Modern Control Systems_P4
where
See the MCS website for a review of complex numbers.
Alternatively, the transfer function can be represented by a magnitude \(|G(j\omega)|\) and a phase \(\phi(j\omega)\) as
where
The graphical representation of the frequency response of the system \(G(j\omega)\) can utilize either Equation (8.8) or Equation (8.9). The polar plot representation of the frequency response is obtained by using Equation (8.8). The coordinates of the polar plot are the real and imaginary parts of \(G(j\omega)\), as shown in Figure 8.1. An example of a polar plot will illustrate this approach.
270. EXAMPLE 8.1 Frequency response of an \(\mathbf{R}C\) filter
A simple \(RC\) filter is shown in Figure 8.2. The transfer function of this filter is
and the sinusoidal steady-state transfer function is
where
FIGURE 8.2 An \(RC\) filter.
FIGURE 8.3 Polar plot for \(RC\) filter.
Then the polar plot is obtained from the relation
The first step is to determine \(R(\omega)\) and \(X(\omega)\) at the two frequencies, \(\omega = 0\) and \(\omega = \infty\). At \(\omega = 0\), we have \(R(\omega) = 1\) and \(X(\omega) = 0\). At \(\omega = \infty\), we have \(R(\omega) = 0\) and \(X(\omega) = 0\). These two points are shown in Figure 8.3. The locus of the real and imaginary parts is also shown in Figure 8.3 and is easily shown to be a circle with the center at \(\left( \frac{1}{2},0 \right)\). When \(\omega = \omega_{1}\), the real and imaginary parts are equal in magnitude, and the angle \(\phi(\omega) = - 45^{\circ}\). The polar plot can also be readily obtained from Equation (8.9) as
where
Hence, when \(\omega = \omega_{1}\), the magnitude is \(\left| G\left( j\omega_{1} \right) \right| = 1/\sqrt{2}\) and the phase \(\phi\left( \omega_{1} \right) = - 45^{\circ}\). Also, when \(\omega\) approaches \(+ \infty\), we have \(|G(j\omega)| \rightarrow 0\) and \(\phi(\omega) = - 90^{\circ}\). Similarly, when \(\omega = 0\), we have \(|G(j\omega)| = 1\) and \(\phi(\omega) = 0\).
271. EXAMPLE 8.2 Polar plot of a transfer function
The polar plot of a transfer function is useful for investigating system stability. Consider a transfer function
FIGURE 8.4
Polar plot for \(G(j\omega) =\) \(K/(j\omega(j\omega\tau + 1))\).
Note that \(\omega = \infty\) is at the origin.
Then the magnitude and phase angle are written as
The phase angle and the magnitude are readily calculated at the frequencies \(\omega = 0,\omega = 1/\tau\), and \(\omega = + \infty\). The polar plot of \(G(j\omega)\) is shown in Figure 8.4.
An alternative solution uses the real and imaginary parts of \(G(j\omega)\) as
where \(R(\omega) = - K\omega^{2}\tau/M(\omega)\) and \(X(\omega) = - \omega K/M(\omega)\), and where \(M(\omega) = \omega^{2} +\) \(\omega^{4}\tau^{2}\). Then when \(\omega = \infty\), we have \(R(\omega) = 0\) and \(X(\omega) = 0\). When \(\omega = 0\), we have \(R(\omega) = - K\tau\) and \(X(\omega) = - \infty\). When \(\omega = 1/\tau\), we have \(R(\omega) = - K\tau/2\) and \(X(\omega) = - K\tau/2\), as shown in Figure 8.4.
Another method of obtaining the polar plot is to evaluate the vector \(G(j\omega)\) graphically at specific frequencies, \(\omega\), along the \(s = j\omega\) axis on the \(s\)-plane. We consider
with the two poles shown on the \(s\)-plane in Figure 8.5.
When \(s = j\omega\), we have
FIGURE 8.5
Two vectors on the \(s\)-plane to evaluate \(G\left( j\omega_{1} \right)\).
where \(p = 1/\tau\). The magnitude and phase of \(G(j\omega)\) can be evaluated at a specific frequency, \(\omega_{1}\), on the \(j\omega\)-axis, as shown in Figure 8.5 . The magnitude and the phase are, respectively,
and
There are several possibilities for coordinates of a graph portraying the frequency response of a system. As we have seen, we may use a polar plot to represent the frequency response (Equation 8.8) of a system. However, the limitations of polar plots are readily apparent. The addition of poles or zeros to an existing system requires the recalculation of the frequency response, as outlined in Examples 8.1 and 8.2. Furthermore, calculating the frequency response in this manner is tedious and does not indicate the effect of the individual poles or zeros.
The introduction of logarithmic plots, often called Bode plots, simplifies the determination of the graphical portrayal of the frequency response. The logarithmic plots are called Bode plots in honor of H. W. Bode, who used them extensively in his studies of feedback amplifiers \(\lbrack 4,5\rbrack\). The transfer function in the frequency domain is
The logarithm of the magnitude is normally expressed in terms of the logarithm to the base 10 , so we use
where the units are decibels (dB). A decibel conversion table is given on the MCS website. The logarithmic gain in \(dB\) and the angle \(\phi(\omega)\) can be plotted versus the frequency \(\omega\) by utilizing several different arrangements. For a Bode diagram, the FIGURE 8.6
Bode plot for \(G(j\omega) = 1/(j\omega\tau + 1)\) :
(a) magnitude plot and (b) phase plot.
(a)
(b)
plot of logarithmic gain in \(dB\) versus \(\omega\) is normally plotted on one set of axes, and the phase \(\phi(\omega)\) versus \(\omega\) on another set of axes, as shown in Figure 8.6. For example, the Bode plot of the transfer function of Example 8.1 can be readily obtained, as we will find in the following example.
272. EXAMPLE 8.3 Bode plot of an \(\mathbf{R}C\) filter
The transfer function of Example 8.1 is
where
the time constant of the network. The logarithmic gain is
For small frequencies - that is, \(\omega \ll 1/\tau\) - the logarithmic gain is
For large frequencies - that is, \(\omega \gg 1/\tau\) - the logarithmic gain is
and at \(\omega = 1/\tau\), we have
The magnitude plot for this network is shown in Figure 8.6(a). The phase angle of the network is
The phase plot is shown in Figure 8.6(b). The frequency \(\omega = 1/\tau\) is often called the break frequency or corner frequency.
A linear scale of frequency is not the most convenient or judicious choice, and we consider the use of a logarithmic scale of frequency. The convenience of a logarithmic scale of frequency can be seen by considering Equation (8.21) for large frequencies \(\omega \gg 1/\tau\), as follows:
Then, on a set of axes where the horizontal axis is \(log\omega\), the asymptotic curve for \(\omega \gg 1/\tau\) is a straight line, as shown in Figure 8.7. The slope of the straight line can be ascertained from Equation (8.21). An interval of two frequencies with a ratio equal to 10 is called a decade, so that the range of frequencies from \(\omega_{1}\) to \(\omega_{2}\), where \(\omega_{2} = 10\omega_{1}\), is called a decade. The difference between the logarithmic gains, for \(\omega \gg 1/\tau\), over a decade of frequency is
that is, the slope of the asymptotic line for this first-order transfer function is \(- 20\text{ }dB/\) decade, and the slope is shown for this transfer function in Figure 8.7. Instead of using a horizontal axis of \(log\omega\) and linear rectangular coordinates, it is easier to use semilog axes with a linear rectangular coordinate for \(dB\) and a logarithmic coordinate for \(\omega\). Alternatively, we could use a logarithmic coordinate for the magnitude as well as for frequency and avoid the necessity of calculating the logarithm of the magnitude.
FIGURE 8.7
Asymptotic curve for \((j\omega\tau + 1)^{- 1}\).
The frequency interval \(\omega_{2} = 2\omega_{1}\) is often used and is called an octave of frequencies. The difference between the logarithmic gains for \(\omega \gg 1/\tau\), for an octave, is
Therefore, the slope of the asymptotic line is \(- 6\text{ }dB/\) octave.
The primary advantage of the logarithmic plot is the conversion of multiplicative factors, such as \((j\omega\tau + 1)\), into additive factors, \(20log(j\omega\tau + 1)\), by virtue of the definition of logarithmic gain. This can be readily ascertained by considering the transfer function
This transfer function includes \(Q\) zeros, \(N\) poles at the origin, \(M\) poles on the real axis, \(P\) pairs of complex conjugate zeros, and \(R\) pairs of complex conjugate poles. The logarithmic magnitude of \(G(j\omega)\) is
and the Bode plot can be obtained by adding the contribution of each individual factor. Furthermore, the separate phase angle plot is obtained as
which is the summation of the phase angles due to each individual factor of the transfer function.
Therefore, the four different kinds of factors that may occur in a transfer function are as follows:
-
Constant gain \(K_{b}\)
-
Poles (or zeros) at the origin \((j\omega)\)
-
Poles (or zeros) on the real axis \((j\omega\tau + 1)\)
-
Complex conjugate poles (or zeros) \(\left\lbrack 1 + \left( 2\zeta/\omega_{n} \right)j\omega + \left( j\omega/\omega_{n} \right)^{2} \right\rbrack\). We can determine the logarithmic magnitude plot and phase angle for these four factors and then use them to obtain a Bode plot for any general form of a transfer function. Typically, the curves for each factor are obtained and then added together graphically to obtain the curves for the complete transfer function. Furthermore, this procedure can be simplified by using the asymptotic approximations to these curves and obtaining the actual curves only at specific important frequencies.
Constant Gain \(K_{b}\). The logarithmic gain for the constant \(K_{b}\) is
and the phase angle is
The gain curve is a horizontal line on the Bode plot.
If the gain is a negative value, \(- K_{b}\), the logarithmic gain remains \(20logK_{b}\). The negative sign is accounted for by the phase angle, \(- 180^{\circ}\).
Poles (or Zeros) at the Origin \((j\omega)\). A pole at the origin has a logarithmic magnitude
and a phase angle
The slope of the magnitude curve is \(- 20\text{ }dB/decade\) for a pole. Similarly, for a multiple pole at the origin, we have
and the phase is
In this case, the slope due to the multiple pole is \(- 20\text{ }N\text{ }dB/\) decade. For a zero at the origin, we have a logarithmic magnitude
FIGURE 8.8
Bode plot for \((j\omega)^{\pm N}\).
where the slope is \(+ 20\text{ }dB/\) decade and the phase angle is
The Bode plot of the magnitude and phase angle of \((j\omega)^{\pm N}\) is shown in Figure 8.8 for \(N = 1\) and \(N = 2\).
Poles or Zeros on the Real Axis. For a pole on the real axis,
The asymptotic curve for \(\omega \ll 1/\tau\) is \(20log1 = 0\text{ }dB\), and the asymptotic curve for \(\omega \gg 1/\tau\) is \(- 20log(\omega\tau)\), which has a slope of \(- 20\text{ }dB/\) decade. The intersection of the two asymptotes occurs when
or when \(\omega = 1/\tau\), the break frequency. The actual logarithmic gain when \(\omega = 1/\tau\) is \(- 3\text{ }dB\). The phase angle is \(\phi(\omega) = - \tan^{- 1}(\omega\tau)\) for the denominator factor. The Bode plot of a pole factor \((1 + j\omega\tau)^{- 1}\) is shown in Figure 8.9.
The Bode plot of a zero factor \(1 + j\omega\tau\) is obtained in the same manner as that of the pole. However, the slope is positive at \(+ 20\text{ }dB/\) decade, and the phase angle is \(\phi(\omega) = + \tan^{- 1}(\omega\tau)\).
A piecewise linear approximation to the phase angle curve can be obtained as shown in Figure 8.9. This linear approximation, which passes through the correct phase at the break frequency, is within \(6^{\circ}\) of the actual phase curve for all frequencies. This approximation will provide a useful means for readily determining the form of the phase angle curves of a transfer function \(G(s)\). However, often the accurate phase angle curves are required, and the actual phase curve for the first-order factor must be obtained via a computer program.
Complex Conjugate Poles or Zeros \(\left\lbrack 1 + \left( 2\zeta/\omega_{n} \right)j\omega + \left( j\omega/\omega_{n} \right)^{2} \right\rbrack\). The quadratic factor for a pair of complex conjugate poles can be written in normalized form as
(a)
Frequency (rad/s)
Bode diagram for \((1 + j\omega\tau)^{- 1}\).
(b)
where \(u = \omega/\omega_{n}\). Then the logarithmic magnitude for a pair of complex conjugate poles is
and the phase angle is
When \(u \ll 1\), the magnitude is
and the phase angle approaches \(0^{\circ}\). When \(u \gg 1\), the logarithmic magnitude approaches
which results in a curve with a slope of \(- 40\text{ }dB/\) decade. The phase angle, when \(u \gg 1\), approaches \(- 180^{\circ}\). The magnitude asymptotes meet at the \(0\text{ }dB\) line when \(u = \omega/\omega_{n} = 1\). However, the difference between the actual magnitude curve and the asymptotic approximation is a function of the damping ratio and must be FIGURE 8.10
Bode diagram for $G(j\omega) = \left\lbrack 1 + \left( 2\zeta/\omega_{n} \right) \right.\ $ \(\left. \ j\omega + \left( j\omega/\omega_{n} \right) \right\rbrack^{- 1}\).
(a)
(b)
accounted for when \(\zeta < 0.707\). The Bode plot of a quadratic factor due to a pair of complex conjugate poles is shown in Figure 8.10. The maximum value \(M_{p\omega}\) of the frequency response occurs at the resonant frequency \(\omega_{r}\). When the damping ratio approaches zero, then \(\omega_{r}\) approaches \(\omega_{n}\), the natural frequency. The resonant frequency is determined by taking the derivative of the magnitude of Equation (8.33) with respect to the normalized frequency, \(u\), and setting it equal to zero. The resonant frequency is given by the relation
FIGURE 8.11
The maximum \(M_{pw}\) of the frequency response and the resonant frequency \(\omega_{r}\) versus \(\zeta\) for a pair of complex conjugate poles.
and the maximum value of the magnitude \(|G(j\omega)|\) is
for a pair of complex poles. The maximum value of the frequency response, \(M_{p\omega}\), and the resonant frequency \(\omega_{r}\) are shown as a function of the damping ratio \(\zeta\) for a pair of complex poles in Figure 8.11. Assuming the dominance of a pair of complex conjugate closed-loop poles, we find that these curves are useful for estimating the damping ratio of a system from an experimentally determined frequency response.
The frequency response curves can be evaluated on the \(s\)-plane by determining the vector lengths and angles at various frequencies \(\omega\) along the \(s = + j\omega\)-axis. For example, considering the second-order factor with complex conjugate poles, we have
FIGURE 8.12
Vector evaluation of the frequency response for selected values of \(\omega\).
(a)
(c)
(b)
(d)
The poles for varying \(\zeta\) lie on a circle of radius \(\omega_{n}\) and are shown for a particular \(\zeta\) in Figure 8.12(a). The transfer function evaluated for real frequency \(s = j\omega\) is written as
where \(s_{1}\) and \({\widehat{s}}_{1}\) are the complex conjugate poles. The vectors \(j\omega - s_{1}\) and \(j\omega - {\widehat{s}}_{1}\) are the vectors from the poles to the frequency \(j\omega\), as shown in Figure 8.12(a). Then the magnitude and phase may be evaluated for various specific frequencies. The magnitude is
and the phase is
The magnitude and phase may be evaluated for three specific frequencies, namely,
as shown in Figure 8.12 in parts (b), (c), and (d), respectively. The magnitude and phase corresponding to these frequencies are shown in Figure 8.13. FIGURE 8.13
Bode diagram for complex conjugate poles.
273. EXAMPLE 8.4 Bode diagram of a twin-T network
As an example of the determination of the frequency response using the pole-zero diagram and the vectors to \(j\omega\), consider the twin-T network shown in Figure 8.14 [6]. The transfer function of this network is
where \(\tau = RC\). The zeros are at \(s = \pm j1/\tau\), and the poles are at \(s = ( - 2 \pm \sqrt{3})/\tau\), in the \(s\)-plane, as shown in Figure 8.15(a). At \(\omega = 0\), we have \(|G(j\omega)| = 1\) and \(\phi(\omega) = 0^{\circ}\). At \(\omega = 1/\tau,|G(j\omega)| = 0\) and the phase angle of the vector from the zero at \(s = j1/\tau\) passes through a transition of \(180^{\circ}\). When \(\omega\) approaches \(\infty,|G(j\omega)| = 1\) and \(\phi(\omega) = 0\) again. The frequency response is shown in Figure 8.15(b).
A summary of the asymptotic curves for basic terms of a transfer function is provided in Table 8.1 .
In the previous examples, the poles and zeros of \(G(s)\) have been restricted to the left-hand plane. However, a system may have zeros located in the right-hand \(s\) plane and may still be stable. Transfer functions with zeros in the right-hand \(s\)-plane are classified as nonminimum phase transfer functions. If the zeros of a transfer function are all reflected about the \(j\omega\)-axis, there is no change in the magnitude of the transfer function, and the only difference is in the phase-shift characteristics. If the phase characteristics of the two system functions are compared, it can be
FIGURE 8.14 Twin-T network. Table 8.1 Asymptotic Curves for Basic Terms of a Transfer Function
Term
- Gain,
- Zero,
- Pole,
- Pole at
the origin,
ranges over less than \(80^{\circ}\), whereas the phase shift of
ranges over \(180^{\circ}\). The meaning of the term minimum phase is illustrated by Figure 8.17 . The range of phase shift of a minimum phase transfer function is the least possible or minimum corresponding to a given amplitude curve, whereas the range of the nonminimum phase curve is the greatest possible for the given amplitude curve. FIGURE 8.16
Pole-zero patterns giving the same amplitude response and different phase characteristics.
FIGURE 8.17
The phase characteristics for the minimum phase and nonminimum phase transfer function.
(a)
(b)
Frequency \((rad/s)\)
A particularly interesting nonminimum phase network is the all-pass network, which can be realized with a symmetrical lattice network [8]. A symmetrical pattern of poles and zeros is obtained as shown in Figure 8.18(a). Again, the magnitude \(|G(j\omega)|\) remains constant; in this case, it is equal to unity. However, the angle varies
(a)
(c)
Frequency (rad/s)
(b)
response, and
(c) a lattice network. from \(0^{\circ}\) to \(- 360^{\circ}\). Because \(\theta_{2} = 180^{\circ} - \theta_{1}\) and \({\widehat{\theta}}_{2} = 180^{\circ} - {\widehat{\theta}}_{1}\), the phase is given by \(\left. \ \phi(\omega) = - 2\left( \theta_{1} + {\widehat{\theta}}_{1} \right\} \right)\). The magnitude and phase characteristic of the all-pass network is shown in Figure 8.18(b). A nonminimum phase lattice network is shown in Figure 8.18(c).
275. EXAMPLE 8.5 Sketching a Bode plot
The Bode plot of a transfer function \(G(s)\), which contains several zeros and poles, is obtained by adding the plot due to each individual pole and zero. The simplicity of this method is illustrated by considering the transfer function
The factors, in order of their occurrence as frequency increases, are:
-
A constant gain \(K = 5\)
-
A pole at the origin
-
A pole at \(\omega = 2\)
-
A zero at \(\omega = 10\)
-
A pair of complex poles at \(\omega = \omega_{n} = 50\).
First, we plot the magnitude characteristic for each individual pole and zero factor and the constant gain:
-
The constant gain is \(20log5 = 14\text{ }dB\), as shown in Figure 8.19.
-
The magnitude of the pole at the origin extends from zero frequency to infinite frequencies and has a slope of \(- 20\text{ }dB/\) decade intersecting the 0 - \(dB\) line at \(\omega = 1\), as shown in Figure 8.19.
-
The asymptotic approximation of the magnitude of the pole at \(\omega = 2\) has a slope of \(- 20\text{ }dB/\) decade beyond the break frequency at \(\omega = 2\). The asymptotic magnitude below the break frequency is \(0\text{ }dB\), as shown in Figure 8.19.
-
The asymptotic magnitude for the zero at \(\omega = + 10\) has a slope of \(+ 20\text{ }dB/\) decade beyond the break frequency at \(\omega = 10\), as shown in Figure 8.19.
FIGURE 8.19 Magnitude asymptotes of poles and zeros used in the example.
- The magnitude for the complex poles is \(- 40\text{ }dB/\) decade. The break frequency is \(\omega = \omega_{n} = 50\), as shown in Figure 8.19. This approximation must be corrected to the actual magnitude because the damping ratio is \(\zeta = 0.3\), and the magnitude differs appreciably from the approximation, as shown in Figure 8.20.
Therefore, the total asymptotic magnitude can be plotted by adding the asymptotes due to each factor, as shown by the solid line in Figure 8.20. Examining the asymptotic curve of Figure 8.20, we note that the curve can be obtained directly by plotting each asymptote in order as frequency increases. Thus, the slope is \(- 20\text{ }dB/\) decade due to \(K(j\omega)^{- 1}\) intersecting \(14\text{ }dB\) at \(\omega = 1\). Then, at \(\omega = 2\), the slope becomes \(- 40\text{ }dB/\) decade due to the pole at \(\omega = 2\). The slope changes to \(- 20\text{ }dB/\) decade due to the zero at \(\omega = 10\). Finally, the slope becomes \(- 60\text{ }dB/\) decade at \(\omega = 50\) due to the pair of complex poles at \(\omega_{n} = 50\).
The exact magnitude curve is then obtained by using Figure 8.9, which provides the difference between the actual and asymptotic curves for a single pole. The single zero follows a similar pattern, but with actual curve \(+ 3\text{ }dB\) at the break frequency. The exact magnitude curve for the pair of complex poles is obtained by utilizing Figure 8.10(a) for the quadratic factor. The exact magnitude curve for \(G(j\omega)\) is shown by a dashed line in Figure 8.20.
The phase characteristic can be obtained by adding the phase due to each individual factor. Usually, the linear approximation of the phase characteristic for a single pole or zero is suitable for the initial analysis. Thus, the individual phase characteristics for the poles and zeros are shown in Figure 8.21 and are:
-
The phase of the constant gain is \(0^{\circ}\).
-
The phase of the pole at the origin is a constant \(- 90^{\circ}\).
-
The linear approximation of the phase characteristic for the pole at \(\omega = 2\) is shown in Figure 8.21 , where the phase shift is \(- 45^{\circ}\) at \(\omega = 2\).
-
The linear approximation of the phase characteristic for the zero at \(\omega = 10\) is also shown in Figure 8.21, where the phase shift is \(+ 45^{\circ}\) at \(\omega = 10\).
FIGURE 8.20 Magnitude characteristic.
FIGURE 8.21
Phase characteristic.
- The actual phase characteristic for the pair of complex poles is obtained from Figure 8.10 and is shown in Figure 8.21.
Therefore, the total phase characteristic, \(\phi(\omega)\), is obtained by adding the phase due to each factor as shown in Figure 8.21. While this curve is an approximation, its usefulness merits consideration as a first attempt to determine the phase characteristic. Thus, a frequency of interest is the frequency for which \(\phi(\omega) = - 180^{\circ}\). The approximate curve indicates that a phase shift of \(- 180^{\circ}\) occurs at \(\omega = 46\). The actual phase shift at \(\omega = 46\) can be readily calculated as
where
Then we find that
and the approximate curve has an error of \(5^{\circ}\) at \(\omega = 46\). However, once the approximate frequency of interest is ascertained from the approximate phase curve, the accurate phase shift for the neighboring frequencies is readily determined by using the exact phase shift relation (Equation 8.43). This approach is usually preferable to the calculation of the exact phase shift for all frequencies over several decades. In summary, we may obtain approximate curves for the magnitude and phase shift of a transfer function \(G(j\omega)\) in order to determine the important frequency ranges. Then, within the relatively small important frequency ranges, the exact magnitude and phase shift can be readily evaluated by using the exact equations, such as Equation (8.43). FIGURE 8.22 The Bode plot of the \(G(j\omega)\) of Equation (8.42).
The Bode plot for the transfer function in Equation 8.42 is shown in Figure 8.22. The plot is generated for four decades, and the \(0 - dB\) line is indicated, as well as the \(- 180^{\circ}\) line. The plot indicates that the magnitude is \(34\text{ }dB\) and that the phase is \(- {92.36}^{\circ}\) at \(\omega = 0.1\). Similarly, the plot indicates that the magnitude is \(- 43\text{ }dB\) and that the phase is \(- 243^{\circ}\) at \(\omega = 100\). Examining the plot, we find that the magnitude is \(0\text{ }dB\) at \(\omega = 3.0\) and the phase is \(- 180^{\circ}\) at \(\omega = 50\).
275.1. FREQUENCY RESPONSE MEASUREMENTS
A sine wave can be used to measure the open-loop frequency response of a system. In practice, a plot of amplitude versus frequency and phase versus frequency will be obtained \(\lbrack 1,3,6\rbrack\). From these two plots, the loop transfer function \(G_{c}(j\omega)G(j\omega)\) can be deduced. Similarly, the closed-loop frequency response of a control system, \(T(j\omega)\), may be obtained and the actual transfer function deduced.
A device called a wave analyzer can be used to measure the amplitude and phase variations as the frequency of the input sine wave is altered. Also, a device called a transfer function analyzer can be used to measure the loop transfer function and closed-loop transfer functions [6].
A typical signal analyzer instrument can perform frequency response measurements from DC to \(100kHz\). Built-in analysis and modeling capabilities can derive poles and zeros from measured frequency responses or construct phase and magnitude responses from user-supplied models. This device can also synthesize the frequency response of a model of a system, allowing a comparison with an actual response.
As an example of determining the transfer function from the Bode plot, consider Figure 8.23. The system is a stable circuit consisting of resistors and capacitors. Because the magnitude declines at about \(- 20\text{ }dB/\) decade as \(\omega\) increases between 100 and 1,000, and because the phase is \(- 45^{\circ}\) and the magnitude is \(- 3\text{ }dB\) at \(300rad/s\), we can deduce that one factor is a pole at \(p_{1} = 300\). Next, we deduce that a pair of quadratic zeros exist at \(\omega_{n} = 2450\). This is inferred by noting that the phase changes abruptly by nearly \(+ 180^{\circ}\), passing through \(0^{\circ}\) at \(\omega_{n} = 2,450\). Also, the slope of the magnitude changes from \(- 20\text{ }dB/\) decade to \(+ 20\text{ }dB/\) decade at \(\omega_{n} = 2,450\). Because the slope of the magnitude returns to \(0\text{ }dB/\) decade as \(\omega\) exceeds 50,000 , we determine that there is a second pole as well as two zeros. This second pole is at FIGURE 8.23
A Bode plot for a system with an unidentified transfer function.
Frequency \((rad/s)\)
(a)
(b)
\(p_{2} = 20,000\), because the magnitude is \(- 3\text{ }dB\) from the asymptote and the phase is \(+ 45^{\circ}\) at this point $\left( - 90^{\circ} \right.\ $ for the first pole, \(+ 180^{\circ}\) for the pair of quadratic zeros, and \(- 45^{\circ}\) for the second pole). We sketch the asymptotes for the poles and the numerator of the proposed transfer function \(T(s)\) of Equation (8.45), as shown in Figure 8.23(a). The equation is
The difference in magnitude from the corner frequency \(\left( \omega_{n} = 2,450 \right)\) of the asymptotes to the minimum response is \(10\text{ }dB\), which, from Equation (8.37), indicates that \(\zeta = 0.16\). (Compare the plot of the quadratic zeros to the plot of the quadratic poles in Figure 8.10. Note that the plots need to be turned "upside down" for the quadratic zeros and that the phase goes from \(0^{\circ}\) to \(+ 180^{\circ}\) instead of \(- 180^{\circ}\).) Therefore, the transfer function is
This frequency response is actually obtained from a bridged-T network.
275.2. PERFORMANCE SPECIFICATIONS IN THE FREQUENCY DOMAIN
We must ask the question: how does the frequency response of a system relate to the expected transient response of the system? In other words, given a set of timedomain (transient performance) specifications, how do we specify the frequency response? For a simple second-order system, we have already answered this question by considering the time-domain performance in terms of overshoot, settling time, and other performance criteria, such as integral squared error. For the second-order system shown in Figure 8.24, the closed-loop transfer function is
The frequency response of the closed-loop system is shown in Figure 8.25. Because this is a second-order system, the damping ratio of the system is related to the maximum magnitude \(M_{p\omega}\), which occurs at the frequency \(\omega_{r}\) as shown in Figure 8.25.
276. At the resonant frequency \(\omega_{r}\) a maximum value \(M_{p\omega}\) of the frequency response is attained.
FIGURE 8.24
A second-order closed-loop system.
FIGURE 8.25
Magnitude characteristic of the second-order system closed-loop transfer function, \(T(s)\).
FIGURE 8.26
Normalized bandwidth, \(\omega_{B}/\omega_{n}\), versus \(\zeta\) for a second-order system (Equation 8.46). The linear approximation \(\omega_{B}/\omega_{n} = - 1.19\zeta + 1.85\) is accurate for \(0.3 \leq \zeta \leq 0.8\).
The bandwidth is the frequency \(\omega_{B}\) at which the frequency response has declined \(3\text{ }dB\) from its low-frequency value. This corresponds to approximately half an octave, or about \(1/\sqrt{2}\) of the low-frequency value.
The resonant frequency \(\omega_{r}\) and the \(- 3\text{ }dB\) bandwidth can be related to the speed of the transient response. Thus, as the bandwidth \(\omega_{B}\) increases, the rise time of the step response of the system will decrease. Furthermore, the overshoot to a step input can be related to \(M_{p\omega}\) through the damping ratio \(\zeta\). The curves of Figure 8.11 relate the resonance magnitude and frequency to the damping ratio of the second-order system. With the damping ratio, the percent overshoot to a unit step can be computed. Thus, we find as the resonant peak \(M_{p\omega}\) increases in magnitude, the percent overshoot to a step input increases. In general, the magnitude \(M_{p\omega}\) indicates the relative stability of a system.
The bandwidth of a system \(\omega_{B}\), as indicated on the frequency response, can be approximately related to the natural frequency of the system. Figure 8.26 shows the normalized bandwidth \(\omega_{B}/\omega_{n}\) versus \(\zeta\) for the second-order system of Equation (8.46). The response of the second-order system to a unit step input is of the form
The greater the magnitude of \(\omega_{n}\) when \(\zeta\) is constant, the more rapidly the response approaches the desired steady-state value. Thus, desirable frequency-domain specifications are as follows:
-
Relatively small resonant magnitudes: \(M_{p\omega} < 1.5\), for example.
-
Relatively large bandwidths so that the system time constant \(\tau = 1/\left( \zeta\omega_{n} \right)\) is sufficiently small.
The usefulness of the frequency response specifications and their relation to the actual transient performance depend upon the approximation of the system by a second-order pair of complex poles, called the dominant roots. If the frequency
response is dominated by a pair of complex poles, the relationships between the frequency response and the time response discussed in this section will be valid. Fortunately, a large proportion of control systems satisfy this dominant second-order approximation in practice.
The steady-state error specification can also be related to the frequency response of a closed-loop system. The steady-state error for a specific test input signal is related to the gain and number of integrations (poles at the origin) of the loop transfer function. Therefore, for the system shown in Figure 8.24, the steady-state error for a ramp input is specified in terms of \(K_{v}\), the velocity constant. The steadystate error for the system is
where \(A\) is the magnitude of the ramp input. The velocity constant for the system of Figure 8.24 without feedback is
The transfer function can be written as
and the gain constant is \(K_{v}\) for this type-one system. For example, reexamining Example 8.5, we had a type-one system with a loop transfer function
where \(u = \omega/\omega_{n}\). Therefore, in this case, we have \(K_{v} = 5\). In general, if the loop transfer function of a feedback system is written as
then the system is type \(N\) and the gain \(K\) is the gain constant for the steady-state error. Thus, for a type-zero system that has two poles, we have
In this equation, \(K = K_{p}\) (the position error constant) that appears as the low-frequency gain on the Bode plot. Furthermore, the gain constant \(K = K_{v}\) for the type-one system appears as the gain of the low-frequency section of the magnitude characteristic. Considering only the pole at the origin and gain of the type-one system of Equation (8.50), we have
and the \(K_{v}\) is equal to the magnitude when this portion of the magnitude characteristic intersects the 0 -dB line. For example, the low-frequency intersection of \(K_{v}/j\omega\) in Figure 8.20 is equal to \(\omega = 5\), as we expect.
Therefore, the frequency response characteristics represent the performance of a system quite adequately, and with some experience, they are quite useful for the analysis and design of feedback control systems.
276.1. LOG-MAGNITUDE AND PHASE DIAGRAMS
There are several alternative methods for presenting the frequency response of a function \(G(j\omega)\). We have seen that suitable graphical presentations of the frequency response are (1) the polar plot and (2) the Bode plot. An alternative approach to portraying the frequency response graphically is to plot the logarithmic magnitude in \(dB\) versus the phase angle for a range of frequencies. Consider the log-magnitude-phase plot for the transfer function
shown in Figure 8.27. The numbers indicated along the curve are for values of frequency \(\omega\).
The log-magnitude-phase curve for the transfer function
is shown in Figure 8.28. This curve can be obtained by utilizing the Bode plots of Figures 8.20 and 8.21 to transfer the frequency response information to the log-magnitude and phase coordinates. The shape of the locus of the frequency response on a log-magnitude-phase diagram is particularly important as the phase approaches \(- 180^{\circ}\) and the magnitude approaches \(0\text{ }dB\). The locus of Equation (8.54) and Figure 8.27 differs substantially from the locus of Equation (8.55) and Figure 8.28. Therefore, as the correlation between the shape of the locus and the transient response of a system is established, we will obtain another useful portrayal of the frequency response of a system. We can establish a stability criterion in the frequency domain for which it will be useful to utilize the log-magnitudephase diagram to investigate the relative stability of closed-loop feedback control systems.
FIGURE 8.27 Log-magnitude-phase curve for \(G_{1}(j\omega)\).
FIGURE 8.28 Log-magnitude-phase curve for \(G_{2}(j\omega)\).
276.2. DESIGN EXAMPLES
In this section, we present two illustrative examples using frequency response methods to design controllers. The first example describes the control of a photovoltaic generator to achieve maximum power delivery as the sunlight varies over time. The second example considers the control of one leg of a six-legged robotic device. In this example, the specifications that must be satisfied include a mix of time-domain specifications (percent overshoot and settling time) and frequency-domain specifications (bandwidth). The design process leads to a viable PID controller meeting all the specifications.
277. EXAMPLE 8.6 Maximum power pointing tracking for photovoltaic generators
One goal of green engineering is to design products that minimize pollution and improve the environment. Using solar energy is one way to provide clean energy using photovoltaic generators converting sunlight to electricity directly. However, the output of a photovoltaic generator is variable and depends on the available sunlight, the temperature, and the attached loads. In this example, we provide a FIGURE 8.29
Photovoltaic generator feedback control system to a track reference input voltage.
discussion on regulating the voltage provided by a photovoltaic generator system using feedback control [24]. We will design a controller to achieve the desired specifications.
Consider the feedback control system in Figure 8.29. The plant transfer function is
where \(K = 300,000\) and \(p = 360\). This model is consistent with a photovoltaic generator with 182 cells generating over \(1,100\text{ }W\) [24]. Assume a controller of the form
where \(K_{c},\tau_{1}\), and \(\tau_{2}\) are to be determined. The controller in Equation (8.56) is a lead or lag compensator depending on \(\tau_{1}\) and \(\tau_{2}\). The controller should minimize the effects of disturbances and plant changes by providing a high gain at low frequencies while minimizing the measurement noise by providing a low gain at high frequencies [24]. To accomplish these goals, the design specifications are:
-
\(\left| G_{c}(j\omega)G(j\omega) \right| \geq 20\text{ }dB\) at \(\omega \leq 10rad/s\)
-
\(\left| G_{c}(j\omega)G(j\omega) \right| \leq - 20\text{ }dB\) at \(\omega \geq 1,000rad/s\)
-
Phase margin P.M. \(\geq 60^{\circ}\)
The phase margin of the uncompensated system is \(P.M. = {36.3}^{\circ}\) implying that the compensated system needs to add approximately P.M. \(= 25^{\circ}\), hence the use of the compensator to add the required phase lead. Also, the magnitude of the uncompensated frequency response at \(\omega = 1000rad/s\) is \(- 11\text{ }dB\) indicating that the gain needs to be further reduced at high frequencies to meet the specifications.
One possible controller is
The compensated phase margin is P.M. \(= {60.4}^{\circ}\). As can be seen in Figure 8.30, the low-frequency, high-gain specification is satisfied, as well as the high-frequency, low-gain specification. The closed-loop step response is shown in Figure 8.31. The settling time is \(T_{s} = 0.11\text{ }s\) and the percent overshoot is \(P.O. = 19.4\%\), both very acceptable for the control of the photovoltaic generator voltage. FIGURE 8.30
Bode plot of compensated system with \(G_{c}(s) =\) \(250\left\lbrack \frac{0.04s + 1}{100s + 1} \right\rbrack\).
FIGURE 8.31
Step response of the closed-loop system.
278. EXAMPLE 8.7 Control of one leg of a six-legged robot
The Ambler is a six-legged walking machine being developed at Carnegie-Mellon University [23]. An artist's conception of the Ambler is shown in Figure 8.32.
In this example we consider the control system design for position control of one leg. The elements of the design process emphasized in this example are highlighted in Figure 8.33. The mathematical model of the actuator and leg is provided. The transfer function is
FIGURE 8.32
An artist's
conception of the six-legged Ambler.
Topics emphasized in this example
FIGURE 8.33
Elements of the control system design process emphasized in this six-legged robot example.
The input is a voltage command to the actuator, and the output is the leg position (vertical position only). A block diagram of the control system is shown in Figure 8.34. The control goal is
279. Control Goal
Control the robot leg position and maintain the position in the presence of unwanted measurement noise.
FIGURE 8.34 Control system for one leg.
The variable to be controlled is
280. Variable to Be Controlled
Leg position, \(Y(s)\).
We want the leg to move to the commanded position as fast as possible but with minimal overshoot. As a practical first step, the design goal will be to produce a system that moves, albeit slowly. In other words, the control system bandwidth will initially be low.
The control design specifications are
281. Control Design Specifications
DS1 Closed-loop bandwidth is \(\omega_{B} \geq 1\text{ }Hz\).
DS2 Percent overshoot is P.O. \(\leq 15\%\) to a step input.
DS3 Zero steady-state tracking error to a step input.
Specifications DS1 and DS2 are intended to ensure acceptable tracking performance. Design specification DS3 is actually a nonissue in our design: the actuator/ leg transfer function is a type-one system so a zero steady-state tracking error to a step input is guaranteed. We need to ensure that \(G_{c}(s)G(s)\) remains at least a typeone system.
Consider the controller
As \(c \rightarrow 0\), a PID controller is obtained with \(K_{P} = K_{a},K_{D} = K\), and \(K_{I} = Kb\). We can let \(c\) be a parameter at this point and see if the additional freedom in selecting \(c \neq 0\) is useful. It may be that we can simply set \(c = 0\) and use the PID form. The key tuning parameters are
282. Select Key Tuning Parameters
\(K,a,b\), and \(c\). The controller in Equation (8.58) is not the only controller that we can consider. For example, we might consider
where \(K,z\), and \(p\) are the key tuning parameters. The design of the type of controller given in Equation (8.59) will be left as a design problem at the end of the chapter.
The response of a closed-loop control system is determined predominantly by the location of the dominant poles. Our approach to the design is to determine appropriate locations for the dominant poles of the closed-loop system. We can determine the locations from the performance specifications by using second-order system approximation formulas. Once the controller parameters are obtained so that the closed-loop system has the desired dominant poles, the remaining poles are located so that their contribution to the overall response is negligible.
Per specification DS1, we want
From the percent overshoot specification, we can determine the minimum value of \(\zeta\). Thus for P.O. \(\leq 15\%\), we require \(\zeta \geq 0.52\); therefore, we will design with \(\zeta = 0.52\). Even though settling time is not a design specification for this problem, we usually attempt to make the system response as fast as possible while still meeting all the design specifications. From Figure 8.26 and Equation (8.60) it follows that
Then with \(\omega_{n} = 5.11rad/s\) and \(\zeta = 0.52\) and using Equation (8.36) we compute \(\omega_{r} = 3.46rad/s\).
So, if we had a second-order system, we would want to determine values of the control gains such that \(\omega_{n} = 5.11rad/s\) and \(\zeta = 0.52\), which yields \(M_{p\omega} = 1.125\) and \(\omega_{r} = 3.46rad/s\).
Our closed-loop system is a fourth-order system and not a second-order system. So, a valid design approach would be to select \(K,a,b\), and \(c\) so that two poles are dominant and located appropriately to meet the design specifications. This will be the approach followed here.
Another valid approach is to develop a second-order approximation of the fourth-order system. In the approximate transfer function, the parameters \(K,a,b\), and \(c\) are left as variables. The objective would be to obtain an approximate transfer function \(T_{L}(s)\) in such a way that the frequency response of \(T_{L}(s)\) is very close to that of the original system.
The loop transfer function is
and the closed-loop transfer function is
The associated characteristic equation is
The desired characteristic polynomial must also be fourth-order, but we want it to be composed of multiple factors, as follows:
where \(\zeta\) and \(\omega_{n}\) are selected to meet the design specifications, and the roots of \(s^{2} + 2\zeta\omega_{n}s + \omega_{n}^{2} = 0\) are the dominant roots. Conversely we want the roots of \(s^{2} + d_{1}s + d_{0} = 0\) to be the nondominant roots. The dominant roots should lie on a vertical line in the complex plane defined by the distance \(s = - \zeta\omega_{n}\) away from the imaginary axis. Let
Then the roots of \(s^{2} + d_{1}s + d_{0} = 0\), when complex or equal, lie on a vertical line in the complex plane defined by \(s = - \alpha\zeta\omega_{n}\). By choosing \(\alpha > 1\), we effectively move the roots to the left of the dominant roots. The larger we select \(\alpha\), the further the nondominant roots lie to the left of the dominant roots. A reasonable value of \(\alpha\) is \(\alpha = 12\). Also, if we select
then we obtain two real roots
Choosing \(d_{0} = \alpha^{2}\zeta^{2}\omega_{n}^{2}\) is not required, but this seems to be a reasonable choice since we would like the contribution of the nondominant roots to the overall response to be quickly fading and nonoscillatory.
The desired characteristic polynomial is then
Equating the coefficients of Equations (8.63) and (8.64) yields four relationships involving \(K,a,b,c\), and \(\alpha\) :
FIGURE 8.35
Step response using the controller in Equation (8.65).
In our case \(\zeta = 0.52,\omega_{n} = 5.11\), and \(\alpha = 12\). Thus we obtain
and the resulting controller is
The step response of the closed-loop system using the controller in Equation (8.65) is shown in Figure 8.35. The percent overshoot is P.O. \(= 14\%\), and the settling time is \(T_{s} = 0.96\text{ }s\).
The magnitude plot of the closed-loop system is shown in Figure 8.36. The bandwidth is \(\omega_{B} = 27.2rad/s = 4.33\text{ }Hz\). This satisfies DS1 but is larger than the \(\omega_{B} = 1\text{ }Hz\) used in the design (due to the fact that our system is not a second-order system). The higher bandwidth leads us to expect a faster settling time. The peak magnitude is \(M_{p\omega} = 1.21\). We were expecting \(M_{p\omega} = 1.125\).
What is the steady-state response of the closed-loop system if the input is a sinusoidal input? From our previous discussions we expect that as the input frequency increases, the magnitude of the output will decrease. Two cases are presented here. In Figure 8.37 the input frequency is \(\omega = 1rad/s\). The output magnitude is approximately equal to 1 in the steady state. In Figure 8.38 the input frequency is \(\omega = 500rad/s\). The output magnitude is less than 0.005 in the steady state. FIGURE 8.36
Magnitude plot of the closedloop system with the controller in Equation (8.65).
Time (s)
FIGURE 8.37
Output response of the closed-loop system when the input is a sinusoidal signal of frequency \(\omega = 1rad/s\).
This verifies our intuition that the system response decreases as the input sinusoidal frequency increases.
Using simple analytic methods, we obtained an initial set of controller parameters for the mobile robot. The controller thus designed proved to satisfy the design requirements. Some fine-tuning would be necessary to meet the design specifications exactly. FIGURE 8.38
Output response of the closed-loop system when the input is a sinusoidal signal of frequency \(\omega = 500rad/s\).
282.1. FREQUENCY RESPONSE METHODS USING CONTROL DESIGN SOFTWARE
In this section, we cover the functions bode and logspace. The bode function is used to generate a Bode plot, and the logspace function generates a logarithmically spaced vector of frequencies utilized by the bode function.
Consider the transfer function
The Bode plot corresponding to Equation (8.66) is shown in Figure 8.39. The plot consists of the logarithmic gain in \(dB\) versus \(\omega\) in one plot and the phase \(\phi(\omega)\) in degrees versus \(\omega\) in \(rad/s\) in a second plot. As with the root locus plots, it will be tempting to rely exclusively on control design software to obtain the Bode plots. The software should be treated as one tool in a tool kit that can be used to design and analyze control systems. It is essential to develop the capability to obtain approximate Bode plots manually. There is no substitute for a clear understanding of the underlying theory.
A Bode plot is obtained with the bode function shown in Figure 8.40. The Bode plot is automatically generated if the bode function is invoked without lefthand arguments. Otherwise, the magnitude and phase characteristics are placed in the workspace through the variables mag and phase. A Bode plot can then be obtained with the plot or semilogx function using mag, phase, and \(\omega\). The vector \(\omega\) contains the values of the frequency in \(rad/s\) at which the Bode plot will be calculated. If \(\omega\) is not specified, the bode function will automatically choose the frequency values by placing more points in regions where the frequency response is changing quickly. If the frequencies are specified explicitly, it is desirable to generate the vector \(\omega\) using the logspace function. The logspace function is shown in Figure 8.41. FIGURE 8.39
The Bode plot associated with Equation (8.66).
FIGURE 8.40
The bode function, given \(G(s)\).
The Bode plot in Figure 8.39 is generated using the script shown in Figure 8.42. The bode function automatically selected the frequency range. This range is user selectable using the logspace function. The bode function can be used with a state variable model, as shown in Figure 8.43. The use of the bode function is exactly the same as with transfer functions, except that the input is a state-space object instead of a transfer function object. FIGURE 8.41 The logspace function.
FIGURE 8.42 The script for the Bode plot in Figure 8.39.
FIGURE 8.43
The bode function with a state variable model.
\(\%\) Bode plot script for Figure 8.39
num \(= 5^{*}\lbrack 0.11\rbrack\)
\(f1 = \lbrack 10\rbrack;f2 = \lbrack 0.51\rbrack;f3 = \lbrack 1/2500.6/50\) 1];
den \(= conv(f1,conv(f2,f3))\);
sys=tf(num,den);
bode(sys)
FIGURE 8.44
(a) The relationship between \(\left( M_{p\omega},\omega_{r} \right)\) and \(\left( \zeta,\omega_{n} \right)\) for a second-order system. (b) m-file script.
(a)
Keep in mind that our goal is to design control systems that satisfy certain performance specifications given in the time domain. Thus, we must establish a connection between the frequency response and the transient time response of a system. The relationship between specifications given in the time domain to those given in the frequency domain depends upon approximation of the system by a second-order system with the poles being the system dominant roots.
Consider the second-order system shown in Figure 8.24. The Bode plot magnitude characteristic associated with the closed-loop transfer function in Equation (8.46) is shown in Figure 8.25. The relationship between the resonant frequency, \(\omega_{r}\), the maximum of the frequency response, \(M_{p\omega}\), and the damping ratio, \(\zeta\), and the natural frequency, \(\omega_{n}\), is shown in Figure 8.44 (and in Figure 8.11). The information in Figure 8.44 will be quite helpful in designing control systems in the frequency domain while satisfying time-domain specifications.
283. EXAMPLE 8.8 Engraving machine system
Engraving machines employ two drive motors and associated lead screws to position the engraving scribe in the desired direction [7]. The block diagram model for the position control system is shown in Figure 8.45. Our objective is to select \(K\) so that the closed-loop system has an acceptable time response to a step command. A functional block diagram describing the frequency-domain design process is shown in Figure 8.46. First, we choose \(K = 2\) and then iterate \(K\) if the performance is unacceptable. The script shown in Figure 8.47 is used in the design. The value of \(K\) is defined at the command level. Then the script is executed and the closed-loop Bode plot is generated. The values of \(M_{p\omega}\) and \(\omega_{r}\) are determined by inspection from the Bode plot. Those values are used in conjunction with Figure 8.44 to determine the corresponding values of \(\zeta\) and \(\omega_{n}\). FIGURE 8.45
Engraving machine block diagram model.
Establish relationship between frequency domain specs and time domain specs.
Determine \(\omega_{n}\) and \(\zeta\).\(\omega_{r}/\omega_{n} = \sqrt{1 - 2\zeta^{2}},\zeta < 0.707\)
FIGURE 8.46
Frequency design functional block diagram for the engraving machine. FIGURE 8.47
Script for the design of an engraving machine. engrave.m
sys=tf(num, den); \(w =\) logspace \(( - 1,1,400)\); [mag,phase,w]=bode(sys,w); \(\lbrack mp,l\rbrack = max(mag);wr = w(l)\);
Closed-loop transfer function. Closed-loop Bode plot. zeta \(= sqrt\left( {0.5}^{\star}\left( 1 - sqrt\left( 1 - 1/{mp}^{\land}2 \right) \right) \right)\); \(wn = wr/sqrt\left( 1 - 2^{\star}z^{2}a^{\land}2 \right)\);
Solving Equations (8.36) ts \(= 4/\) zeta/wn po $= 100^{*}exp\left( - \right.\ $ zeta \(\ ^{*}\) pi/sqrt $\left( 1 - \right.\ $ zeta^\(\left. \ \left. \ \ ^{\land}2 \right) \right)\)
\(> K = 2\); engrave
ts \(=\)
15.7962
39.4570
Check specifications and iterate, if necessary.
Given the damping ratio, \(\zeta\), and the natural frequency, \(\omega_{n}\), the settling time and percent overshoot can be estimated. If the time-domain specifications are not satisfied, then we adjust \(K\) and iterate.
The values for \(\zeta\) and \(\omega_{n}\) corresponding to \(K = 2\) are \(\zeta = 0.29\) and \(\omega_{n} = 0.88\). This leads to a prediction of P.O. \(= 37\%\) and \(T_{s} = 15.7\text{ }s\). The step response, shown in Figure 8.48, is a verification that the performance predictions are quite accurate and that the closed-loop system performs adequately.
In this example, the second-order system approximation is reasonable and leads to an acceptable design. However, the second-order approximation may not always lead directly to a good design. Fortunately, the control design software allows us to construct an interactive design facility to assist in the design process by reducing the manual computational loads while providing easy access to a host of classical and modern control tools.
283.1. SEQUENTIAL DESIGN EXAMPLE: DISK DRIVE READ SYSTEM
The disk drive uses a flexure suspension to hold the reader head mount. This flexure may be modeled by a spring and mass. In this chapter, we will include the effect of the flexure within the model of the motor-load system [22].
We model the flexure with the mounted head as a mass \(M\), a spring \(k\), and a sliding friction \(b\), as shown in Figure 8.49. Here, we assume that the force \(u(t)\) is exerted on the flexure by the arm. The transfer function of a spring-massdamper is
FIGURE 8.48
(a) Engraving machine step response for \(K = 2\). (b) m-file script.
FIGURE 8.49
Spring, mass, friction model of flexure and head.
(a)
\(y = step(sys,t);plot(t,y);\) grid
xlabel('Time (s)'), ylabel('y(t)')
(b)
A typical flexure and head has \(\zeta = 0.3\) and a natural resonance at \(f_{n} = 3000\text{ }Hz\). Therefore, \(\omega_{n} = 18.85 \times 10^{3}\) as shown in the model of the system (see Figure 8.50).
First, we sketch the magnitude characteristics for the open-loop Bode diagram. The Bode plot of the loop transfer function with \(K = 400\) is shown in Figure 8.51. Note that the actual plot has a \(10 - dB\) gain (over the asymptotic plot) at the resonance \(\omega = \omega_{n}\), as shown in the sketch. Note the resonance at \(\omega_{n}\). Clearly, we wish to avoid exciting this resonance.
Plots of the magnitude of the loop transfer function and the closed-loop transfer function are shown in Figure 8.52. The bandwidth of the closed-loop system
FIGURE 8.50 Disk drive head position control, including effect of flexure head mount.
FIGURE 8.51
Sketch of the Bode diagram magnitude for the system of Figure 8.50.
Frequency (rad/s)
is \(\omega_{B} = 2000rad/s\). We can estimate the settling time (with a \(2\%\) criterion) of this system where \(\zeta \simeq 0.8\) and \(\omega_{n} \simeq \omega_{B} = 2000rad/s\). Therefore, we expect \(T_{s} = 2.5\text{ }ms\) for the system of Figure 8.50. As long as \(K \leq 400\), the resonance is outside the bandwidth of the system.
283.2. SUMMARY
In this chapter, we considered the representation of a feedback control system by its frequency response characteristics. The frequency response of a system was defined as the steady-state response of the system to a sinusoidal input signal. Several alternative forms of frequency response plots were considered. They included the polar plot of the frequency response and logarithmic plots, often called Bode plots. The value of the logarithmic measure was also illustrated. The ease of obtaining a Bode
(a)
Frequency \((rad/s)\)
(b)
FIGURE 8.52 The magnitude Bode plot for (a) the loop transfer function and (b) the closed-loop system.
plot for the various factors of \(G(j\omega)\) was noted. The asymptotic approximation for sketching the Bode plot simplifies the computation considerably. A summary of fifteen typical Bode plots is shown in Table 8.2. Several performance specifications in the frequency domain were discussed; among them were the maximum magnitude \(M_{p\omega}\) and the resonant frequency \(\omega_{r}\). The relationship between the Bode plot and the system error constants $\left( K_{p} \right.\ $ and \(\left. \ K_{v} \right)\) was noted. Finally, the log-magnitude versus phase diagram was considered for graphically representing the frequency response of a system. Section 8.9 Summary
593
594 Chapter 8 Frequency Response Methods
284. SKILLS CHECK
In this section, we provide three sets of problems to test your knowledge: True or False, Multiple Choice, and Word Match. To obtain direct feedback, check your answers with the answer key provided at the conclusion of the end-of-chapter problems. Use the block diagram in Figure 8.53 as specified in the various problem statements.
FIGURE 8.53 Block diagram for the Skills Check.
In the following True or False and Multiple Choice problems, circle the correct answer.
- The frequency response represents the steady-state response of a stable system to a sinusoidal input signal at various frequencies.
True or False
- A plot of the real part of \(G(j\omega)\) versus the imaginary part of \(G(j\omega)\) is called a Bode plot.
True or False
- A transfer function is termed minimum phase if all its zeros lie in the right-hand s-plane.
True or False
- The resonant frequency and bandwidth can be related to the speed of the transient response.
True or False
-
One advantage of frequency response methods is the ready availability of sinusoidal test signals for various ranges of frequencies and amplitudes. True or False
-
Consider the stable system represented by the differential equation
Determine the phase of this system at the frequency \(\omega = 3rad/s\).
a. \(\phi = 0^{\circ}\)
b. \(\phi = - 45^{\circ}\)
c. \(\phi = - 60^{\circ}\)
d. \(\phi = - 180^{\circ}\)
In Problems 7 and 8, consider the feedback system in Figure 8.53 with the loop transfer function
-
The Bode plot of this system corresponds to which plot in Figure 8.54?
-
Determine the frequency at which the gain has unit magnitude and compute the phase angle at that frequency:
a. \(\omega = 1rad/s,\phi = - 82^{\circ}\)
b. \(\omega = 1.26rad/s,\phi = - 133^{\circ}\)
c. \(\omega = 1.26rad/s,\phi = 133^{\circ}\)
d. \(\omega = 4.2rad/s,\phi = - 160^{\circ}\)
(a)
(c)
(b)
(d)
FIGURE 8.54 Bode plot selections.
In Problems 9 and 10, consider the feedback system in Figure 8.53 with the loop transfer function
-
The break frequency on the Bode plot is
a. \(\omega = 1rad/s\)
b. \(\omega = 4.47rad/s\)
c. \(\omega = 8.94rad/s\)
d. \(\omega = 10rad/s\) -
The slope of the asymptotic plot at very low \((\omega \ll 1)\) and high \((\omega \gg 10)\) frequencies are, respectively:
a. At low frequency: slope \(= 20\text{ }dB/\) decade and at high frequency: slope \(= 20\text{ }dB/\) decade
b. At low frequency: slope \(= 0\text{ }dB/\) decade and at high frequency: slope \(= - 20\text{ }dB/\) decade
c. At low frequency: slope \(= 0\text{ }dB/\) decade and at high frequency: slope \(= - 40\text{ }dB/\) decade
d. At low frequency: slope \(= - 20\text{ }dB/\) decade and at high frequency: slope \(= - 20\text{ }dB/\) decade 11. Consider the Bode plot in Figure 8.55.
FIGURE 8.55 Bode plot for unknown system.
Which loop transfer function \(L(s) = G_{c}(s)G(s)\) corresponds to the Bode plot in Figure 8.55?
a. \(L(s) = G_{c}(s)G(s) = \frac{100}{s(s + 5)(s + 6)}\)
b. \(L(s) = G_{c}(s)G(s) = \frac{24}{s(s + 2)(s + 6)}\)
c. \(L(s) = G_{c}(s)G(s) = \frac{24}{s^{2}(s + 6)}\)
d. \(L(s) = G_{c}(s)G(s) = \frac{10}{s^{2} + 0.5s + 10}\)
-
Suppose that one design specification for a feedback control system requires that the percent overshoot to a step input be \(P.O. \leq 10\%\). The corresponding specification in the frequency domain is
a. \(M_{p\omega} \leq 0.55\)
b. \(M_{p\omega} \leq 0.59\)
c. \(M_{p\omega} \leq 1.05\)
d. \(M_{p\omega} \leq 1.27\) -
Consider the feedback control system in Figure 8.53 with loop transfer function
The resonant frequency, \(\omega_{r}\), and the bandwidth, \(\omega_{b}\), are:
a. \(\omega_{r} = 1.59rad/s,\omega_{b} = 1.86rad/s\)
b. \(\omega_{r} = 3.26rad/s,\omega_{b} = 16.64rad/s\)
c. \(\omega_{r} = 12.52rad/s,\omega_{b} = 3.25rad/s\)
d. \(\omega_{r} = 5.51rad/s,\omega_{b} = 11.6rad/s\)
For Problems 14 and 15, consider the frequency response of a process \(G(j\omega)\) depicted in Figure 8.56.
FIGURE 8.56 Bode plot for \(G(j\omega)\).
-
Determine the system type (that is, the number of integrators, \(N\) ):
a. \(N = 0\)
b. \(N = 1\)
c. \(N = 2\)
d. \(N > 2\) -
The transfer function corresponding to the Bode plot in Figure 8.56 is:
a. \(G(s) = \frac{100(s + 10)(s + 5000)}{s(s + 5)(s + 6)}\)
b. \(G(s) = \frac{100}{(s + 1)(s + 20)}\)
c. \(G(s) = \frac{100}{(s + 1)(s + 50)(s + 200)}\)
d. \(G(s) = \frac{100(s + 20)(s + 5000)}{(s + 1)(s + 50)(s + 200)}\) In the following Word Match problems, match the term with the definition by writing the correct letter in the space provided.
a. Laplace transform pair
b. Decibel (dB)
c. Fourier transform
d. Bode plot
e. Transfer function in the frequency domain
f. Decade
g. Dominant roots
h. All-pass network
i. Logarithmic magnitude
j. Natural frequency
k. Fourier transform pair
l. Minimum phase
m. Bandwidth
n. Frequency response
o. Resonant frequency
p. Break frequency
q. Polar plot
r. Maximum value of the frequency response
s. Nonminimum phase
The logarithm of the magnitude of the transfer function and the phase are plotted versus the logarithm of \(\omega\), the frequency.
The logarithm of the magnitude of the transfer function, \(20\log_{10}|G(j\omega)|\).
A plot of the real part of \(G(j\omega)\) versus the imaginary part of \(G(j\omega)\).
The steady-state response of a system to a sinusoidal input signal.
All the zeros of a transfer function lie in the left-hand side of the \(s\)-plane.
The frequency at which the frequency response has declined \(3\text{ }dB\) from its low-frequency value.
The frequency at which the maximum value of the frequency response of a complex pair of poles is attained.
The frequency of natural oscillation that would occur for two complex poles if the damping were equal to zero.
Transfer functions with zeros in the right-hand \(s\)-plane.
The frequency at which the asymptotic approximation of the frequency response for a pole (or zero) changes slope.
The transformation of a function of time into the frequency domain.
The ratio of the output to the input signal where the input is a sinusoid.
The units of the logarithmic gain.
A pair of complex poles will result in a maximum value for the frequency response occurring at the resonant frequency.
A nonminimum phase system that passes all frequencies with equal gain.
A factor of ten in frequency.
The roots of the characteristic equation that represent or dominate the closed-loop transient response.
A pair of functions, one in the time domain, and the other in the frequency domain, and both related by the Fourier transform.
A pair of functions, one in the time domain, and the other in the frequency domain, and both related by the Laplace transform.
285. EXERCISES
E8.1 Increased track densities for computer disk drives necessitate careful design of the head positioning control [1]. The loop transfer function is
Plot the frequency response for this system when \(K = 10\). Calculate the phase and magnitude at \(\omega = 0,0.5,1,2,4\), and \(\infty\).
Answer: \(|L(j0.5)| = 8\) and \(/L(j0.5) = - {53.13}^{\circ}\)
E8.2 A tendon-operated robotic hand can be implemented using a pneumatic actuator [8]. The actuator can be represented by
Plot the frequency response of \(G(j\omega)\). Show that the magnitude of \(G(j\omega)\) is \(- 17\text{ }dB\) at \(\omega = 10\) and \(- 27.1\text{ }dB\) at \(\omega = 200\). Show also that the phase is \(- {138.7}^{\circ}\) at \(\omega = 700\).
E8.3 A robotic arm has a joint-control loop transfer function
Show that the frequency equals \(\omega = 11.7rad/s\) when the phase angle of \(L(j\omega)\) is \(- 135^{\circ}\). Find the magnitude of \(L(j\omega)\) at \(\omega = 11.7rad/s\).
Answer: \(|L(j11.7)| = - 5.1\text{ }dB\)
E8.4 The frequency response for the system
is shown in Figure E8.4. Estimate \(K\) and \(a\) by examining the frequency response curves.
E8.5 The magnitude plot of a transfer function
is shown in Figure E8.5. Estimate \(K^{'},a,b\), and \(c\) from the plot.
Answer: \(K = 6,a = 1/3,b = 1/12,c = 1/6\)
E8.6 Several studies have proposed an extravehicular robot that could move around in a NASA space station and perform physical tasks at various worksites [9]. The arm is controlled by a unity feedback control with loop transfer function
Sketch the Bode plot for \(K = 30\), and determine the frequency when \(20log|L(j\omega)|\) is \(0\text{ }dB\).
E8.7 Consider a system with a closed-loop transfer function
This system will have a steady-state error for a step input. (a) Plot the frequency response, noting the two peaks in the magnitude response. (b) Predict the time response to a step input, noting that the system has four poles and cannot be represented as a dominant second-order system. (c) Plot the step response.
E8.8 Two feedback systems with their respective loop transfer function are represented as:
(i) \(T(s) = \frac{100(s - 1)}{\left( s^{2} + 25s + 100 \right)}\)
FIGURE E8.4
Bode plot.
FIGURE E8.5
Bode plot.
FIGURE E8.9
Bode plot.
Frequency \((rad/s)\)
(ii) \(T_{1}(s) = \frac{100(s + 1)}{\left( s^{2} + 25s + 100 \right)}\)
For each transfer function, (a) determine the break frequencies for the Bode plot. (b) Determine the slope of the asymptotic plot at very low frequencies and at high frequencies. (c) Sketch the Bode magnitude plot and compare them.
E8.9 The Bode plot of a system is shown in Figure E8.9. Estimate the transfer function \(G(s)\).
E8.10 The dynamic analyzer shown in Figure E8.10(a) can be used to display the frequency response of a system. Figure E8.10(b) shows the actual frequency response of a system. Estimate the poles and zeros of the device. Note \(X = 1.37kHz\) at the first cursor, and \(\Delta X = 1.257kHz\) to the second cursor.
E8.11 Consider the feedback control system in Figure E8.11. Sketch the Bode plot of \(G(s)\), and determine the crossover frequency, that is, the frequency when \(20\log_{10}|G(j\omega)| = 0\text{ }dB\).
E8.12 Consider the system represented in state variable form
(a) Determine the transfer function representation of the system. (b) Sketch the Bode plot.
E8.13 Determine the bandwidth of the feedback control system in Figure E8.13.
E8.14 Consider the nonunity feedback system in Figure E8.14, where the controller gain is \(K = 2\). Sketch the Bode plot of the loop transfer function. Determine the phase of the loop transfer function when the magnitude \(20log|L(j\omega)| = 0\text{ }dB\). Recall that the loop transfer function is \(L(s) = G_{c}(s)G(s)H(s)\).
(a)
(b)
FIGURE E8.10 (a) Photo showing a typical signal analyzer. (b) Frequency response. (Courtesy of the Syafiq Adnan/Shutterstock.)
FIGURE E8.11
Unity feedback system.
FIGURE E8.13
Third-order
feedback system.
FIGURE E8.14
Nonunity feedback system with controller gain \(K\).
E8.15 Consider the single-input, single-output system described by
where
Compute the bandwidth of the system for \(K = 1,2\), and 10 . As \(K\) increases, does the bandwidth increase or decrease?
286. PROBLEMS
P8.1 Sketch the polar plot for the following loop transfer functions:
(a) \(L(s) = G_{c}(s)G(s) = \frac{1}{(1 + 0.5s)(1 + 2s)}\)
(b) \(L(s) = G_{c}(s)G(s) = \frac{3\left( s^{2} + 1.5s + 1 \right)}{(s - 2)^{2}}\)
(c) \(L(s) = G_{c}(s)G(s) = \frac{s - 6}{s^{2} + 5s + 6}\)
(d) \(L(s) = G_{c}(s)G(s) = \frac{10(s + 6)}{s(s + 1)(s + 3)}\)
P8.2 Sketch the Bode plot representation of the frequency response for the transfer functions given in Problem P8.1.
P8.3 A rejection network is the bridged-T network shown in Figure P8.3. The transfer function of this network is
FIGURE P8.3 Bridged-T network.
where \(\omega_{n}^{2} = 2/LC,Q = \omega_{n}L/R_{1}\), and \(R_{2}\) is adjusted so that \(R_{2} = \left( \omega_{n}L \right)^{2}/4R_{1}\) [3]. (a) Determine the poles and zeros. (b) Sketch the Bode plot. FIGURE P8.4
(a) Pressure controller. (b) Block diagram model.
P8.4 A control system for controlling the pressure in a closed chamber is shown in Figure P8.4. Sketch the Bode plot of the loop transfer function.
P8.5 The global robot industry is growing rapidly [8]. A typical industrial robot has multiple degrees of freedom. A unity feedback position control system for a force-sensing joint has a loop transfer function.
\(G_{c}(s)G(s) = \frac{K}{(1 + s/4)(1 + s/8)(1 + s/16)(1 + s/32)}\),
where \(K = 20\). Sketch the Bode plot of this system.
P8.6 The asymptotic log-magnitude curves for two loop transfer functions are given in Figure P8.6. Sketch the corresponding asymptotic phase shift curves for each system. Estimate the transfer function for each system. Assume that the systems have minimum phase transfer functions.
P8.7 Driverless vehicles can be used in warehouses, airports, and many other applications. These vehicles follow a wire embedded in the floor and adjust the steerable front wheels in order to maintain proper direction, as shown in Figure P8.7(a) [10]. The sensing coils, mounted on the front wheel assembly, detect an error in the direction of travel and adjust the steering. The overall control system is shown in Figure P8.7(b). The loop transfer function is
(a)
(b)
FIGURE P8.6 Log-magnitude curves.
(a) Set \(K_{v} = \pi\) and sketch the Bode plot. (b) Using the Bode plot, determine the phase at the crossover frequency. FIGURE P8.7
Steerable wheel control.
FIGURE P8.8
Second-order unity feedback system.
Energized guidepath wire
(a)
(b)
P8.8 A feedback control system is shown in Figure P8.8. The specification for the closed-loop system requires that the percent overshoot to a step input be P.O. \(\leq 10\%\). (a) Determine the corresponding specification \(M_{p\omega}\) in the frequency domain for the closed-loop transfer function. (b) Determine the resonant frequency \(\omega_{r}\). (c) Determine the bandwidth of the closed-loop system \(\omega_{n}\).
P8.9 Sketch the logarithmic-magnitude versus phase angle curve for the transfer functions (a) and (b) of Problem P8.1.
FIGURE P8.10
Linear actuator control.
P8.10 A linear actuator is used in the system shown in Figure P8.10 to position a mass \(M\). The actual position of the mass is measured by a slide wire resistor, and thus \(H(s) = 1.0\). The amplifier gain is selected so that the steady-state error of the system is less than \(1\%\) of the magnitude of the position reference \(R(s)\). The actuator has a field coil with a resistance \(R_{f} = 0.1\Omega\) and \(L_{f} = 0.2H\). The mass of the load is \(M = 0.1\text{ }kg\), and the friction is \(b = 0.2\text{ }N\text{ }s/m\). The
FIGURE P8.11
Frequency response of ship control system.
spring constant is \(k = 0.4\text{ }N/m\). (a) Determine the gain \(K\) necessary to maintain a steady-state error for a step input less than 1%. (b) Sketch the Bode plot of the loop transfer function. (c) Sketch the Bode plot for the closed-loop transfer function. Determine \(M_{p\omega},\omega_{r}\), and the bandwidth.
P8.11 Automatic steering of a ship is a particularly useful application of feedback control theory [20]. In the case of heavily traveled seas, it is important to maintain the motion of the ship along an accurate track. An automatic system is more likely to maintain a smaller error from the desired heading than a helmsman who recorrects at infrequent intervals. A mathematical model of the steering system has been developed for a ship moving at a constant velocity and for small deviations from the desired track. For a large tanker, the transfer function of the ship is
where \(E(s)\) is the Laplace transform of the deviation of the ship from the desired heading and \(\delta(s)\) is the Laplace transform of the angle of deflection of the steering rudder. Verify that the Bode plot of \(G(j\omega)\) is that shown in Figure P8.11.
P8.12 The block diagram of a feedback control system is shown in Figure P8.12(a). The transfer functions of the blocks are represented by the frequency response curves shown in Figure P8.12(b). (a) When \(G_{3}\) is disconnected from the system, determine the damping ratio \(\zeta\) of the system. (b) Connect \(G_{3}\) and determine the damping ratio \(\zeta\). Assume that the systems have minimum phase transfer functions.
P8.13 A position control system may be constructed by using an AC motor and AC components, as shown in Figure P8.13. The syncro and control transformer may be considered to be a transformer with a rotating winding. The syncro position detector rotor turns with the load through an angle \(\theta_{0}\). The syncro motor is energized with an AC reference voltage, for example, 115 volts, \(60\text{ }Hz\). The input signal or command is \(R(s) = \theta_{\text{in}\text{~}}(s)\) and is applied by turning the rotor of the control transformer. The AC two-phase motor operates as a result of the amplified error signal. The advantages of an AC control system are (1) freedom from DC drift effects and (2) the simplicity and accuracy of AC components. To measure the open-loop frequency response, we simply disconnect \(X\) from \(Y\) and \(X^{'}\) from \(Y^{'}\) and then apply a sinusoidal modulation signal generator to the \(Y - Y^{'}\) terminals and measure the response at \(X - X^{'}\). (The error \(\left( \theta_{0} - \theta_{i} \right)\) will be adjusted to zero before applying the AC generator.) The resulting frequency response of the loop transfer function \(L(j\omega)\) is shown in Figure P8.13(b). Determine the loop transfer function. Assume that the system has a minimum phase transfer function.
P8.14 A bandpass amplifier may be represented by the circuit model shown in Figure P8.14 [3]. When \(R_{1} = R_{2} = 10k,C_{1} = 1\mu F,C_{2} = 10\mu F\), and \(K = 100\), show that
(a) Sketch the Bode plot of \(G(j\omega)\). (b) Find the mid band gain (in \(dB\) ). (c) Find the high and low frequency \(- 3\text{ }dB\) points.
(a)
FIGURE P8.12
Feedback system.
P8.15 To determine the transfer function of a process \(G(s)\), the frequency response may be measured using a sinusoidal input. One system yields the data in the following table:
| \(\omega\), rad/s | $$ | G(j\omega) |
| 0.1 | 50 | -90 |
| 1 | 5.02 | -92.4 |
| 2 | 2.57 | -96.4 |
| 4 | 1.36 | -100 |
| 5 | 1.17 | -104 |
| 6.3 | 1.03 | -110 |
| 8 | 0.97 | -120 |
| 10 | 0.97 | -143 |
| 12.5 | 0.74 | -169 |
| 20 | 0.13 | -145 |
| 31 | 0.026 | -158 |
Determine the transfer function \(G(s)\).
P8.16 A space shuttle was used to repair satellites. Figure P8.16 illustrates how a crew member, with her feet strapped to the platform on the end of the shuttle's robotic arm, used her arms to stop the satellite's spin. The control system of the robotic arm has a closed-loop transfer function
(a) Determine the response \(y(t)\) to a unit step input, \(R(s) = 1/s\). (b) Determine the bandwidth of the system.
P8.17 The experimental Oblique Wing Aircraft (OWA) has a wing that pivots, as shown in Figure P8.17. The wing is in the normal unskewed position for low speeds and can move to a skewed position for improved supersonic flight [11]. The aircraft control system loop transfer function is
FIGURE P8.13
(a) AC motor control. (b) Bode plot of the loop transfer function.
(a)
(b)
rich sensory information acquired by the robot to the operator with a sensation of presence. This concept is called tele-existence or telepresence [9].
The tele-existence system consists of a system with a visual and auditory sensation of presence, a computer control system, and an anthropomorphic robot mechanism with an arm having seven degrees of freedom and a locomotion mechanism. The operator's head movement, right arm movement, right hand movement, and other auxiliary motion are measured. A specially designed stereo visual and auditory input system mounted on the neck mechanism of the robot gathers visual and auditory information from the remote environment. These pieces of information are fed back and are applied to the specially designed stereo display system to evoke the sensation of presence of the operator. The locomotion control system has the loop transfer function (a) Sketch the Bode plot. (b) Find the crossover frequency. Find the frequency when the phase is \(\phi(\omega) = - 180^{\circ}\).
P8.18 Remote operation plays an important role in hostile environments. Research engineers have been trying to improve teleoperations by feeding back
FIGURE P8.16 Satellite repair.
Obtain the Bode plot for the loop transfer function, and determine the crossover frequency.
FIGURE P8.17 The Oblique Wing Aircraft, top and side views.
P8.19 A DC motor controller used extensively in automobiles is shown in Figure P8.19(a). The measured plot of \(\Theta(s)/I(s)\) is shown in Figure P8.19(b). Determine the transfer function of \(\Theta(s)/I(s)\).
P8.20 For the successful development of space projects, robotics and automation will be a key technology. Autonomous and dexterous space robots can reduce
(a)
(b)
FIGURE P8.19
(a) Motor controller.
(b) Bode plot.
FIGURE P8.20 A space robot with three arms, shown capturing a satellite.
the workload of astronauts and increase operational efficiency in many missions. Figure P8.20 shows a concept called a free-flying robot \(\lbrack 9,13\rbrack\). A major characteristic of space robots, which clearly distinguishes them from robots operated on earth, is the lack of a fixed base. Any motion of the manipulator arm will induce reaction forces and moments in the base, which disturb its position and attitude.
The control of one of the joints of the robot can be represented by the loop transfer function
(a) Sketch the Bode plot of \(L(j\omega)\). (b) Determine the maximum value of \(L(j\omega)\), the frequency at which it occurs, and the phase at that frequency.
P8.21 Low-altitude wind shear is a major cause of air carrier accidents in the United States. Most of these accidents have been caused by either microbursts (small-scale, low-altitude, intense thunderstorm downdrafts that impact the surface and cause strong divergent outflows of wind) or by the gust front at the leading edge of expanding thunderstorm outflows. A microburst encounter is a serious problem for either landing or departing aircraft, because the aircraft is at low altitudes and is traveling at just over \(25\%\) above its stall speed [12].
The design of the control of an aircraft encountering wind shear after takeoff may be treated as a problem of stabilizing the climb rate about a desired value of the climb rate. The resulting controller uses only climb rate information.
The standard negative unity feedback system of Figure 8.24 has a loop transfer function
Note the negative gain in loop transfer function. This system represents the control system for the climb rate. Sketch the Bode plot and determine gain (in \(dB\) ) when the phase is \(\phi(\omega) = - 180^{\circ}\).
P8.22 The frequency response of a process \(G(j\omega)\) is shown in Figure P8.22. Determine \(G(s)\).
P8.23 The frequency response of a process \(G(j\omega)\) is shown in Figure P8.23. Deduce the type number (number of integrations) for the system. Determine the transfer function of the system, \(G(s)\). Calculate the error to a unit step input.
P8.24 The Bode plot of a closed-loop film transport system is shown in Figure P8.24 [17]. Assume that the system transfer function \(T(s)\) has two dominant complex
FIGURE P8.22 Bode plot of \(G(s)\).
FIGURE P8.23 Frequency response of \(G(j\omega)\).
FIGURE P8.24
Bode plot of a closed-film transport system.
conjugate poles. (a) Determine the best secondorder model for the system. (b) Determine the system bandwidth. (c) Predict the percent overshoot and settling time (with a \(2\%\) criterion) for a step input.
P8.25 A unity feedback closed-loop system has a steady-state error equal to \(A/10\), where the input is \(r(t) = At^{2}/2\). The Bode plot is shown in Figure P8.25 for \(G(j\omega)\). Determine the transfer function \(G(s)\).
P8.26 Determine the transfer function of the opamp circuit shown in Figure P8.26. Assume an ideal op-amp. Plot the frequency response when \(R = 10k\Omega,R_{1} = 9k\Omega,R_{2} = 1k\Omega\), and \(C = 1\mu F\).
P8.27 A unity feedback system has the loop transfer function
Sketch the Bode plot of the loop transfer function, and indicate how the magnitude \(20log|L(j\omega)|\) plot varies as \(K\) varies. Develop a table for \(K = 10,20\), and 30 , and for each \(K\) determine the crossover frequency ( \(\omega_{c}\) for \(20log|L(j\omega)| = 0\text{ }dB\) ), the magnitude at low frequency \((20log|L(j\omega)|\) for \(\omega \ll 1)\), and for the closed-loop system determine the bandwidth for each \(K\).
FIGURE P8.25 Bode plot of a unity feedback system.
FIGURE P8.26
An op-amp circuit.
287. ADVANCED PROBLEMS
AP8.1 A high pass amplifier may be represented by the circuit model shown in Figure AP8.1. When \(R_{1} = 100\Omega,R_{2} = 100k\Omega,C_{1} = 100\mu F,C_{2} = 10\mu F\), and \(K = 1000\), show that
FIGURE AP8.1
A high pass amplifier
(a) Sketch the Bode plot of \(G(j\omega)\). (b) Find the pass band gain (in \(dB\) ). (c) Find the low frequency \(- 3\text{ }dB\) points.
FIGURE AP8.2 System with parameter \(b\) and controller gain \(K\).
AP8.2 A system is shown in Figure AP8.2. The nominal value of the parameter \(b\) is 4.0 and \(K = 5.0\). Determine the sensitivity \(S_{b}^{T}\), and plot \(20log\left| S_{b}^{T}(j\omega) \right|\).
AP8.3 As an automobile moves along the road, the vertical displacements at the tires act as the motion excitation to the automobile suspension system [16].
FIGURE AP8.3 Auto suspension system model.
Figure AP8.3 is a schematic diagram of a simplified automobile suspension system, for which we assume the input is sinusoidal. Determine the transfer function \(X(s)/R(s)\), and sketch the Bode plot when \(M = 1\text{ }kg,b = 4\text{ }N\text{ }s/m\), and \(k = 18\text{ }N/m\).
AP8.4 A helicopter with a load on the end of a cable is shown in Figure AP8.4(a). The position control system is shown in Figure AP8.4(b), where the visual feedback is represented by \(H(s)\). Sketch the Bode plot of the loop transfer function. Determine the crossover frequency, that is, where \(20\log_{10}|H(j\omega)G(j\omega)| = 0\text{ }dB\).
AP8.5 A closed-loop system with unity feedback has a transfer function
(a) Determine the loop transfer function. (b) Plot the logarithmic-magnitude versus phase curve, and identify the frequency points for \(\omega\) equal to \(1,10,50,110\), and 500 . (c) Is the open-loop system stable? Is the closed-loop system stable?
AP8.6 Consider the spring-mass system depicted in Figure AP8.6. Develop a transfer function model
FIGURE AP8.4
A helicopter feedback control system.
(a)
FIGURE AP8.6 Suspended spring-mass system with parameters \(k\) and \(b\).
to describe the motion of the mass \(M = 2\text{ }kg\), when the input is \(u(t)\) and the output is \(x(t)\). Assume that the initial conditions are \(x(0) = 0\) and \(x(0) = 0\). Determine values of \(k\) and \(b\) such that the maximum
FIGURE AP8.7 Op-amp lead compensator circuit.
steady-state response of the system to a sinusoidal input \(u(t) = sin(\omega t)\) is less than 1 for all \(\omega\). For the values you selected for \(k\) and \(b\), what is the frequency at which the peak response occurs?
AP8.7 An op-amp circuit is shown in Figure AP8.7. The circuit represents a lead compensator.
(a) Determine the transfer function of this circuit.
(b) Sketch the Bode plot of the circuit when \(R_{1} =\) \(10k\Omega,R_{2} = 10\Omega,C_{1} = 0.1\mu F\), and \(C_{2} = 1mF\).
288. DESIGN PROBLEMS
CDP8.1 In this chapter, we wish to use a PD controller such that
The tachometer is not used (see Figure CDP4.1). Obtain the Bode plot for the system when \(M_{pw}\) Determine the step response of this system and estimate the overshoot and settling time (with a \(2\%\) criterion).
DP8.1 Understanding the behavior of a human steering an automobile remains an interesting subject \(\lbrack 14,15\), 16, 21]. The design and development of systems for four-wheel steering, active suspensions, active, independent braking, and "drive-by-wire" steering provide the engineer with considerably more freedom in altering vehicle-handling qualities than existed in the past.
The vehicle and the driver are represented by the model in Figure DP8.1, where the driver develops anticipation of the vehicle deviation from the center line. For \(K = 1\), obtain the Bode plot of (a) the loop transfer function \(L(s) = G_{c}(s)G(s)\) and (b) the closed-loop transfer function \(T(s)\). (c) Repeat parts (a) and (b) when \(K = 50\). (d) A driver can select the gain \(K\). Determine the appropriate gain so that \(M_{p\omega} \leq 2\), and the bandwidth is the maximum attainable for the closed-loop system. (e) Determine the steady-state error of the system for a ramp input \(r(t) = t\).
DP8.2 The unmanned exploration of planets requires a high level of autonomy because of the communication delays between robots in space and their Earthbased stations. This affects all the components of the system: planning, sensing, and mechanism. In particular, such a level of autonomy can be achieved only if each robot has a perception system that can reliably build and maintain models of the environment. The perception system is a major part of the development
FIGURE DP8.1
Human steering control system.
(a)
FIGURE DP8.2
(a) The Mars-bound Spider-bot. (Photo courtesy of NASA.) (b) Block diagram of the control system for one leg.
(b) of a complete system that includes planning and mechanism design. The target vehicle is the Spiderbot, a four-legged walking robot shown in Figure DP8.2(a), being developed at NASA Jet Propulsion Laboratory [18]. The control system of one leg is shown in Figure DP8.2(b).
(a) Sketch the Bode plot for the loop transfer function when \(K = 20\). Determine (1) the frequency when the phase is \(\phi(\omega) = - 180^{\circ}\) and (2) the crossover frequency. (b) Obtain the Bode plot for the closed-loop transfer function \(T(s)\) when \(K = 20\). (c) Determine \(M_{p\omega},\omega_{r}\), and \(\omega_{B}\) for the closed-loop system when \(K = 22\) and \(K = 25\). (d) Select the best gain of the two specified in part (c) when it is desired that the percent overshoot of the system to a step input \(r(t)\) be P.O. \(\leq 5\%\) and the settling time be as short as possible.
DP8.3 A table is used to position vials under a dispenser head, as shown in Figure DP8.3(a). The objective is speed, accuracy, and smooth motion in order to eliminate spilling. The position control system is shown in Figure DP8.3(b). Determine a \(K\) such that the bandwidth is maximized while keeping P.O. \(\leq 20\%\) to a unit step input. What is the maximum bandwidth, \(\omega_{b}\), when \(P.O\). \(\leq 20\) ? Determine the range of \(K\) such that the closed-loop system is stable.
DP8.4 Anesthesia can be administered automatically by a control system. To ensure adequate operating conditions for the surgeon, muscle relaxant drugs, which block involuntary muscle movements, are administered.
A conventional method used by anesthesiologists for muscle relaxant administration is to inject a bolus dose whose size is determined by experience and to inject supplements as required. Significant improvements may be achieved by introducing the concept of automatic control, which results in a considerable reduction in the total relaxant drug consumed [19].
A model of the anesthesia process is shown in Figure DP8.4. Select a gain \(K\) so that the bandwidth of the closed-loop system is maximized while \(M_{p\omega} \leq 1.5\). Determine the bandwidth attained for your design.
DP8.5 Consider the control system depicted in Figure DP8.5(a) where the plant is a "black box" for which little is known in the way of mathematical models. The only information available on the plant is the frequency response shown in Figure DP8.5(b). Design a controller \(G_{c}(s)\) to meet the following specifications: (i) the crossover frequency is between \(10rad/s\) and \(50rad/s\); (ii) the magnitude of the loop transfer function is greater than \(20\text{ }dB\) for \(\omega < 0.1rad/s\).
DP8.6 A single-input, single-output system is described by
FIGURE DP8.3
Automatic table and dispenser.
(a)
(b)
FIGURE DP8.4 Model of an anesthesia control system.
DP8.7 Consider the system of Figure DP8.7. Consider the controller to be similar to a proportional plus derivative \((PD)\) given by
Design the PD controller gains to achieve (a) a velocity constant \(Kv \geq 1\), (b) a phase margin of P.M. \(\geq 60^{\circ}\), and (c) a bandwidth \(\omega_{b} \geq 2.0\). Plot the response of the closed-loop system to a unit step input.
(a)
FIGURE DP8.5
(a) Feedback system with
"black box" plant.
(b) Frequency response plot of the "black box" represented by \(G(s)\).
(b)
FIGURE DP8.7
Closed-loop
feedback system.
289. COMPUTER PROBLEMS
CP8.1 Consider the closed-loop transfer function
Develop an m-file to obtain the Bode plot, and verify that the resonant frequency is \(5rad/s\) and that the peak magnitude \(M_{p\omega}\) is \(14\text{ }dB\).
CP8.2 For the following transfer functions, sketch the Bode plots, then verify with the bode function:
(a) \(G(s) = \frac{1000}{(s + 10)(s + 100)}\) (b) \(G(s) = \frac{s + 100}{(s + 2)(s + 25)}\)
(c) \(G(s) = \frac{100}{s^{2} + 2s + 50}\)
(d) \(G(s) = \frac{s - 6}{(s + 3)\left( s^{2} + 12s + 50 \right)}\)
CP8.3 For each of the following transfer functions, sketch the Bode plot and determine the crossover frequency:
(a) \(G(s) = \frac{2500}{(s + 10)(s + 100)}\) (b) \(G(s) = \frac{50}{(s + 1)\left( s^{2} + 10s + 2 \right)}\)
(c) \(G(s) = \frac{30(s + 100)}{(s + 1)(s + 30)}\)
(d) \(G(s) = \frac{100\left( s^{2} + 14s + 50 \right)}{(s + 1)(s + 2)(s + 200)}\)
CP8.4 A unity negative feedback system has the loop transfer function
Determine the closed-loop system bandwidth. Using the bode function, obtain the Bode plot and label the plot with the bandwidth.
CP8.5 A block plot of a second-order system is shown in Figure CP8.5.
(a) Determine the resonant peak \(M_{p\omega}\) the resonant frequency \(\omega_{r}\), and the bandwidth \(\omega_{B}\), of the system from the closed-loop Bode plot. Generate the Bode plot with an m-file for \(\omega = 0.1\) to \(\omega = 1,000\) \(rad/s\) using the logspace function. (b) Estimate the system damping ratio, \(\zeta\), and natural frequency \(\omega\). (c) From the closed-loop transfer function, compute the actual \(\zeta\) and \(\omega_{n}\) and compare with your results in part (b).
FIGURE CP8.5 A second-order feedback control system.
CP8.6 Consider the feedback system in Figure CP8.6. Obtain the Bode plots of the loop transfer function and the closed-loop transfer function using an \(m\)-file.
FIGURE CP8.6 Closed-loop feedback system.
CP8.7 A unity feedback system has the loop transfer function
Generate a plot of the bandwidth versus the parameter \(p\) as \(0.1 < p < 5\).
CP8.8 Consider the problem of controlling an inverted pendulum on a moving base, as shown in Figure CP8.8(a). The transfer function of the system is
The design objective is to balance the pendulum (i.e., \(\theta(t) \approx 0\) ) in the presence of disturbance inputs. A block diagram representation of the system is depicted in Figure CP8.8(b). Let \(M_{s} = 10\text{ }kg,M_{b} =\) \(100\text{ }kg,L = 1\text{ }m,g = 9.81\text{ }m/s^{2},a = 5\), and \(b = 10\). The design specifications, based on a unit step disturbance, are as follows:
-
settling time (with a \(2\%\) criterion) of \(T_{s} \leq 10\text{ }s\),
-
percent overshoot of P.O. \(\leq 40\%\), and
-
steady-state tracking error less than \({0.1}^{\circ}\) in the presence of the disturbance.
Develop a set of interactive \(m\)-file scripts to aid in the control system design. The first script should accomplish at least the following:
-
Compute the closed-loop transfer function from the disturbance to the output with \(K\) as an adjustable parameter.
-
Draw the Bode plot of the closed-loop system.
-
Automatically compute and output \(M_{p\omega}\) and \(\omega_{r}\).
As an intermediate step, use \(M_{p\omega}\) and \(\omega_{r}\) and Equations (8.36) and (8.37) in Section 8.2 to estimate \(\zeta\) and \(\omega_{n}\). The second script should at least estimate the settling time and percent overshoot using \(\zeta\) and \(\omega_{n}\) as input variables.
If the performance specifications are not satisfied, change \(K\) and iterate on the design using the first two scripts. After completion of the first two steps, the final step is to test the design by simulation. The functions of the third script are as follows:
-
plot the response, \(\theta(t)\), to a unit step disturbance with \(K\) as an adjustable parameter, and
-
label the plot appropriately. FIGURE CP8.8
(a) An inverted pendulum on a moving base.
(b) A block diagram representation.
(a)
(b)
Utilizing the interactive scripts, design the controller to meet the specifications using frequency response Bode methods. To start the design process, use analytic methods to compute the minimum value of \(K\) to meet the steady-state tracking error specification. Use the minimum \(K\) as the first guess in the design iteration.
CP8.9 Design a filter, \(G(s)\), with the following frequency response:
-
For \(\omega < 1rad/s\), the magnitude \(20\log_{10}|G(j\omega)|\) \(< 0\text{ }dB\)
-
For \(1 < \omega < 1000rad/s\), the magnitude \(20\log_{10}\) \(|G(j\omega)| \geq 0\text{ }dB\)
-
For \(\omega > 1000rad/s\), the magnitude \(20\log_{10}\) \(|G(j\omega)| < 0\text{ }dB\)
Try to maximize the peak magnitude as close to \(\omega = 40rad/s\) as possible.
290. ANSWERS TO SKILLS CHECK
True or False: (1) True; (2) False; (3) False; (4) True; (5) True
Multiple Choice: (6) b; (7) a; (8) b; (9) b; (10) c; (11) b; (12) c; (13) d; (14) a; (15) d
Word Match (in order, top to bottom): d, i, q, n, l, m, o, j, s, p, c, e, b, r, h, f, g, k, a
291. TERMS AND CONCEPTS
All-pass network A nonminimum phase system that passes all frequencies with equal gain.
Bandwidth The frequency at which the frequency response has declined \(3\text{ }dB\) from its low-frequency value.
Bode plot The logarithm of the magnitude of the transfer function is plotted versus the logarithm of \(\omega\), the frequency. The phase \(\phi\) of the transfer function is separately plotted versus the logarithm of the frequency.
Break frequency The frequency at which the asymptotic approximation of the frequency response for a pole (or zero) changes slope.
Corner frequency See Break frequency. Decade A factor of 10 in frequency (e.g., the range of frequencies from \(1rad/s\) to \(10rad/s\) is one decade).
Decibel (dB) The units of the logarithmic gain.
Dominant roots The roots of the characteristic equation that represent or dominate the closed-loop transient response.
Fourier transform The transformation of a function of time \(f(t)\) into the frequency domain.
Fourier transform pair A pair of functions, one in the time domain, denoted by \(f(t)\), and the other in the frequency domain, denoted by \(F(\omega)\), related by the Fourier transform as \(F(\omega) = \mathcal{F}\{ f(t)\}\), where \(\mathcal{F}\) denotes the Fourier transform.
Frequency response The steady-state response of a system to a sinusoidal input signal.
Laplace transform pair A pair of functions, one in the time domain, denoted by \(f(t)\), and the other in the frequency domain, denoted by \(F(\text{ }s)\), related by the Laplace transform as \(F(s) = \mathcal{L}\{ f(t)\}\), where \(\mathcal{L}\) denotes the Laplace transform.
Logarithmic magnitude The logarithm of the magnitude of the transfer function, usually expressed in units of \(20\text{ }dB\), thus \(20\log_{10}|G|\).
Logarithmic plot See Bode plot.
Maximum value of the frequency response \(A\) pair of complex poles will result in a maximum value for the frequency response occurring at the resonant frequency.
Minimum phase transfer function All the zeros of a transfer function lie in the left-hand side of the \(s\)-plane.
Natural frequency The frequency of natural oscillation that would occur for two complex poles if the damping were equal to zero.
Nonminimum phase transfer function Transfer functions with zeros in the right-hand \(s\)-plane.
Octave The frequency interval \(\omega_{2} = 2\omega_{1}\) is an octave of frequencies (e.g., the range of frequencies from \(\omega_{1} = 100rad/s\) to \(\omega_{2} = 200rad/s\) is one octave).
Polar plot A plot of the real part of \(G(j\omega)\) versus the imaginary part of \(G(j\omega)\).
Resonant frequency The frequency \(\omega_{r}\) at which the maximum value of the frequency response of a complex pair of poles is attained.
Transfer function in the frequency domain The ratio of the output to the input signal where the input is a sinusoid. It is expressed as \(G(j\omega)\).
Stability in the Frequency
Domain
9.1 Introduction 623
9.2 Mapping Contours in the s-Plane 624
9.3 The Nyquist Criterion 630
9.4 Relative Stability and the Nyquist Criterion 641
9.5 Time-Domain Performance Criteria in the Frequency Domain 648
9.6 System Bandwidth 655
9.7 The Stability of Control Systems with Time Delays 655
9.8 Design Examples 659
9.9 PID Controllers in the Frequency Domain 677
9.10 Stability in the Frequency Domain Using Control Design Software 678
9.11 Sequential Design Example: Disk Drive Read System 686
9.12 Summary 689
292. PREVIEW
In previous chapters, we discussed stability and developed various tools to determine stability and to assess relative stability. We continue that discussion in this chapter by showing how frequency response methods can be used to investigate stability. The important concepts of gain margin, phase margin, and bandwidth are developed in the context of Bode plots, Nyquist plots, and Nichols charts. A frequency response stability result - known as the Nyquist stability criterion - is presented and its use illustrated through several interesting examples. The implications of having pure time delays in the system on both stability and performance are discussed. We will see that the phase lag introduced by the time delay can destabilize an otherwise stable system. The chapter concludes with a frequency response analysis of the Sequential Design Example: Disk Drive Read System.
293. DESIRED OUTCOMES
Upon completion of Chapter 9, students should be able to:
\(\square\) Explain the Nyquist stability criterion and the role of the Nyquist plot.
\(\square\) Identify time-domain performance specifications in the frequency domain.
\(\square\) Describe the importance of considering time delays in feedback control systems.
$\square\ $ Analyze the relative stability and performance of feedback control systems using frequency response methods considering phase and gain margin, and system bandwidth with Bode plots, Nyquist plots, and Nichols charts.
293.1. INTRODUCTION
Stability is a key characteristic of a feedback control system. Furthermore, if the system is stable, it is possible to investigate the relative stability. There are several methods of determining the absolute and relative stability of a system. The Routh-Hurwitz method is useful for investigating the characteristic equation expressed in terms of the complex variable \(s = \sigma + j\omega\). The relative stability of a system can be investigated utilizing the root locus method, which is also expressed in terms of the complex variable \(s\). In this chapter, we are concerned with investigating the stability of a system in the frequency domain, that is, in terms of the frequency response.
The frequency response of a system represents the sinusoidal steady-state response of a system and provides sufficient information for the determination of the relative stability of the system. The frequency response of a system can readily be obtained experimentally by exciting the system with sinusoidal input signals; therefore, it can be utilized to investigate the relative stability of a system when the system parameter values have not been determined. Furthermore, a frequency-domain stability criterion would be useful for determining suitable approaches to adjusting the parameters of a system in order to increase its relative stability.
A frequency domain stability criterion was developed by H. Nyquist in 1932, and it remains a fundamental approach to the investigation of the stability of linear control systems \(\lbrack 1,2\rbrack\). The Nyquist stability criterion is based on a theorem in the theory of the function of a complex variable due to Cauchy. Cauchy's theorem is concerned with mapping contours in the complex s-plane, and fortunately the theorem can be understood without a formal proof requiring complex variable theory.
To determine the relative stability of a closed-loop system, we must investigate the characteristic equation of the system:
For the unity feedback control system of Figure 9.1, the loop transfer function is \(L(s) = G_{c}(s)G(s)\). For a multiloop system, in terms of signal-flow graphs, the characteristic equation is
where \(\Delta(s)\) is the graph determinant. Therefore, we can represent the characteristic equation of single-loop or multiple-loop systems by Equation (9.1), where \(L(s)\) is a rational function of \(s\). To ensure stability, we must ascertain that all the zeros of \(F(s)\) lie in the left-hand \(s\)-plane. Nyquist thus proposed a mapping of the right-hand \(s\)-plane into the \(F(s)\)-plane. Therefore, to use and understand Nyquist's criterion, we shall first consider briefly the mapping of contours in the complex plane.
(a)
FIGURE 9.1
Unity feedback control system.
(b)
293.2. MAPPING CONTOURS IN THE s-PLANE
Consider the mapping of contours in the \(s\)-plane by a function \(F(s)\). A contour map is a contour or trajectory in one plane mapped or translated into another plane by a relation. Since \(s\) is a complex variable, \(s = \sigma + j\omega\), the function \(F(s)\) is itself complex; it can be defined as \(F(s) = u + jv\) and can be represented on a complex \(F(s)\)-plane with coordinates \(u\) and \(v\). As an example, let us consider a function \(F(s) = 2s + 1\) and a contour in the \(s\)-plane, as shown in Figure 9.2(a).
FIGURE 9.2
Mapping a square contour by \(F(s) = 2s + 1 =\) \(2(s + 1/2)\).
(a)
(b) The mapping of the \(s\)-plane unit square contour to the \(F(s)\)-plane is accomplished through the relation \(F(s)\), and so
Therefore, in this case, we have
and
Thus, the contour has been mapped by \(F(s)\) into a contour of an identical form, a square, with the center shifted by one unit and the magnitude of a side multiplied by two. This type of mapping, which retains the angles of the \(s\)-plane contour on the \(F(s)\)-plane, is called a conformal mapping. We also note that a closed contour in the \(s\)-plane results in a closed contour in the \(F(s)\)-plane.
The points \(A,B,C\), and \(D\), as shown in the \(s\)-plane contour, map into the points \(A,B,C\), and \(D\) shown in the \(F(s)\)-plane. Furthermore, a direction of traversal of the \(s\)-plane contour can be indicated by the direction \(ABCD\) and the arrows shown on the contour. Then a similar traversal occurs on the \(F(s)\)-plane contour as we pass \(ABCD\) in order, as shown by the arrows. By convention, the area within a contour to the right of the traversal of the contour is considered to be the area enclosed by the contour. Therefore, we will assume clockwise traversal of a contour to be positive and the area enclosed within the contour to be on the right. This convention is opposite to that usually employed in complex variable theory, but is equally applicable and is generally used in control system theory. We might consider the area on the right as we walk along the contour in a clockwise direction and call this rule "clockwise and eyes right."
Typically, we are concerned with an \(F(s)\) that is a rational function of \(s\). Therefore, it will be worthwhile to consider another example of a mapping of a contour. Let us again consider the unit square contour for the function
Several values of \(F(s)\) as \(s\) traverses the square contour are given in Table 9.1, and the resulting contour in the \(F(s)\)-plane is shown in Figure 9.3(b). The contour in the \(F(s)\)-plane encloses the origin of the \(F(s)\)-plane because the origin lies within the enclosed area of the contour in the \(F(s)\)-plane.
294. Table 9.1 Values of \(F(s)\)
| Point \(\mathbf{A}\) | Point \(\mathbf{B}\) | Point \(\mathbf{C}\) | Point \(\mathbf{D}\) | |||||
| $$s = \sigma + j\omega$$ | $$1 + j1$$ | 1 | $$1 - j1$$ | $$- j1$$ | $$- 1 - j1$$ | -1 | $$- 1 + j1$$ | $$j1$$ |
| $$F(s) = u + jv$$ | $$\frac{4 + 2j}{10}$$ | $$\frac{1}{3}$$ | $$\frac{4 - 2j}{10}$$ | $$\frac{1 - 2j}{5}$$ | $$- j$$ | -1 | $$+ j$$ | $$\frac{1 + 2j}{5}$$ |
FIGURE 9.3
Mapping for
\(F(s) = s/(s + 2)\).
(a)
(b)
Cauchy's theorem is concerned with mapping a function \(F(s)\) that has a finite number of poles and zeros within the contour, so that we may express \(F(s)\) as
where \(- z_{i}\) are the zeros of the function \(F(s)\) and \(- p_{k}\) are the poles of \(F(s)\). The function \(F(s)\) is the characteristic equation, and so
where
Therefore, we have
and the poles of \(L(s)\) are the poles of \(F(s)\). However, it is the zeros of \(F(s)\) that are the characteristic roots of the system and that indicate its response. This is clear if we recall that the output of the system is
where \(P_{k}\) and \(\Delta_{k}\) are the path factors and cofactors as defined in Section 2.7. Reexamining the example when \(F(s) = 2(s + 1/2)\), we have one zero of \(F(s)\) at \(s = - 1/2\), as shown in Figure 9.2. The contour that we chose (that is, the unit square) enclosed and encircled the zero once within the area of the contour. Similarly, for the function \(F(s) = s/(s + 2)\), the unit square encircled the zero at the origin but did not encircle the pole at \(s = - 2\). The encirclement of the poles and zeros of \(F(s)\) can be related to the encirclement of the origin in the \(F(s)\)-plane by Cauchy's theorem, commonly known as the principle of the argument, which states [3, 4]:
If a contour \(\Gamma_{s}\) in the \(s\)-plane encircles \(Z\) zeros and \(P\) poles of \(F(s)\) and does not pass through any poles or zeros of \(F(s)\) and the traversal is in the clockwise direction along the contour, the corresponding contour \(\Gamma_{F}\) in the \(F(s)\)-plane encircles the origin of the \(F(s)\)-plane \(N = Z - P\) times in the clockwise direction.
Thus, for the examples shown in Figures 9.2 and 9.3, the contour in the \(F(s)\) plane encircles the origin once, because \(N = Z - P = 1\), as we expect. As another example, consider the function \(F(s) = s/(s + 1/2)\). For the unit square contour shown in Figure 9.4(a), the resulting contour in the \(F(s)\) plane is shown in Figure 9.4(b). In this case, \(N = Z - P = 0\), as is the case in Figure 9.4(b), since the contour \(\Gamma_{F}\) does not encircle the origin.
Cauchy's theorem can be best comprehended by considering \(F(s)\) in terms of the angle due to each pole and zero as the contour \(\Gamma_{s}\) is traversed in a clockwise direction. Thus, let us consider the function
where \(- z_{i}\) is a zero of \(F(s)\), and \(- p_{k}\) is a pole of \(F(s)\). Equation (9.10) can be written as
Now, considering the vectors as shown for a specific contour \(\Gamma_{s}\) (Figure 9.5a), we can determine the angles as \(s\) traverses the contour. Clearly, the net angle change as \(s\) traverses along \(\Gamma_{s}\) (a full rotation of \(360^{\circ}\) for \(\phi_{p_{1}},\phi_{p_{2}}\), and \(\phi_{z_{2}}\) ) is zero degrees. However, for \(\phi_{z_{1}}\) as \(s\) traverses \(360^{\circ}\) around \(\Gamma_{s}\), the angle \(\phi_{z_{1}}\) traverses a full \(360^{\circ}\) clockwise. Thus, as \(\Gamma_{s}\) is completely traversed, the net angle increase of \(F(s)\) is equal to \(360^{\circ}\), since only one zero is enclosed. If \(Z\) zeros were enclosed within \(\Gamma_{s}\), then the net angle increase would be equal to \(\phi_{z} = 2\pi Z\) rad. Following this reasoning, if \(Z\) zeros and \(P\) poles are encircled as \(\Gamma_{S}\) is traversed, then \(2\pi Z - 2\pi P\) is the net resultant angle increase of \(F(s)\). Thus, the net angle increase of \(\Gamma_{F}\) of the contour in the \(F(s)\)-plane is simply FIGURE 9.4
Mapping for
\(F(s) = s/(s + 1/2)\).
(a)
(b)
or
and the net number of encirclements of the origin of the \(F(s)\)-plane is \(N = Z - P\). Thus, for the contour shown in Figure 9.5(a), which encircles one zero, the contour \(\Gamma_{F}\) shown in Figure 9.5(b) encircles the origin once in the clockwise direction.
(a)
(b)
Evaluation of the net angle of \(\Gamma_{F}\). FIGURE 9.6
Example of Cauchy's theorem with three zeros and one pole within \(\Gamma_{s}\).
(a)
(b)
As an example of the use of Cauchy's theorem, consider the pole-zero pattern shown in Figure 9.6(a) with the contour \(\Gamma_{s}\) to be considered. The contour encloses and encircles three zeros and one pole. Therefore, we obtain
and \(\Gamma_{F}\) completes two clockwise encirclements of the origin in the \(F(s)\)-plane, as shown in Figure 9.6(b).
For the pole and zero pattern shown and the contour \(\Gamma_{s}\) as shown in Figure 9.7(a), one pole is encircled and no zeros are encircled. Therefore, we have
and we expect one encirclement of the origin by the contour \(\Gamma_{F}\) in the \(F(s)\)-plane. However, since the sign of \(N\) is negative, we find that the encirclement moves in the counterclockwise direction, as shown in Figure 9.7(b).
(a)
(b)
FIGURE 9.7
Example of Cauchy's theorem \(\Gamma_{S}\). Now that we have developed and illustrated the concept of mapping of contours through a function \(F(s)\), we are ready to consider the stability criterion proposed by Nyquist.
294.1. THE NYQUIST CRITERION
To investigate the stability of a control system, we consider the characteristic equation
For a system to be stable, all the zeros of \(F(s)\) must lie in the left-hand \(s\)-plane. Thus, we find that the roots of a stable system (the zeros of \(F(s)\) ) must lie to the left of the \(j\omega\)-axis in the \(s\)-plane. Therefore, we choose a contour \(\Gamma_{s}\) in the \(s\)-plane that encloses the entire right-hand \(s\)-plane, and we determine whether any zeros of \(F(s)\) lie within \(\Gamma_{s}\) by utilizing Cauchy's theorem. That is, we plot \(\Gamma_{F}\) in the \(F(s)\)-plane and determine the number of encirclements of the origin \(N\). Then the number of zeros of \(F(s)\) within the \(\Gamma_{s}\) contour (and therefore, the unstable zeros of \(F(s)\) ) is
Thus, if \(P = 0\), as is usually the case, we find that the number of unstable roots of the system is equal to \(N\), the number of encirclements of the origin of the \(F(s)\)-plane.
The Nyquist contour that encloses the entire right-hand \(s\)-plane is shown in Figure 9.8. The contour \(\Gamma_{s}\) passes along the \(j\omega\)-axis from \(- j\infty\) to \(+ j\infty\), and this part of the contour provides the familiar \(F(j\omega)\). The contour is completed by a
FIGURE 9.8
Nyquist contour is shown as the heavy line.
semicircular path of radius \(r\), where \(r\) approaches infinity so this part of the contour typically maps to a point. This contour \(\Gamma_{F}\) is known as the Nyquist plot.
The Nyquist criterion is concerned with the mapping of the function
and the number of encirclements of the origin of the \(F(s)\)-plane. Alternatively, we may define the function
The change of functions represented by Equation (9.16) is very convenient because the loop transfer function \(L(s)\) is typically available in factored form, while \(1 + L(s)\) is not. Then, the mapping of \(\Gamma_{s}\) in the \(s\)-plane will be through the function \(F^{'}(s) = L(s)\) into the \(L(s)\)-plane. In this case, the number of clockwise encirclements of the origin of the \(F(s)\)-plane becomes the number of clockwise encirclements of the -1 point in the \(F^{'}(s) = L(s)\)-plane because \(F^{'}(s) = F(s) - 1\). Therefore, the Nyquist stability criterion can be stated as follows:
295. A feedback system is stable if and only if the contour \(\Gamma_{L}\) in the \(L(s)\)-plane does not encircle the \(( - 1,0)\) point when the number of poles of in the right-hand \(s\)-plane is zero \((P = 0)\).
When the number of poles of \(L(s)\) in the right-hand \(s\)-plane is other than zero, the Nyquist criterion is stated as follows:
A feedback control system is stable if and only if, for the contour \(\Gamma_{L}\), the number of counterclockwise encirclements of the \(( - 1,0)\) point is equal to the number of poles of with positive real parts.
The basis for the two statements is the fact that, for the \(F^{'}(s) = L(s)\) mapping, the number of roots (or zeros) of \(1 + L(s)\) in the right-hand \(s\)-plane is represented by the expression
Clearly, if the number of poles of \(L(s)\) in the right-hand \(s\)-plane is zero \((P = 0)\), we require for a stable system that \(N = 0\), and the contour \(\Gamma_{p}\) must not encircle the -1 point. Also, if \(P\) is other than zero and we require for a stable system that \(Z = 0\), then we must have \(N = - P\), or \(P\) counterclockwise encirclements.
296. EXAMPLE 9.1 System with two real poles
A unity feedback control system is shown in Figure 9.1, where
In this case, \(L(s) = G_{c}(s)G(s)\), and we use a contour \(\Gamma_{L}\) in the \(L(s)\)-plane. The contour \(\Gamma_{s}\) in the \(s\)-plane is shown in Figure 9.9(a), and the contour \(\Gamma_{L}\) is shown in Figure 9.9(b) for \(\tau_{1} = 1,\tau_{2} = 1/10\), and \(K = 100\).
The \(+ j\omega\)-axis is mapped into the solid line, as shown in Figure 9.9. The \(- j\omega\)-axis is mapped into the dashed line, as shown in Figure 9.9. The semicircle with \(r \rightarrow \infty\) in the \(s\)-plane is mapped into the origin of the \(L(s)\)-plane.
We note that the number of poles of \(L(s)\) in the right-hand \(s\)-plane is zero, and thus \(P = 0\). Therefore, for this system to be stable, we require \(N = Z = 0\), and the contour must not encircle the -1 point in the \(L(s)\)-plane. Examining Figure 9.9(b) and Equation (9.17), we find that, irrespective of the value of \(K\), the contour does not encircle the -1 point, and the system is always stable for all \(K\) greater than zero.
297. EXAMPLE 9.2 System with a pole at the origin
A unity feedback control system is shown in Figure 9.1, where
In this single-loop case, \(L(s) = G_{c}(s)G(s)\), and we determine the contour \(\Gamma_{L}\) in the \(L(s)\)-plane. The contour \(\Gamma_{s}\) in the \(s\)-plane is shown in Figure 9.10(a), where an infinitesimal detour around the pole at the origin is effected by a small semicircle of radius \(\varepsilon\), where \(\varepsilon \rightarrow 0\). This detour is a consequence of the condition of Cauchy's
FIGURE 9.9
Nyquist contour and mapping for \(L(s) =\) \(\frac{100}{(s + 1)(s/10 + 1)}\).
(a)
(b) FIGURE 9.10
Nyquist contour and mapping for \(L(s) =\) \(K/(s(\tau s + 1))\).
(a)
(b)
theorem, which requires that the contour cannot pass through the pole at the origin. A sketch of the contour \(\Gamma_{L}\) is shown in Figure 9.10(b). Clearly, the portion of the contour \(\Gamma_{L}\) from \(\omega = 0^{+}\)to \(\omega = + \infty\) is a plot of the real and imaginary components of \(L(j\omega) = u(\omega) + jv(\omega)\). Let us consider each portion of the Nyquist contour \(\Gamma_{s}\) in detail and determine the corresponding portions of the \(L(s)\)-plane contour \(\Gamma_{L}\).
(a) The Origin of the s-Plane. The small semicircular detour around the pole at the origin can be represented by setting \(s = \varepsilon e^{j\phi}\) and allowing \(\phi\) to vary from \(- 90^{\circ}\) at \(\omega = 0^{-}\)to \(+ 90^{\circ}\) at \(\omega = 0^{+}\). Because \(\varepsilon\) approaches zero, the mapping for \(L(s)\) is
Therefore, the angle of the contour in the \(L(s)\)-plane changes from \(90^{\circ}\) at \(\omega = 0_{-}\) to \(- 90^{\circ}\) at \(\omega = 0_{+}\), passing through \(0^{\circ}\) at \(\omega = 0\). The radius of the contour in the \(L(s)\)-plane for this portion of the contour is infinite, and this portion of the contour is shown in Figure 9.10(b). The points denoted by \(A,B\), and \(C\) in Figure 9.10(a) map to \(A,B\), and \(C\), respectively, in Figure 9.10(b).
(b) The Portion from \(\omega = \mathbf{0}_{+}\)to \(\omega = + \infty\). The portion of the contour \(\Gamma_{s}\) from \(\omega = 0_{+}\)to \(\omega = + \infty\) is mapped by the function \(L(j\omega)\) where
for this part of the contour. This results in the plot with \(\omega = 0_{+}\)to \(\omega = + \infty\) shown in Figure 9.10(b). When \(\omega\) approaches \(+ \infty\), we have
Therefore, the magnitude approaches zero at an angle of \(- 180^{\circ}\). (c) The Portion from \(\omega = + \infty\) to \(\omega = - \infty\). The portion of \(\Gamma_{s}\) from \(\omega = + \infty\) to \(\omega = - \infty\) is mapped into the point zero at the origin of the \(L(s)\)-plane by the function \(L(s)\). The mapping is represented by
as \(\phi\) changes from \(\phi = + 90^{\circ}\) at \(\omega = + \infty\) to \(\phi = - 90^{\circ}\) at \(\omega = - \infty\). Thus, the contour moves from an angle of \(- 180^{\circ}\) at \(\omega = + \infty\) to an angle of \(+ 180^{\circ}\) at \(\omega = - \infty\). The magnitude of the \(L(s)\) contour when \(r\) is infinite is always zero or a constant.
(d) The Portion from \(\omega = - \infty\) to \(\omega = \mathbf{0}_{-}\). The portion of the contour \(\Gamma_{s}\) from \(\omega = - \infty\) to \(\omega = 0_{-}\)is mapped by the function \(L( - j\omega)\) where
Thus, we obtain the complex conjugate of \(L(j\omega)\), and the plot for the portion of the plot from \(\omega = - \infty\) to \(\omega = 0_{-}\)is symmetrical to the plot from \(\omega = + \infty\) to \(\omega = 0_{+}\). This symmetrical plot is shown on the \(L(s)\)-plane in Figure 9.10(b).
To investigate the stability of this second-order system, we first note that the number of poles, \(P\), within the right-hand \(s\)-plane is zero. Therefore, for this system to be stable, we require \(N = Z = 0\), and the contour \(\Gamma_{L}\) must not encircle the -1 point in the \(L(s)\)-plane. Examining Figure 9.10(b), we find that irrespective of the value of the gain \(K\) and the time constant \(\tau\), the contour does not encircle the -1 point, and the system is always stable. We are considering positive values of gain \(K\). If negative values of gain are to be considered, we should use \(- K\), where \(K \geq 0\).
We may draw two general conclusions from this example:
-
The plot of the contour \(\Gamma_{L}\) for the range \(- \infty < \omega < 0\) - will be the complex conjugate of the plot for the range \(0_{+} < \omega < + \infty\), and the Nyquist plot of \(L(s) = G_{c}(s)G(s)\) will be symmetrical in the \(L(s)\)-plane about the \(u\)-axis. Therefore, it is sufficient to construct the contour \(\mathbf{\Gamma}_{\mathbf{L}}\) for the frequency range \(\mathbf{0}_{+} < \omega < + \infty\) in order to investigate the stability (keeping in mind the detour around the origin).
-
The magnitude of \(L(s) = G_{c}(s)G(s)\) as \(s = re^{j\phi}\) and \(r \rightarrow \infty\) will normally approach zero or a constant.
298. EXAMPLE 9.3 System with three poles
Consider the unity feedback system shown in Figure 9.1 with the loop transfer function
The Nyquist contour \(\Gamma_{S}\) is shown in Figure 9.10(a). This mapping is symmetrical for \(L(j\omega)\) and \(L( - j\omega)\) so it is sufficient to investigate the \(L(j\omega)\)-locus. The small semicircle around the origin of the \(s\)-plane maps into a semicircle of infinite radius, as in Example 9.2. Also, the semicircle \(re^{j\phi}\) in the \(s\)-plane as \(r \rightarrow \infty\) maps into the point \(L(j\omega) = 0\), as we expect. Therefore, to investigate the stability of the system, it is sufficient to plot the portion of the contour \(\Gamma_{L}\) that is the magnitude and phase of \(L(j\omega)\) for \(0_{+} < \omega < + \infty\). Thus, when \(s = + j\omega\), we have
When \(\omega = 0_{+}\), the magnitude of the locus is infinite at an angle of \(- 90^{\circ}\) in the \(L(s)\) plane. When \(\omega\) approaches \(+ \infty\), we have
Therefore, \(L(j\omega)\) approaches a magnitude of zero at an angle of \(- 270^{\circ}\) [29]. To approach at an angle of \(- 270^{\circ}\), the locus must cross the \(u\)-axis in the \(L(s)\)-plane, as shown in Figure 9.11. Thus, it is possible to encircle the -1 point. The number of encirclements when the -1 point lies within the locus, as shown in Figure 9.11, is equal to two, and the system is unstable with two roots in the right-hand \(s\)-plane. The
FIGURE 9.11 Nyquist plot for \(L(s) =\) $K/\left( s\left( \tau_{1}s + 1 \right) \right.\ $ \(\left. \ \left( \tau_{2}s + 1 \right) \right)\). The tic mark shown to the left of the origin is the -1 point.
point where the \(L(s)\)-locus intersects the real axis can be found by setting the imaginary part of \(L(j\omega) = u + jv\) equal to zero. We then have, from Equation (9.24),
Thus, \(v = 0\) when \(1 - \omega^{2}\tau_{1}\tau_{2} = 0\) or \(\omega = 1/\sqrt{\tau_{1}\tau_{2}}\). The magnitude of the real part of \(L(j\omega)\) at this frequency is
Therefore, the system is stable when
or
Consider the case where \(\tau_{1} = \tau_{2} = 1\), so that
Using Equation (9.28), we expect stability when
The Nyquist plots for three values of \(K\) are shown in Figure 9.12.
(a)
(b)
(c)
FIGURE 9.12 Nyquist plot for \(L(s) = G_{c}(s)G(s) = \frac{K}{s(s + 1)^{2}}\) when (a) \(K = 1\), (b) \(K = 2\), and (c) \(K = 3\).
299. EXAMPLE 9.4 System with two poles at the origin
Consider the unity feedback system shown in Figure 9.1 when
When \(s = j\omega\), we have
We note that the angle of \(L(j\omega)\) is always \(- 180^{\circ}\) or less, and the locus of \(L(j\omega)\) is above the \(u\)-axis for all values of \(\omega\). As \(\omega\) approaches \(0_{+}\), we have
As \(\omega\) approaches \(+ \infty\), we have
At the small semicircular detour at the origin of the \(s\)-plane where \(s = \epsilon e^{j\phi}\), we have
where \(- \pi/2 \leq \phi \leq \pi/2\). Thus, the contour \(\Gamma_{L}\) ranges from an angle of \(+ \pi\omega = 0_{-}\)to \(- \pi\) at \(\omega = 0_{+}\)and passes through a full circle of \(2\pi\) rad as \(\omega\) changes from \(\omega = 0_{-}\) to \(\omega = 0_{+}\). The complete contour plot of \(\Gamma_{L}\) is shown in Figure 9.13. Because the contour encircles the -1 point twice, there are two roots of the closed-loop system in the right-hand plane, and the system, irrespective of the gain \(K\), is unstable.
300. EXAMPLE 9.5 System with a pole in the right-hand s-plane
Consider the control system shown in Figure 9.14 and determine the stability of the system. First, consider the system without derivative feedback, so that \(K_{2} = 0\). We then have the loop transfer function
Thus, the loop transfer function has one pole in the right-hand \(s\)-plane, and therefore \(P = 1\). For this system to be stable, we require \(N = - P = - 1\), one counterclockwise encirclement of the -1 point. At the semicircular detour at the origin of the \(s\)-plane, we let \(s = \epsilon e^{j\phi}\) when \(- \pi/2 \leq \phi \leq \pi/2\). Then, when \(s = \epsilon e^{j\phi}\), we have FIGURE 9.13
Nyquist contour plot for \(L(s) =\) \(K/\left( s^{2}(\tau s + 1) \right)\).
FIGURE 9.14
Second-order feedback control system. (a) Signalflow graph.
(b) Block diagram.
(a)
(b)
Therefore, this portion of the contour \(\Gamma_{L}\) is a semicircle of infinite magnitude in the left-hand \(L(s)\)-plane, as shown in Figure 9.15. When \(s = j\omega\), we have
FIGURE 9.15
Nyquist plot for
\(L(s) = K_{1}/(s(s - 1))\).
Finally, for the semicircle of radius \(r\) as \(r\) approaches infinity, we have
where \(\phi\) varies from \(\pi/2\) to \(- \pi/2\) in a clockwise direction. Therefore, the contour \(\Gamma_{L}\), at the origin of the \(L(s)\)-plane, varies \(2\pi rad\) in a counterclockwise direction. The contour \(\Gamma_{L}\) in the \(L(s)\)-plane encircles the -1 point once in the clockwise direction so \(N = + 1\), and there is one pole \(s = 1\) in the right-hand plane so \(P = 1\). Hence,
and the system is unstable because two roots of the characteristic equation, irrespective of the value of the gain \(K_{1}\), lie in the right half of the s-plane.
Let us now consider again the system when the derivative feedback is included in the system shown in Figure \(9.14\left( K_{2} > 0 \right)\). Then the loop transfer function is
The portion of the contour \(\Gamma_{L}\) when \(s = \varepsilon e^{j\phi}\) is the same as the system without derivative feedback, as shown in Figure 9.16. However, when \(s = re^{j\phi}\) as \(r\) approaches infinity, we have
FIGURE 9.16
Nyquist plot
for \(L(s) =\)
\((s(s - 1))\).
and the \(\Gamma_{L}\)-contour at the origin of the \(L(s)\)-plane varies \(\pi\) rad in a counterclockwise direction. The locus \(L(j\omega)\) crosses the \(u\)-axis at a point determined by considering
The \(L(j\omega)\)-locus intersects the \(u\)-axis at a point where the imaginary part of \(L(j\omega)\) is zero. Therefore,
at this point, or \(\omega^{2} = 1/K_{2}\). The value of the real part of \(L(j\omega)\) at the intersection is then
Therefore, when \(- K_{1}K_{2} < - 1\) or \(K_{1}K_{2} > 1\), the contour \(\Gamma_{L}\) encircles the -1 point once in a counterclockwise direction, and therefore \(N = - 1\). Then the number of zeros of the system in the right-hand plane is
Thus, the system is stable when \(K_{1}K_{2} > 1\). Often, it may be useful to utilize a computer to plot the Nyquist plot [5].
301. EXAMPLE 9.6 System with a zero in the right-hand \(s\)-plane
Consider the feedback control system shown in Figure 9.1 when
FIGURE 9.17 Nyquist plot for Example 9.6 for \(L(j\omega)/K\).
We have
As \(\omega\) approaches \(+ \infty\) on the \(+ j\omega\) axis, we have
When \(\omega = \sqrt{5}\), we have \(L(j\omega) = K/2\). At \(\omega = 0_{+}\), we have \(L(j\omega) = - 2K\). The Nyquist plot for \(L(j\omega)/K\) is shown in Figure 9.17. \(L(j\omega)\) intersects the \(- 1 + j0\) point when \(K = 1/2\). Thus, the system is stable for the limited range of gain \(0 < K \leq 1/2\). When \(K > 1/2\), the number of encirclements of the -1 point is \(N = 1\). The number of poles of \(L(s)\) in the right half \(s\)-plane is \(P = 0\). Therefore, we have
and the system is unstable. Examining the Nyquist plot of Figure 9.17 we conclude that the system is unstable for all \(K > 1/2\).
301.1. RELATIVE STABILITY AND THE NYQUIST CRITERION
For the \(s\)-plane, we defined the relative stability of a system as the property measured by the relative settling time of each root or pair of roots. Therefore, a system with a shorter settling time is considered relatively more stable. We would like to determine a similar measure of relative stability useful for the frequency response method. The Nyquist criterion provides us with suitable information concerning the absolute stability and, furthermore, can be utilized to define and ascertain the relative stability of a system.
The Nyquist stability criterion is defined in terms of the \(( - 1,0)\) point on the Nyquist plot or the \(0 - dB, - 180^{\circ}\) point on the Bode plot. The proximity of the \(L(j\omega)\)-locus to this stability point is a measure of the relative stability of a system. The critical section of the Nyquist plot for the loop transfer function for several values of \(K\) with
is shown in Figure 9.18. As \(K\) increases, the Nyquist plot approaches the -1 point and eventually encircles the -1 point for a gain \(K = K_{3}\). We determined in Section 9.3 that the locus intersects the \(u\)-axis at a point
Therefore, the system has roots on the \(j\omega\)-axis when
As \(K\) is decreased below this marginal value, the stability is increased, and the margin between the critical gain \(K = \left( \tau_{1} + \tau_{2} \right)/\tau_{1}\tau_{2}\) and a gain \(K = K_{2}\) is a measure of the relative stability. This measure of relative stability is called the gain margin and is defined as the reciprocal of the gain \(|L(j\omega)|\) at the frequency at which the phase angle reaches \(- \mathbf{180}^{\circ}\) (that is, \(v = 0\) ). The gain margin is a measure of the factor by which the system gain would have to be increased for the \(L(j\omega)\) locus to pass through the \(u = - 1\) point. Thus, for a gain \(K = K_{2}\) in Figure 9.18, the gain margin
is equal to the reciprocal of \(L(j\omega)\) when \(v = 0\). Because \(\omega = 1/\sqrt{\tau_{1}\tau_{2}}\) when the phase shift is \(- 180^{\circ}\), we have a gain margin equal to
The gain margin can be defined in terms of a logarithmic (decibel) measure as
For example, when \(\tau_{1} = \tau_{2} = 1\), the system is stable when \(K \leq 2\). Thus, when \(K = K_{2} = 0.5\), the gain margin is equal to
or, in logarithmic measure,
Therefore, the gain margin indicates that the system gain can be increased by a factor of four \((12\text{ }dB)\) before the stability boundary is reached.
The gain margin is the increase in the system gain when phase \(= - 180^{\circ}\) that will result in a marginally stable system with intersection of the \(- 1 + j0\) point on the Nyquist plot.
An alternative measure of relative stability can be defind in terms of the phase margin between a specific and a system that is marginally stable. The phase margin is defined as the phase angle through which the \(\mathbf{L}(\mathbf{j}\omega)\) locus must be rotated so that the unity magnitude \(|L(j\omega)| = 1\) point will pass through the \(( - 1,0)\) point in the \(\mathbf{L}(\mathbf{j\omega})\) plane. This measure of relative stability is equal to the additional phase lag required before the system becomes unstable. This information can be determined from the Nyquist plot shown in Figure 9.18. For a gain \(K = K_{2}\), an additional phase angle, \(\phi_{2}\), may be added to the system before the system becomes unstable. Similarly, for the gain \(K_{1}\), the phase margin is equal to \(\phi_{1}\), as shown in Figure 9.18.
The phase margin is the amount of phase shift of the \(L(j\omega)\) at unity magnitude that will result in a marginally stable system with intersection of the \(- 1 + j0\) point on the Nyquist plot.
The gain phase margins are easily evaluated from the Bode plot. The critical point for stability is \(u = - 1,v = 0\) in the \(L(j\omega)\) - plane, which is equivalent to a logarithmic magnitude of \(0\text{ }dB\) and a phase angle of \(180^{\circ}\) (or \(- 180^{\circ}\) ) on the Bode plot. FIGURE 9.19
Bode plot for
\((0.2j\omega + 1))\).
It is relatively straightforward to examine the Nyquist plot of a minimum-phase system. Special care is required with a nonminimum-phase system, however, and the complete Nyquist plot should be studied to determine stability.
The Bode plot associated with the loop transfer function
is shown in Figure 9.19. The phase angle when the logarithmic magnitude is \(0\text{ }dB\) is equal to \(- 137^{\circ}\). Thus, the phase margin is \(180^{\circ} - 137^{\circ} = 43^{\circ}\). The logarithmic magnitude when the phase angle is \(- 180^{\circ}\) is \(- 15\text{ }dB\), and therefore the gain margin is \(G \cdot M. = 15\text{ }dB\).
The frequency response of a system can be graphically portrayed on the logarithmic-magnitude-phase-angle diagram. For the log-magnitude-phase diagram, the critical stability point is the \(0 - dB, - 180^{\circ}\) point, and the gain margin and phase margin can be easily determined and indicated on the diagram. The log-magnitude-phase locus of the loop transfer function, \(L(j\omega)\), in Equation (9.50) is shown in Figure 9.20. The indicated phase margin is \(P.M. = 43^{\circ}\), and the gain margin is G.M. \(= 15\text{ }dB\). For comparison, the locus for
is also shown in Figure 9.20. The gain margin for \(L_{2}(j\omega)\) is \(G.M. = 5.7\text{ }dB\), and the phase margin for \(L_{2}\) is \(P.M. = 20^{\circ}\). Clearly, the feedback system \(L_{2}(j\omega)\) is relatively less stable than the system \(L_{1}(j\omega)\). However, the question still remains: How much less stable is the system \(L_{2}(j\omega)\) in comparison to the system \(L_{1}(j\omega)\) ? In the following, we answer this question for a second-order system, and the general usefulness of the relation that we develop will depend on the presence of dominant roots. FIGURE 9.20
Log-magnitudephase curves for \(L_{1}\) and \(L_{2}\).
Let us now determine the phase margin of a second-order system and relate the phase margin to the damping ratio \(\zeta\) of an underdamped system. Consider the loop-transfer function of the system shown in Figure 9.1, where
The characteristic equation for the closed-loop second-order system is
Therefore, the closed-loop roots are
The frequency domain form of Equation (9.52) is
The magnitude of the frequency response is equal to 1 at the crossover frequency \(\omega_{c}\); thus,
Rearranging Equation (9.55), we obtain
Solving for \(\omega_{c}\) yields
The phase margin for this system is
Equation (9.57) is the relationship between the damping ratio \(\zeta\) and the phase margin \(\phi_{pm}\), which provides a correlation between the frequency response and the time response. A plot of \(\zeta\) versus \(\phi_{pm}\) is shown in Figure 9.21. The actual curve of \(\zeta\) versus \(\phi_{\text{pm}\text{~}}\) can be approximated by the dashed line shown in Figure 9.21. The slope of the linear approximation is equal to 0.01 , and therefore an approximate linear relationship between the damping ratio and the phase margin is
where the phase margin is measured in degrees. This approximation is reasonably accurate for \(\zeta \leq 0.7\) and is a useful index for correlating the frequency response with the transient performance of a system. Equation (9.58) is a suitable approximation for a second-order system and may be used for higher-order systems if we can assume that the transient response of the system is primarily due to a pair of dominant underdamped roots. The approximation of a higher-order system by a dominant second-order system is a useful approximation. Although it must be used
FIGURE 9.21
Damping ratio versus phase margin for a second-order system.
with care, control engineers find this approach to be a simple, yet fairly accurate, technique of setting the specifications of a control system.
Therefore, for the system with a loop transfer function, \(L(j\omega)\), in Equation (9.50) we found the phase margin \(P.M. = 43^{\circ}\). Thus, the damping ratio is approximately
Then the percent overshoot to a step input for this system is approximately
It is possible to calculate and plot the phase margin and gain margin versus the gain \(K\) for a specified \(L(j\omega)\). Consider the system of Figure 9.1 with the loop transfer function
The gain for which the system is marginally stable is \(K = K^{*} = 128\). The gain margin and the phase margin plotted versus \(K\) are shown in Figures 9.22(a) and (b), respectively. The gain margin is plotted versus the phase margin, as shown in Figure 9.22(c). The phase margin and the gain margin are suitable measures of the performance of the system. We will normally emphasize phase margin as a frequency-domain specification.
(a)
(b)
b)
(c) (a) Gain margin versus gain \(K\). (b) Phase margin versus gain \(K\). (c) Gain margin versus phase margin. The phase margin is a frequency response measure for indicating the expected transient performance of a system. Another useful index of performance in the frequency domain is \(M_{p\omega}\), the maximum magnitude of the closed-loop frequency response, and we shall now consider this practical index.
301.2. TIME-DOMAIN PERFORMANCE CRITERIA IN THE FREQUENCY DOMAIN
The transient performance of a feedback system can be estimated from the closedloop frequency response. The closed-loop frequency response is the frequency response of the closed-loop transfer function \(T(j\omega)\). The open- and closed-loop frequency responses for a single-loop system are related. Consider the closed-loop system:
The Nyquist criterion and the phase margin index are defined for the loop transfer function \(L(j\omega) = G_{c}(j\omega)G(j\omega)\). However, as we found in Section 8.2, the maximum magnitude of the closed-loop frequency response can be related to the damping ratio of a second-order system of
Because this relationship between the closed-loop frequency response and the transient response is a useful one, we would like to be able to determine \(M_{p\omega}\) from the plots completed for the investigation of the Nyquist criterion. That is, we want to be able to obtain the closed-loop frequency response from the open-loop frequency response. Of course, we could determine the closed-loop roots of \(1 + L(s)\) and plot the closed-loop frequency response. However, once we have invested all the effort necessary to find the closed-loop roots of a characteristic equation, then a closedloop frequency response is not necessary.
The relation between the closed-loop and open-loop frequency response is illuminated on the magnitude-phase plot when considering unity feedback systems. In the unity feedback case, key performance indicators such as \(M_{p\omega}\) and \(\omega_{r}\) can be determined from the magnitude-phase plot using circles of constant magnitude of the closed-loop transfer function. These circles are known as constant \(M\)-circles.
The relationship between \(T(j\omega)\) and \(L(j\omega)\) is readily obtained in terms of complex variables in the \(L(s)\)-plane. The coordinates of the \(L(s)\)-plane are \(u\) and \(v\), and we have
Therefore, the magnitude of the closed-loop transfer function is
Squaring Equation (9.65) and rearranging, we obtain
Dividing Equation (9.66) by \(1 - M^{2}\) and adding the term \(\left\lbrack M^{2}/\left( 1 - M^{2} \right) \right\rbrack^{2}\) to both sides, we have
Rearranging, we obtain
which is the equation of a circle on the \((u,v)\)-plane with the center at
The radius of the circle is equal to \(\left| M/\left( 1 - M^{2} \right) \right|\). Therefore, we can plot several circles of constant magnitude \(M\) in the \(L(s)\)-plane. Several constant \(M\) circles are shown in Figure 9.23. The circles to the left of \(u = - 1/2\) are for \(M > 1\), and the circles to the right of \(u = - 1/2\) are for \(M < 1\). When \(M = 1\), the circle becomes the straight line \(u = - 1/2\), which is evident from inspection of Equation (9.66).
The frequency response for a system is shown in Figure 9.24 for two gain values where \(K_{2} > K_{1}\). The frequency response curve for the system with gain \(K_{1}\) is tangent to magnitude circle \(M_{1}\) at a frequency \(\omega_{r1}\). Similarly, the frequency response curve for gain \(K_{2}\) is tangent to magnitude circle \(M_{2}\) at the frequency \(\omega_{r2}\). Therefore, the closed-loop frequency response magnitude curves are estimated as shown in Figure 9.25. Hence, we can obtain the closed-loop frequency response of a system
FIGURE 9.23 Constant \(M(\omega)\) circles.
FIGURE 9.24 The frequency response plot of \(G_{c}(j\omega)G(j\omega)\) for two values of a gain \(\left( K_{2} > K_{1} \right)\).
FIGURE 9.25 Closed-loop frequency response of \(T(j\omega) = G_{c}(j\omega)G(j\omega)/\left( 1 + G_{c}(j\omega)G(j\omega) \right)\). Note that \(K_{2} > K_{1}\).
from the \(L(s)\)-plane. If the maximum magnitude, \(M_{p\omega}\), is the only information desired, then it is sufficient to read this value directly from the Nyquist plot. The maximum magnitude of the closed-loop frequency response, \(M_{p\omega}\), is the value of the \(M\) circle that is tangent to the \(L(j\omega)\)-locus. The point of tangency occurs at the frequency \(\omega_{r}\), the resonant frequency. The complete closed-loop frequency response of a system can be obtained by reading the magnitude \(M\) of the circles that the \(L(j\omega)\)-locus intersects at several frequencies. Therefore, the system with a gain \(K = K_{2}\) has a closed-loop magnitude \(M_{1}\) at the frequencies \(\omega_{1}\) and \(\omega_{2}\). This magnitude is read from Figure 9.24 and is shown on the closed-loop frequency response in Figure 9.25. The bandwidth for \(K_{1}\) is shown as \(\omega_{B1}\).
It may be empirically shown that the crossover frequency \(\omega_{c}\) on the open-loop Bode plot is related to the closed-loop system bandwidth \(\omega_{B}\) by the approximation for \(\zeta\) in the range 0.2 to 0.8 :
In a similar manner, we can obtain circles of constant closed-loop phase angles. Thus, for Equation (9.62), the angle relation is
Taking the tangent of both sides and rearranging, we have
where \(N = tan\phi\). Adding the term \(1/4\left\lbrack 1 + 1/N^{2} \right\rbrack\) to both sides of the equation and simplifying, we obtain
which is the equation of a circle with its center at \(u = - 1/2\) and \(v = + 1/(2N)\). The radius of the circle is equal to \(1/2\left\lbrack 1 + 1/N^{2} \right\rbrack^{1/2}\). Therefore, the constant phase angle curves can be obtained for various values of \(N\) in a manner similar to the \(M\) circles.
The constant \(M\) and \(N\) circles can be used for analysis and design in the \(L(s)\) plane. However, it is much easier to obtain the Bode plot for a system, and it would be preferable if the constant \(M\) and \(N\) circles were translated to a logarithmic gain phase. N. B. Nichols transformed the constant \(M\) and \(N\) circles to the log-magnitudephase diagram, and the resulting chart is called the Nichols chart [3, 7]. The \(M\) and \(N\) circles appear as contours on the Nichols chart shown in Figure 9.26. The coordinates of the log-magnitude-phase diagram are the same as those used in Section 8.5. However, superimposed on the log-magnitude-phase plane we find constant \(M\) and \(N\) lines. The constant \(M\) lines are given in decibels and the \(N\) lines in degrees. An example will illustrate the use of the Nichols chart to determine the closed-loop frequency response.
302. EXAMPLE 9.7 Stability using the Nichols chart
Consider a unity feedback system with a loop transfer function in Equation (9.50). The frequency response of \(L(j\omega)\) is plotted on the Nichols chart and is shown in Figure 9.27. The maximum magnitude, \(M_{p\omega}\), is equal to \(+ 2.5\text{ }dB\) and occurs at a frequency \(\omega_{r} = 0.8\). The closed-loop phase angle at \(\omega_{r}\) is equal to \(- 72^{\circ}\). The \(3 - dB\) closed-loop bandwidth, where the closed-loop magnitude is \(- 3\text{ }dB\), is equal to \(\omega_{B} = 1.33\), as shown in Figure 9.27. The closed-loop phase angle at \(\omega_{B}\) is equal to \(- 142^{\circ}\).
303. EXAMPLE 9.8 Third-order system
Let us consider a unity feedback system with a loop transfer function
where \(\zeta = 0.5\) for the complex poles. The Nichols chart for this system is shown in Figure 9.28. The phase margin for this system as it is determined from the Nichols chart is \(30^{\circ}\). On the basis of the phase, we use Equation (9.58) to estimate the system damping ratio as \(\zeta = 0.30\). The maximum magnitude is equal to \(+ 9\text{ }dB\) occurring at a frequency \(\omega_{r} = 0.88\). Therefore,
FIGURE 9.26 Nichols chart. The phase curves for the closed-loop system are shown as heavy curves. FIGURE 9.27 Nichols chart for \(G_{c}(j\omega)G(j\omega) =\) \(1/(j\omega(j\omega + 1)\) \((0.2j\omega + 1))\). Three points on curve are shown for \(\omega = 0.5\), 0.8 , and 1.35 , respectively.
Solving Equation (9.63), we find that \(\zeta = 0.18\). We are confronted with two conflicting damping ratios, where one is obtained from a phase margin measure and another from a peak frequency response measure. In this case, we have discovered an example in which the correlation between the frequency domain and the time domain is unclear and uncertain. This apparent conflict is caused by the nature of the frequency response which slopes rapidly toward the \(180^{\circ}\) line from the 0 - \(dB\) axis. If we determine the roots of the characteristic equation for \(1 + L(s)\), we obtain
The damping ratio of the complex conjugate roots is equal to 0.124 , where the complex roots do not dominate the response of the system. Therefore, the real root will add some damping to the system, and we might estimate the damping ratio to be approximately the value determined from the \(M_{p\omega}\) index; that is, \(\zeta = 0.18\). A designer must use the frequency-domain-to-time-domain correlations with caution. However, we are usually safe if the lower value of the damping ratio resulting from the phase margin and the \(M_{p\omega}\) relation is used for analysis and design purposes. FIGURE 9.28 Nichols chart for \(G_{c}(j\omega)G(j\omega) =\) $0.64/\left( j\omega\left\lbrack (j\omega)^{2} + \right.\ \right.\ $ \(j\omega + 1)\).
The Nichols chart can be used for design purposes by altering the frequency response of the loop transfer fraction, \(L(s) = G_{c}(s)G(s)\), so we can obtain a desirable phase margin and \(M_{p\omega}\). The system gain \(K\) is readily adjusted to provide a suitable phase margin and \(M_{p\omega}\) by inspecting the Nichols chart. For example, let us consider again Example 9.8, where
The \(G_{c}(j\omega)G(j\omega)\)-locus on the Nichols chart for \(K = 0.64\) is shown in Figure 9.28. Let us determine a suitable value for \(K\) so that the system damping ratio is greater than 0.30. From Equation (9.63), we find that it is required that \(M_{p\omega}\) be less than \(1.75(4.9\text{ }dB)\). From Figure 9.28, we find that the \(G_{c}(j\omega)G(j\omega)\)-locus will be tangent to the \(4.9 - dB\) curve if the magnitude is lowered by a factor of \(2.2\text{ }dB\). Therefore, \(K\) should be reduced by 1.28 . Thus, the gain \(K\) must be less than \(0.64/1.28 = 0.50\) if the system damping ratio is to be greater than 0.30 .
303.1. SYSTEM BANDWIDTH
The bandwidth of the closed-loop control system is an excellent measure of the response of the system. In systems where the low-frequency magnitude is \(0\text{ }dB\) on the Bode plot, the bandwidth is measured at the \(- 3 - dB\) frequency. The speed of response to a step input will be roughly proportional to \(\omega_{B}\), and the settling time is inversely proportional to \(\omega_{B}\). Thus, we seek a large bandwidth consistent with reasonable system components [12].
Consider the two second-order systems with closed-loop transfer functions
Both systems have \(\zeta = 0.5\). The frequency response of both closed-loop systems is shown in Figure 9.29(a). The natural frequency is \(\omega_{n_{1}} = 10\) and \(\omega_{n_{2}} = 30\) for systems \(T_{1}(s)\) and \(T_{2}(s)\), respectively. The bandwidth is \(\omega_{B_{1}} = 12.7\) and \(\omega_{B_{2}} = 38.1\) for systems \(T_{1}(s)\) and \(T_{2}(s)\), respectively. Both systems have a \(P.O. = 16\%\), but \(T_{1}(s)\) has a peak time of \(T_{p} = 0.12\text{ }s\) compared to \(T_{p} = 0.36\) for \(T_{2}(\text{ }s)\), as shown in Figure 9.29(b). Also, note that the settling time for \(T_{2}(s)\) is \(T_{s} = 0.27\text{ }s\), while the settling time for \(T_{1}(s)\) is \(T_{s} = 0.8\text{ }s\). The system with a larger bandwidth provides a faster response.
303.2. THE STABILITY OF CONTROL SYSTEMS WITH TIME DELAYS
Many control systems have a time delay within the closed loop of the system that affects the stability. A time delay is the time interval between the start of an event at one point in a system and its resulting action at another point in the system. Fortunately, the Nyquist criterion can be utilized to determine the effect of the time delay on the relative stability of the feedback system. A pure time delay, without attenuation, is represented by the transfer function
where \(T\) is the delay time. The Nyquist criterion remains valid for a system with a time delay because the factor \(e^{- sT}\) does not introduce any additional poles or zeros within the contour. The factor adds a phase shift to the frequency response without altering the magnitude curve.
This type of time delay occurs in systems that have a movement of a material that requires a finite time to pass from an input or control point to an output or measured point \(\lbrack 8,9\rbrack\). For example, a steel rolling mill control system is shown in Figure 9.30. The motor adjusts the separation of the rolls so that the thickness error is minimized. If the steel is traveling at a velocity \(v\), then the time delay between the roll adjustment and the measurement is
Therefore, to have a negligible time delay, we must decrease the distance to the measurement and increase the velocity of the flow of steel. Usually, we cannot eliminate the effect of time delay; thus, the loop transfer function is [10]
FIGURE 9.29
Response of two second-order systems.
FIGURE 9.30 Steel rolling mill control system.
Frequency \((rad/s)\)
(a)
(b)
The frequency response of this system is obtained from
The frequency response of the loop transfer function is graphed on the Nyquist plot and the stability ascertained relative to the -1 point. Alternatively, we can obtain the Bode plot including the delay factor and investigate the stability relative to the \(0 - dB, - 180^{\circ}\) point. The delay factor \(e^{- j\omega T}\) results in a phase shift
and is readily added to the phase shift resulting from \(G_{c}(j\omega)G(j\omega)\). Note that the angle is in radians in Equation (9.80). An example will show the simplicity of this approach on the Bode diagram.
304. EXAMPLE 9.9 Liquid level control system
A level control system is shown in Figure 9.31(a) and the block diagram in Figure 9.31(b) [11]. The time delay between the valve adjustment and the fluid output is \(T = d/v\). Therefore, if the flow rate is \(v = 5{\text{ }m}^{3}/s\), and the distance is equal to \(d = 5\text{ }m\), then we have a time delay \(T = 1\text{ }s\). The loop transfer function is then
The Bode plot for this system is shown in Figure 9.32. The phase angle is shown both for the denominator factors alone and with the additional phase lag due to the time delay. The logarithmic gain curve crosses the 0 - \(dB\) line at \(\omega = 0.8\). Therefore, the phase margin of the system without the pure time delay would be P.M. \(= 40^{\circ}\). However, with the time delay added, we find that the phase margin is equal to \(P.M. = - 3^{\circ}\), and the system is unstable. Consequently, the system gain must be reduced in order to provide a reasonable phase margin. To provide a phase margin of \(P.M. = 30^{\circ}\), the gain would have to be decreased by a factor of \(5\text{ }dB\), to \(K = 31.5/1.78 = 17.7\).
A time delay \(e^{- sT}\) in a feedback system introduces an additional phase lag and results in a less stable system. Therefore, as pure time delays are unavoidable in many systems, it is often necessary to reduce the loop gain in order to obtain a stable response. However, the cost of stability is the resulting increase in the steadystate error of the system as the loop gain is reduced.
The systems considered by most analytical tools are described by rational functions (that is, transfer functions) or by a finite set of ordinary constant coefficient differential equations. Since the time-delay is given by \(e^{- sT}\), where \(T\) is the delay, we see that the time delay is nonrational. It would be helpful if we could obtain a rational function approximation of the time-delay. Then it would be more convenient to incorporate the delay into the block diagram for analysis and design purposes.
The Padé approximation uses a series expansion of the transcendental function \(e^{- sT}\) and matches as many coefficients as possible with a series expansion of a rational function of specified order. For example, to approximate the function \(e^{- sT}\) with a first-order rational function, we begin by expanding both functions in a series (actually a Maclaurin series \(\ ^{1}\) ),
FIGURE 9.31
(a) Liquid level control system.
(b) Block diagram.
(a)
(b)
Frequency (rad/s)
FIGURE 9.32
Bode plot for the liquid level control system. and
For a first-order approximation, we want to find \(n_{0},n_{1},d_{0}\), and \(d_{1}\) such that
Equating the corresponding coefficients of the terms in \(s\), we obtain the relationships
Solving for \(n_{0},d_{0},n_{1}\), and \(d_{1}\) yields
Setting \(d_{0} = 1\) and solving yields
A series expansion of Equation (9.83) yields
Comparing Equation (9.84) to Equation (9.82), we verify that the first three terms match. So for small \(s\), the Padé approximation is a reasonable representation of the time-delay. Higher-order rational functions can be obtained.
304.1. DESIGN EXAMPLES
In this example, we present three illustrative examples. The first example we consider is a design problem that supports green engineering and involves controlling the pitch angles of blades on large-scale wind turbines. The wind speeds are assumed to be high enough so that the pitch angle of the turbine blades can be prescribed properly to shed excess power to regulate the generated wind power at desired levels. The second example is a remotely controlled reconnaissance vehicle control design. The Nichols chart is illustrated as a key element of the design of a controller gain to meet time-domain specifications. The third example considers the control of a hot ingot robot used in manufacturing. The goal is to minimize the tracking error in the presence of disturbances and a known time-delay. The design process is illustrated, leading to a PI controller that meets a mixture of time-domain and frequency-domain performance specifications.
305. EXAMPLE 9.10 PID control of wind turbines for clean energy
Wind energy is currently the fastest-growing energy source in the world. It is a cost-effective, environmentally friendly solution to energy needs. Modern wind turbines are large, flexible structures operating in uncertain environments as wind direction and flow constantly changes. There are many controls challenges associated with efficient energy capture and delivery for wind turbines. In this design problem, we consider the so-called "above-rated" operational mode of the wind turbine. In this mode, the wind speeds are high enough that the pitch angle of the turbine blades needs to be prescribed properly to shed excess power so that the generated wind power is regulated at desired levels. This mode of operation readily permits the application of linear control theory.
Wind turbines are generally constructed in either a vertical axis configuration or a horizontal axis configuration, as shown in Figure 9.33. The horizontal axis configuration is the most common for energy production today. A horizontal axis wind turbine is mounted on a tower with two or three blades rotating placed atop a tall tower and driving an electric generator. The high placement of the blades takes advantage of the higher wind velocities. The vertical axis wind turbines are generally smaller and present a reduced noise footprint.
When there is sufficient wind, in order to regulate the rotor speed of the turbine shaft and thus the generator, the pitch of the wind turbine blades is collectively adjusted using a blade pitch motor, as illustrated in Figure 9.34(a). A simplified model of the turbine from the pitch command to the rotor speed is obtained by including a generator mode represented by a first-order transfer function in series with the
FIGURE 9.33
(a) Vertical axis wind turbine (photo courtesy of Visions of America/ SuperStock), and (b) horizontal axis wind turbine (photo courtesy of David Williams/Alamy).
(a)
(b) drive train compliance represented by a second-order transfer function [32]. The third-order transfer function of the turbine is given by
where \(K = - 7000,\tau_{g} = 5\text{ }s,\zeta_{g} = 0.005\), and \(\omega_{n_{g}} = 20rad/s\). The input to the turbine model is the commanded pitch angle (in radians) plus disturbances and the output is the rotor speed (in rpm). For commercial wind turbines, pitch control is often achieved using a PID controller, as shown in Figure 9.34(b). Selecting a PID controller requires selecting the coefficients of the controller \(K_{P},K_{I}\), and \(K_{D}\). The objective is to design the PID system for fast and accurate control. The control specifications are gain margin G.M. \(\geq 6\text{ }dB\) and phase margin \(30^{\circ} \leq\) P.M. \(\leq 60^{\circ}\). The specifications for the transient response are rise time \(T_{r_{1}} < 4\text{ }s\) and time to peak \(T_{P} < 10\text{ }s\).
The output \(\omega(s)\) shown in Figure 9.34 is actually the deviation from the rated speed of the turbine. At the rated speed, the pitch control of the blades is used to regulate the rotor speed. In the linear setting described by Figure 9.34, the input desired rotor speed \(\omega_{d}(s) = 0\) and the goal is to regulate the output to zero in the presence of disturbances.
The loop transfer function is
The objective is to determine the gains \(K_{P},K_{I}\), and \(K_{D}\) to meet the control design specifications. The phase margin specification can be used to determine a target damping of the dominant roots yielding
FIGURE 9.34
(a) Block diagram model of the wind turbine system.
(b) Block diagram for control system design.
(a)
(b) where we target for a phase margin P.M. \(= 30^{\circ}\). Then we utilize the rise time design formula to obtain a target natural frequency of the dominant roots. To this end, we use the design formula
to obtain \(\omega_{n} > 0.31\) when \(\zeta = 0.3\). For design purposes, we choose \(\omega_{n} = 0.4\) and \(\zeta = 0.3\) for the dominant poles. As a final check on the target damping and natural frequency, we verify that the time to peak specification is reachable with \(\omega_{n} = 0.4\) and \(\zeta = 0.3\). The rise time and time to peak are estimated to be
which meet the design specification. First we locate the PID zeros in the left halfplane in the desired performance region defined by \(\omega_{n}\) and \(\zeta\) by specifying the ratios \(K_{P}/K_{D}\) and \(K_{I}/K_{D}\) and select the gain \(K_{D}\) to meet the phase margin and gain margin specifications using frequency response plots (that is, Bode plot).
The Bode plot is shown in Figure 9.35, where \(K_{P}/K_{D} = 5\) and \(K_{I}/K_{D} = 20\). The value of \(K_{D} = - 6.22 \times 10^{- 6}\) was determined by observing the effects of varying the gain on the phase and gain margins and selecting the gain that satisfied the specifications as closely as possible. The PID controller is then given by
The final design results in a phase margin of \(P.M. = {32.9}^{\circ}\) and a gain margin of \(G.M. = 13.9\text{ }dB\). The step response is shown in Figure 9.36. The rise time \(T_{r_{1}} = 3.2\text{ }s\) and the time to peak \(T_{P} = 7.6\text{ }s\). All the specifications are satisfied.
FIGURE 9.35
Bode plot with \(K_{P}/K_{D} = 5\), \(K_{1}/K_{D} = 20\), and \(K_{D} = - 6.22 \times 10^{- 6}\).
FIGURE 9.36
Closed-loop step response to a unit step showing rise time and time to peak specifications are satisfied.
FIGURE 9.37
Disturbance response showing the rotor speed deviation from the rated speed.
The dominant poles of the closed-loop feedback control system are \(\omega_{n} = 0.41\) and \(\zeta = 0.29\). This is very close to the design values which demonstrates the effectiveness of the design formulas even when the system under consideration is not a second-order system.
The response of the wind turbine to an impulsive disturbance is shown in Figure 9.37. In this numerical experiment, the disturbance (possibly a wind gust) imparts a step change in the wind turbine blade pitch angle. In practice, the disturbance would lead to varying pitch angle disturbances on the each blade, but for purposes of demonstration, we model this as a single step disturbance input. The result of the disturbance is a change on the rotor speed from the nominal that is brought back to zero in about 25 seconds.
306. EXAMPLE 9.11 Remotely controlled vehicle
One concept of a remotely controlled vehicle is shown in Figure 9.38(a), \(R(s)\) and a proposed speed control system is shown in Figure 9.38(b). The desired speed is transmitted by radio to the vehicle; the disturbance \(T_{d}(s)\) represents hills and rocks. The goal is to achieve good overall control with a low steady-state error and a low-overshoot response to step commands, \(R(s)\) [13].
First, to achieve a low steady-state error for a unit step command, we calculate
where \(L(s) = G_{c}(s)G(s)\). If we select \(K = 20\), we will obtain a steady-state error of \(9\%\) of the magnitude of the input command. Using \(K = 20\), we reformulate \(L(s) = G_{c}(s)G(s)\) for the Bode plot calculations, obtaining
FIGURE 9.38
(a) Remotely controlled reconnaissance vehicle. (b) Speed control system. (a)
(b) The Nichols chart for \(K = 20\) is shown in Figure 9.39. Examining the Nichols chart, we find that \(M_{p\omega} = 12\text{ }dB\) and the phase margin is \(P.M. = 15^{\circ}\). The step response of this system is underdamped, and we use Equation (9.58) to compute \(\zeta \simeq 0.15\) to predict an excessive overshoot of approximately P.O. \(= 61\%\).
To reduce the percent overshoot to a step input, we can reduce the gain to achieve a predicted overshoot. To limit the percent overshoot to P.O. \(= 25\%\), we select a desired \(\zeta\) of the dominant roots as 0.4 and thus require \(M_{p\omega} = 1.35\) (from Equation (9.63)) or \(20logM_{p\omega} = 2.6\text{ }dB\). To lower the gain, we will move the frequency response vertically down on the Nichols chart, as shown in Figure 9.39. At \(\omega_{1} = 2.8\), we just intersect the \(2.6 - dB\) closed-loop curve. The reduction (vertical drop) in gain is equal to \(13\text{ }dB\), or a factor of 4.5. Thus, \(K = 20/4.5 = 4.44\). For this reduced gain, the steady-state error is
so that we have a \(e_{ss} = 31\%\) steady-state error.
FIGURE 9.39
Nichols chart for the design example when \(K = 20\) and for two reduced gains.
FIGURE 9.40
The response of the system for three values of \(K\) for a unit step input \(r(t)\).
The actual step response when \(K = 4.44\), as shown in Figure 9.40, has an overshoot of P.O. \(= 32\%\). If we use a gain of 10 , we have a percent overshoot of P.O. \(= 48\%\) with a steady-state error of \(e_{ss} = 17\%\). The performance of the system is summarized in Table 9.2. As a suitable compromise, we select \(K = 10\) and draw the frequency response on the Nichols chart by moving the response for \(K = 20\) down by \(20log2 = 6\text{ }dB\), as shown in Figure 9.39.
Examining the Nichols chart for \(K = 10\), we have \(M_{p\omega} = 7\text{ }dB\), and a phase margin of P.M. \(= 26^{\circ}\). Thus, we estimate a \(\zeta\) for the dominant roots of \(\zeta = 0.26\) which should result in an overshoot to a step input of \(P.O. = 43\%\). The actual response is recorded in Table 9.2. The bandwidth of the system is \(\omega_{B} = 5.4rad/s\). Therefore, we predict a settling time (with a \(2\%\) criterion) of
Table 9.2 Actual Response for Selected Gains
Percent overshoot (%)
Settling time (seconds)
Peak time (seconds)
The actual settling time is approximately \(T_{s} = 5.4\text{ }s\), as shown in Figure 9.41. The steady-state effect of a unit step disturbance can be determined by using the final-value theorem with \(R(s) = 0\), as follows:
Thus, the unit disturbance is reduced by the factor \(4 + 2K\). For \(K = 10\), we have \(y(\infty) = 1/24\), or the steady-state disturbance is reduced to \(4\%\) of the disturbance magnitude. Thus we have achieved a reasonable result with \(K = 10\).
The best compromise design would be \(K = 10\), since we achieve a compromise steady-state error of \(e_{ss} = 16.7\%\). If the percent overshoot and settling time are excessive, then we need to reshape the frequency response on the Nichols chart.
307. EXAMPLE 9.12 Hot ingot robot control
The hot ingot robot mechanism is shown in Figure 9.41. The robot picks up hot ingots and sets them in a quenching tank. A vision sensor is in place to provide a measurement of the ingot position. The controller uses the sensed position information to orient the robot over the ingot (along the \(x\)-axis). The vision sensor provides the desired position input \(R(s)\) to the controller. The block diagram depiction of the closed-loop system is shown in Figure 9.42. More information on robots and robot vision systems can be found in \(\lbrack 15,30,31\rbrack\).
The position of the robot along the track is also measured (by a sensor other than the vision sensor) and is available for feedback to the controller. We assume that the position measurement is noise free. This is not a restrictive assumption FIGURE 9.42
Hot ingot robot control system block diagram.
since many accurate position sensors are available today. For example some laser diode systems are self-contained (including the power supply, optics, and laser diode) and provide position accuracy of over \(99.9\%\).
The robot dynamics are modeled as a second-order system with two poles at \(s = - 1\) and include a time delay of \(T = \pi/4\text{ }s\). Therefore,
where \(T = \pi/4\text{ }s\). The elements of the design process emphasized in this example are highlighted in Figure 9.43. The control goal is as follows:
308. Control Goal
Minimize the tracking error \(E(s) = R(s) - Y(s)\) in the presence of external disturbances while accounting for the known time-delay.
To this end the following control specifications must be satisfied:
309. Design Specifications
DS1 Achieve a steady-state tracking error \(e_{ss} \leq 10\%\) for a step input.
DS2 Phase margin P.M. \(\geq 50^{\circ}\) with the time-delay \(T = \pi/4\text{ }s\).
DS3 Percent overshoot P.O. \(\leq 10\%\) for a step input.
Our design method is first to consider a proportional controller. We will show that the design specifications cannot be simultaneously satisfied with a proportional controller; however, the feedback system with proportional control provides a useful vehicle to discuss in some detail the effects of the time-delay. In particular, we consider the effects of the time-delay on the Nyquist plot. The final design uses a PI controller, which is capable of providing adequate performance (that is, it satisfies all design specifications).
As a first attempt, we consider a simple proportional controller:
Then ignoring the time-delay for the moment, we have the loop gain
Topics emphasized in this example
FIGURE 9.43 Elements of the control system design process emphasized in the hot ingot robot control example.
FIGURE 9.44
Hot ingot robot control system block diagram with the proportional controller and no time-delay.
The feedback control system is shown in Figure 9.44 with a proportional controller and no time-delay. The system is a type-zero system, so we expect a nonzero steadystate tracking error to a step input. The closed-loop transfer function is
With the tracking error defined as
and with \(R(s) = a/s\), where \(a\) is the input magnitude, we have
Using the final value theorem (which is possible since the system is stable for all positive values of \(K\) ) yields
Per specification DS1, we require the steady-state tracking error be less than \(10\%\). Therefore,
Solving for the appropriate gain \(K\) yields \(K \geq 9\). With \(K = 9\), we obtain the Bode plot shown in Figure 9.45.
If we raise the gain above \(K = 9\), we find that the crossover moves to the right (that is, \(\omega_{c}\) increases) and the corresponding phase margin decreases. Is a \(P.M. = {38.9}^{\circ}\) at \(\omega = 2.8rad/s\) sufficient for stability in the presence of a time-delay of \(T = \pi/4\text{ }s\) ? The addition of the time-delay term causes a phase lag without changing the magnitude plot. The amount of time-delay that our system can withstand while remaining stable is \(\phi = - \omega T\) which implies that
FIGURE 9.45 Bode plot with \(K = 9\) and no timedelay showing gain margin G.M. \(= \infty\) and phase margin P.M. \(= {38.9}^{\circ}\).
Solving for \(T\) yields \(T = 0.24\text{ }s\). Thus for time-delays less than \(T = 0.24\text{ }s\), our closed-loop system remains stable. However, the time-delay \(T = \pi/4\text{ }s\) will cause instability. Raising the gain only exacerbates matters, since the phase margin goes down further. Lowering the gain raises the phase margin, but the steady-state tracking error exceeds the \(10\%\) limit. A more complex controller is necessary. Before proceeding, let us consider the Nyquist plot and see how it changes with the addition of the time-delay. The Nyquist plot for the system (without the time-delay)
is shown in Figure 9.46, where we use \(K = 9\). The number of open-loop poles of \(G_{c}(s)G(s)\) in the right half-plane is \(P = 0\). From Figure 9.46 we see that there are no encirclements of the -1 point, thus, \(N = 0\).
By the Nyquist theorem, we know that the net number of encirclements \(N\) equals the number of zeros \(Z\) (or closed-loop system poles) in the right half-plane minus the number of open-loop poles \(P\) in the right half-plane. Therefore,
Since \(Z = 0\), the closed-loop system is stable. More importantly, even when the gain \(K\) is increased (or decreased), the -1 point is never encircled-the gain margin is \(\infty\). Similarly when the time-delay is absent, the phase margin is always positive. The value of the P.M. varies as \(K\) varies, but the P.M. is always greater than zero.
With the time-delay in the loop, we can rely on analytic methods to obtain the Nyquist plot. The loop transfer function with the time-delay is
FIGURE 9.46 Nyquist plot with \(K = 9\) and no time-delay showing no encirclements of the minus 1 point.
Using the Euler identity
\(L(s)\) and substituting \(s = j\omega\) into \(L(s)\) yields
where
Generating a plot of \(Re(L(j\omega))\) versus \(Im(L(j\omega))\) yields the plot shown in Figure 9.47. With \(K = 9\), the number of encirclements of the -1 point is \(N = 2\). Therefore, the system is unstable since \(Z = N + P = 2\).
Figure 9.48 shows the Nyquist plot for four values of time-delay: \(T = 0,0.1\), 0.24 , and \(\pi/4 = 0.78\text{ }s\). For \(T = 0\) there is no possibility of an encirclement of the -1 point as \(K\) varies (see the upper left graph of Figure 9.48). We have stability (that is, \(N = 0\) ) for \(T = 0.1\text{ }s\) (upper right graph), marginal stability for \(T = 0.24\text{ }s\) (lower left graph), and for \(T = \pi/4 = 0.78\text{ }s\) we have \(N = 1\) (lower right graph), thus the closed-loop system is unstable.
Since we know that \(T = \pi/4\) in this example, the proportional gain controller is not a viable controller. With it we cannot meet the steady-state error specifications and have a stable closed-loop system in the presence of the time-delay \(T = \pi/4\).
FIGURE 9.47 Nyquist plot with \(K = 9\) and \(T = \pi/4\) showing two encirclements of the -1 point, \(N = 2\).
FIGURE 9.48 Nyquist plot with \(K = 9\) and various time-delays.
However, before proceeding with the design of a controller that meets all the specifications, let us take a closer look at the Nyquist plot with a time-delay.
Suppose we have the case where \(K = 9\) and \(T = 0.1\text{ }s\). The associated Nyquist plot is shown in the upper right of Figure 9.48. The Nyquist plot intersects (or crosses over) the real axis whenever the imaginary part of \(G_{c}(j\omega)G(j\omega) = 0\), or
Thus we obtain the relation that describes the frequencies \(\omega\) at which crossover occurs:
Equation (9.90) has an infinite number of solutions. The first real-axis crossing (farthest in the left half-plane) occurs when \(\omega = 4.43rad/s\).
The magnitude of \(|L(j4.43)|\) is equal to \(0.0484\text{ }K\). For stability we require that \(|L(j\omega)| < 1\) when \(\omega = 4.43\) (to avoid an encirclement of the -1 point). Thus, for stability we find
when \(T = 0.1\). When \(K = 9\), the closed-loop system is stable, as we already know. If the gain \(K = 9\) increases by a factor of 2.3 to \(K = 20.67\), we will be on the border of instability. This factor \(\delta\) is the gain margin:
Consider the PI controller
The loop system transfer function is
The system type is now equal to 1; thus we expect a zero steady-state error to a step input. The steady-state error specification DS1 is satisfied. We can now concentrate on meeting specification DS3, P.O. \(< 10\%\) and DS2, the requirement for stability in the presence of the time-delay \(T = \pi/4\text{ }s\).
From the percent overshoot specification we can determine a desired system damping ratio. Thus we determine for P.O. \(\leq 10\%\) that \(\zeta \geq 0.59\). Due to the PI controller, the system now has a zero at \(s = - K_{I}/K_{P}\). The zero will not affect the closed-loop system stability, but it will affect the performance. Using the approximation (valid for small \(\zeta,P.M\). expressed in degrees)
we determine a good target phase margin (since we want \(\zeta \geq 0.59\) ) to be \(60\%\). We can rewrite the PI controller as
where \(1/\tau = K_{I}/K_{P}\) is the break frequency of the controller. The PI controller is essentially a low-pass filter and adds phase lag to the system below the break frequency. We would like to place the break frequency below the crossover frequency so that the phase margin is not reduced significantly due to the presence of the PI zero.
The uncompensated Bode plot is shown in Figure 9.49 for
where \(T = \pi/4\). The uncompensated system phase margin is \(P.M. = - {88.34}^{\circ}\) at \(\omega_{c} = 2.83rad/s\). Since we want P.M. \(= 60^{\circ}\), we need the phase to be minus \(120^{\circ}\) at the crossover frequency. In Figure 9.49 we can estimate the phase \(\phi = - 120^{\circ}\) at \(\omega \approx 0.87rad/s\). This is an approximate value but is sufficiently accurate for the design procedure. At \(\omega = 0.87\) the magnitude is about \(14.5\text{ }dB\). If we want the FIGURE 9.49
Uncompensated Bode plot with \(K = 9\) and \(T = \pi/4\).
crossover to be \(\omega_{c} = 0.87rad/s\), the controller needs to attenuate the system gain by \(14.5\text{ }dB\), so that the magnitude is \(0\text{ }dB\) at \(\omega_{c} = 0.87\). With
we can consider \(K_{P}\) to be the gain of the compensator (a good approximation for large \(\omega)\). Therefore,
Finally we need to select \(K_{I}\). Since we want the break frequency of the controller to be below the crossover frequency (so that the phase margin is not reduced significantly due to the presence of the PI zero), a good rule-of-thumb is to select \(1/\tau = K_{I}/K_{P} = 0.1\omega_{c}\). To make the break frequency of the controller zero one decade below the crossover frequency. The final value of \(K_{I}\) is computed to be \(K_{I} = 0.1\omega_{c}K_{P} = 0.0164\), where \(\omega_{c} = 0.87rad/s\). Thus the PI controller is
The Bode plot of \(G_{c}(s)G(s)e^{- sT}\) is shown in Figure 9.50, where \(T = \pi/4\). The gain and phase margins are G.M. \(= 5.3\text{ }dB\) and \(P.M. = {56.5}^{\circ}\).
We consider whether the design specifications have been met. The steadystate tracking specification (DS1) is certainly satisfied since our system is type one; FIGURE 9.50
Compensated Bode plot with \(K = 9\) and \(T = \pi/4\) and with its \(PI\) controller.
FIGURE 9.51 Hot ingot robot control step response with the PI controller.
the PI controller introduced an integrator. The phase margin (with the time-delay) is \(P.M. = {56.5}^{\circ}\), so the phase margin specification, DS2, is satisfied. The unit step response is shown in Figure 9.51. The percent overshoot is approximately P.O. \(\approx 4.2\%\). The target percent overshoot was P.O. \(= 10\%\), so DS3 is satisfied. Overall the design specifications are satisfied.
309.1. PID CONTROLLERS IN THE FREQUENCY DOMAIN
The PID controller provides a proportional term, an integral term, and a derivative term. We then have the PID controller transfer function as
In general, we note that PID controllers are particularly useful for reducing the steady-state error and improving the transient response when \(G(s)\) has one or two poles (or may be approximated by a second-order process).
We may use frequency response methods to represent the addition of a PID controller. The PID controller, Equation (9.93), may be rewritten as
The Bode plot of Equation (9.94) is shown in Figure 9.52 for \(K_{I} = 2,\tau = 1\), and \(\alpha = 10\). The PID controller is a form of a notch (or bandstop) compensator with a variable gain, \(K_{I}\). Of course, it is possible that the controller will have complex zeros and a Bode plot that will be dependent on the \(\zeta\) of the complex zeros. The PID controller with complex zeros is
Typically, we choose \(0.9 > \zeta > 0.7\).
FIGURE 9.52
Bode plot for a PID controller using the asymptomatic approximation for the magnitude curve with \(K_{1} = 2,\alpha = 10\), and \(\tau = 1\).
309.2. STABILITY IN THE FREQUENCY DOMAIN USING CONTROL DESIGN SOFTWARE
We now approach the issue of stability using the computer as a tool. This section revisits the Nyquist plot, the Nichols chart, and the Bode plot in our discussions on relative stability. Two examples will illustrate the frequency-domain design approach. We will make use of the frequency response of the closed-loop transfer function \(T(j\omega)\) as well as the loop transfer function \(L(j\omega)\). We also present an illustrative example that shows how to deal with a time delay in the system by utilizing a Padé approximation [6]. The functions covered in this section are nyquist, nichols, margin, pade, and ngrid.
It is generally more difficult to manually generate the Nyquist plot than the Bode plot. However, we can use the control design software to generate the Nyquist plot. The Nyquist plot is generated with the nyquist function, as shown in Figure 9.53. When nyquist is used without left-hand arguments, the Nyquist plot is automatically generated; otherwise, the real and imaginary parts of the frequency response (along with the frequency vector \(\omega\) ) is returned. An illustration of the nyquist function is given in Figure 9.54.
As discussed in Section 9.4, relative stability measures of gain margin and phase margin can be determined from both the Nyquist plot and the Bode plot. The gain margin is a measure of how much the system gain would have to be increased for the \(L(j\omega)\) locus to pass through the \(- 1 + j0\) point, thus resulting in an unstable system. The phase margin is a measure of the additional phase lag required before the system becomes unstable. Gain and phase margins can be determined from both the Nyquist plot and the Bode plot.
FIGURE 9.53 The nyquist function. \(> >\) num \(= \lbrack 0.5\rbrack;\) den=[1 \(\left. \ \begin{matrix} 1 & 1 & 0.5 \end{matrix} \right\rbrack\);
\(> >\) sys=tf(num,den);
nyquist(sys)
FIGURE 9.54 An example of the nyquist function.
FIGURE 9.55 A closed-loop control system example for Nyquist and Bode with relative stability.
Consider the system shown in Figure 9.55. Relative stability can be determined from the Bode plot using the margin function, which is shown in Figure 9.56. If the margin function is invoked without left-hand arguments, the Bode plot is automatically generated with the gain and phase margins labeled on the plot. This is illustrated in Figure 9.57 for the system shown in Figure 9.55.
The script to generate the Nyquist plot for the system in Figure 9.55 is shown in Figure 9.58. In this case, the number of poles of \(L(s) = G_{c}(s)G(s)\) with positive real parts is zero, and the number of counterclockwise encirclements of -1 is zero; hence, the closed-loop system is stable. We can also determine the gain margin and phase margin, as indicated in Figure 9.58.
Nichols charts can be generated using the nichols function, shown in Figure 9.59. If the nichols function is invoked without left-hand arguments, the Nichols chart is automatically generated; otherwise the nichols function returns the magnitude and FIGURE 9.56
The margin function.
Example
num \(= \lbrack 0.5\rbrack;\) den \(= \begin{bmatrix} 1 & 2 & 1 & 0.5 \end{bmatrix}\) sys=tf(num, den); margin(sys);
\(Gm =\) gain margin \((dB)\)
\(Pm =\) phase margin \((deg)\)
\(Wcg =\) freq. for phase \(= - 180\)
\(Wcp =\) freq. for gain \(= 0\text{ }dB\)
Frequency ( \(rad/s)\)
FIGURE 9.57
The Bode plot for the system in Figure 9.55 with the gain margin and the phase margin indicated on the plots.
(a)
(b)
FIGURE 9.58 (a) The Nyquist plot for the system in Figure 9.55 with gain and phase margins.
(b) m-file script.
phase in degrees (along with the frequency \(\omega\) ). A Nichols chart grid is drawn on the existing plot with the ngrid function. The Nichols chart, shown in Figure 9.60, is for the system
FIGURE 9.59
The nichols function.
Set up to generate
Figure 9.27
num=[1]; den=[0.2 1.210 ]; sys=tf(num,den);
\(w =\) logspace \(( - 1,1,400)\);
nichols(sys, w);
ngrid
Plot Nichols chart and add grid lines.
310. EXAMPLE 9.13 Liquid level control system
Consider a liquid level control system described by the block diagram shown in Figure 9.31. Note that this system has a time delay. The loop transfer function is given by
We first change Equation (9.97) in such a way that \(L(s)\) has a transfer function form with polynomials in the numerator and the denominator. To do this, we can make an approximation to \(e^{- sT}\) with the pade function, shown in Figure 9.61. For example, suppose our time delay is \(T = 1\text{ }s\), and we want a second-order approximation \(n = 2\). Using the pade function, we find that
Substituting Equation (9.98) into Equation (9.97) yields
Now we can build a script to investigate the relative stability of the system using the Bode plot. Our goal is to have a phase margin of \(P.M. = 30^{\circ}\). The associated script is shown in Figure 9.62. To make the script interactive, we let the gain \(K\) (now set at \(K = 31.5\) ) be adjustable and defined outside the script at the command level. Then we set \(K\) and run the script to check the phase margin and iterate if necessary. The final selected gain is \(K = 16\). Remember that we have utilized a second-order Padé approximation of the time delay in our analysis.
311. EXAMPLE 9.14 Remotely controlled vehicle
Consider the speed control system for a remotely controlled vehicle shown in Figure 9.38. The design objective is to achieve good control with a low steady-state error and a low overshoot to a step command. Building a script will allow us to
(a)
(a) Bode plot for the liquid level control system. (b) m-file script.
(b)
perform many design iterations quickly and efficiently. First, we investigate the steady-state error specification. The steady-state error to a unit step command is
The effect of the gain \(K\) on the steady-state error is clear from Equation (9.99): If \(K = 20\), the error is \(e_{ss} = 9\%\) of the input magnitude; if \(K = 10\), the error is \(e_{ss} = 17\%\) of the input magnitude. Now we can investigate the overshoot specification in the frequency domain. Suppose we require that the percent overshoot is less than \(50\%\). Solving
for \(\zeta\) yields \(\zeta \geq 0.215\). Referring to Equation (9.63), we find that \(M_{p\omega} \leq 2.45\). We must keep in mind that Equation (9.63) is for second-order systems only and can be used here only as a guideline. We now compute the closed-loop Bode plot and check the values of \(M_{p\omega}\). Any gain \(K\) for which \(M_{p\omega} \leq 2.45\) may be a valid gain for our design, but we will have to investigate step responses further to check the actual overshoot. The script in Figure 9.63 aids us in this task. We further investigate the gains \(K = 20,10\), and 4.44 (even though \(M_{p\omega} > 2.45\) for \(K = 20\) ).
We can plot the step responses to quantify the percent overshoot as shown in Figure 9.64. Additionally, we could have used a Nichols chart to aid the design process, as shown in Figure 9.65.
(a)
FIGURE 9.63
Remotely controlled vehicle. (a) Closed-loop system Bode plot. (b) m-file script.
FIGURE 9.64
Remotely controlled vehicle. (a) Step response. (b) \(m\)-file script.
(a)
(b)
The results of the analysis are summarized in Table 9.3 for \(K = 20,10\), and 4.44. We choose \(K = 10\) as our design gain. Then we obtain the Nyquist plot and check relative stability, as shown in Figure 9.66. The gain margin is G.M. \(= \infty\) and the phase margin is \(P.M. = {26.1}^{\circ}\).
311.1. SEQUENTIAL DESIGN EXAMPLE: DISK DRIVE READ SYSTEM
In this chapter, we will examine the disk drive read system including the effect of the flexure resonance and incorporating a PD controller with a zero at \(s = - 1\). We will determine the system gain margin and phase margin when \(K = 400\).
The Bode plot for the system when \(K = 400\) is shown in Figure 9.67. The gain margin is G.M. \(= 22.9\text{ }dB\), and the phase margin is \(P.M. = {37.2}^{\circ}\). The plot of the step response of this system is shown in Figure 9.68. The settling time of this design is \(T_{s} = 9.6\text{ }ms\).
(a)
FIGURE 9.65
Remotely controlled vehicle. (a) Nichols chart. (b) m-file script.
(b)
(a)
(a) Nyquist plot for the remotely controlled vehicle with \(K = 10\).
(b) m-file script.
(b)
FIGURE 9.67
Bode plot of the disk drive read system.
Frequency \((rad/s)\) FIGURE 9.68
Response of the disk drive read system to a step input.
311.2. SUMMARY
The stability of a feedback control system can be determined in the frequency domain by utilizing Nyquist's criterion. Furthermore, Nyquist's criterion provides us with two relative stability measures: (1) gain margin and (2) phase margin. These relative stability measures can be utilized as indices of the transient performance on the basis of correlations established between the frequency domain and the time-domain transient response. The magnitude and phase of the closed-loop system can be determined from the frequency response of the open-loop transfer function by utilizing constant magnitude and phase circles on the polar plot. Alternatively, we can utilize a log-magnitudephase diagram with closed-loop magnitude and phase curves superimposed (called the Nichols chart) to obtain the closed-loop frequency response. A measure of relative stability, the maximum magnitude of the closed-loop frequency response, \(M_{p\omega}\), is available from the Nichols chart. The frequency response, \(M_{p\omega}\), can be correlated with the damping ratio of the time response and is a useful index of performance. Finally, a control system with a pure time delay can be investigated in a manner similar to that for systems without time delay. A summary of the Nyquist criterion, the relative stability measures, and the Nichols chart is given in Table 9.3 for several transfer functions.
Table 9.3 is very useful and important to the designer and analyst of control systems. If we have the model of a process \(G(s)\) and a controller \(G_{c}(s)\), then we can determine \(L(s) = G_{c}(s)G(s)\). With this loop transfer function, we can examine the transfer function table in column 1. This table contains fifteen typical transfer functions. For a selected transfer function, the table gives the Bode plot, the Nichols chart, and the root locus. With this information, the designer can determine or estimate the performance of the system and consider the addition or alteration of the controller \(G_{c}(s)\). Table 9.3 Key Plots for Typical Loop Transfer Functions
-
\(\frac{K}{s\tau_{1} + 1}\)
-
\(\frac{K}{\left( s\tau_{1} + 1 \right)\left( s\tau_{2} + 1 \right)}\)
-
\(\frac{K}{\left( s\tau_{1} + 1 \right)\left( s\tau_{2} + 1 \right)\left( s\tau_{3} + 1 \right)}\)
-
\(\frac{K}{S}\)
Table 9.3 (continued)
Nichols Chart
312. Comments
Stable; gain margin \(= \infty\)
Elementary regulator; stable; gain margin \(= \infty\)
Regulator with additional energystorage component; unstable, but can be made stable by reducing gain
Ideal integrator; stable Nyquist Plot
-
\(\frac{K}{s\left( s\tau_{1} + 1 \right)}\)
-
\(\frac{K}{s\left( s\tau_{1} + 1 \right)\left( s\tau_{2} + 1 \right)}\)
-
\(\frac{K}{s^{2}}\)
313. Bode Plot
Table 9.3 (continued)
Nichols Chart
314. Root Locus
315. Comments
Elementary instrument servo; inherently stable; gain margin \(= \infty\)
Instrument servo with field control motor or power servo with elementary Ward-Leonard drive; stable as shown, but may become unstable with increased gain
Elementary instrument servo with phase-lead (derivative) compensator; stable
Inherently marginally stable; must be compensated
316. Nyquist Plot
-
\(\frac{K}{s^{2}\left( s\tau_{1} + 1 \right)}\)
-
\(\frac{K\left( s\tau_{a} + 1 \right)}{s^{3}}\)
317. Bode Plot
Table 9.3 (continued)
Nichols Chart
Root Locus
318. Comments
Inherently unstable; must be compensated
Stable for all gains
Inherently unstable
Inherently unstable 13. \(\frac{K\left( s\tau_{a} + 1 \right)\left( s\tau_{b} + 1 \right)}{s^{3}}\)
-
\(\frac{K\left( s\tau_{a} + 1 \right)\left( s\tau_{b} + 1 \right)}{s\left( s\tau_{1} + 1 \right)\left( s\tau_{2} + 1 \right)\left( s\tau_{3} + 1 \right)\left( s\tau_{4} + 1 \right)}\)
-
\(\frac{K\left( s\tau_{a} + 1 \right)}{s^{2}\left( s\tau_{1} + 1 \right)\left( s\tau_{2} + 1 \right)}\)
Table 9.3 (continued)
Nichols Chart
319. Comments
Conditionally stable; becomes unstable if gain is too low
Conditionally stable; stable at low gain, becomes unstable as gain is raised, again becomes stable as gain is further increased, and becomes unstable for very high gains
Conditionally stable; becomes unstable at high gain
320. SKILLS CHECK
In this section, we provide three sets of problems to test your knowledge: True or False, Multiple Choice, and Word Match. To obtain direct feedback, check your answers with the answer key provided at the conclusion of the end-of-chapter problems. Use the block diagram in Figure 9.69 as specified in the various problem statements.
FIGURE 9.69 Block diagram for the Skills Check.
In the following True or False and Multiple Choice problems, circle the correct answers.
- The gain margin of a system is the increase in the system gain when the phase is \(- 180^{\circ}\) that will result in a marginally stable system.
True or False
- A conformal mapping is a contour mapping that retains the angles on the \(s\)-plane on the transformed \(F(s)\)-plane.
True or False
- The gain and phase margin are readily evaluated on either a Bode plot or a Nyquist plot.
True or False
- A Nichols chart displays curves describing the relationship between the open-loop and closed-loop frequency responses.
True or False
- The phase margin of a second-order system (with no zeros) is a function of both the damping ratio \(\zeta\) and the natural frequency, \(\omega_{n}\).
True or False
- Consider the closed-loop system in Figure 9.69 where
The crossover frequency and the phase margin are:
a. \(\omega = 2.0rad/s,P.M. = {37.2}^{\circ}\)
b. \(\omega = 2.6rad/s,P.M. = {54.9}^{\circ}\)
c. \(\omega = 5.3rad/s,P.M. = {68.1}^{\circ}\)
d. \(\omega = 10.7rad/s,P.M. = {47.9}^{\circ}\)
- Consider the block diagram in Figure 9.69. The plant transfer function is
and the controller is
Utilize the Nyquist stability criterion to characterize the stability of the closed-loop system.
a. The closed-loop system is stable.
b. The closed-loop system is unstable.
c. The closed-loop system is marginally stable.
d. None of the above.
For Problems 8 and 9, consider the block diagram in Figure 9.69 where
and the controller is the proportional-plus-derivative (PD) controller
-
When \(T_{d} = 0\), the PD controller reduces to a proportional controller, \(G_{c}(s) = K\). In this case, use the Nyquist plot to determine the limiting value of \(K\) for closed-loop stability.
a. \(K = 0.5\)
b. \(K = 1.6\)
c. \(K = 2.4\)
d. \(K = 4.3\) -
Using the value of \(K\) in Problem 8, compute the gain and phase margins when \(T_{d} = 0.2\).
a. \(G \cdot M. = 14\text{ }dB,P.M. = 27^{\circ}\)
b. \(G.M. = 20\text{ }dB,P.M. = {64.9}^{\circ}\)
c. \(G.M. = \infty dB,P.M. = 60^{\circ}\)
d. Closed-loop system is unstable -
Determine whether the closed-loop system in Figure 9.69 is stable or not, given the loop transfer function
In addition, if the closed-loop system is stable, compute the gain and phase margins.
a. Stable, G.M. \(= 24\text{ }dB,P.M. = {2.5}^{\circ}\)
b. Stable, G.M. \(= 3\text{ }dB,P.M. = 24^{\circ}\)
c. Stable, G.M. \(= \infty dB,P.M. = 60^{\circ}\)
d. Unstable
- Consider the closed-loop system in Figure 9.69, where the loop transfer function is
Determine the value of the gain \(K\) such that the phase margin is P.M. \(= 40^{\circ}\).
a. \(K = 1.64\)
b. \(K = 2.15\)
c. \(K = 2.63\)
d. Closed-loop system is unstable for all \(K > 0\) 12. Consider the feedback system in Figure 9.69 , where
Notice that the plant contains a time-delay of \(T = 0.2\) seconds. Determine the gain \(K\) such that the phase margin of the system is \(P.M. = 50^{\circ}\). What is the gain margin for the same gain \(K\) ?
a. \(K = 8.36\), G.M. \(= 2.6\text{ }dB\)
b. \(K = 2.15,G \cdot M. = 10.7\text{ }dB\)
c. \(K = 5.22,G.M. = \infty dB\)
d. \(K = 1.22,G \cdot M. = 14.7\text{ }dB\)
- Consider the control system in Figure 9.69, where the loop transfer function is
The value of the resonant peak, \(M_{p_{\omega}}\) and the damping factor, \(\zeta\), for the closed-loop system are:
a. \(M_{p_{\omega}} = 0.37,\zeta = 0.707\)
b. \(M_{p_{\omega}} = 1.15,\zeta = 0.5\)
c. \(M_{p_{\omega}} = 2.55,\zeta = 0.5\)
d. \(M_{p_{\omega}} = 0.55,\zeta = 0.25\)
- A feedback model of human reaction time used in analysis of vehicle control can use the block diagram model in Figure 9.69 with
A typical driver has a reaction time of \(T = 0.3\text{ }s\). Determine the bandwidth of the closedloop system.
a. \(\omega_{b} = 0.5rad/s\)
b. \(\omega_{b} = 10.6rad/s\)
c. \(\omega_{b} = 1.97rad/s\)
d. \(\omega_{b} = 200.6rad/s\)
- Consider a control system with unity feedback as in Figure 9.69 with loop transfer function
The gain and phase margin are:
a. \(G \cdot M. = \infty dB,P.M. = {58.1}^{\circ}\)
b. \(G.M. = 20.4\text{ }dB,P.M. = {47.3}^{\circ}\)
c. \(G.M. = 6.6\text{ }dB,P.M. = {60.4}^{\circ}\)
d. Closed-loop system is unstable In the following Word Match problems, match the term with the definition by writing the correct letter in the space provided.
a. Time delay
The frequency response of the closed-loop transfer function \(T(j\omega)\).
b. Cauchy's theorem
A chart displaying the curves for the relationship between the open-loop and closed-loop frequency response.
c. Bandwidth
A contour mapping that retains the angles on the \(s\)-plane on the \(F(s)\)-plane.
d. Contour map
e. Nichols chart
If a contour encircles \(Z\) zeros and \(P\) poles of \(F(s)\) traversing clockwise, the corresponding contour in the \(F(s)\)-plane encircles the origin of the \(F(s)\)-plane \(N = Z - P\) times clockwise.
The amount of phase shift of \(G_{c}(j\omega)G(j\omega)\) at unity magnitude that will result in a marginally stable system with intersections of the point \(- 1 + j0\) on the Nyquist diagram.
f. Closed-loop frequency response
Events occurring at time \(t\) at one point in the system occur at another point in the system at a later time, \(t + T\).
g. Logarithmic (decibel) measure
A feedback system is stable if and only if the contour in the \(G(s)\)-plane does not encircle the \(( - 1,0)\) point when the number of poles of \(G(s)\) in the right-hand \(s\)-plane is zero. If \(G(s)\) has \(P\) poles in the right-hand plane, then the number of counterclockwise encirclements of the \(( - 1,0)\) point must be equal to for a stable system.
h. Gain margin
i. Nyquist stability criterion
A contour or trajectory in one plane is mapped into another plane by a relation \(F(s)\).
The increase in the system gain when plane \(= - 180^{\circ}\) that will result in a marginally stable system with intersection of the \(- 1 + j0\) point on the Nyquist diagram.
j. Phase margin
The frequency at which the frequency response has declined \(3\text{ }dB\) from its low-frequency value.
k. Conformal mapping
A measure of the gain margin.
321. EXERCISES
E9.1 A system has the open-loop transfer function
Obtain the Bode plot. Show that the phase margin is \(P.M. = {20.1}^{\circ}\) and that the gain margin is \(G.M. =\) \(6.61\text{ }dB\).
E9.2 A system has the loop transfer function
where \(K = 4\). Show that the system crossover frequency is \(\omega_{c} = 3.51rad/s\) and that the phase margin is P.M. \(= {56.9}^{\circ}\). E9.3 An integrated circuit is available to serve as a feedback system to regulate the output voltage of a power supply. The Bode plot of the loop transfer function is shown in Figure E9.3 Estimate the phase margin of the regulator.
Answer: P.M. \(= 75^{\circ}\)
FIGURE E9.3 Power supply regulator.
E9.4 Consider a system with a loop transfer function
We wish to obtain a resonant peak \(M_{p\omega} = 3.0\text{ }dB\) for the closed-loop system. The peak occurs between 6 and \(9rad/s\) and is only \(1.25\text{ }dB\). Plot the Nichols chart for the range of frequency from 6 to \(15rad/s\). Show that the system gain needs to be raised by \(4.6\text{ }dB\) to 171. Determine the resonant frequency for the adjusted system.
Answer: \(\omega_{r} = 11rad/s\)
E9.5 An integrated CMOS digital circuit can be represented by the Bode plot shown in Figure E9.5. (a) Find the gain and phase margins of the circuit. (b) Estimate how much we would need to reduce the system gain (dB) to obtain a phase margin of P.M. \(= 60^{\circ}\).
E9.6 A system has a loop transfer function
Determine the range of \(K\) for closed-loop stability. Find the gain margin and phase margin of the system with \(K = 40\).
E9.7 A unity feedback system has a loop transfer function
Determine the range of \(K\) for which the system is stable using the Nyquist plot.
(a)
FIGURE E9.5 Bode plot of the CMOS circuit.
E9.8 Consider a unity feedback with the loop transfer function
(a) For \(K = 27\), show that the gain margin is G.M. \(= 11\text{ }dB\).
(b) To achieve a gain margin G.M. \(= 28\text{ }dB\), determine the value of the gain \(K\).
Answer: (b) \(K = 3.8\)
E9.9 Consider a unity feedlack system with loop transfer function
Compute the phase margin and gain margin.
E9.10 Consider a system with the loop transfer function
Obtain the Bode plot, and show that the P.M. \(= {16.9}^{\circ}\) and that the G.M. \(= 9.63\text{ }dB\). Also, show that the bandwidth of the closed-loop system is \(\omega_{B} = 4.57rad/s\).
E9.11 Consider a unity feedback system with the loop transfer function
(a) Obtain the Bode plot. (b) Find the gain margin and the phase margin. E9.12 A unity feedback system with the loop transfer function
where \(\tau_{1} = 0.01\) and \(\tau_{2} = 0.3\text{ }s\). (a) Select a gain \(K\) so that the steady-state error for a ramp input is \(20\%\) of the magnitude of the ramp function \(A\), where \(r(t) = A(t),t \geq 0\). (b) Obtain the Bode plot of loop transfer function, and determine the phase and gain margins. (c) Determine the bandwidth \(\omega_{B}\), the resonant peak \(M_{p\omega}\), and the resonant frequency \(\omega_{r}\) of the closed-loop system.
322. Answer:
(a) \(K = 20\)
(b) P.M. \(= 19^{\circ},G \cdot M. = 14.6\text{ }dB\)
(c) \(\omega_{B} = 12.4,M_{p\omega} = 9.82,\omega_{r} = 7.92\)
E9.13 A unity feedback system has a loop transfer function
(a) Find the maximum magnitude of the closed-loop frequency response. (b) Find the bandwidth and the resonant frequency of this system. (c) Use these frequency measures to estimate the overshoot of the system to a step response.
Answers:(a) \(11.1\text{ }dB,(\text{ }b)\omega_{B} = 21.8rad/s,\omega_{r} = 13.9\)
E9.14 A Nichols chart is given in Figure E9.14 for a system with \(G_{c}(j\omega)G(j\omega)\). Using the following table, find (a) the peak resonance \(M_{p\omega}\) in \(dB\); (b) the resonant frequency \(\omega_{r}\); (c) the 3-dB bandwidth; and (d) the phase margin of the system.
| $$\omega_{1}$$ | $$\omega_{2}$$ | $$\omega_{3}$$ | $$\omega_{4}$$ | |
| $$\mathbf{rad}/\mathbf{s}$$ | 1 | 3 | 6 | 10 |
E9.15 Consider a unity feedback system with the loop transfer function
Find the bandwidth of the closed-loop system.
Answer: \(\omega_{B} = 6.4rad/s\)
FIGURE E9.14 Nichols chart for \(G_{c}(j\omega)G(j\omega)\).
E9.16 The pure time delay \(e^{- sT}\) may be approximated by a transfer function as
Obtain the Bode plot for the actual transfer function and the approximation for \(T = 0.05\) for \(0 < \omega < 100\).
(a) Obtain the Bode plot, and (b) determine the gain \(K\) required to obtain a phase margin of \(P \cdot M. = 56^{\circ}\). What is the steady-state error for a ramp input for the gain of part (b)?
E9.17 A unity feedback system has a loop transfer E9.18 An actuator for a disk drive uses a shock mount function
to absorb vibrational energy at approximately \(60\text{ }Hz\)
(a)
RE E9.18 FIGURE E9.20
Automobile control system.
[14]. The Bode plot of the loop transfer function of the control system is shown in Figure E9.18. (a) Find the expected percent overshoot for a step input for the closed-loop system, (b) estimate the bandwidth of the closed-loop system, and (c) estimate the settling time (with a \(2\%\) criterion) of the system.
E9.19 A unity feedback system with \(G_{c}(s) = K\) has
Select a gain \(K\) so that the phase margin of the system is \(P.M. = 40^{\circ}\). Determine the gain margin for the selected gain, \(K\).
E9.20 Consider a simple model of an automobile driver following another car on the highway at high speed. The model shown in Figure E9.20 incorporates the driver's reaction time, \(T\). One driver has \(T = 1\text{ }s\), and another has \(T = 1.5\text{ }s\). Determine the time response \(y(t)\) of the system for both drivers for a step change in the command signal \(R(s) = - 1/s\), due to the braking of the lead car.
E9.21 A unity feedback control system has a loop transfer function
Determine the phase margin, the crossover frequency, and the gain margin when \(K = 1300\).
E9.22 A unity feedback system has a loop transfer function
(a) Using a Bode plot for \(K = 40\), determine the system phase margin. (b) Select a gain \(K\) so that the phase margin is \(P.M. \geq 55^{\circ}\).
E9.23 Consider again the system of E9.21 when \(K = 100\). Determine the closed-loop system bandwidth, resonant frequency, and \(M_{p\omega}\).
Answers: \(\omega_{B} = 5.44rad/s,\omega_{r} = 3.61rad/s\), \(M_{p\omega} = 5.75\)
E9.24 A unity feedback system has a loop transfer function
where \(K = 0.4\) and \(\tau = 1\). The Nyquist plot for \(G_{c}(j\omega)G(j\omega)\) is shown in Figure E9.24. Determine whether the system is stable by using the Nyquist criterion.
FIGURE E9.24 Nyquist plot for \(G_{c}(s)G(s) = K/( - 1 + \tau s)\) where \(K = 0.4\).
E9.25 A unity feedback system has a loop transfer function
Determine the phase margin and the crossover frequency.
Answer: P.M. \(= 28^{\circ},\omega_{c} = 8.31rad/s\)
E9.26 For the system of E9.25, determine \(M_{p\omega},\omega_{r}\), and \(\omega_{B}\) for the closed-loop frequency response by using the Nichols chart.
E9.27 A unity feedback system has a loop transfer function
(a) Determine the maximum gain \(K\) for which the phase margin is \(P.M. \geq 30^{\circ}\), and the gain margin is G.M. \(= 8\text{ }dB\). (b) Determine the value of gain \(K\) and cross over frequency for marginal stability.
E9.28 A unity feedback system has the loop transfer function
(a) Determine the phase margin for the system. (b) Use the phase margin to estimate the damping ratio, and predict the percent overshoot. (c) Calculate the actual response for this system, and compare the result with the part (b) estimate.
E9.29 A loop transfer function is
Using the contour in the \(s\)-plane shown in Figure E9.29, determine the corresponding contour in the \(F(s)\)-plane \((B = - 1 + j)\).
FIGURE E9.29 Contour in the s-plane.
E9.30 A system is represented in state variable form
where
Sketch the Bode plot.
E9.31 A closed-loop feedback system is shown in Figure E9.31. Sketch the Bode plot, and determine the phase margin.
FIGURE E9.31 Nonunity feedback system.
E9.32 Consider the system described in state variable form by
where
Compute the phase margin.
E9.33 Consider the system shown in Figure E9.33. Compute the loop transfer function \(L(s)\), and sketch the Bode plot. Determine the phase margin and gain margin when the controller gain \(K = 5\).
FIGURE E9.33
Nonunity feedback system with proportional controller \(K\).
323. PROBLEMS
P9.1 For the Nyquist plots of Problem P8.1, use the Nyquist criterion to ascertain the stability of the various systems. In each case, specify the values of \(N,P\), and \(Z\).
P9.2 Sketch the Nyquist plots of the following loop transfer functions \(L_{1}(s) = Gc_{1}(s)G_{1}(s)\), and determine whether the system is stable by applying the Nyquist criterion:
(a) \(L(s) = G_{c}(s)G(s) = \frac{K}{s\left( s^{2} + 2s + 5 \right)}\).
(b) \(L(s) = G_{c}(s)G(s) = \frac{K(s + 2)}{s^{2}(s + 5)}\).
If the system is stable, find the maximum value for \(K\) by determining the point where the Nyquist plot crosses the \(u\)-axis.
P9.3 (a) Find a suitable contour \(\Gamma_{s}\) in the \(s\)-plane that can be used to determine whether all roots of the characteristic equation have damping ratios greater than \(\zeta_{1}\). (b) Find a suitable contour \(\Gamma_{s}\) in the \(s\)-plane that can be used to determine whether all the roots of the characteristic equation have real parts less than \(s = - \sigma_{1}\). (c) Using the contour of part (b) and Cauchy's theorem, determine whether the following characteristic equation has roots with real parts less than \(s = - 1\) :
P9.4 The Nyquist plot of a conditionally stable system is shown in Figure P9.4 for a specific gain \(K\). (a) Determine
whether the system is stable, and find the number of roots (if any) in the right-hand \(s\)-plane. The system has no poles of \(G_{c}(s)G(s)\) in the right half-plane. (b) Determine whether the system is stable if the -1 point lies at the dot on the axis.
P9.5 A speed control for a gasoline engine is shown in Figure P9.5. Because of the restriction at the carburetor intake and the capacitance of the reduction manifold, the lag \(\tau_{t}\) occurs and is equal to 1.5 seconds. The engine time constant \(\tau_{e}\) is equal to \(J/b = 4\text{ }s\). The speed measurement time constant is \(\tau_{m} = 0.6\text{ }s\). (a) Determine the necessary gain \(K\) if the steadystate speed error is required to be less than \(20\%\) of the speed reference setting. (b) With the gain determined from part (a), apply the Nyquist criterion to investigate the stability of the system. (c) Determine the phase and gain margins of the system.
P9.6 A direct-drive arm is an innovative mechanical arm in which no reducers are used between motors and their loads. Because the motor rotors are directly coupled to the loads, the drive systems have no backlash, small friction, and high mechanical stiffness, which are all important features for fast and accurate positioning and dexterous handling using sophisticated torque control.
The goal of the MIT direct-drive arm project is to achieve arm speeds of \(10\text{ }m/s\) [15]. The arm has torques of up to \(660\text{ }N\text{ }m(475ftlb)\). Feedback and a set of position and velocity sensors are used with each motor. The frequency response of one joint of the arm is shown in Figure P9.6(a). The two poles appear at \(3.7\text{ }Hz\) and \(68\text{ }Hz\). Figure P9.6(b) shows the step response with position and velocity feedback used. The time constant of the closed-loop system is \(82\text{ }ms\). Develop the block diagram of the drive system and prove that \(82\text{ }ms\) is a reasonable result.
P9.7 A vertical takeoff (VTOL) aircraft is an inherently unstable vehicle and requires an automatic stabilization system. An attitude stabilization system for the K-16B U.S. Army VTOL aircraft has been designed and is shown in block diagram form in Figure P9.7 [16]. (a) Obtain the Bode plot of the loop transfer function \(L(s)\) when the gain is \(K = 2\). (b) Determine the
FIGURE P9.4 Nyquist plot of conditionally
stable system.
FIGURE P9.5
Engine speed control.
FIGURE P9.6
The MIT arm:
(a) frequency response, and (b) position response.
(a)
(b)
FIGURE P9.7
VTOL aircraft stabilization system. thus controlling the hydraulic fluid flow to the actuator. The block diagram of a closed-loop electrohydraulic servomechanism using pressure feedback to obtain damping is shown in Figure P9.8(b) [17, 18]. Typical values for this system are \(\tau = 0.02\text{ }s\); for the hydraulic system they are \(\omega_{2} = 7(2\pi)\) and \(\zeta_{2} = 0.05\). The structural resonance \(\omega_{1}\) is equal to \(10(2\pi)\), and the damping is \(\zeta_{1} = 0.05\). The loop gain is \(K_{A}K_{1}K_{2} = 1.0\). (a) Sketch the Bode plot and determine the phase margin of the system. (b) The damping of the system can be increased by drilling a small hole in the piston so that \(\zeta_{2} = 0.25\). Sketch the Bode plot and determine the phase margin of this system.
P9.9 The space shuttle, shown in Figure P9.9(a), carried large payloads into space and returned them to Earth FIGURE P9.8
(a) A servovalve and actuator
(b) Block diagram.
(a)
(b)
edge of the wing and a brake on the tail to control the flight during entry. The block diagram of a pitch rate control system is shown in Figure P9.9(b).
(a) Sketch the Bode plot of the system when \(G_{c}(s) = 2\) and determine the stability margin. (b) Sketch the Bode plot of the system when
The gain \(K_{P}\) should be selected so that the gain margin is \(10\text{ }dB\).
P9.10 Machine tools are often automatically controlled as shown in Figure P9.10. These automatic systems are often called numerical machine controls [9]. On each axis, the desired position of the machine tool is compared with the actual position and is used to actuate a solenoid coil and the shaft of a hydraulic actuator. The transfer function of the actuator is where \(K_{a} = 1\) and \(\tau_{a} = 0.4\text{ }s\). The output voltage of the difference amplifier is
where \(x_{d}(t)\) is the desired position input. The force on the shaft is proportional to the current \(i(t)\), so that \(F = K_{2}i(t)\), where \(K_{2} = 3.0\). The spring constant \(K_{s}\) is equal to \(1.5,R = 0.1\), and \(L = 0.2\).
(a) Determine the gain \(K_{1}\) that results in a system with a phase margin of P.M. \(= 30^{\circ}\). (b) For the gain \(K_{1}\) of part (a), determine \(M_{p\omega},\omega_{r}\), and the closed-loop system bandwidth. (c) Estimate the percent overshoot of the transient response to a step input \(X_{d}(s) = 1/s\), and the settling time (to within \(2\%\) of the final value).
(a)
FIGURE P9.9
(a) The Earthorbiting space shuttle against the blackness of space. The remote manipulator robot is shown with the cargo bay doors open in this top view, taken by a satellite. (b) Pitch rate control system. (Courtesy of NASA.)
(b)
GURE P9.10
Machine tool control.
P9.11 A control system for a chemical concentration control system is shown in Figure P9.11. The system receives a granular feed of varying composition, and we want to maintain a constant composition of the output mixture by adjusting the feed-flow valve.
The transport of the feed along the conveyor requires a transport (or delay) time, \(T = 1.5\text{ }s\). (a) Sketch the Bode plot when \(K_{1} = K_{2} = 1\), and investigate the stability of the system. (b) Sketch the Bode plot when \(K_{1} = 0.1\) and \(K_{2} = 0.04\), and investigate FIGURE P9.11
Chemical concentration control.
the stability of the system. (c) When \(K_{1} = 0\), use the Nyquist criterion to calculate the maximum allowable gain \(K_{2}\) for the system to remain stable.
P9.12 A simplified model of the control system for regulating the pupillary aperture in the human eye is shown in Figure P9.12 [20]. The gain \(K\) represents the pupillary gain, and \(\tau\) is the pupil time constant, which is \(0.75\text{ }s\). The time delay \(T\) is equal to \(0.6\text{ }s\). The pupillary gain is \(K = 2.5\).
(a) Assuming the time delay is negligible, sketch the Bode plot for the system. Determine the phase margin of the system. (b) Include the effect of the time delay by adding the phase shift due to the delay.
Determine the phase margin of the system with the time delay included.
P9.13 A controller is used to regulate the temperature of a mold for plastic part fabrication, as shown in Figure P9.13. The value of the delay time is estimated as \(1.2\text{ }s\). (a) Using the Nyquist criterion, determine the stability of the system for \(K_{a} = K = 1\). (b) Determine a suitable value for \(K_{a}\) for a stable system that will yield a phase margin P.M. \(\geq 50^{\circ}\) when \(K = 1\).
P9.14 Electronics and computers are being used to control automobiles. Figure P9.14 is an example of an automobile control system, the steering control for a research automobile. The control stick is used
FIGURE P9.12
Human pupil aperture control.
FIGURE P9.13
Temperature controller.
FIGURE P9.14
Automobile steering control. for steering. A typical driver has a reaction time of \(T = 0.2\text{ }s\).
(a) Using the Nichols chart, determine the magnitude of the gain \(K\) that will result in a system with a peak magnitude of the closed-loop frequency response \(M_{p\omega} \leq 2\text{ }dB\).
(b) Estimate the damping ratio of the system based on (1) \(M_{p\omega}\) and (2) the phase margin. Compare the results and explain the difference, if any.
(c) Determine the closed-loop bandwidth of the system.
P9.15 Consider the automatic ship-steering system transfer function.
The deviation of the tanker from the straight track is measured by radar and is used to generate the error signal, as shown in Figure P9.15. This error signal is used to control the rudder angle \(\delta(s)\).
(a) Is this system stable? Discuss what an unstable ship-steering system indicates in terms of the transient response of the system. Recall that the system under consideration is a ship attempting to follow a straight track.
(b) Is it possible to stabilize this system by lowering the gain of the transfer function \(G(s)\) ?
(c) Is it possible to stabilize this system with a derivative feedback controller?
(d) Suggest a suitable feedback controller.
(e) Repeat part (a), (b), and (c) when switch \(S\) is closed.
P9.16 An electric carrier that automatically follows a tape track laid out on a factory floor is shown in Figure P9.16(a) [15]. Closed-loop feedback systems are used
FIGURE P9.15 Automatic ship steering.
to control the guidance and speed of the vehicle. The cart senses the tape path by means of an array of 16 phototransistors. The block diagram of the steering system is shown in Figure P9.16(b). Select a gain \(K\) so that the phase margin is \(P.M. = 30^{\circ}\).
P9.17 The primary objective of many control systems is to maintain the output variable at the desired or reference condition when the system is subjected to a disturbance [22]. A typical chemical reactor control scheme is shown in Figure P9.17. The disturbance is represented by \(U(s)\), and the chemical process by \(G_{3}\) and \(G_{4}\). The controller is represented by \(G_{1}\) and the valve by \(G_{2}\). The feedback sensor is \(H(s)\) and will be assumed to be equal to 1 . We will assume that \(G_{2},G_{3}\), and \(G_{4}\) are all of the form
where \(\tau_{3} = \tau_{4} = 4\text{ }s\) and \(K_{3} = K_{4} = 0.1\). The valve constants are \(K_{2} = 20\) and \(\tau_{2} = 0.5\text{ }s\). We want to FIGURE P9.16
(a) An electric carrier vehicle (photo courtesy of Control Engineering Corporation).
(b) Block diagram.
(a)
(b)
FIGURE P9.17
Chemical reactor control. a percent overshoot of P.O. \(\leq 30\%\), but P.O. \(\geq\) \(5\%\). For parts (a) and (b), use the approximation of the damping ratio as a function of phase margin that yields \(\zeta = 0.01\phi_{pm}\). For these calculations, assume that \(U(s) = 0\).
(c) Estimate the settling time (with a \(2\%\) criterion) of the step response of the system for the controller of parts (a) and (b). (b) If the controller has a proportional term plus an integral term so that \(G_{1}(s) = K_{1}(1 + 1/s)\), determine a suitable gain to yield a system with maintain a steady-state error \(e_{ss} = 5\%\) of the desired
(a) When \(G_{1}(s) = K_{1}\), find the necessary gain to satisfy the error-constant requirement. For this condition, determine the expected percent overshoot to a step change in the reference signal \(r(t)\). (d) The system is expected to be subjected to a step disturbance \(U(s) = A/s\). For simplicity, assume that the desired reference is \(r(t) = 0\) when the system has settled. Determine the response of the system of part (b) to the disturbance.
P9.18 A model of an automobile driver attempting to steer a course is shown in Figure P9.18, where \(K = 2.0\). (a) Find the frequency response and the gain and phase margins when the reaction time \(T = 0\). (b) Find the phase margin when the reaction time is \(T = 0.3\text{ }s\). (c) Find the reaction time that will cause the system to be borderline stable \(\left( P.M. = 0^{\circ} \right)\).
P9.19 In the United States, billions of dollars are spent annually for solid waste collection and disposal. One system, which uses a remote-control pick-up arm for collecting waste bags, is shown in Figure P9.19. The loop transfer function of the remote pick-up arm is
(a) Plot Nichols chart, and show that the gain margin G.M. is \(33.5\text{ }dB\). (b) Determine the phase margin and the \(M_{p\omega}\) for the closed loop. Also, determine the closed-loop bandwidth.
P9.20 The Bell-Boeing V-22 Osprey Tiltrotor is both an airplane and a helicopter. Its advantage is the ability to rotate its engines to a vertical position, as shown in Figure P7.33(a), for takeoffs and landings and then switch the engines to a horizontal position for cruising as an airplane. The altitude control system in the helicopter mode is shown in Figure P9.20. (a) Obtain the frequency response of the system for \(K = 100\). (b) Find the gain margin and the phase margin for this system. (c) Select a suitable gain \(K\) so that the phase margin is P.M. \(= 40^{\circ}\). (Decrease the gain above \(K = 100\).)
(d) Find the response \(y(t)\) of the system for the gain selected in part (c).
P9.21 Consider a unity feedback system with the loop transfer function
FIGURE P9.18
Automobile and driver control.
FIGURE P9.19
Waste collection system.
FIGURE P9.20
Tiltrotor aircraft control. (a) Sketch the Bode plot for \(K = 1\). Determine P9.23 A closed-loop system has a loop transfer function (b) the gain margin, (c) the value of \(K\) required to provide a gain margin equal to \(20\text{ }dB\), and (d) the value of \(K\) to yield a steady-state error of \(10\%\) of the magnitude \(A\) for the ramp input \(r(t) = At,t > 0\). Can this gain be utilized and achieve acceptable performance?
P9.22 The Nichols chart for \(G_{c}(j\omega)G(j\omega)\) of a closed-loop system is shown in Figure P9.22. The frequency for each point on the graph is given in the following table:
| Point | $$\mathbf{1}$$ | $$\mathbf{2}$$ | $$\mathbf{3}$$ | $$\mathbf{4}$$ | $$\mathbf{5}$$ | $$\mathbf{6}$$ | $$\mathbf{7}$$ | $$\mathbf{8}$$ | $$\mathbf{9}$$ |
| 1 | 2.0 | 2.6 | 3.4 | 4.2 | 5.2 | 6.0 | 7.0 | 8.0 |
Determine (a) the resonant frequency, (b) the bandwidth, (c) the phase margin, and (d) the gain margin. (e) Estimate the overshoot and settling time (with a \(2\%\) criterion) of the response to a step input.
(a) Determine the gain \(K\) so that the phase margin is \(P.M. = 40^{\circ}\). (b) For the gain \(K\) selected in part (a), determine the gain margin of the system.
P9.24 A closed-loop system with unity feedback has a loop transfer function
(a) Determine the gain \(K\) so that the phase margin is \(P.M. = 35^{\circ}\). (b) For the gain \(K\) selected in part (a) determine the gain margin of the system. (c) Predict the bandwidth of the closed-loop system.
FIGURE P9.22 Nichols chart.
P9.25 A closed-loop system has the loop transfer function
(a) Determine the gain \(K\) so that the phase margin is \(P.M. = 60^{\circ}\) when \(T = 0.15\). (b) Plot the phase margin versus the time delay \(T\) for \(K\) as in part (a).
P9.26 A specialty machine shop is improving the efficiency of its surface-grinding process [21]. The existing machine is mechanically sound, but manually operated. Automating the machine will free the operator for other tasks and thus increase overall throughput of the machine shop. The grinding machine is shown in Figure P9.26(a) with all three axes automated with motors and feedback systems. The control system for the \(y\)-axis is shown in Figure P9.26(b). To achieve a low steady-state error to a ramp command, we choose \(K = 2\). Sketch the Bode plot and Nichols chart.
Determine the gain margin and phase margin of the system and the bandwidth of the closed-loop system. Estimate the damping ratio of the system and the predicted percent overshoot and settling time.
P9.27 Consider the system shown in Figure P9.27. Determine the maximum value of \(K = K_{\max}\) for which the closed-loop system is stable. Plot the phase margin as a function of the gain \(1 \leq K \leq K_{\max}\). Explain what happens to the phase margin as \(K\) approaches \(K_{\max}\).
P9.28 Consider the feedback system shown in Figure P9.28.
(a) Determine the value of \(K_{P}\) such that the phase margin is P.M. \(= 60^{\circ}\).
(b) Using the P.M. obtained, predict the percent overshoot of the closed-loop system to a unit step input.
(c) Plot the step response, and compare the actual percent overshoot with the predicted percent overshoot.
(a)
(b)
FIGURE P9.26 Surface-grinding wheel control system.
FIGURE P9.27
Nonunity feedback system with proportional controller \(K\).
FIGURE P9.28 A unity feedback system with a proportional controller in the loop.
324. ADVANCED PROBLEMS
AP9.1 For positive constants of \(K,T_{1}\), and \(T_{2}\), a control system is described by its loop transfer function as
Considering gain \(K = 0.06\), compute the phase margin and gain margin for (a) \(T_{1} = 5\) and \(T_{2} = 2\). (b) \(T_{1} = 2\) and \(T_{2} = 5\). (c) Comment on the stability.
AP9.2 Anesthesia is used in surgery to induce unconsciousness. One problem with drug-induced unconsciousness is differences in patient responsiveness. Furthermore, the patient response changes during an operation. A model of drug-induced anesthesia control is shown in Figure AP9.2. The proxy for unconsciousness is the arterial blood pressure.
(a) Obtain the Bode plot and determine the gain margin and the phase margin when \(T = 0.05\text{ }s\).
(b) Repeat part (a) when \(T = 0.1\text{ }s\). Describe the effect of the \(100\%\) increase in the time delay \(T\). (c) Using the phase margin, predict the percent overshoot for a step input for parts (a) and (b).
AP9.3 Figure AP9.3 shows an automatic water treatment plant. It is typically a mechanical-chemical arrangement that uses reagent to purify water. The plant comprises an input pipeline embedded with a flow meter and sensors. The temperature and the flow rate of the boiler and the cooling water are regulated based on the input-output concentration of the chemical water.
The transfer function of the system is given by
\(L(s) = G(s)H(s) = \frac{0.2s^{3} + 0.06s^{2} + 0.03s + 0.04}{s^{3} + 1.12s^{2} + 0.8s + 0.4}\).
(a) Draw the Bode plot indicating the gain margin and phase margin. (b) Find the peak overshoot of the system for unit step input.
FIGURE AP9.2
Control of blood pressure with anesthesia.
FIGURE AP9.3
An automatic water treatment plant.
AP9.4 The loop transfer function of a system is described as
Find the value of \(K\) for marginal stability. Find the gain margin and the phase margin for \(K = 22\) and \(K = 42\).
AP9.5 A unity feedback control system given by
Determine the (a) phase margin for \(T = 0\) and (b) limiting value of \(T\) for stability.
AP9.6 The acidity of water draining from a coal mine is often controlled by adding lime to the water. A valve controls the lime addition and a sensor is downstream. For the model of the system shown in Figure AP9.6, determine \(K\) and the distance \(D\) to maintain stability. We require \(D > 2\) meters in order to allow full mixing before sensing.
AP9.7 Building elevators are limited to about 800 meters. Above that height, elevator cables become too thick and too heavy for practical use. One solution is to eliminate the cable. The key to the cordless elevator is the linear motor technology now being applied to the development of magnetically levitated rail transportation systems. Under consideration is a linear synchronous motor that propels a passenger car along the track like guideway running the length of the elevator shaft. The motor works by the interaction of an electromagnetic field from electric coils on the guideway with magnets on the car [28].
If we assume that the motor has negligible friction, the system may be represented by the model shown in Figure AP9.7. Determine \(K\) so that the phase margin of the system is \(P.M. = 60^{\circ}\). For the gain \(K\) selected, determine the system bandwidth. Also calculate the maximum value of the output for a unit step disturbance for the selected gain.
FIGURE AP9.6
Mine water acidity control.
FIGURE AP9.7
Elevator position
AP9.8 A control system is shown in Figure AP9.8. The gain \(K\) is greater than 500 and less than 4000 . Select a gain that will cause the system step response to have a percent overshoot of P.O. \(\leq 20\%\). Plot the Nichols chart and calculate the phase margin.
AP9.9 Consider a unity feedback system with
and
Let
and determine the gain \(K_{P}\) that provides the maximum phase margin.
AP9.10 A multiloop block diagram is shown in Figure AP9.10.
(a) Compute the transfer function \(T(s) = Y(s)/R(s)\). (b) Determine \(K\) such that the steady-state tracking error to a unit step input \(R(s) = 1/s\) is zero. Plot the unit step response.
(c) Using \(K\) from part (b), compute the system bandwidth \(\omega_{b}\).
AP9.11 Patients with a cardiological illness and less than normal heart muscle strength can benefit from an assistance device. An electric ventricular assist device (EVAD) converts electric power into blood flow by moving a pusher plate against a flexible blood sac. The pusher plate reciprocates to eject blood in systole and to allow the sac to fill in diastole. The EVAD will be implanted in tandem or in parallel with the intact natural heart as shown in Figure AP9.11(a). The EVAD is driven by rechargeable batteries, and the electric power is transmitted inductively across the skin through a transmission system. The batteries and the transmission system limit the electric energy storage and the transmitted peak power. We desire to drive the EVAD in a fashion that minimizes its electric power consumption [33].
The EVAD has a single input, the applied motor voltage, and a single output, the blood flow rate. The control system of the EVAD performs two main
FIGURE AP9.8
Gain selection.
FIGURE AP9.10
Multiloop feedback control system.
FIGURE AP9.11
(a) An electric ventricular assist device for cardiology patients.
(b) Feedback control system.
(a)
(b) tasks: It adjusts the motor voltage to drive the pusher plate through its desired stroke, and it varies the EVAD blood flow to meet the body's cardiac output demand. The blood flow controller adjusts the blood flow rate by varying the EVAD beat rate. A model of the feedback control system is shown in Figure AP9.11(b). The motor, pump, and blood sac can be modeled by a nominal time delay with \(T = 2\text{ }s\). The goal is to achieve a step response with zero steady-state error and percent overshoot P.O. \(\leq 15\%\).
Consider the controller
For the nominal time delay of \(T = 2\text{ }s\), plot the step response and verify that steady-state tracking error and percent overshoot specifications are satisfied. Determine the maximum time delay, \(T\), possible with the controller that continues to stabilize the closed-loop system. Plot the phase margin as a function of time delay up to the maximum allowed for stability.
325. DESIGN PROBLEMS
CDP9.1 The system of Figure CDP4.1 uses a controller \(G_{c}(s) = K_{a}\). Determine the value of \(K_{a}\) so that the phase margin is \(P.M. = 70^{\circ}\). Plot the response of this system to a step input.
DP9.1 A mobile robot for toxic waste cleanup is shown in Figure DP9.1(a) [23]. The closed-loop speed control is a unity feedback system. The Nichols chart in Figure DP9.1(b) shows the plot of \(G_{c}(j\omega)G(j\omega)/K\) versus \(\omega\).
(a)
cleanup. (b) Nichols chart.
(b) The value of the frequency at the points indicated is DP9.4 A robot tennis player is shown in Figure DP9.4(a), recorded in the following table:
| Point | $$\mathbf{1}$$ | $$\mathbf{2}$$ | $$\mathbf{3}$$ | $$\mathbf{4}$$ | $$\mathbf{5}$$ |
| 2 | 5 | 10 | 20 | 50 |
and a simplified control system for \(\theta_{2}(t)\) is shown in Figure DP9.4(b). The goal of the control system is to attain the best step response while attaining a high \(K_{v}\) for the system. Select \(K_{v1} = 0.35\) and \(K_{v2} = 0.65\), and determine the phase margin, gain margin, bandwidth, percent overshoot, and settling time for each case. Obtain the step response for each case, and select the best value for \(K\).
closed-loop system when \(K = 1\). (b) Determine the resonant peak in \(dB\) and the resonant frequency for \(K = 1\). (c) Determine the system bandwidth and estimate the settling time (with a \(2\%\) criterion) and percent overshoot of this system for a step input. (d) Determine the appropriate gain \(K\) so that the percent overshoot to a step input is P.O. \(= 30\%\), and estimate the settling time of the system.
DP9.2 Flexible-joint robotic arms are constructed of lightweight materials and exhibit lightly damped open-loop dynamics [15]. A feedback control system for a flexible arm is shown in Figure DP9.2. Select \(K\) so that the system has maximum phase margin. Predict the percent overshoot for a step input based on the phase margin attained, and compare it to the actual overshoot for a step input. Determine the bandwidth of the closed-loop system. Predict the settling time (with a \(2\%\) criterion) of the system to a step input and compare it to the actual settling time. Discuss the suitability of this control system.
DP9.3 An automatic drug delivery system is used in the regulation of critical care patients suffering from cardiac failure [24]. The goal is to maintain stable patient status within narrow bounds. Consider the use of a drug delivery system for the regulation of blood pressure by the infusion of a drug. The feedback control system is shown in Figure DP9.3. Select an appropriate gain \(K\) that maintains narrow deviation for blood pressure while achieving a good dynamic response.
DP9.5 An electrohydraulic actuator is used to actuate large loads for a robot manipulator, as shown in Figure DP9.5 [17]. The system is subjected to a step input, and we desire the steady-state error to be minimized. However, we wish to keep the percent overshoot P.O. \(\leq 10\%\). Let \(T = 0.8\text{ }s\).
(a) Select the gain \(K\) when \(G_{c}(s) = K\), and determine the resulting percent overshoot, settling time (with a \(2\%\) criterion), and steady-state error. (b) Repeat part (a) when \(G_{c}(s) = K_{1} + K_{2}/s\) by selecting \(K_{1}\) and \(K_{2}\). Sketch the Nichols chart for the selected gains \(K_{1}\) and \(K_{2}\).
DP9.6 The physical representation of a steel strip-rolling mill is a damped-spring system [8]. The output thickness sensor is located a negligible distance from the output of the mill, and the objective is to keep the thickness as close to a reference value as possible. Any change of the input strip thickness is regarded as a disturbance. The system is a nonunity feedback system, as shown in Figure DP9.6. Depending on the maintenance of the mill, the parameter varies as \(50 \leq b < 400\).
Determine the phase margin and gain margin for the two extreme values of \(b\) when the normal value of the gain is \(K = 100\). Recommend a reduced value for \(K\) so that the phase margin is \(P.M. \geq 45^{\circ}\) and the gain margin is G.M. \(\geq 6\text{ }dB\) for the range of \(b\).
FIGURE DP9.2
Control of a flexible robot arm.
FIGURE DP9.3
Automatic drug delivery.
(a)
FIGURE DP9.4
(a) An articulated two-link tennis player robot.
(b) Unity feedback control system.
FIGURE DP9.5 Electrohydraulic actuator.
FIGURE DP9.6 Steel strip-rolling mill.
(b)
DP9.7 Vehicles for lunar construction and exploration work will face conditions unlike anything found on Earth. Furthermore, they will be controlled via remote control. A block diagram of such a vehicle and the control are shown in Figure DP9.7. Select the gain \(K\) to have a percent overshoot of P.O. \(= 10\%\). For this \(K\), what is the maximum allowed time delay \(T\) for stability?
DP9.8 The control of a high-speed steel-rolling mill is a challenging problem. The goal is to keep the strip thickness accurate and readily adjustable. The model of the control system is shown in Figure DP9.8. Design a control system by selecting \(K\) so that the step response of the system is as fast as possible with a percent overshoot of P.O. \(\leq 0.7\%\), and a settling time (with a \(2\%\) criterion) of \(T_{s} \leq 4\text{ }s\). Use the root locus to select \(K\), and calculate the roots for the selected \(K\). Describe the dominant root(s) of the system.
DP9.9 A two-tank system containing a heated liquid has the model shown in Figure DP9.9(a), where \(T_{0}\) is the temperature of the fluid flowing into the first tank FIGURE DP9.7
Lunar vehicle
control.
FIGURE DP9.8
Steel-rolling mill control.
(a)
(b) and \(T_{2}\) is the temperature of the liquid flowing out of the second tank. The block diagram model is shown in Figure DP9.9(b). The system of the two tanks has a heater in tank 1 with a controllable heat input \(Q\). The time constants are \(\tau_{1} = 10\text{ }s\) and \(\tau_{2} = 50\text{ }s\).
(a) Determine \(T_{2}(s)\) in terms of \(T_{0}(s)\) and \(T_{2d}(s)\).
(b) If \(T_{2d}(s)\), the desired output temperature, is changed instantaneously from \(T_{2d}(s) = A/s\) to \(T_{2d}(s) = 2A/s\), determine the transient response of \(T_{2}(t)\) when \(G_{c}(s) = K = 500\). Assume that, prior to the abrupt temperature change, the system is at steady state. (c) Find the steady-state error \(e_{ss}\) for the system of part (b), where \(E(s) = T_{2d}(s) - T_{2}(s)\).
(d) Let \(G_{c}(s) = K/s\) and repeat parts (b) and (c). Use a gain \(K\) such that the percent overshoot is P.O. \(\leq 10\%\).
(e) Design a controller that will result in a system with a settling time (with a \(2\%\) criterion) of \(T_{s} \leq 150\text{ }s\) and a percent overshoot of P.O. \(\leq 10\%\), while maintaining a zero steady-state error when
FIGURE DP9.11
Nuclear reactor control.
(f) Prepare a table comparing the percent overshoot, settling time, and steady-state error for the designs of parts (b) through (e).
DP9.10 Consider the system is described in state variable form by
where
Assume that the input is a linear combination of the states, that is,
where \(r(t)\) is the reference input and the gain matrix is \(K = \begin{bmatrix} K_{1} & K_{2} \end{bmatrix}\). Substituting \(u(t)\) into the state variable equation yields the closed-loop system
(a) Obtain the characteristic equation associated with A-BK. (b) Design the gain matrix \(\mathbf{K}\) to meet the following specifications: (i) the closed-loop system is stable; (ii) the system bandwidth \(\omega_{b} \geq 1rad/s\); and (iii) the steady-state error to a unit step input \(R(s) = 1/s\) is zero.
DP9.11 The primary control loop of a nuclear power plant includes a time delay due to the need to transport the fluid from the reactor to the measurement point as shown in Figure DP9.11. The transfer function of the controller is
The transfer function of the reactor and time delay is
where \(T = 0.5\text{ }s\) and \(\tau = 0.3\text{ }s\). Using frequency response methods, design the controller so that the percent overshoot of the system is P.O. \(\leq 20\%\). With this controller in the loop, estimate the percent overshoot and settling time (with a \(2\%\) criterion) to a unit step. Determine the actual overshoot and settling time, and compare with the estimated values.
326. COMPUTER PROBLEMS
CP9.1 Consider a unity negative feedback control system with
Verify that the gain margin is \(\infty\) and that the phase margin is \(10^{\circ}\). CP9.2 Using the nyquist function, obtain the Nyquist plot CP9.6 A block diagram of the yaw acceleration control for the following transfer functions:
(a) \(G(s) = \frac{15}{s + 20}\);
(b) \(G(s) = \frac{40}{s^{2} + 6s + 25}\);
(c) \(G(s) = \frac{12}{s^{3} + 4s^{2} + 4s + 1}\).
CP9.3 Using the nichols function, obtain the Nichols chart with a grid for the following transfer functions:
(a) \(G(s) = \frac{1}{s + 0.2}\);
(b) \(G(s) = \frac{1}{s^{2} + 2s + 1}\);
(c) \(G(s) = \frac{12}{s^{3} + 6s^{2} + 11s + 6}\).
Determine the approximate phase and gain margins from the Nichols charts and label the charts accordingly.
CP9.4 A negative feedback control system has the loop transfer function
(a) When \(T = 0.05\text{ }s\), find \(K\) such that the phase margin is \(P.M. \geq 55^{\circ}\) using the margin function. (b) Obtain a plot of phase margin versus \(T\) for \(K\) as in part (a), with \(0 \leq T \leq 0.4\text{ }s\).
CP9.5 Consider a unity feedlach system with the loop transfer function
Develop an m-file to plot the bandwidth of the closedloop system as \(K\) varies in the interval \(1 \leq K \leq 80\). system for a bank-to-turn missile is shown in Figure CP9.6. The input is yaw acceleration command (in g's), and the output is missile yaw acceleration (in g's). The controller is specified to be a proportional, integral
(PI) controller. The nominal value of \(b_{0}\) is 0.5 .
(a) Using the margin function, compute the phase margin, gain margin, and system crossover frequency assuming the nominal value of \(b_{0}\).
(b) Using the gain margin from part (a), determine the maximum value of \(b_{0}\) for a stable system. Verify your answer with a Routh-Hurwitz analysis of the characteristic equation.
CP9.7 An engineering laboratory has presented a plan to operate an Earth-orbiting satellite that is to be controlled from a ground station. A block diagram of the proposed system is shown in Figure CP9.7. It takes \(T\) seconds for a signal to reach the spacecraft from the ground station and the identical delay for a return signal. The proposed ground-based controller is a proportional-derivative (PD) controller, where
(a) Assume no transmission time delay (i.e., \(T = 0\) ), and design the controller to the following specifications: (1) percent overshoot P.O. \(\leq 20\%\) to a unit step input and (2) time to peak \(T_{p} \leq 30\text{ }s\).
(b) Compute the phase margin with the controller in the loop but assuming a zero transmission time delay. Estimate the amount of allowable time delay for a stable system from the phase margin calculation.
(c) Using a second-order Padé approximation to the time delay, determine the maximum allowable delay \(T_{\max}\) for system stability by developing a
FIGURE CP9.6 A feedback control system for the yaw acceleration control of a bank-to-turn missile.
327. FIGURE CP9.7
A block diagram of a ground-controlled satellite.
\(m\)-file script that employs the pade function and computes the closed-loop system poles as a function of the time delay \(T\). Compare your answer with the one obtained in part (b).
CP9.8 Consider the system represented in state variable
Using the nyquist function, obtain the Nyquist plot.
CP9.9 For the system in CP9.8, use the nichols function to obtain the Nichols chart and determine the phase margin and gain margin.
CP9.10 A closed-loop feedback system is shown in Figure CP9.10. (a) Obtain the Nyquist plot, and determine the phase margin. Assume that the time delay \(T = 0\text{ }s\). (b) Compute the phase margin when \(T = 0.05\text{ }s\). (c) Determine the minimum time delay that destabilizes the closed-loop system.
328. ANSWERS TO SKILLS CHECK
True or False: (1) True; (2) True; (3) True; (4) True; (5) False
Multiple Choice: (6) b; (7) a; (8) d; (9) a; (10) d; (11) b; (12) a; (13) b; (14) c; (15) a
FIGURE CP9.10 Nonunity feedback system with a time delay.
329. TERMS AND CONCEPTS
Bandwidth The frequency at which the frequency response has declined \(3\text{ }dB\) from its low-frequency value.
Cauchy's theorem If a contour encircles \(Z\) zeros and \(P\) poles of \(F(s)\) traversing clockwise, the corresponding contour in the \(F(s)\)-plane encircles the origin of the \(F(s)\)-plane \(N = Z - P\) times clockwise.
Closed-loop frequency response The frequency response of the closed-loop transfer function \(T(j\omega)\).
Conformal mapping A contour mapping that retains the angles on the \(s\)-plane on the \(F(s)\)-plane.
Contour map A contour or trajectory in one plane is mapped into another plane by a relation \(F(s)\).
Gain margin The increase in the system gain when phase \(= - 180^{\circ}\) that will result in a marginally stable system with intersection of the \(- 1 + j0\) point on the Nyquist diagram.
Logarithmic (decibel) measure A measure of the gain margin defined as \(20\log_{10}(1/d)\), where \(\frac{1}{d} = \frac{1}{|L(j\omega)|}\) when the phase shift is \(- 180^{\circ}\).
Word Match (in order, top to bottom): f, e, k, b, j, a, i, d, h, c, g
Nichols chart A chart displaying the curves for the relationship between the open-loop and closed-loop frequency response.
Nyquist stability criterion A feedback system is stable if, and only if, the contour in the \(L(s)\)-plane does not encircle the \(( - 1,0)\) point when the number of poles of \(L(s)\) in the right-hand \(s\)-plane is zero. If \(L(s)\) has \(P\) poles in the right-hand plane, then the number of counterclockwise encirclements of the \(( - 1,0)\) point must be equal to \(P\) for a stable system.
Phase margin The amount of phase shift of the \(L(j\omega)\) at unity magnitude that will result in a marginally stable system with intersections of the \(- 1 + j0\) point on the Nyquist diagram.
Principle of the argument See Cauchy's theorem.
Time delay A time delay \(T\), so that events occurring at time \(t\) at one point in the system occur at another point in the system at a later time \(t + T\).
330. CHAPTER 10
The Design of Feedback Control Systems
10.1 Introduction 729
10.2 Approaches to System Design 730
10.3 Cascade Compensators 731
10.4 Phase-Lead Design Using the Bode Plot 735
10.5 Phase-Lead Design Using the Root Locus 741
10.6 System Design Using Integration Compensators 747
10.7 Phase-Lag Design Using the Root Locus 750
10.8 Phase-Lag Design Using the Bode Plot 753
10.9 Design on the Bode Plot Using Analytical Methods 758
10.10 Systems with a Prefilter 759
10.11 Design for Deadbeat Response 762
10.12 Design Examples 764
10.13 System Design Using Control Design Software 774
10.14 Sequential Design Example: Disk Drive Read System 781
10.15 Summary 783
331. PREVIEW
In this chapter, we address the central issue of the design of compensators. Using the methods of the previous chapters, we develop several design techniques in the frequency domain that enable us to achieve the desired system performance. The powerful lead and lag controllers are used in several design examples. Phase-lead and phase-lag control design approaches using both root locus plots and Bode plots are presented. The proportional-integral (PI) controller is revisited in the context of achieving high steady-state tracking accuracies. The chapter concludes with a proportional-derivative (PD) controller design with prefiltering for the Sequential Design Example: Disk Drive Read System.
332. DESIRED OUTCOMES
Upon completion of Chapter 10, students should be able to:
$\square\ $ Explain the design of lead and lag compensators using root locus and Bode plot methods.
$\square\ $ Identify the value of prefilters and design for deadbeat response.
\(\square\) Distinguish between the varied approaches available for control system design.
332.1. INTRODUCTION
The performance of a feedback control system is of primary importance. A suitable control system is stable and results in an acceptable response to input commands, is less sensitive to system parameter changes, results in a minimum steady-state error for input commands, and is able to reduce the effect of undesirable disturbances. A feedback control system that provides an optimum performance without any necessary adjustments is rare. Usually, we find it necessary to compromise among the many conflicting and demanding specifications and to adjust the system parameters to provide a suitable and acceptable performance when it is not possible to obtain all the desired optimum specifications.
It is generally possible to adjust the system parameters in order to provide the desired system response. However, we may find that it is not sufficient to adjust a single system parameter and obtain the desired performance. Rather, we may be required to consider the structure of the system and redesign the system in order to obtain a suitable one. That is, sometimes we must examine the scheme or plan of the system and obtain a new design or plan that results in a suitable system. Thus, the design of a control system is concerned with the arrangement, or the plan, of the system structure and the selection of suitable components and parameters. For example, if we desire a set of performance measures to be less than some specified values, we often encounter a conflicting set of requirements. Hence, if we wish a system to have a percent overshoot less P.O. \(\leq 20\%\) and \(\omega_{n}T_{p} = 3.3\), we obtain a conflicting requirement on the system damping ratio \(\zeta\). If we are unable to relax these two performance requirements, we must alter the system in some way. The alteration or adjustment of a control system in order to provide a suitable performance is called compensation; that is, compensation is the adjustment of a system in order to make up for deficiencies or inadequacies.
In redesigning a control system to alter the system response, an additional component is inserted within the structure of the feedback system. It is this additional component or device that equalizes or compensates for the performance deficiency. The compensating device is often called a compensator.
333. A compensator is an additional component that is inserted into a control system to compensate for a deficient performance.
The transfer function of a compensator is designated as \(G_{c}(s) = E_{o}(s)/E_{\text{in}\text{~}}(s)\), and the compensator can be placed in a suitable location within the structure of the system. Several types of compensation are shown in Figure 10.1 for a simple, single-loop feedback control system. The compensator placed in the feedforward path is called a cascade compensator (Figure 10.1a). Similarly, the other compensation schemes are called feedback, output (or load), and input compensation, as shown in Figures 10.1(b), (c), and (d), respectively. The selection of the compensation scheme depends upon a consideration of the specifications, the power levels at various signal nodes in the system, and the networks available for use. Usually, the output \(Y(s)\) is a direct output of the process \(G(s)\) and the output compensation of Figure \(10.1(c)\) is not physically realizable. FIGURE 10.1
Types of compensation.
(a) Cascade
compensation.
(b) Feedback
compensation.
(c) Output, or load, compensation.
(d) Input
compensation.
(a)
(c)
(b)
(d)
333.1. APPROACHES TO SYSTEM DESIGN
The performance of a control system can be described in terms of the time-domain performance measures or the frequency-domain performance measures. The performance of a system can be specified by requiring a certain peak time, \(T_{p}\), maximum percent overshoot, P.O., and settling time, \(T_{s}\), for a step input. Furthermore, it is usually necessary to specify the maximum allowable steady-state error for several test signal inputs and disturbance inputs. These performance specifications can be defined in terms of the desirable location of the poles and zeros of the closed-loop system transfer function \(T(s)\). Thus, the location of the \(s\)-plane poles and zeros of \(T(s)\) can be specified. The locus of the roots of the closed-loop system can be readily obtained for the variation of one system parameter. However, when the locus of roots does not result in a suitable root configuration, we must add a compensator to alter the locus of the roots as the parameter is varied. Therefore, we can use the root locus method and determine a suitable compensator so that the resultant root locus yields the desired closed-loop root configuration.
Alternatively, we can describe the performance of a feedback control system in terms of frequency performance measures. Then a system can be described in terms of the peak of the closed-loop frequency response, \(M_{p\omega}\), the resonant frequency, \(\omega_{r}\), the bandwidth, \(\omega_{B}\), the gain margin, G.M., and the phase margin, P.M., of the system. We can add a suitable compensator, if necessary, in order to satisfy the system specifications. The design of the compensator, represented by \(G_{c}(s)\), is developed in terms of the frequency response as portrayed on the Nyquist plot, the Bode plot, or the Nichols chart. Because a cascade transfer function is readily accounted for on a Bode plot by adding the frequency response of the network, we often prefer to approach the frequency response methods.
Thus, the design of a system is concerned with the alteration of the frequency response or the root locus of the system in order to obtain a suitable system performance. For frequency response methods, we are concerned with altering the system so that the frequency response of the compensated system will satisfy the system specifications. Hence, in the frequency response approach, we use compensators to alter and reshape the system characteristics represented on the Nyquist plot, Bode plot, or Nichols chart.
Alternatively, the design of a control system can be accomplished in the \(s\)-plane by root locus methods. For the case of the \(s\)-plane, the designer wishes to alter and reshape the root locus so that the roots of the system lie in the desired locations in the s-plane.
When possible, one way to improve the performance of a control system is to alter the process itself. That is, if the system designer is able to specify and alter the design of the process represented by the transfer function \(G(s)\), then the performance of the system may be improved. For example, to improve the transient behavior of a servomechanism position controller, we might choose a better motor for the system. In the case of an airplane control system, we might be able to alter the aerodynamic design of the airplane and thus improve the flight transient characteristics. Thus, a control system designer should recognize that an alteration of the process may result in an improved system. However, the process is often unalterable or has been altered as much as possible and still results in unsatisfactory performance. Then the addition of compensators becomes imperative for improving the performance of the system.
In the following sections, we will assume that the process has been improved as much as possible and that the \(G(s)\) representing the process is unalterable. First, we shall consider the addition of a phase-lead compensator and describe the design of the network by root locus and frequency response techniques. Then, using both the root locus and frequency response techniques, we will describe the design of the integration compensators in order to obtain a suitable system performance.
333.2. CASCADE COMPENSATORS
In this section, we consider the design of a cascade compensator, as shown in Figures 10.1(a) and (b), respectively. The compensator, \(G_{c}(s)\), is cascaded with the process \(G(s)\) to provide a suitable loop transfer function \(L(s) = G_{c}(s)G(s)H(s)\). The compensator \(G_{c}(s)\) can be chosen to alter either the shape of the root locus or the frequency response. In either case, the compensator may be chosen to have a transfer function
Then the problem reduces to the judicious selection of the poles and zeros of the compensator. To illustrate the properties, we consider a first-order compensator. The compensation approach developed on the basis of a first-order compensator can then be extended to higher-order compensators, for example, by cascading several first-order compensators.
A compensator \(G_{c}(s)\) is used with a process \(G(s)\) so that the overall loop gain can be set to satisfy the steady-state error requirement, and then \(G_{c}(s)\) is used to adjust the system dynamics favorably without affecting the steady-state error. FIGURE 10.2
Pole-zero diagram of the phase-lead compensator.
Consider the first-order compensator with the transfer function
The design problem then becomes the selection of \(z,p\), and \(K\) in order to provide a suitable performance. When \(|z| < |p|\), the compensator is called a phase-lead compensator and has a pole-zero \(s\)-plane configuration, as shown in Figure 10.2. If the pole was negligible, that is, \(|p| \gg |z|\), and the zero occurred at the origin of the \(s\)-plane, we would have a differentiator so that
Thus, a compensator of the form of Equation (10.2) is a differentiator-type compensator. The differentiator compensator of Equation (10.3) has the frequency characteristic
and a phase angle of \(+ 90^{\circ}\). Similarly, the frequency response of the differentiating compensator of Equation (10.2) is
where \(\tau = 1/p\) and \(p = \alpha z\). The frequency response of this phase-lead compensator is shown in Figure 10.3. The angle of the frequency characteristic is
Because the zero occurs first on the frequency axis, we obtain a phase-lead characteristic, as shown in Figure 10.3. The slope of the asymptotic magnitude curve is \(+ 20\text{ }dB/\) decade.
The phase-lead compensator transfer function can be written as
FIGURE 10.3
Bode plot of the phase-lead compensator.
where \(\tau = 1/p\) and \(\alpha = p/z > 1\). The maximum value of the phase lead occurs at a frequency \(\omega_{m}\), where \(\omega_{m}\) is the geometric mean of \(p = 1/\tau\) and \(z = 1/(\alpha\tau)\); that is, the maximum phase lead occurs halfway between the pole and zero frequencies on the logarithmic frequency scale. Therefore,
To obtain an equation for the maximum phase-lead angle, we rewrite the phase angle of Equation (10.5) as
Then, substituting the frequency for the maximum phase angle, \(\omega_{m} = 1/(\tau\sqrt{a})\), we have
We use the trigonometric relationship \(sin\phi = tan\phi/\sqrt{1 + \tan^{2}\phi}\) and obtain
Equation (10.11) is very useful for calculating a necessary \(\alpha\) ratio between the pole and zero of a compensator in order to provide a required maximum phase lead. A plot of \(\phi_{m}\) versus \(\alpha\) is shown in Figure 10.4. The phase angle readily obtainable from this compensator is not much greater than \(70^{\circ}\). Also, there are practical limitations on the maximum value of \(\alpha\) that we should attempt to FIGURE 10.4
Maximum phase angle \(\phi_{m}\) versus \(\alpha\) for a phase-lead compensator.
FIGURE 10.5
Pole-zero diagram of the phase-lag compensator.
obtain. Therefore, if we required a maximum angle greater than \(70^{\circ}\), two cascade compensators could be used.
It is often useful to add a cascade compensator that provides a phase-lag characteristic. The phase-lag compensator transfer function is
where \(\tau = 1/z\) and \(\alpha = z/p > 1\). The pole lies closest to the origin of the \(s\)-plane, as shown in Figure 10.5. This type of compensator is often called an integrating compensator because it has a frequency response like an integrator over a finite range of frequencies. The Bode plot of the phase-lag compensator is obtained from the transfer function
and is shown in Figure 10.6. The form of the Bode plot of the lag compensator is similar to that of the phase-lead compensator; the difference is the resulting attenuation and phase-lag angle instead of amplification and phase-lead angle. However, note that the shapes of the diagrams of Figures 10.3 and 10.6 are similar. Therefore, we can show that the maximum phase lag occurs at \(\omega_{m} = \sqrt{zp}\).
In the succeeding sections, we wish to utilize these compensation networks to obtain a desired system frequency response or \(s\)-plane root location. The lead compensator can provide a phase-lead angle and thus a satisfactory phase margin for a system. Alternatively, the phase-lead compensator can enable us to reshape the root locus and thus provide the desired root locations. The phase-lag compensator
FIGURE 10.6 Bode plot of the phase-lag compensator.
is used, not to provide a phase-lag angle, which is normally a destabilizing influence, but rather to provide an attenuation and to increase the steady-state error constant [3].
333.3. PHASE-LEAD DESIGN USING THE BODE PLOT
The Bode plot is used to design a suitable phase-lead compensator in preference to other frequency response plots. The frequency response of the cascade compensator is added to the frequency response of the uncompensated system. That is, because the total loop transfer function of Figure 10.1(a) is \(L(j\omega) = G_{c}(j\omega)G(j\omega)H(j\omega)\), we will first plot the Bode plot for \(G(j\omega)H(j\omega)\). Then we can examine the plot for \(G(j\omega)H(j\omega)\) and determine a suitable location for \(p\) and \(z\) of \(G_{c}(j\omega)\) in order to satisfactorily reshape the frequency response. The uncompensated \(G(j\omega)H(j\omega)\) is plotted with the desired gain to allow an acceptable steady-state error. Then the phase margin and the expected \(M_{p\omega}\) are examined to find whether they satisfy the specifications. If the phase margin is not sufficient, phase lead can be added to the phase-angle curve of the system by placing the \(G_{c}(j\omega)\) in a suitable location. To obtain maximum additional phase lead, we adjust the network so that the frequency \(\omega_{m}\) is located at the frequency where the magnitude of the compensated magnitude curve crosses the 0 - \(dB\) axis. The value of the added phase lead required allows us to determine the necessary value for \(\alpha\) from Equation (10.11) or Figure 10.4. The zero \(z = 1/(\alpha\tau)\) is located by noting that the maximum phase lead should occur at \(\omega_{m} = \sqrt{zp}\), halfway between the pole and the zero. Because the total magnitude gain for the compensator is \(20log\alpha\), we expect a gain of \(10log\alpha\) at \(\omega_{m}\). Thus, we determine the compensator by completing the following steps:
-
Evaluate the uncompensated system phase margin when the error constants are satisfied.
-
Allowing for a small amount of safety, determine the necessary additional phase lead \(\phi_{m}\).
-
Evaluate \(\alpha\) from Equation (10.11).
-
Assume \(K/\alpha = 1\) in \(G_{s}(s)\) in Equation (10.7). This gain will be adjusted in step 8 .
-
Evaluate \(10log\alpha\) and determine the frequency where the uncompensated magnitude curve is equal to \(- 10log\alpha dB\). Because the compensator provides a gain of \(10log\alpha\) at \(\omega_{m}\), this frequency is the new 0 - \(dB\) crossover frequency and \(\omega_{m}\) simultaneously.
-
Calculate the pole \(p = \omega_{m}\sqrt{a}\) and \(z = p/\alpha\).
-
Draw the compensated frequency response, check the resulting phase margin, and repeat the steps if necessary.
-
Finally, for an acceptable design, raise the gain, \(K\), compensator in order to account for the attenuation \((1/\alpha)\).
334. EXAMPLE 10.1 A lead compensator for a type-two system
Let us consider a single-loop feedback control system as shown in Figure 10.1(a), where
and \(H(s) = 1\). The uncompensated system is a type-two system and at first appears to possess a satisfactory steady-state error for both step and ramp input signals. However, the response of the uncompensated system is an undamped oscillation because
Therefore, the compensator is added so that the loop transfer function is \(L(s) = G_{c}(s)G(s)\). The specifications for the system are
Settling time, \(T_{s} \leq 4\text{ }s\);
System damping constant \(\zeta \geq 0.45\).
The settling time (with a \(2\%\) criterion) requirement is
therefore,
FIGURE 10.7
Bode plot for Example 10.1.
Perhaps the easiest way to check the value of \(\omega_{n}\) for the frequency response is to relate \(\omega_{n}\) to the bandwidth \(\omega_{B}\), and evaluate the \(- 3 - dB\) bandwidth of the closedloop system. For a closed-loop system with \(\zeta = 0.45\), we estimate the closed-loop bandwidth \(\omega_{B} = ( - 1.19\zeta + 1.85)\omega_{n} = 3.00\). The bandwidth can be checked following compensation by utilizing the Nichols chart. The Bode plot of
is shown as solid lines in Figure 10.7. The phase margin of the system is required to be approximately
The phase margin of the uncompensated system is \(0^{\circ}\) because the double integration results in a constant \(180^{\circ}\) phase lag. Therefore, we must add a \(45^{\circ}\) phase-lead angle at the crossover \((0 - dB)\) frequency of the compensated magnitude curve. Evaluating the value of \(\alpha\), we have
and thus \(\alpha = 5.8\). To provide a margin of safety, we will use \(\alpha = 6\). The value of \(10log\alpha\) is then equal to \(7.78\text{ }dB\). Then the lead compensator will add an additional gain of \(7.78\text{ }dB\) at the frequency \(\omega_{m}\), and we want to have \(\omega_{m}\) equal to the compensated slope near the 0 - \(dB\) axis (the dashed line) so that the new crossover is \(\omega_{m}\) and the dashed magnitude curve is \(7.78\text{ }dB\) above the uncompensated curve at the crossover frequency. Thus, the compensated crossover frequency is located by evaluating the frequency where the uncompensated magnitude curve is equal to \(- 7.78\text{ }dB\), which in this case is \(\omega = 4.95\). Then the maximum phase-lead angle is added to \(\omega = \omega_{m} = 4.95\), as shown in Figure 10.7. Using step 6, we determine the pole \(p = \omega_{m}\sqrt{\alpha} = 12.0\) and the zero \(z = p/\alpha = 2.0\).
The transfer function of the compensator is
in the form of Equation (10.8). We select \(K = 6\) so that the total loop gain is still equal to 10 . When we add the compensated Bode plot to the uncompensated Bode plot, as in Figure 10.7, we assume that we can raise the gain to account for the \(1/\alpha\) attenuation.
The total loop transfer function is
The closed-loop transfer function is
and the effects of the zero at \(s = - 2\) and the third pole at \(s = - 6\) will affect the transient response. The percent overshoot is \(P.O. = 34\%\), the settling time is \(T_{s} = 1.3\text{ }s\), the bandwidth is \(\omega_{B} = 8.4rad/s\), and the phase margin is \(P.M. = {45.6}^{\circ}\).
335. EXAMPLE 10.2 A lead compensator for a second-order system
A unity feedback control system has a loop transfer function
where \(L(s) = G_{c}(s)G(s)\). We want to have a steady-state error for a ramp input of \(e_{ss} = 5\%\) of the velocity of the ramp. Therefore, we require that
Furthermore, we desire that the phase margin of the system be at least P.M. \(= 40^{\circ}\). The first step is to obtain the Bode plot of the uncompensated transfer function
where \(K = K_{v}\), as shown in Figure 10.8(a). The frequency at which the magnitude curve crosses the \(0 - dB\) line is \(6.2rad/s\), and the phase margin at this frequency is determined readily from
336. क्र है \(\theta\)
(a)
(a) Bode plot for Example 10.2.
(b) Nichols chart for Example 10.2.
(b) At the crossover frequency \(\omega = \omega_{c} = 6.2rad/s\), we have
and therefore the phase margin is \(P.M. = 18^{\circ}\). We need to add a phase-lead compensator so that the phase margin is raised to \(P.M. = 40^{\circ}\) at the new crossover \((0 - dB)\) frequency. Because the compensation crossover frequency is greater than the uncompensated crossover frequency, the phase lag of the uncompensated system is also greater. We shall account for this additional phase lag by attempting to obtain a maximum phase lead of \(40^{\circ} - 18^{\circ} = 22^{\circ}\), plus a small increment of phase lead to account for the added lag. Thus, we will design a compensator with a maximum phase lead equal to \(22^{\circ} + 8^{\circ} = 30^{\circ}\). Then, calculating \(\alpha\), we obtain
and therefore \(\alpha = 3\).
The maximum phase lead occurs at \(\omega_{m}\), and this frequency will be selected so that the new crossover frequency and \(\omega_{m}\) coincide. The lead compensator will add an additional \(10log\alpha = 10log3 = 4.8\text{ }dB\) at \(\omega_{m}\). The compensated crossover frequency is then evaluated where the magnitude of \(G(j\omega)\) is \(- 4.8\text{ }dB\), and thus \(\omega_{m} = \omega_{c} = 8.4\). Drawing the compensated magnitude line so that it intersects the 0 - \(dB\) axis at \(\omega = \omega_{c} = 8.4\), we find that \(z = \omega_{m}/\sqrt{\alpha} = 4.8\) and \(p = \alpha z = 14.4\). Therefore, the compensator is
The total loop gain must be raised by a factor of three in order to account for the factor \(1/\alpha\). With \(K = 3\), the compensated loop transfer function is
To verify the final phase margin, we can evaluate the phase of \(G_{c}(j\omega)G(j\omega)\) at \(\omega = \omega_{c} = 8.4\) and thus obtain the phase margin. The phase angle is then
Therefore, the phase margin for the compensated system is \(P.M. = {43.7}^{\circ}\). The step response of this system yields \(P.O. = 28\%\) with a settling time of \(T_{s} = 0.9\text{ }s\). The compensated system has a steady-state error of \(5\%\) to a ramp, as desired.
The Nichols chart for the compensated and uncompensated system is shown in Figure 10.8(b). The reshaping of the frequency response locus is clear on this chart. Note the increased phase margin for the compensated system as well as the reduced magnitude of \(M_{p\omega}\), the maximum magnitude of the closed-loop frequency response. In this case, \(M_{p\omega}\) has been reduced from an uncompensated value of \(+ 12\text{ }dB\) to a compensated value of approximately \(+ 3.2\text{ }dB\). Also, we note that the closed-loop \(3 - dB\) bandwidth of the compensated system is equal to \(12rad/s\) compared with 9.5 \(rad/s\) for the uncompensated system.
336.1. PHASE-LEAD DESIGN USING THE ROOT LOCUS
The design of the phase-lead compensator can also be readily accomplished using the root locus. The locations of the compensator zero and pole are selected so as to result in a satisfactory root locus for the compensated system. The specifications of the system are used to specify the desired location of the dominant roots of the system. The \(s\)-plane root locus method is as follows:
-
List the system specifications and translate them into a desired root location for the dominant roots.
-
Sketch the root locus with a constant gain controller, \(G_{c}(s) = K\), and determine whether the desired root locations can be realized.
-
If a compensator is necessary, place the zero of the phase-lead compensator directly below the desired root location (or to the left of the first two real poles).
-
Determine the pole location so that the total angle at the desired root location is \(180^{\circ}\) and therefore is on the compensated root locus.
-
Evaluate the total system gain at the desired root location and then calculate the error constant.
-
Repeat the steps if the error constant is not satisfactory.
Therefore, we first locate our desired dominant root locations so that the dominant roots satisfy the specifications in terms of \(\zeta\) and \(\omega_{n}\), as shown in Figure 10.9(a). The root locus of the system with \(G_{c}(s) = K\) is sketched as illustrated in Figure 10.9(b). Then the zero is added to provide a phase lead by placing it to the left of the first two real poles. Some caution is necessary because the zero must not alter the dominance of the desired roots; that is, the zero should not be placed closer to the origin than the second pole on the real axis, or a real root near the origin will result and will dominate the system response. Thus, in Figure 10.9(c), we note that the desired root is directly above the second pole, and we place the zero \(z\) somewhat to the left of the second real pole.
Consequently, the real root may be near the real zero, and the coefficient of this term of the partial fraction expansion may be relatively small. Hence, the response due to this real root may have very little effect on the overall system response. Nevertheless, the designer must be continually aware that the compensated system response will be influenced by the roots and zeros of the system and that the dominant roots will not by themselves dictate the response. It is usually wise to allow for some margin of error in the design and to test the compensated system using a computer simulation.
Because the desired root is a point on the root locus when the final compensation is accomplished, we expect the algebraic sum of the vector angles to be \(180^{\circ}\) at that point. Thus, we calculate the angle \(\theta_{p}\) from the pole of the compensator in FIGURE 10.9
Compensation on the s-plane using a phase-lead compensator.
(a) Desired root location
(c) Addition of zero
(b) Root locus with \(G_{c}(s) = K\)
(d) Location of new pole
order to result in a total angle of \(180^{\circ}\). Then, locating a line at an angle \(\theta_{p}\) intersecting the desired root, we are able to evaluate the compensator pole \(p\), as shown in Figure 10.9(d).
The advantage of the root locus method is the ability of the designer to specify the location of the dominant roots and therefore the dominant transient response. The disadvantage of the method is that we cannot directly specify an error constant (for example, \(K_{v}\) ) as in the Bode plot approach. After the design is complete, we evaluate the gain of the system at the root location, which depends on \(p\) and \(z\), and then calculate the error constant for the compensated system. If the error constant is not satisfactory, we must repeat the design steps and alter the location of the desired root as well as the location of the compensator pole and zero.
337. EXAMPLE 10.3 Lead compensator using the root locus
Consider again the system of Example 10.1 where the loop transfer function is
The characteristic equation of the closed-loop system is
and the root locus is the \(j\omega\)-axis. Therefore, we propose to compensate this system with the compensator
where \(|z| < |p|\). The specifications for the system are
Settling time (with a \(2\%\) criterion), \(T_{s} \leq 4\text{ }s\);
Percent overshoot for a step input P.O. \(\leq 35\%\).
Therefore, the damping ratio should be \(\zeta \geq 0.32\). The settling time requirement is
so \(\zeta\omega_{n} = 1\). We will choose a desired dominant root location as
as shown in Figure 10.10 (hence, \(\zeta = 0.45\) ).
Now we place the zero of the compensator directly below the desired location at \(s = - z = - 1\), as shown in Figure 10.10. Measuring the angle at the desired root, we have
FIGURE 10.10
Phase-lead compensator design for Example 10.3.
Therefore, to have a total of \(180^{\circ}\) at the desired root, we evaluate the angle from the undetermined pole, \(\theta_{p}\), as
or \(\theta_{p} = 38^{\circ}\). Then a line is drawn at an angle \(\theta_{p} = 38^{\circ}\) intersecting the desired root location and the real axis, as shown in Figure 10.10. The point of intersection with the real axis is then \(s = - p = - 3.6\). Therefore, the compensator is
and the compensated loop transfer function for the system is
The gain \(K\) is evaluated by measuring the vector lengths from the poles and zeros to the root location. Hence,
Finally, the error constants of this system are evaluated. We find that this system with two open-loop integrations will result in a zero steady-state error for a step and ramp input signal. The acceleration constant is
The steady-state performance of this system is quite satisfactory, and therefore the compensation is complete. When we compare the compensator evaluated by the \(s\)-plane method with the compensator obtained by using the Bode plot approach, we find that the magnitudes of the poles and zeros are different. However, the resulting system will have the same performance, and we need not be concerned with the difference. In fact, the difference arises from the design step (number 3), which places the zero directly below the desired root location. If we placed the zero at \(s = - 2.0\), we would find that the pole evaluated by the \(s\)-plane method is approximately equal to the pole evaluated by the Bode plot approach.
The specifications for the transient response of this system were originally expressed in terms of the percent overshoot and the settling time. These specifications were translated, on the basis of an approximation of the system by a second-order system, to an equivalent \(\zeta\) and \(\omega_{n}\) and therefore a desired root location. However, the original specifications will be satisfied only if the selected roots are dominant. The zero of the compensator and the root resulting from the addition of the compensator pole result in a third-order system with a zero. The validity of approximating this system with a second-order system without a zero is dependent upon the validity of the dominance assumption. Often, the designer will simulate the final design and obtain the actual transient response of the system. The actual percent overshoot is P.O. \(= 46\%\) and a settling time (to within \(2\%\) of the final value) is \(T_{s} = 3.8\text{ }s\) for a step input. These values compare moderately well with the specified values of \(P.O. = 35\%\) and \(T_{S} = 4\text{ }s\), and they justify the use of the dominant root specifications. The difference in the percent overshoot from the specified value is due to the zero, which is not negligible. Thus, again we find that the specification of dominant roots is a useful approach but must be utilized with caution and understanding. A second attempt to obtain a compensated system with a percent overshoot of P.O. \(= 30\%\) would use a prefilter to eliminate the effect of the zero in the closed-loop transfer function.
338. EXAMPLE 10.4 Lead compensator for a type-one system
Consider the system of Example 10.2 and design a compensator based on the root locus approach. The system loop transfer function is
when \(G_{c}(s) = K\). We want the damping ratio of the dominant roots of the system to be \(\zeta = 0.4\) and the velocity error constant to be \(K_{v} \geq 20\).
To achieve a rapid settling time, we will select the real part of the desired roots as \(\zeta\omega_{n} = 4\), and therefore \(T_{s} = 1\text{ }s\). This implies the natural frequency of these roots is fairly large, \(\omega_{n} = 10\); hence, the velocity constant should be reasonably large. The location of the desired roots is shown in Figure 10.11(a) for \(\zeta\omega_{n} = 4,\zeta = 0.4\), and \(\omega_{n} = 10\).
The zero of the compensator is placed at \(s = - z = - 4\), directly below the desired root location. Then the angle at the desired root location is
Therefore, the angle from the undetermined pole is determined from
and thus \(\theta_{p} = 54^{\circ}\). This angle is drawn to intersect the desired root location, and \(p\) is evaluated as \(s = - p = - 10.6\), as shown in Figure 10.11(a). The gain of the compensated system is then
The compensated system loop transfer function is then
Therefore, the velocity constant of the compensated system is
(a)
FIGURE 10.11
(a) Design of a phase-lead compensator on the s-plane for Example 10.4. (b) Step response of the compensated system of Example 10.4.
(b)
The velocity constant of the compensated system meets the requirement \(K_{v} \geq 20\).
The step response of the compensated system yields a percent overshoot of P.O. \(= 34\%\) with a settling time of \(T_{s} = 1.06\text{ }s\), as shown in Figure 10.11(b). The phase margin is \(P.M. = {38.4}^{\circ}\).
The phase-lead compensator is useful for altering the performance of a control system. The phase-lead compensator adds a phase-lead angle to provide an adequate phase margin. Using an \(s\)-plane design approach, we can choose the phase-lead compensator in order to alter the system root locus and place the roots of the system in a desired position in the \(s\)-plane. When the design specifications include an error constant requirement, the Bode plot method is more suitable, because the error constant of a system designed on the \(s\)-plane must be ascertained following the choice of a compensator pole and zero. Therefore, the root locus method often results in an iterative design procedure when the error constant is specified. On the other hand, the root locus is a very satisfactory approach when the specifications are given in terms of percent overshoot and settling time, thus specifying the \(\zeta\) and \(\omega_{n}\) of the desired dominant roots in the \(s\)-plane. The use of a lead compensator extends the bandwidth of a feedback system, which may be objectionable for systems subjected to large amounts of noise. Also, lead compensators are not suitable for providing high steady-state accuracy in systems requiring very high error constants. To provide large error constants, typically \(K_{p}\) and \(K_{v}\), we must consider the use of integration-type compensators.
338.1. SYSTEM DESIGN USING INTEGRATION COMPENSATORS
For many control systems, the primary objective is obtaining a high steady-state accuracy. Another goal is maintaining the transient performance of these systems within reasonable limits. The steady-state accuracy of many feedback systems can be improved by increasing the gain in the forward channel. However, the resulting transient response may be unacceptable-even unstable. Therefore, it is often necessary to introduce a compensator in the forward path of a feedback control system in order to provide a sufficient steady-state accuracy.
Consider the single-loop control system shown in Figure 10.12. The compensator is chosen to provide a large error constant. With \(G_{p}(s) = 1\), the steady-state error of this system is
The steady-state error of a system depends on the number of poles at the origin for \(L(s) = G_{c}(s)G(s)H(s)\). A pole at the origin can be considered an integration, and therefore the steady-state accuracy of a system ultimately depends on the number of integrations in the loop transfer function. If the steady-state accuracy is not sufficient, we will introduce an integration-type compensator \(G_{c}(s)\) in order to compensate for the lack of integration in the uncompensated loop transfer function \(G_{c}(s)H_{c}(s)\).
One widely used form of controller is the proportional plus integral (PI) controller, which has a transfer function
FIGURE 10.12
Single-loop feedback control system.
As an example, consider a control system where the transfer function \(H(s) = 1\), and the transfer function of the process is [28]
The steady-state error of the uncompensated system is
where \(R(s) = A/s\), and \(K = \lim_{s \rightarrow 0}\mspace{2mu} G(s)\). To obtain a small steady-state error, the magnitude of the gain \(K\) must be quite large. However, when \(K\) is quite large, the transient performance of the system will very likely be unacceptable. Therefore, we must consider the addition of a compensator \(G_{c}(s)\), as shown in Figure 10.12. To eliminate the steady-state error of this system, we might choose
The steady-state error for a step input of the system is always zero, because
The transient performance can be adjusted to satisfy the system specifications by adjusting the constants \(K,K_{P}\), and \(K_{I}\). The adjustment of the transient response is perhaps best accomplished by using root locus methods and drawing a root locus for the gain \(K_{P}K\) after locating the zero \(s = - K_{I}/K_{P}\) on the \(s\)-plane.
The addition of an integration as \(G_{c}(s) = K_{P} + K_{I}/s\) can also be used to reduce the steady-state error for a ramp input \(r(t) = t,t \geq 0\). For example, if the uncompensated system \(G(s)\) possessed one integration, the additional integration due to \(G_{c}(s)\) would result in a zero steady-state error for a ramp input.
339. EXAMPLE 10.5 Temperature control system
The transfer function of a temperature control system process is
To maintain zero steady-state error for a step input, we will add the PI compensation compensator
FIGURE 10.13
The s-plane design of an integration compensator.
Therefore, the loop transfer function is
The transient response of the system is required to have a percent overshoot less than or equal to P.O. \(\leq 20\%\). Since the PI compensator introduces a zero that will interact with the dominant poles, we will target a slightly higher damping ratio of the dominant poles to increase the likelihood of achieving the desired percent overshoot. Therefore, the dominant complex roots will be placed on the \(\zeta = 0.6\) line, as shown in Figure 10.13. We will adjust the compensator zero so that the negative real part of the complex roots is \(\zeta\omega_{n} = 0.75\), and thus the settling time (with a \(2\%\) criterion) is \(T_{s} = 4/\left( \zeta\omega_{n} \right) = \frac{16}{3}\text{ }s\). We determine the location of the zero \(z = - K_{I}/K_{P}\) by ensuring that the angle at the desired root is \(- 180^{\circ}\). Therefore, the sum of the angles at the desired root is
where \(\theta_{z}\) is the angle from the undetermined zero. Consequently, we find that \(\theta_{z} = + 89^{\circ}\), and the location of the zero is \(z = - 0.75\). Finally, to determine the gain at the desired root, we evaluate the vector lengths from the poles and zeros and obtain
The compensated root locus and the location of the zero are shown in Figure 10.13. Note that the zero \(z = - K_{I}/K_{P}\) should be placed to the left of the pole at \(s = - 0.5\) to ensure that the complex roots dominate the transient response. In fact, the third root of the compensated system of Figure 10.13 can be determined as \(s = - 1.0\), and therefore this real root is only \(\frac{4}{3}\) times the real part of the complex roots. Although complex roots dominate the response of the system, the equivalent damping of the system is somewhat less than \(\zeta = 0.60\) due to the real root and zero.
The closed-loop transfer function is
The effect of the zero is to increase the overshoot to a step input. The percent overshoot is P.O. \(= 16\%\), the setting time is \(T_{s} = 4.9\text{ }s\), and the steady-state error to a unit step is zero, as desired.
339.1. PHASE-LAG DESIGN USING THE ROOT LOCUS
The phase-lag compensator is an integration-type compensator and can be used to increase the error constant of a feedback control system. The transfer function of the phase-lag compensator is of the form
where
Begin by supposing that the controller is a constant gain controller, \(G_{c}(s) = K\). We refer to the system with loop transfer function \(L(s) = KG(s)\) as the uncompensated system. Then, for example, the velocity error constant of a type-one uncompensated system is
If we add the phase-lag compensator in Equation (10.55), we have
or
Now, if the pole and zero of the compensator are chosen so that \(|z| = \alpha|p| < 1\), the resultant \(K_{v\text{,comp}\text{~}}\) will be increased at the desired root location by \(\alpha\). Then, for example, if \(z = 0.1\) and \(p = 0.01\), the velocity constant of the desired root location will be increased by a factor of 10 . If the compensator pole and zero appear relatively close together on the \(s\)-plane, their effect on the location of the desired root will be negligible. Therefore, the compensator pole-zero combination near the origin of the \(s\)-plane can be used to increase the error constant of a feedback system by the factor \(\alpha\) while altering the root location very slightly.
The steps necessary for the design of a phase-lag compensator on the s-plane are as follows:
-
Obtain the root locus of the uncompensated system with a constant gain controller, \(G_{c}(s) = K\).
-
Determine the transient performance specifications for the system and locate suitable dominant root locations on the uncompensated root locus that will satisfy the specifications. 3. Calculate the loop gain at the desired root location and thus the uncompensated system error constant.
-
Compare the uncompensated error constant with the desired error constant, and calculate the necessary increase that must result from the pole-zero ratio \(\alpha\) of the compensator.
-
With the known ratio of the pole-zero combination of the compensator, determine a suitable location of the pole and zero of the compensator so that the compensated root locus will still pass through the desired root location. Locate the pole and zero near the origin of the \(s\)-plane.
The fifth requirement can be satisfied if the magnitudes of the pole and zero are significantly less than \(\omega_{n}\) of the dominant roots and they appear to merge as measured from the desired root location. The pole and zero will appear to merge at the root location if the angles from the compensator pole and zero are essentially equal as measured to the root location. One method of locating the zero and pole of the compensator is based on the requirement that the difference between the angle of the pole and the angle of the zero as measured at the desired root is less than \(2^{\circ}\).
340. EXAMPLE 10.6 Design of a phase-lag compensator
Consider a unity feedback system where the uncompensated loop transfer function is
We require the damping ratio of the dominant complex roots to be \(\zeta \geq 0.45\), with a system velocity constant \(K_{v} \geq 20\). The uncompensated root locus is a vertical line at \(s = - 1\) and results in a root on the \(\zeta = 0.45\) line at \(s = - 1 \pm j2\), as shown in Figure 10.14. Measuring the gain at this root, we have \(K = (2.24)^{2} = 5\). Therefore, the velocity constant of the uncompensated system is
FIGURE 10.14
Root locus of the uncompensated system of Example 10.6.
FIGURE 10.15
Root locus of the compensated system of Example 10.6. Note that the actual root will differ from the desired root by a slight amount. The vertical portion of the locus leaves the \(\sigma\) axis at \(\sigma = - 0.95\)
Thus, the required ratio of the zero to the pole of the compensator is
Examining Figure 10.15, we find that we might set \(z = 0.1\) and then \(p = 0.1/8\). The difference of the angles from \(p\) and \(z\) at the desired root is approximately \(1^{\circ}\); therefore, \(s = - 1 \pm j2\) is still the location of the dominant roots. The compensated root locus is shown as a heavy line in Figure 10.15. Thus, the compensated system loop transfer function is
341. EXAMPLE 10.7 Design of a phase-lag compensator
Consider a system that is difficult to design using a phase-lead compensator. The loop transfer function of the uncompensated unity feedback system is
It is specified that the velocity constant of this system be \(K_{v} \geq 20\), while the damping ratio of the dominant roots is equal to \(\zeta = 0.707\). The gain necessary for a \(K_{v} = 20\) is
or \(K = 2000\). However, using Routh's criterion, we find that the roots of the characteristic equation lie on the \(j\omega\)-axis at \(\pm j10\) when \(K = 2000\). The roots of the system when the \(K_{v}\) requirement is satisfied are a long way from satisfying the damping ratio specification, and it would be difficult to bring the dominant roots from the \(j\omega\)-axis FIGURE 10.16 Design of a phaselag compensator on the s-plane.
to the \(\zeta = 0.707\) line by using a phase-lead compensator. Therefore, we will attempt to satisfy the \(K_{v}\) and \(\zeta\) requirements by using a phase-lag compensator. The uncompensated root locus of this system is shown in Figure 10.16, and the roots are shown when \(\zeta = 0.707\) and \(s = - 2.9 \pm j2.9\). Measuring the gain at these roots, we find that \(K = 242\). Therefore, the necessary ratio of the zero to the pole of the compensator is
Thus, we will choose \(z = 0.1\) and \(p = 0.1/9\) in order to allow a small margin of safety. Examining Figure 10.16, we find that the difference between the angle from the pole and zero of \(G_{c}(s)\) is negligible. Therefore, the compensated system loop transfer function is
where \(G_{c}(s) = \frac{242(s + 0.1)}{(s + 0.0111)}\)
341.1. PHASE-LAG DESIGN USING THE BODE PLOT
The design of a phase-lag compensator can be readily accomplished on the Bode plot.
The transfer function of the phase-lag compensator, written in Bode plot form, is
The Bode plot of the phase-lag compensator is shown in Figure 10.6. On the Bode plot, the pole and the zero of the compensator have a magnitude much smaller than the smallest pole of the uncompensated system. Thus, the phase lag is not the useful effect of the compensator; it is the attenuation \(- 20log\alpha\) that is the useful effect for compensation. The phase-lag compensator is used to provide an attenuation and therefore to lower the \(0 - dB\) (crossover) frequency of the system. However, at lower crossover frequencies, we usually find that the phase margin of the system is increased, and our specifications can be satisfied. The design procedure for a phaselag compensator on the Bode plot is as follows:
-
Obtain the Bode plot of the uncompensated system with the constant gain controller, \(G_{c}(s) = K\), and with the gain adjusted for the desired error constant.
-
Determine the phase margin of the uncompensated system and, if it is insufficient, proceed with the following steps.
-
Determine the frequency where the phase margin requirement would be satisfied if the magnitude curve crossed the \(0 - dB\) line at this frequency, \(\omega_{c}^{'}\). (Allow for \(5^{\circ}\) phase lag from the phase-lag compensator when determining the new crossover frequency.)
-
Place the zero of the compensator one decade below the new crossover frequency \(\omega_{c}^{'}\), and thus ensure only \(5^{\circ}\) of additional phase lag at \(\omega_{c}^{'}\) (see Figure 10.8) due to the lag network.
-
Measure the necessary attenuation at \(\omega_{c}^{'}\) to ensure that the magnitude curve crosses at this frequency.
-
Calculate \(\alpha\) by noting that the attenuation introduced by the phase-lag compensator is \(- 20log\alpha\) at \(\omega_{c}^{'}\).
-
Calculate the pole as \(\omega_{p} = 1/(\alpha\tau) = \omega_{z}/\alpha\), and the design is completed.
An example of this design procedure will illustrate that the method is simple to carry out in practice.
342. EXAMPLE 10.8 Design of a phase-lag compensator
Consider the unity feedback system of Example 10.6 and design a phase-lag compensator so that the desired phase margin is obtained. The uncompensated loop transfer function is
where \(K_{v} = K/2\). We want \(K_{v} \geq 20\) with a phase margin of \(P.M. = 45^{\circ}\). The uncompensated Bode plot is shown as a solid line in Figure 10.17. The uncompensated system has a phase margin of P.M. \(= 18^{\circ}\), and the phase margin must be increased. Allowing \(5^{\circ}\) for the phase-lag compensator, we locate the frequency \(\omega\) where \(\phi(\omega) = - 130^{\circ}\), which is to be our new crossover frequency \(\omega_{c}^{'}\). In this case, we find that \(\omega_{c}^{'} = 1.66\). We select \(\omega_{c}^{'} = 1.5\) to allow for a small margin of safety. The attenuation necessary to cause \(\omega_{c}^{'}\) to be the new crossover frequency is equal to \(20\text{ }dB\). Both the compensated and uncompensated magnitude curves are an asymptotic approximation. Thus, \(\omega_{c}^{'} = 1.5\), and the required attenuation is \(20\text{ }dB\).
(a)
(b)
FIGURE 10.17 (a) Design of a phase-lag compensator on the Bode plot for Example 10.8. (b) Time response to a step input for the uncompensated system (solid line) and the compensated system (dashed line) of Example 10.8. Then we find that \(20\text{ }dB = 20log\alpha\), or \(\alpha = 10\). Therefore, the zero is one decade below the crossover, or \(\omega_{z} = \omega_{c}^{'}/10 = 0.15\), and the pole is at \(\omega_{p} = \omega_{z}/10 = 0.015\). The compensated system is then
and the phase-lag compensator is
The frequency response of the compensated system is shown in Figure 10.17(a) with dashed lines. It is evident that the phase lag introduces an attenuation that lowers the crossover frequency and therefore increases the phase margin. Note that the phase angle of the lag compensator has almost totally disappeared at the crossover frequency \(\omega_{c}^{'}\). As a final check, we numerically evaluate the phase margin and find that P.M. \(= {46.9}^{\circ}\) at \(\omega_{c}^{'} = 1.58\) which is the desired result. Using the Nichols chart, we find that the closed-loop bandwidth of the system has been reduced from \(\omega = 10rad/s\) for the uncompensated system to \(\omega = 2.5rad/s\) for the compensated system. Due to the reduced bandwidth, we expect a slower time response to a step command.
The time response of the system is shown in Figure 10.17(b). Note that the percent overshoot is P.O. \(= 25\%\) and the peak time is \(T_{p} = 1.84\text{ }s\). Thus, the response is within the specifications.
343. EXAMPLE 10.9 Design of a phase-lag compensator
Consider the unity feedback system of Example 10.7 with
where \(K_{v} = K/100\). A velocity constant of \(K_{v} \geq 20\) is specified. Furthermore, we aim for a phase margin P.M. \(= 70^{\circ}\). The frequency response of the uncompensated system is shown in Figure 10.18. The phase margin of the uncompensated system is \(0^{\circ}\). Allowing \(5^{\circ}\) for the phase-lag compensator, we locate the frequency where the phase is \(\phi(\omega) = - 105^{\circ}\). This frequency is equal to \(\omega = 1.3\), and therefore we will attempt to locate the new crossover frequency at \(\omega_{c}^{'} = 1.3\). Measuring the necessary attenuation at \(\omega = \omega_{c}\ ^{'}\), we find that \(24\text{ }dB\) is required; then \(24 = 20log\alpha\) gives \(\alpha = 16\). The zero of the compensator is located one decade below the crossover frequency, and thus
The pole is then
Therefore, the compensated system is
FIGURE 10.18
Design of a phaselag compensator on the Bode plot for Example 10.9.
where
The compensated frequency response is shown in Figure 10.18. As a final check, we evaluate the phase margin at \(\omega_{c}^{'} = 1.24\) and find that \(P.M. = {70.3}^{\circ}\), which is within the specifications.
We have seen that a phase-lag compensator can be used to alter the frequency response of a feedback control system in order to attain satisfactory system performance. The system design is satisfactory when the asymptotic curve for the magnitude of the compensated system crosses the 0-dB line with a slope of \(- 20\text{ }dB/\) decade. The attenuation of the phase-lag compensator reduces the magnitude of the crossover \((0 - dB)\) frequency to a point where the phase margin of the system is satisfactory. Thus, in contrast to the phase-lead compensator, the phaselag compensator reduces the closed-loop bandwidth of the system as it maintains a suitable error constant.
The phase-lead compensator alters the frequency response of a system by adding a positive (leading) phase angle and therefore increases the phase margin at the crossover \((0 - dB)\) frequency. It becomes evident that a designer might wish to consider using a compensator that provides the attenuation of a phase-lag compensator and the lead-phase angle of a phase-lead compensator. Such a network does exist. It is called a lead-lag network. The transfer function of this compensator is
The first factors in the numerator and denominator, which are functions of \(\tau_{1}\), provide the phase-lead portion of the compensator. The second factors, which are functions of \(\tau_{2}\), provide the phase-lag portion of the compensator. The parameter \(\beta\) is adjusted to provide suitable attenuation of the low-frequency portion of the frequency response, and the parameter \(\alpha\) is adjusted to provide an additional phase lead at the new crossover \((0 - dB)\) frequency. Alternatively, the compensation can be designed on the \(s\)-plane by placing the lead pole and zero compensation in order to locate the dominant roots in a desired location. Then the phase-lag compensator is used to raise the error constant at the dominant root location. The design of a phase lead-lag compensator follows the procedures already discussed. Other literature will further illustrate the utility of lead-lag compensation [2, 3, 25].
343.1. DESIGN ON THE BODE PLOT USING ANALYTICAL METHODS
An analytical technique of selecting the parameters of a single-stage compensator has been developed for the Bode plot [3-5]. For a single-stage compensator,
where \(\alpha < 1\) yields a phase lag and \(\alpha > 1\) yields phase lead. The phase contribution of the compensator at the desired crossover frequency \(\omega_{c}\) (see Equation 10.9) is given by
The magnitude \(M\) (in \(dB\) ) of the compensator in Equation (10.70) at \(\omega_{c}\) is
Eliminating \(\omega_{c}\tau\) from Equations (10.71) and (10.72), we obtain the nontrivial solution equation for \(\alpha\) as
For a single-stage compensator, it is necessary that \(c > p^{2} + 1\). If we solve for \(\alpha\) from Equation (10.73), we can obtain \(\tau\) from
The design steps for adding phase lead are:
-
Select the desired \(\omega_{c}\).
-
Determine the phase margin desired and therefore the required phase \(\phi\) for Equation (10.71).
-
Verify that the phase lead is applicable: \(\phi > 0\) and \(M > 0\).
-
Determine whether a single stage will be sufficient by testing \(c > p^{2} + 1\). 5. Determine \(\alpha\) from Equation (10.73).
-
Determine \(\tau\) from Equation (10.74).
If we need to design a single-stage, phase-lag compensator, then \(\phi < 0\) and \(M < 0\) (step 3). Step 4 will require \(c < 1/\left( 1 + p^{2} \right)\). Otherwise the method is similar.
344. EXAMPLE 10.10 Design using an analytical technique
Consider the system of Example 10.1 using the analytical technique. Examine the uncompensated curves in Figure 10.7. We select \(\omega_{c} = 5\). Then, as before, we desire a phase margin of P.M. \(= 45^{\circ}\). The compensator must yield this phase, so
The required magnitude contribution is \(8\text{ }dB\), or \(M = 8\), so that
Using \(c\) and \(p\), we obtain
Solving for \(\alpha\), we obtain \(\alpha = 5.84\). Solving Equation (10.74), we obtain \(\tau = 0.087\). Therefore, the compensator is
The pole is equal to 11.5 , and the zero is 1.94 . This can be written in phase-lead compensator form as
344.1. SYSTEMS WITH A PREFILTER
In the earlier sections of this chapter, we utilized compensators of the form
that alter the roots of the characteristic equation of the closed-loop system. However, the closed-loop transfer function \(T(s)\) will contain the zero of \(G_{c}(s)\) as a zero of \(T(s)\). This zero will significantly affect the response of the system \(T(s)\).
Let us consider the system shown in Figure 10.19, where
We will introduce a PI compensator, so that
FIGURE 10.19
Control system with a prefilter \(G_{p}(s)\).
The closed-loop transfer function of the system with a prefilter is
For illustrative purposes, the specifications require a settling time (with a \(2\%\) criterion) of \(T_{S} = 0.5\text{ }s\) and a percent overshoot of approximately P.O. \(= 4\%\). We use \(\zeta = 1/\sqrt{2}\) and note that
Thus, we require that \(\zeta\omega_{n} = 8\) or \(\omega_{n} = 8\sqrt{2}\). We now obtain
The closed-loop transfer function when \(G_{p}(s) = 1\) is then
The effect of the zero on the step response is significant. The percent overshoot to a step is \(P.O. = 21\%\).
We use a prefilter \(G_{p}(s)\) to eliminate the zero from \(T(s)\) while maintaining the \(DC\) gain of 1 , thus requiring that
Then we have
and the percent overshoot of this system is P.O. \(= 4.5\%\), as expected.
Let us now consider again Example 10.3, which includes the design of a lead compensator. The resulting closed-loop transfer function can be determined to be (using Figure 10.22)
If \(G_{p}(s) = 1\) (no prefilter), then we obtain a response with a percent overshoot of P.O. \(= 46.6\%\) and a settling time of \(T_{s} = 3.8\text{ }s\). If we use a prefilter,
we obtain a percent overshoot of P.O. \(= 6.7\%\) and a settling time of \(T_{S} =\) \(3.8\text{ }s\). The real root at \(s = - 1.66\) helps to damp the step response. The prefilter is very useful in permitting the designer to introduce a compensator with a zero to adjust the root locations (poles) of the closed-loop transfer function while eliminating the effect of the zero incorporated in \(T(s)\).
In general, we will add a prefilter for systems with lead compensators or PI compensators. Typically, we will not use a prefilter for a system with a lag compensator, since we expect the effect of the zero to be insignificant. To check this assertion, let us consider again the design obtained in Example 10.6. The system with a phase-lag compensator is
The closed-loop transfer function is then
since the zero at \(s = - 0.1\) and the pole at \(s = - 0.104\) approximately cancel. We expect a percent overshoot of \(P.O. = 20\%\) and a settling time (with a \(2\%\) criterion) of \(T_{s} = 4.0\text{ }s\) for the design parameters \(\zeta = 0.45\) and \(\zeta\omega_{n} = 1\). The actual response has a percent overshoot of P.O. \(= 26\%\) and a settling time of \(T_{s} = 5.8\text{ }s\). Thus, we usually do not use a prefilter with systems that utilize lag compensators.
345. EXAMPLE 10.11 Design of a third-order system
Consider a system of the form shown in Figure 10.19 with
Design a system that will yield a step response with a percent overshoot \(P.O. \leq 2\%\) and a settling time \(T_{s} \leq 3\text{ }s\) by using both \(G_{c}(s)\) and \(G_{p}(s)\) to achieve the desired response.
Consider the lead compensator
and select \(K\) to find the complex roots with \(\zeta = 1/\sqrt{2}\). Then, with \(K = 78.7\), the closed-loop transfer function is
If we choose
Table 10.1 Effect of a Prefilter on the Step Response
| $$G_{p}(\text{ }s)$$ | $$\mathbf{p}\mathbf{= 1}$$ | $$\mathbf{p} = \mathbf{1.20}$$ | $$p = 2.4$$ |
| Percent overshoot | $$0%$$ | $$0%$$ | $$5%$$ |
| \(90\%\) rise time (seconds) | 2.6 | 2.2 | 1.60 |
| Settling time (seconds) | 4.0 | 3.0 | 3.2 |
the closed-loop transfer function is
If \(p = 1.2\), we cancel the effect of the zero. The response of the system with a prefilter is summarized in Table 10.1. We choose the appropriate value for \(p\) to achieve the response desired. Note that \(p = 2.40\) will provide a response that may be desirable, since it effects a faster rise time than \(p = 1.20\). The prefilter provides an additional parameter to select for design purposes.
345.1. DESIGN FOR DEADBEAT RESPONSE
Often, the goal for a control system is to achieve a fast response to a step command with minimal overshoot. We define a deadbeat response as a response that proceeds rapidly to the desired level and holds at that level with minimal overshoot. We use the \(\pm 2\%\) band at the desired level as the acceptable range of variation from the desired response. Then, if the response enters the band at time \(T_{s}\), it has satisfied the settling time \(T_{s}\) upon entry to the band, as illustrated in Figure 10.20. A deadbeat response has the following characteristics:
-
Steady-state error \(= 0\)
-
Fast response \(\rightarrow\) minimum \(T_{r}\) and \(T_{s}\)
-
\(0.1\% \leq P.O. < 2\%\)
-
Percent undershoot P.O. \(< 2\%\).
Characteristics (3) and (4) require that the response remain within the \(\pm 2\%\) band so that the entry to the band occurs at the settling time.
Consider the transfer function \(T(s)\) of a closed-loop system. To determine the coefficients that yield the optimal deadbeat response, the standard transfer function is first normalized. An example of this for a third-order system is
Dividing the numerator and denominator by \(\omega_{n}^{3}\) yields
FIGURE 10.20 The deadbeat response. \(A\) is the magnitude of the step input.
Let \(\bar{s} = s/\omega_{n}\) to obtain
Equation (10.83) is the normalized, third-order, closed-loop transfer function. For a higher order system, the same method is used to derive the normalized equation. The coefficients of the equation \(- \alpha,\beta,\gamma\), and so on - are then assigned the values necessary to meet the requirement of deadbeat response. The coefficients recorded in Table 10.2 were selected to achieve deadbeat response and minimize settling time and rise time \(T_{r}\). The form of Equation (10.83) is normalized since \(\bar{s} = s/\omega_{n}\). Thus, we choose \(\omega_{n}\) based on the desired settling time or rise time. Therefore, if we have a third-order system with a required settling time of \(T_{s} = 1.2\text{ }s\), we note from Table 10.2 that the normalized settling time is
| $$\begin | |||||||||
| \text{}\text{System}\text{} \ | |||||||||
| \text{}\text{Order}\text{} | |||||||||
| \end{matrix}$$ | Coefficients | $$\begin |
\text{~}\text{Percent}\text{~} \\
\text{~}\text{Overshoot P.O.}\text{~}
\end{matrix}$$ | $$\begin{matrix}
\text{~}\text{Percent}\text{~} \\
\text{~}\text{Overshoot P.U.}\text{~}
\end{matrix}$$ | $$\begin{matrix}
90\%\text{~}\text{Rise}\text{~} \\
\text{~}\text{Time}\text{~}T_{r}
\end{matrix}$$ | $$\begin{matrix}
\text{~}\text{Settling}\text{~} \\
\text{~}\text{Time}\text{~}T_{s}
\end{matrix}$$ |
| | $$\alpha$$ | $$\beta$$ | $$\gamma$$ | $$\delta$$ | $$\in$$ | | | | |
| 2nd | 1.82 | | | | | $$0.10%$$ | $$0.00%$$ | 3.47 | 4.82 |
| $$3rd$$ | 1.90 | 2.20 | | | | $$1.65%$$ | $$1.36%$$ | 3.48 | 4.04 |
| 4 th | 2.20 | 3.50 | 2.80 | | | $$0.89%$$ | $$0.95%$$ | 4.16 | 4.81 |
| 5 th | 2.70 | 4.90 | 5.40 | 3.40 | | $$1.29%$$ | $$0.37%$$ | 4.84 | 5.43 |
| 6th | 3.15 | 6.50 | 8.70 | 7.55 | 4.05 | $$1.63%$$ | $$0.94%$$ | 5.49 | 6.04 |
346. Table 10.2 Coefficients and Response Measures of a Deadbeat System
Note: All times are normalized. Therefore, we require that
Once \(\omega_{n}\) is chosen, the complete closed-loop transfer function is known, having the form of Equation (10.81). When designing a system to obtain a deadbeat response, the compensator is chosen, and the closed-loop transfer function is found. This compensated transfer function is then set equal to Equation (10.81), and the required compensator can be determined.
347. EXAMPLE 10.12 Design of a system with a deadbeat response
Consider a unity feedback system with a compensator \(G_{c}(s)\) and a prefilter \(G_{p}(s)\). The process is
and the compensator is
Using the necessary prefilter yields
The closed-loop transfer function is
We use Table 10.2 to determine the required coefficients, \(\alpha = 1.90\) and \(\beta = 2.20\). If we select a settling time (with a \(2\%\) criterion) of \(T_{s} = 2\text{ }s\), then \(\omega_{n}T_{s} = 4.04\), and thus \(\omega_{n} = 2.02\). The required closed-loop system has the characteristic equation
Then, we determine that \(p = 2.84,z = 1.34\), and \(K = 6.14\). The response of this system will have \(T_{s} = 2\text{ }s\), and \(T_{r} = 1.72\text{ }s\).
347.1. DESIGN EXAMPLES
In this section we present two illustrative examples. The first example is a rotor winder control system where both a lead and lag compensator are designed using root locus methods. In the second example, precise control of a milling machine used in manufacturing is employed to illustrate the design process. A lag compensator is designed using root locus methods to meet steady-state tracking error and percent overshoot specifications. FIGURE 10.21
(a) Rotor winder control system.
(b) Block diagram.
(a)
(b)
348. EXAMPLE 10.13 Rotor winder control system
Our goal is to replace a manual operation using a machine to wind copper wire onto the rotors of small motors. Each motor has three separate windings of several hundred turns of wire. It is important that the windings be consistent and that the throughput of the process be high. The operator simply inserts an unwound rotor, pushes a start button, and then removes the completely wound rotor. The DC motor is used to achieve accurate rapid windings. Thus, the goal is to achieve high steady-state accuracy for both position and velocity. The control system is shown in Figure 10.21(a) and the block diagram in Figure 10.21(b). This system has zero steady-state error for a step input, and the steady-state error for a ramp input is
where
When \(G_{c}(s) = K\), we have \(K_{v} = K/50\). If we select \(K = 500\), we will have \(K_{v} = 10\), but the percent overshoot to a step is \(P.O. = 70\%\), and the settling time is \(T_{S} = 8\text{ }s\).
We first try a lead compensator so that
FIGURE 10.22
Root locus for lead compensator.
Selecting \(z_{1} = 4\) and the pole \(p_{1}\) so that the complex roots have a \(\zeta = 0.6\), we have (see Figure 10.22)
We find the response to a step input has a P.O. \(= 3\%\) and a settling time of \(T_{s} = 1.5\text{ }s\). However, the velocity constant is
which is inadequate.
If we use a phase-lag compensator, we select
in order to achieve \(K_{v} = 38\). Thus, the velocity constant of the phase-lag compensated system is
Using a root locus, we select \(K = 105\) in order to achieve a reasonable uncompensated step response with a percent overshoot of P.O. \(\leq 10\%\). We select \(\alpha = z/p\) to achieve the desired \(K_{v}\). We then have
Table 10.3 Design Example Results
| Controller | Gain, \(K\) | $$\begin |
\text{~}\text{Lead}\text{~} \\
\text{~}\text{Compensator}\text{~}
\end{matrix}$$ | $$\begin{matrix}
\text{~}\text{Lag}\text{~} \\
\text{~}\text{Compensator}\text{~}
\end{matrix}$$ | $$\begin{matrix}
\text{~}\text{Lead-Lag}\text{~} \\
\text{~}\text{Compensator}\text{~}
\end{matrix}$$ |
| Step overshoot | $$70%$$ | $$3%$$ | $$12%$$ | $$5%$$ |
| Settling time (seconds) | 8 | 1.5 | 2.5 | 2.0 |
| Steady-state error for ramp | $$10%$$ | $$48%$$ | $$2.6%$$ | $$4.8%$$ |
| $$K_{v}$$ | 10 | 2.1 | 38 | 21 |
(a)
(b)
FIGURE 10.23 (a) Step response and (b) ramp response for rotor winder system.
Selecting \(z_{2} = 0.1\) to avoid affecting the uncompensated root locus, we have \(p_{2} = 0.0055\). We then obtain a step response with a P.O. \(= 12\%\) and a settling time of \(T_{s} = 2.5\text{ }s\). The results for the simple gain, the lead network, and the lag network are summarized in Table 10.3.
Let us return to the phase-lead compensator system and add a cascade phaselag compensator, so that the lead-lag compensator is
The lead compensator of Equation (10.86) requires \(K = 191.2,z_{1} = 4\), and \(p_{1} = 7.3\). The root locus for the system is shown in Figure 10.22. We recall that this lead compensator resulted in \(K_{v} = 2.1\) (see Table 10.3). To obtain \(K_{v} = 21\), we use \(\alpha = 10\) and select \(z_{2} = 0.1\) and \(p_{2} = 0.01\). Then the compensated loop transfer function is
The step response and ramp response of this system are shown in Figure 10.23 in parts (a) and (b), respectively, and are summarized in Table 10.3. Clearly, the leadlag design is suitable for satisfaction of the design goals. FIGURE 10.24
A depiction of the milling machine.
349. EXAMPLE 10.14 Milling machine control system
Smaller, lighter, less costly sensors are being developed by engineers for machining and other manufacturing processes. A milling machine table is depicted in Figure 10.24. This particular machine table has a new sensor that obtains information about the cutting process (that is, the depth-of-cut) from the acoustic emission (AE) signals. Acoustic emissions are low-amplitude, high-frequency stress waves that originate from the rapid release of strain energy in a continuous medium. The AE sensors are commonly piezoelectric amplitude sensitive in the \(100kHz\) to \(1MHz\) range; they are cost effective and can be mounted on most machine tools.
There is a relationship between the sensitivity of the AE power signal and small depth-of-cut changes \(\lbrack 15,18,19\rbrack\). This relationship can be exploited to obtain a feedback signal or measurement of the depth-of-cut. A simplified block diagram of the feedback system is shown in Figure 10.25. The elements of the design process emphasized in this example are highlighted in Figure 10.26.
Since the acoustic emissions are sensitive to material, tool geometry, tool wear, and cutting parameters such as cutter rotational speed, the measurement of the depth-of-cut is modeled as being corrupted by noise, denoted by \(N(s)\) in Figure 10.25. Also disturbances to the process, denoted by \(T_{d}(s)\), are modeled. These could represent external disturbances resulting in unwanted motion of the cutter, fluctuations in the cutter rotation speed, and so forth.
noise
FIGURE 10.25
A simplified block diagram of the milling machine feedback system.
FIGURE 10.26 Elements of the control system design process emphasized in this milling machine control system design example.
The process model \(G(s)\) is given by
and represents the model of the cutter apparatus and the AE sensor dynamics. The input to \(G(s)\) is a control signal to actuate an electromechanical device, which then applies downward pressure on the cutter.
There are a variety of methods available to obtain the model represented by Equation (10.88). One approach would be to use basic principles to obtain a mathematical model in the form of a nonlinear differential equation, which can then be linearized about an operating point leading to a linear model (or equivalently, a transfer function). The basic principles include Newton's laws, the various conservation laws, and Kirchhoff's laws. Another approach would be to assume a form of the model (such as a second-order system) with unknown parameters (such as \(\omega_{n}\) and \(\zeta\) ), and then experimentally obtain good values of the unknown parameters. FIGURE 10.27
Hypothetical impulse response of the milling machine.
A third approach is to conduct a laboratory experiment to obtain the step or impulse response of the system. In other words we can apply an input (in this case, a voltage) to the system and measure the output - the depth-of-cut into the desired workpiece. Suppose, for example, we have the impulse response data shown in Figure 10.27 (the small circles on the graph represent the data). If we had access to the function \(C_{imp}(t)\)-the impulse response function of the milling machine-we could take the Laplace transform to obtain the transfer function model. There are many methods available for curve fitting the data to obtain the function \(C_{imp}(t)\). We will not cover curve fitting here, but we can say a few words regarding the basic structure of the function.
From Figure 10.27 we see that the response approaches a steady-state value:
So we expect that
where \(\Delta C_{\text{imp}\text{~}}(t)\) is a function that goes to zero as \(t\) gets large. This leads us to consider \(\Delta C_{imp}(t)\) as a sum of stable exponentials. Since the response does not oscillate, we might expect that the exponentials are, in fact, real exponentials,
where \(\tau_{i}\) are positive real numbers. The data in Figure 10.27 can be fitted by the function
for which the Laplace transform is
Thus we can obtain the transfer function model of the milling machine.
The control goal is to develop a feedback system to track a desired step input. In this case the reference input is the desired depth-of-cut. The control goal is stated as
350. Control Goal
Control the depth-of-cut to the desired value.
The variable to be controlled is the depth-of-cut, or
351. Variable to Be Controlled
Depth-of-cut \(y(t)\).
Since we are focusing on lead and lag controllers in this chapter, the key tuning parameters are the parameters associated with the compensator given in Equation (10.89).
352. Select Key Tuning Parameters
Compensator variables: \(p,z\), and \(K\).
The control design specifications are
353. Control Design Specifications
DS1 Track a ramp input, \(R(s) = a/s^{2}\), with a steady-state tracking error less than \(a/8\), where \(a\) is the ramp velocity.
DS2 Percent overshoot to a step input of P.O. \(\leq 20\%\).
The phase-lag compensator is given by
where \(\alpha = z/p > 1\) and \(\tau = 1/z\). The tracking error is
where
Therefore,
With \(R(s) = a/s^{2}\) and using the final value theorem, we find that
or equivalently,
According to DS1, we require that
or
Substituting for \(G(s)\) and \(G_{c}(s)\) from Equations (10.88) and (10.89), respectively, we obtain the compensated velocity constant
The compensated velocity constant is the velocity constant of the system when the phase-lag compensator is in the loop.
The loop transfer function is
We separate the phase-lag compensator from the process and obtain the uncompensated root locus by considering the feedback loop with the gain \(K\), but not the phase-lag compensator zero and pole factors. The uncompensated root locus for the characteristic equation
is shown in Figure 10.28.
From DS2 we determine that the target damping ratio of the dominant roots is \(\zeta > 0.45\). We find that \(K \leq 2.09\) at \(\zeta \geq 0.45\). Then with \(K = 2.0\) the uncompensated velocity constant is
FIGURE 10.28
Root locus for the uncompensated system.
