Modern Control Systems_P5
The compensated velocity constant is
Therefore with \(\alpha = z/p\), we obtain the relationship
We require \(K_{v\text{,comp}\text{~}} > 8\). A possible choice is \(K_{v\text{,comp}\text{~}} = 10\) as the desired velocity constant. Then
But \(\alpha = z/p\), thus our phase-lag compensator should have \(p = 0.08z\). If we select \(z = 0.01\) then \(p = 0.0008\).
The compensated loop transfer function is given by
The phase-lag compensator with \(z\) and \(p\) as above is determined to be
The step response is shown in Figure 10.29. The percent overshoot is P.O. \(= 22\%\). The velocity error constant is \(K_{v} = 10\), which satisfies DS1. FIGURE 10.29
Step response for the compensated system.
353.1. SYSTEM DESIGN USING CONTROL DESIGN SOFTWARE
We want to use computers, when appropriate, to assist the designer in the selection of the parameters of a compensator. The development of algorithms for computer-aided design is an important alternative approach to the trial-and-error methods considered in earlier sections. Computer programs have been developed for the selection of suitable parameter values for compensators based on satisfaction of frequency response criteria such as the phase margin \(\lbrack 3,4\rbrack\).
In this section, the compensation of control systems is illustrated using frequency response and \(s\)-plane methods. We will consider again the rotor winder design. To illustrate the use of \(m\)-file scripts in designing and developing control systems with good performance characteristics. We examine both the phase-lead and phase-lag compensators for this design example and obtain the system response using computer-based analysis tools.
354. EXAMPLE 10.15 Rotor winder control system
Consider again the rotor winder control system shown in Figure 10.21. The design objective is to achieve high steady-state accuracy to a ramp input. The steady-state error to a unit ramp input \(R(s) = 1/s^{2}\) is
where
The performance specification of percent overshoot and settling time must be considered, as must the steady-state tracking error. In all likelihood, a simple gain will not be satisfactory, so we will also consider compensation utilizing phase-lead and phase-lag compensators, using both Bode plot and root locus plot design methods. Our approach is to develop a series of \(m\)-file scripts to aid in the compensator designs.
Consider a simple gain controller
Then the steady-state error is
The larger we make \(K\), the smaller is the steady-state error \(e_{ss}\). However, we must consider the effect that increasing \(K\) has on the transient response, as shown in Figure 10.30 . When \(K = 500\), our steady-state error for a ramp is \(10\%\), but the percent overshoot is \(P.O. = 70\%\), and the settling time is approximately \(T_{s} = 8\text{ }s\) for a step input. We consider this to be unacceptable performance and thus turn
(a)
FIGURE 10.30
(a) Transient response for simple gain controller. (b) m-file script.
Compute response for four gains.
numg=[1]; deng=[1 1550 0]; sysg=tf(numg, deng); \(t = \lbrack 0:0.1:5\rbrack\); \(\%\)
for \(i = 1:4\) sys=feedback(K(i)*sysg,[1]); \(y =\) step(sys,t);
Ys(:,i)=y; end \(\%\) \(plot(t,Ys(:,1),t,Ys(:,2),t,Ys(:,3),t,Ys(:,4))\) xlabel('Time \((s)\) '), ylabel('y(t)')
Store response for \(i\) th gain in \(Ys\).
Closed-loop transfer function. to compensation. The two important compensator types that we consider are phase-lead and phase-lag compensators.
First, we try a phase-lead compensator
where \(|z| < |p|\) The phase-lead compensator will give us the capability to improve the transient response. We will use a frequency-domain approach to design the phase-lead compensator.
We want a steady-state error of \(e_{ss} \leq 10\%\) to a ramp input and \(K_{v} = 10\). In addition to the steady-state specifications, we want to meet certain performance specifications: (1) settling time (with a \(2\%\) criterion) \(T_{s} \leq 3\text{ }s\), and (2) percent overshoot for a step input P.O. \(\leq 10\%\). Solving for \(\zeta\) and \(\omega_{n}\) using
yields \(\zeta = 0.59\) and \(\omega_{n} = 2.26\). We thus obtain the phase margin requirement:
The steps leading to the final design are as follows:
-
Obtain the uncompensated system Bode plot with \(K = 500\), and compute the phase margin.
-
Determine the amount of necessary phase lead \(\phi_{m}\).
-
Evaluate \(\alpha\) from \(sin\phi_{m} = (\alpha - 1)/(\alpha + 1)\).
-
Compute \(10log\alpha\) and find the frequency \(\omega_{m}\) on the uncompensated Bode plot where the magnitude curve is equal to \(- 10log\alpha\).
-
In the neighborhood of \(\omega_{m}\) on the uncompensated Bode plot draw a line through the 0 -dB point at \(\omega_{m}\) with slope equal to the current slope plus \(20\text{ }dB/decade\). Locate the intersection of the line with the uncompensated Bode plot to determine the phase-lead compensation zero location. Then calculate the phase-lead compensator pole location as \(p = \alpha z\).
-
Obtain the compensated Bode plot and check the phase margin. Repeat any steps if necessary.
-
Raise the gain to account for the attenuation \(1/\alpha\).
-
Verify the final design with simulation using step function inputs, and repeat any design steps if necessary.
We use three scripts in the design. The design scripts are shown in Figures 10.31-10.33. The script in Figure 10.31 is for the Bode plot of the uncompensated system. The script in Figure 10.32 is for the detailed Bode plot of the compensated system. The script in Figure 10.33 is for the step response analysis. The final phase-lead compensator design is
where \(K = 1800\) was selected after iteratively using the m-file script. FIGURE 10.31
(a) Bode plot.
(b) m-file script.
(a)
Additional phase lead.
Compute \(\alpha\).
Plot \(- 10log(\alpha)\) line to aid in locating \(\omega_{m}\).
(b)
The settling time and percent overshoot specifications are satisfied, but \(K_{v} = 5\), resulting in a \(20\%\) steady-state error to a ramp input. It is possible to continue the design iteration and refine the compensator somewhat, although it should be clear that the phase-lead compensator has added phase margin and improved the transient response as anticipated.
To reduce the steady-state error, we can consider the phase-lag compensator, which has the form
where \(|p| < |z|\). We will use a root locus approach to design the phase-lag compensator, although it can be done using a Bode plot as well. The desired root location region of the dominant roots is specified by
(a)
FIGURE 10.32
Phase-lead compensator:
(a) compensated Bode plot, (b) m-file script.
FIGURE 10.33
Phase-lead compensator:
(a) step response,
(b) m-file script.
355. \(K = 1800\);
numg=[1]; deng=[1 1550 0];
numgc \(= K^{*}\begin{bmatrix} 1 & 3.5 \end{bmatrix}\); dengc \(= \begin{bmatrix} 1 & 25 \end{bmatrix}\)
sysg=tf(numg,deng);
sysgc=tf(numgc, dengc);
sys=series(sysgc,sysg);
margin(sys)
Increase \(K\) to account for attenuation of \(1/\alpha\).
Lead compensator.
(a)
\(K = 1800\);
numg=[1]; deng=[1 1550 0]; sysg=tf(numg,deng);
numgc \(= K^{\star}\begin{bmatrix} 13.5 \end{bmatrix};\) dengc \(= \begin{bmatrix} 1 & 25 \end{bmatrix};\) sysgc \(= tf(\) numgc, dengc \()\);
syso \(=\) series \((\) sysgc, sysg);
sys=feedback(syso,[1]);
\(t = \lbrack 0:0.01:2\rbrack\);
step(sys,t)
ylabel ('y(t)') (b)
356. FIGURE 10.34
Phase-lag compensator: (a) uncompensated root locus, (b) m-file script.
(a)
(b)
The steps in the design are as follows:
-
Obtain the root locus of the uncompensated system.
-
Locate suitable root locations on the uncompensated system that lie in the region defined by \(\zeta = 0.59\) and \(\omega_{n} = 2.26\).
-
Calculate the loop gain at the desired root location and the system error constant, \(K_{v,\text{~}\text{unc}\text{~}}\).
-
With \(\alpha\) known, determine suitable locations of the compensator pole and zero so that the compensated root locus still passes through the desired location.
-
Verify with simulation and repeat any steps if necessary.
The design methodology is illustrated in Figures 10.34-10.36. Using the rlocfind function, we can compute the gain \(K\) associated with the roots of our choice on the uncompensated root locus that lie in the performance region. We then compute \(\alpha\) to ensure that we achieve the desired \(K_{v}\). We place the lag compensator pole and zero to avoid affecting the uncompensated root locus. In Figure 10.35, the phase-lag compensator pole and zero are very near the origin, at \(z = - 0.1\) and \(p = - 0.01\). FIGURE 10.35
Phase-lag compensator:
(a) compensated root locus, (b) m-file script.
(a)
(b)
357. Table 10.4 Compensator Design Results
| Controller | Gain, \(\mathbf{K} = \mathbf{500}\) | Lead | Lag |
| Step overshoot | $$70%$$ | $$8%$$ | $$13%$$ |
| Settling time (seconds) | 8 | 1 | 9 |
| Steady-state error for ramp | $$10%$$ | $$20%$$ | $$10%$$ |
| $$K_{v}$$ | 10 | 5 | 10 |
The settling time and percent overshoot specifications are not satisfied, but \(K_{v} = 10\), as desired. It is possible to continue the design iteration and refine the compensator somewhat, although it should be clear that the phase-lag compensator has improved the steady-state errors to a ramp input relative to the phase-lead compensator design. The final phase-lag compensator design is
The resulting performance is summarized in Table 10.4. FIGURE 10.36
Phase-lag compensator:
(a) step response,
(b) m-file response.
(a)
(b)
357.1. SEQUENTIAL DESIGN EXAMPLE: DISK DRIVE READ SYSTEM
In this chapter, we design a PD controller to achieve the specified response to a unit step input. The specifications are given in Table 10.5. The closed-loop system is shown in Figure 10.37. A prefilter is used to eliminate any undesired effects of the term \(s + z\) introduced in the closed-loop transfer function. We will use the deadbeat system, where the desired closed-loop transfer function is
358. Table 10.5 Disk Drive Control System Specifications and Actual Performance
| Performance Measure | Desired Value | Actual Response |
| Percent overshoot | Less than \(5\%\) | $$0.1%$$ |
| Settling time | Less than \(250\text{ }ms\) | $$40\text{ }ms$$ |
| $$\begin | ||
| \text{~}\text{Maximum response}\text{~} \ | ||
| \text{~}\text{to a unit disturbance}\text | ||
| \end{matrix}$$ | Less than \(5 \times 10^{- 3}\) | $$6.9 \times 10^{- 5}$$ |
FIGURE 10.37 Disk drive control system with PD controller (second-order model).
For the second-order model shown in Figure 10.37, we require \(\alpha = 1.82\) (see Table 10.2). Then the settling time is
Since we want a settling time \(T_{s} \leq 50\text{ }ms\), we will use \(\omega_{n} = 120\). Then we expect \(T_{s} = 40\text{ }ms\). Therefore, the denominator of Equation (10.91) is
The characteristic equation of the closed-loop system of Figure 10.37 is
Equating Equations (10.92) and (10.93), we have
and
Therefore, \(K_{P} = 2880\) and \(K_{D} = 39.68\). Then we note that
The prefilter will then be
The model neglected the motor field. Nevertheless, this design will be very accurate. The actual response is given in Table 10.5. All the specifications are satisfied.
358.1. SUMMARY
In this chapter, we have considered several alternative approaches to the design of feedback control systems. In the first two sections, we discussed the concepts of design and compensation and noted the several design cases. Then we examined the possibility of introducing cascade compensators within the feedback loops of control systems. The cascade compensators are useful for altering the shape of the root locus or frequency response of a system. The phase-lead compensator and the phase-lag compensator were considered in detail as candidates for system compensators. Then system compensation was studied by using a phase-lead compensator on the Bode plot and the root locus. We noted that the phase-lead compensator increases the phase margin of the system and thus provides additional stability. When the design specifications include an error constant, the design of a phase-lead compensator is more readily accomplished on the Bode plot. Alternatively, when an error constant is not specified but the settling time and percent overshoot for a step input are specified, the design of a phase-lead compensator is more readily carried out on the \(s\)-plane. When large error constants are specified for a feedback system, it is usually easier to compensate the system by using integration (phase-lag) compensators. We also noted that the phase-lead compensation increases the system bandwidth, whereas the phase-lag compensation decreases the system bandwidth. The bandwidth may often be an important factor when noise is present at the input and generated within the system. Also, we noted that a satisfactory system is obtained when the asymptotic course for magnitude of the compensated system crosses the 0 - \(dB\) line with a slope of \(- 20\text{ }dB/\) decade. The characteristics of the phase-lead and phase-lag compensators are summarized in Table 10.6. Operational amplifier circuits for phase-lead and phase-lag and for PI and PD compensators are summarized in Table 10.7 [1].
359. Table 10.6 A Summary of the Characteristics of Phase-Lead and Phase-Lag Compensators
| Compensation | ||
| Phase-Lead | Phase-Lag | |
| Approach | $$\begin |
\text{~}\text{Addition of phase-lead angle near cross-}\text{~} \\
\text{~}\text{over frequency on Bode plot. Add lead}\text{~} \\
\text{~}\text{compensator to yield desired dominant}\text{~} \\
\text{~}\text{roots in}\text{~}s\text{-plane.}\text{~}
\end{matrix}$$ | $$\begin{matrix}
\text{~}\text{Addition of phase-lag to yield an increased}\text{~} \\
\text{~}\text{error constant while maintaining desired}\text{~} \\
\text{~}\text{dominant roots in}\text{~}s\text{-plane or phase margin}\text{~} \\
\text{~}\text{on Bode plot}\text{~}
\end{matrix}$$ |
| Results | $$\begin{matrix}
\text{~}\text{1. Increases system bandwidth}\text{~} \
\text{~}\text{2. Increases gain at higher frequencies}\text{~}
\end{matrix}$$ | 1. Decreases system bandwidth |
| Advantages | $$\begin{matrix}
\text{~}\text{1. Yields desired response}\text{~} \
\text{~}\text{2. Improves dynamic response}\text{~}
\end{matrix}$$ | $$\begin{matrix}
\text{~}\text{1. Suppresses high-frequency noise}\text{~} \
\text{~}\text{2. Reduces steady-state error}\text{~}
\end{matrix}$$ |
| Disadvantages | $$\begin{matrix}
\text{~}\text{1. Requires additional amplifier gain}\text{~} \
\text{~}\text{2. Increases bandwidth and thus}\text{~} \
\text{~}\text{susceptibility to noise}\text{~}
\end{matrix}$$ | 1. Slows down transient response |
| Applications | 1. When fast transient response is desired | 1. When error constants are specified |
| $$\begin{matrix}
\text{~}\text{Situations not}\text{~} \
\text{}\text{applicable}\text{}
\end{matrix}$$ | $$\begin{matrix}
\text{~}\text{1. When phase decreases rapidly near}\text{~} \
\text{~}\text{crossover frequency}\text{~}
\end{matrix}$$ | $$\begin{matrix}
\text{~}\text{1. When no low-frequency range exists where}\text{~} \
\text{~}\text{phase is equal to desired phase margin}\text{~}
\end{matrix}$$ |
Table 10.7 Operational Amplifier Circuits for Compensators
Type of
Controller
360. PD
PI
Lead or lag
Lead if
Lag if
361. 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 10.38 as specified in the various problem statements.
FIGURE 10.38 Block diagram for the Skills Check. In the following True or False and Multiple Choice problems, circle the correct answer.
- A cascade compensator is a compensator that is placed in parallel with the system process.
True or False
-
Generally, a phase-lag compensator speeds up the transient response.
-
The arrangement of the system and the selection of suitable components and parameters is part of the process of control system design.
True or False
- A deadbeat response of a system is a rapid response with minimal percent overshoot and zero steady-state error to a step input.
True or False
-
A phase-lead compensator can be used to increase the system bandwidth. True or False
-
Consider the feedback system in Figure 10.38, where
A phase-lag compensator is designed for the system:
When compared with the uncompensated system (that is, \(G_{c}(s) = 1\) ), the compensated system utilizing the phase-lag compensator:
a. Increases the phase lag near the cross-over frequency.
b. Increases the phase margin.
c. Provides additional attenuation at higher frequencies.
d. All of the above.
- A position control system can be analyzed using the feedback system in Figure 10.38, where the process transfer function is
A phase-lag compensator that provides a phase margin of \(P.M. \approx 30^{\circ}\) is:
a. \(G_{c}(s) = \frac{1 + s}{1 + 106s}\)
b. \(G_{c}(s) = \frac{1 + 26s}{1 + 115s}\)
c. \(G_{c}(s) = \frac{1 + 106s}{1 + 118s}\)
d. None of the above
- Consider a unity feedback system in Figure 10.38, where
A phase-lead compensator is introduced into the feedback loop, where
The peak magnitude and the bandwidth of the closed-loop frequency response are:
a. \(M_{p_{\omega}} = 1.9\text{ }dB;\omega_{b} = 12.1rad/s\)
b. \(M_{p_{\omega}} = 12.8\text{ }dB;\omega_{b} = 14.9rad/s\)
c. \(M_{p_{\omega}} = 5.3\text{ }dB;\omega_{b} = 4.7rad/s\)
d. \(M_{p_{\omega}} = 4.3\text{ }dB;\omega_{b} = 24.2rad/s\) 9. Consider the feedback system in Figure 10.38, where the plant model is
and the controller is a proportional-plus-integral (PI) controller given by
Selecting \(K_{I} = 1\), determine a suitable value of \(K_{P}\) for a percent overshoot of P. \(O = 20\%\).
a. \(K_{P} = 0.5\)
b. \(K_{P} = 1.5\)
c. \(K_{P} = 2.5\)
d. \(K_{P} = 5.0\)
- Consider the feedback system in Figure 10.38, where
The design specifications are: \(K_{v} \geq 100,G.M. \geq 10\text{ }dB,P.M. \geq 45^{\circ}\), and the crossover frequency, \(\omega_{c} \geq 10rad/s\). Which of the following controllers meets these specifications?
a. \(G_{c}(s) = \frac{(1 + s)(1 + 20s)}{(1 + s/0.01)(1 + s/50)}\)
b. \(G_{c}(s) = \frac{100(1 + s)(1 + s/5)}{(1 + s/0.1)(1 + s/50)}\)
c. \(G_{c}(s) = \frac{1 + 100s}{1 + 120s}\)
d. \(G_{c}(s) = 100\)
- Consider a feedback system in which a phase-lead compensator
is placed in series with the plant
The feedback system is a negative unity feedback control system shown in Figure 10.38. Compute the gain and phase margin.
a. \(G \cdot M. = \infty dB,P.M. = 60^{\circ}\)
b. \(G.M. = 20.5\text{ }dB,P.M. = {47.8}^{\circ}\)
c. \(G.M. = 8.6\text{ }dB,P.M. = {33.7}^{\circ}\)
d. Closed-loop system is unstable.
- Consider the feedback system in Figure 10.38, where
Which of the following represents a suitable phase-lag compensator that achieves a steady-state \(e_{ss} \leq 0.1\) for a unit ramp input and a damping ratio of the closed-loop system dominant roots of \(\zeta \approx 0.707\).
a. \(G_{c}(s) = \frac{2850(s + 1)}{(10s + 1)}\)
b. \(G_{c}(s) = \frac{100(s + 1)(s + 5)}{(s + 10)(s + 50)}\)
c. \(G_{c}(s) = \frac{10}{s + 1}\)
d. Closed-loop system cannot track a ramp input for any \(G_{c}(s)\).
- A viable phase-lag compensator for a unity negative feedback system with plant transfer function
that satisfies the design specifications: (i) no percent overshoot; (ii) rise time \(T_{r} < 5\text{ }s\), and (iii) position error constant \(K_{p} > 6\), is which of the following:
a. \(G_{c}(s) = \frac{s + 1}{s + 0.074}\)
b. \(G_{c}(s) = \frac{s + 0.074}{s + 1}\)
c. \(G_{c}(s) = \frac{20s + 1}{100s + 1}\)
d. \(G_{c}(s) = 20\)
- Consider the feedback system depicted in Figure 10.38, where
A suitable compensator for \(G_{c}(s)\) this system that satisfies the specifications:
(i) P.O. \(\leq 20\%\), and (ii) velocity error constant \(K_{v} \geq 10\), is which of the following:
a. \(G_{c}(s) = \frac{s + 4}{(s + 1)}\)
b. \(G_{c}(s) = \frac{160(10s + 1)}{200s + 1}\)
c. \(G_{c}(s) = \frac{24(s + 1)}{s + 4}\)
d. None of the above
- Using a Nichols chart, determine the gain and phase margin of the system in Figure 10.38 with loop gain transfer function
a. \(G.M. = 20.4\text{ }dB,P.M. = {58.1}^{\circ}\)
b. \(G \cdot M. = \infty dB,P \cdot M. = 47^{\circ}\)
c. \(G \cdot M. = 6\text{ }dB,P \cdot M. = 45^{\circ}\)
d. \(G \cdot M. = \infty dB,P.M. = 23^{\circ}\) In the following Word Match problems, match the term with the definition by writing the correct letter in the space provided.
a. Deadbeat response A system with a rapid response, minimal overshoot, an zero steady-state error for a step input.
b. Phase lead A network that provides a positive phase angle over the compensation
c. PI controller
d. Lead-lag compensator frequency range of interest.
A compensator hat acts, in part, like an integrator.
A compensator with the characteristics of both a lead compensator and a lag compensator.
e. Design of a control system
A compensator that provides a negative phase angle and a significant attenuation over the frequency range of interest.
f. Phase lag compensation
g. Integration network
h. Compensator
i. Compensation
An additional component or circuit that is inserted into the system to compensate for a performance deficiency.
A compensator placed in cascade or series with the system process.
Controller with a proportional term and an integral term.
A transfer function, \(G_{P}(s)\), that filters the input signal \(R(s)\) prior to calculating the error signal.
j. Phase-lag network
k. Cascade compensation network
- Phase-lead network A widely-used compensator that possesses one zero and one pole with the pole closer to the origin of the \(s\)-plane.
m. Prefilter
A widely-used compensator that possesses one zero and one pole with the zero closer to the origin of the \(s\)-plane.
362. EXERCISES
E10.1 A negative feedback control system has a transfer E10.2 A control system with negative unity feedback has function
We select a compensator
in order to achieve zero steady-state error for a step input. Select \(a\) and \(K\) to target a percent overshoot to a step of P.O. \(\leq 10\%\) and the settling time (with a \(2\%\) criterion) of \(T_{s} \leq 1\text{ }s\). Does the actual P.O. and \(T_{s}\) match the targeted values? If not, explain why not.
Answer: \(K = 3,a = 15.27\) a process
and we select a proportional plus integral compensation, where
Note that the steady-state error of this system for a ramp input is zero. (a) Set \(K_{I} = 1\) and find a suitable value of \(K_{P}\) so the step response will have a percent overshoot of P.O. \(\leq 20\%\). (b) What is the expected settling time (with a \(2\%\) criterion) of the compensated system?
Answer: \(K_{P} = 0.5\)
E10.3 A unity feedback control system in a manufacturing system has a process transfer function
and it is proposed to use a compensator to achieve a percent overshoot P.O. \(\leq 5\%\) to a step input. The compensator is [4]
which provides proportional plus integral control. Show that one solution is \(K = 0.44\) and \(\tau = 1.5\).
E10.4 Consider a unity feedback system with
where \(K\) is set equal to 100 in order to achieve a specified \(K_{v} = 2\). We wish to add a lead-lag compensator
Show that the gain margin of the compensated system is \(G \cdot M. = 28.6\text{ }dB\) and that the phase margin is P.M. \(= {75.4}^{\circ}\).
E10.5 Consider a unity feedback system with the transfer function
We desire to obtain the dominant roots with \(\omega_{n} = 2\) and \(\zeta = 0.55\). The compensator is
Determine the value of \(K\) that should be selected.
Answer: \(K = 42\)
E10.6 Consider the system with the loop transfer functions
When \(K = 10\), find \(T(s)\) and estimate the expected percent overshoot and settling time (with a \(2\%\) criterion). Compare your estimates with the actual percent overshoot of P.O. \(= 47.5\%\) and a settling time of \(T_{s} = 32.1\text{ }s\).
E10.7 NASA astronauts retrieved a satellite and brought it into the cargo bay of the space shuttle, as shown in Figure E10.7(a). A model of the feedback control system is shown in Figure E10.7(b). Determine the value of \(K\) that will result in a phase margin of P.M. \(= 40^{\circ}\) when \(T = 0.6\text{ }s\).
Answer: \(K = 34.15\)
(a)
FIGURE E10.7
Retrieval of a satellite. (Photo courtesy of NASA.)
Visual feedback
(b) E10.8 A unity feedback system has a plant
where \(\tau = 2.8\text{ }ms\). Select a compensator
so that the dominant roots of the characteristic equation have damping ratio equal to \(\zeta = 1/\sqrt{2}\). Plot \(y(t)\) for a step input.
E10.9 A control system with a controller is shown in Figure E10.9. Select \(K_{P}\) and \(K_{I}\) so that the percent overshoot to a step input is P.O. \(= 4\%\), and the velocity constant \(K_{v}\) is equal to 10 . Verify the results of your design.
E10.10 A control system with a controller is shown in Figure E10.10. Select \(K_{I} = 2\) in order to provide a reasonable steady-state error to a step [8]. Find \(K_{P}\) to obtain a phase margin of P.M. \(= 60^{\circ}\). Find the peak time and percent overshoot of this system.
E10.11 A unity feedback system has
A lead network is selected so that
Determine the peak magnitude, \(M_{p\omega}\), and the bandwidth, \(\omega_{b}\), of the closed-loop frequency response. From \(M_{p\omega}\), estimate the percent overshoot, P.O., to a unit step. Compare with the actual P.O. and comment.
E10.12 The control of an automobile ignition system has unity feedback and a loop transfer function \(L(s) = G_{c}(s)G(s)\), where
Let \(K_{I}/K_{P} = 1\) and determine \(KK_{P}\) so that the complex roots have a damping ratio of \(\zeta = 1/\sqrt{2}\).
E10.13 The design of Example 10.3 determined a lead network in order to obtain desirable dominant root locations using a cascade compensator \(G_{c}(s)\) in the system configuration shown in Figure 10.1(a). The same lead network would be obtained if we used the feedback compensation configuration of Figure 10.1(b). Determine the closed-loop transfer function \(T(s) = Y(s)/R(s)\) of both the cascade and feedback configurations, and show how the transfer function of each configuration differs. Explain how the response to a step \(R(s)\) will be different for each system.
E10.14 A robot will be operated by NASA to build a permanent lunar station. The unity feedback position control system for the gripper tool has the process transfer function
Determine a phase-lag compensator \(G_{c}(s)\) that will provide a phase margin of \(P.M. = 40^{\circ}\).
Answer: \(G(s) = \frac{(1 + 7.69)}{(1 + 28.92s)}\)
E10.15 A unity feedback control system has a plant transfer function
We desire to attain a steady-state error to a ramp \(r(t) = At\) of less than \(0.16A\) and a phase margin of \(P.M. = 35^{\circ}\). We desire to have a crossover frequency \(\omega_{c} = 20rad/s\). Determine whether a phase-lead or a phase-lag compensator is required.
FIGURE E10.9
Design of a PI controller.
FIGURE E10.10 Design of a PI controller.
E10.16 Consider again the system and specifications of Exercise E10.15 when the required crossover frequency is \(\omega_{c} = 3rad/s\).
E10.17 Consider again the system of Exercise 10.9. Select \(K_{P}\) and \(K_{I}\) so that the step response is deadbeat and the settling time (with a \(2\%\) criterion) is \(T_{s} \leq 2\text{ }s\).
E10.18 The nonunity feedback control system shown in Figure E10.18 has the transfer functions
Design a compensator \(G_{c}(s)\) and prefilter \(G_{p}(s)\) so that the closed-loop system is stable and meets the following specifications: (i) a percent overshoot to a unit step input of \(P.O. \leq 10\%\), (ii) a settling time of \(T_{s} \leq 2\text{ }s\), and (iii) zero steady-state tracking error to a unit step.
E10.19 A unity feedback control system has the plant transfer function
Design a PID controller of the form
so that the closed-loop system has a settling time of \(T_{s} \leq 0.5\text{ }s\) to a unit step input.
E10.20 Consider the system shown in Figure E10.20. Design the proportional-derivative controller \(G_{c}(s)\) such that the system has a phase margin of \(40^{\circ} \leq\) P.M. \(\leq 60^{\circ}\).
E10.21 Consider the unity feedback system shown in Figure E10.21. Design the controller gain, \(K\), such that the maximum value of the output \(y(t)\) in response to a unit step disturbance \(T_{d}(s) = 1/s\) is less than 0.1.
FIGURE E10.18 Nonunity feedback system with a prefilter.
FIGURE E10.20 Unity feedback system with \(PD\) controller.
FIGURE E10.21 Closed-loop feedback system with a disturbance input.
363. PROBLEMS
P10.1 The design of a lunar excursion module is an interesting control problem. The attitude control system for the lunar vehicle is shown in Figure P10.1. The vehicle damping is negligible, and the attitude is controlled by gas jets. The torque, as a first approximation, will be considered to be proportional to the signal \(V(s)\) so that \(T(s) = K_{2}V(s)\). The loop gain may be selected by the designer in order to provide a suitable damping. A damping ratio of \(\zeta = 0.6\) with a settling time (with a \(2\%\) criterion) of less than 2.5 seconds is required. Using a lead network compensation, select the necessary compensator \(G_{c}(s)\) by using (a) frequency response techniques and (b) root locus methods.
P10.2 A magnetic tape recorder transport for modern computers requires a high-accuracy, rapid-response control system. The requirements for a specific transport are as follows: (1) The tape must stop or start in \(10\text{ }ms\), and (2) it must be possible to read 45,000 characters per second. This system is illustrated in Figure P10.2. We will use a tachometer in this case and set \(K_{a} = 50000\) and \(K_{2} = 1\). To provide a suitable performance, a compensator \(G_{c}(s)\) is inserted immediately following the photocell transducer. Select a compensator \(G_{c}(s)\) so that the percent overshoot of the system for a step input is \(P.O \leq 25\%\). We assume that \(\tau_{1} = 0.1\text{ }ms,\tau_{a} = 0.1\text{ }ms,K_{1} = 2,R/L = 0.5\text{ }ms,K_{b} =\) \(0.4,r = 0.2,K_{T}/LJ = 2.0\), and \(K_{p} = 1\).
P10.3 A simplified version of the attitude rate control for a supersonic aircraft is shown in Figure P10.3. When the vehicle is flying at four times the speed of sound (Mach 4) at an altitude of \(100,000ft\), the parameters are [26]
FIGURE P10.1
Attitude control system for a lunar excursion module.
FIGURE P10.2 Block diagram of a tape control system.
FIGURE P10.3
Aircraft attitude control.
Design a compensator \(G_{c}(s)\) so that the response to a step input has a percent overshoot of \(P.O \leq 10\%\).
P10.4 Magnetic particle clutches are useful actuator devices for high power requirements because they can typically provide a \(200 - W\) mechanical power output. The particle clutches provide a high torque-to-inertia ratio and fast time-constant response. A particle clutch positioning system for nuclear reactor rods is shown in Figure P10.4. The motor drives two counterrotating clutch housings. The clutch housings are geared through parallel gear trains, and the direction of the servo output is dependent on the clutch that is energized. The time constant of a \(200 - W\) clutch is \(\tau = 1/10\text{ }s\). The constants are such that \(K_{T}n/J = 1\). We want the percent overshoot for a step input to be P.O. \(\leq 20\%\). Design a compensator so that the system is adequately stabilized. The settling time (with a \(2\%\) criterion) of the system should be \(T_{s} \leq 7\text{ }s\).
P10.5 A stabilized precision rate table uses a precision tachometer and a DC direct-drive torque motor, as shown in Figure P10.5. We want to maintain a high steady-state accuracy for the speed control. To obtain a zero steady-state error for a step command design, select a proportional plus integral compensator. Select the appropriate gain constants so that the system has a percent overshoot of P.O. \(= 15\%\) and a settling time (with a \(2\%\) criterion) of \(T_{s} \leq 2\text{ }s\).
P10.6 Repeat Problem P10.5 by using a phase-lead compensator and compare the results.
P10.7 A chemical reactor process whose production rate is a function of catalyst addition is shown in block diagram form in Figure P10.7 [10]. The time delay is \(T = 50\text{ }s\), and the time constant \(\tau\) is approximately 40 s. The gain of the process is \(K = 1\). Design a compensator using Bode plot methods in order to provide a suitable system response. We want to have a steadystate error less than \(0.10A\) for a step input \(R(s) = A/s\). For the system with the compensation added, estimate the settling time of the system.
P10.8 A numerical path-controlled machine turret lathe is an interesting problem in attaining sufficient accuracy \(\lbrack 2,23\rbrack\). A block diagram of a turret lathe control system is shown in Figure P10.8. The gear ratio is \(n = 0.2,J = 10^{- 3}\), and \(b = 2.0 \times 10^{- 2}\). It is necessary
FIGURE P10.4 Nuclear reactor rod control.
FIGURE P10.5 Stabilized rate table.
FIGURE P10.7 Chemical reactor control.
FIGURE P10.8 Path-controlled turret lathe. to attain an accuracy of \(5 \times 10^{- 4}\) in., and therefore a steady-state position accuracy of \(2.5\%\) is specified for a ramp input. Design a cascade compensator to be inserted before the silicon-controlled rectifiers in order to provide a response to a step command with a percent overshoot of \(P.O. \leq 5\%\). A suitable damping ratio for this system is \(\zeta = 0.7\). The gain of the silicon-controlled rectifiers is \(K_{R} = 5\). Design a suitable phase-lag compensator.
P10.9 The Avemar ferry, shown in Figure P10.9(a), is a large 670-ton ferry hydrofoil built for Mediterranean ferry service. It is capable of 45 knots \((52mph)\) [29]. The boat's appearance, like its performance, derives from the innovative design of the narrow "wavepiercing" hulls which move through the water like racing shells. Between the hulls is a third quasihull which gives additional buoyancy in rough seas. Loaded with 900 passengers and crew, and a mix of cars, buses, and freight cars trucks, one of the boats can carry almost its own weight. The Avemar is capable of operating in seas with waves up to \(8ft\) in amplitude at a speed of 40 knots as a result of an automatic stabilization control system. Stabilization is achieved by means of flaps on the main foils and the adjustment of the aft foil. The stabilization control system maintains a level flight through rough seas. Thus, a system that minimizes deviations from a constant lift force or, equivalently, that minimizes the pitch angle \(\theta(t)\) has been designed. A block diagram of the lift control system is shown in Figure P10.9(b). The desired response of the system to wave disturbance is a constant-level travel of the craft. Establish a set of reasonable specifications and design a compensator \(G_{c}(s)\) so that the performance of the system is suitable. Assume that the disturbance is due to waves with a frequency \(\omega = 6rad/s\).
P10.10 A unity feedback system has the loop transfer function
(a) Determine the step response when \(G_{c}(s) = 1\), and calculate the settling time and steady state for a ramp input \(r(t) = t,t > 0\). (b) Design a phase-lag compensator using the root locus method so that the velocity constant is increased to 10 . Determine the settling time (with a \(2\%\) criterion) of the compensated system.
P10.11 A unity feedback control system has the loop transfer function
Select a lead-lag compensator so that the percent overshoot for a step input is P.O. \(\leq 5\%\) and the settling time (with a \(2\%\) criterion) is \(T_{s} \leq 1\text{ }s\). It also is desired that the acceleration constant \(K_{a}\) be greater than 7500 .
P10.12 A unity feedback control system has a plant
FIGURE P10.9
(a) The Avemar ferry built for ferry service between Barcelona and the Balearic Islands. (b) A block diagram of the lift control system. (a)
(b) Select a compensator \(G_{c}(s)\) so that the phase margin is \(P.M. \geq 80^{\circ}\). Use a two-stage lead compensator
It is required that the error for a ramp input be \(1\%\) of the magnitude of the ramp input \(\left( K_{v} = 100 \right)\).
P10.13 Materials testing requires the design of control systems that can faithfully reproduce normal specimen operating environments over a range of specimen parameters [23]. From the control system design viewpoint, a materials-testing machine system can be considered a servomechanism in which we want to have the load waveform track the reference signal. The system is shown in Figure P10.13.
(a) With \(G_{c}(s) = K\), choose \(K\) so that a phase margin of P.M. \(= 45^{\circ}\) is achieved. Determine the system bandwidth for this design.
(b) The additional requirement introduced is that the velocity constant \(K_{v}\) be equal to 1 . Design a lag compensator so that the phase margin is \(P.M. = 45^{\circ}\) and \(K_{v} = 1\).
P10.14 For the system described in Problem P10.13, the goal is to achieve a phase margin of \(P.M. = 45^{\circ}\) with the additional requirement that the time to settle (to within \(2\%\) of the final value) is \(T_{s} \leq 10\text{ }s\). Design a phase-lead compensator to meet the specifications. As before, we require \(K_{v} = 1\).
P10.15 A robot with an extended arm has a heavy load, whose effect is a disturbance, as shown in Figure P10.15 [22]. Let \(R(s) = 0\) and design \(G_{c}(s)\) so that the maximum value of the disturbance response is less than 0.25 and the steady-state error to a unit step disturbarce is zero.
P10.16 A driver and car may be represented by the simplified model shown in Figure P10.16 [17]. The goal is to have the speed adjust to a step input with a percent overshoot of P.O. \(\leq 10\%\) and a settling time (with a \(2\%\) criterion) of \(T_{s} = 1\text{ }s\). Select a proportional plus integral (PI) controller to yield these specifications. For the selected controller, determine the actual response (a) for \(G_{p}(s) = 1\) and (b) with a prefilter \(G_{p}(s)\) that removes the zero from the closed-loop transfer function \(T(s)\).
P10.17 A unity feedback control system for a robot submarine has a plant with a third-order transfer function [20]:
We want the percent overshoot to be P.O. \(= 7.5\%\) for a step input and the settling time (with a \(2\%\) criterion) of the system be \(T_{s} = 400\text{ }ms\). Find a suitable lead compensator by using root locus
FIGURE P10.13 Materials testing machine system.
FIGURE P10.15 Robot control.
FIGURE P10.16 Speed control of an automobile.
(a)
FIGURE P10.18
(a) Conceptual diagram of a remote manipulator on the Moon controlled by a person on the Earth. (b) Feedback diagram of the remote manipulator control system with \(\tau =\) transmission time delay of the video signal.
(b) methods. Let the zero of the compensator be located at \(s = - 15\), and determine the compensator pole. Determine the resulting system \(K_{v}\).
P10.18 NASA is developing remote manipulators that can be used to extend the hand and the power of humankind through space by means of radio. A concept of a remote manipulator is shown in Figure P10.18(a) \(\lbrack 11,22\rbrack\). The closed-loop control is shown schematically in Figure P10.18(b). Assuming an average distance of 238,855 miles from Earth to the Moon, the time delay \(T\) in transmission of a communication signal is \(1.28\text{ }s\). The operator uses a control stick to control remotely the manipulator placed on the Moon to assist in geological experiments, and the TV display to access the response of the manipulator. The time constant of the manipulator is \(1/4\) second.
(a) Set the gain \(K_{1}\) so that the system has a phase margin of P.M. \(= 30^{\circ}\). Evaluate the percentage steady-state error for this system for a step input. (b) To reduce the steady-state error for a position command input to \(5\%\), add a lag compensation network in cascade with \(K_{1}\). Plot the step response.
P10.19 There have been significant developments in the application of robotics technology to nuclear power plant maintenance problems. Thus far, robotics technology in the nuclear industry has been used primarily on spent-fuel reprocessing and waste management. The industry is applying the technology to such areas as primary containment inspection, reactor maintenance, facility decontamination, and accident recovery activities. These developments suggest that the application of remotely operated devices can significantly reduce radiation exposure to personnel and improve maintenance-program performance.
Currently, an operational robotic system is under development to address particular operational problems within a nuclear power plant. This device, IRIS (Industrial Remote Inspection System), is a general-purpose surveillance system that conducts particular inspection and handling tasks with the goal of significantly reducing personnel exposure to high radiation fields [12]. The device is shown in Figure P10.19. The open-loop transfer function is
(a) Determine a suitable gain \(K\) for the system when \(T = 0.5\text{ }s\), so that the percent overshoot to a step FIGURE P10.19
Remotely controlled robot for nuclear plants.
input is P.O. \(\leq 30\%\). Determine the steady-state error. (b) Design a compensator
to improve the step response for the system in part (a) so that the steady-state error is less than \(12\%\). Assume a unity feedback control system.
P10.20 An uncompensated control system with unity feedback has a plant transfer function
We want to have a velocity error constant of \(K_{v} = 20\). We also want to have a phase margin of P.M. \(= 45^{\circ}\) and a closed-loop bandwidth \(\omega_{B} \geq 4rad/s\). Use two identical cascaded phase-lead compensators to compensate the system.
P10.21 For the system of Problem P10.20, design a phaselag compensator to yield the desired specifications, with the exception that a bandwidth \(\omega_{B} \geq 2rad/s\) will be acceptable.
P10.22 For the system of Problem P10.20, we wish to achieve the same phase margin and \(K_{v}\), but in addition, we wish to limit the bandwidth to \(2rad/s \leq \omega_{B} \leq 10rad/s\). Use a lead-lag compensation to compensate the system. The compensator could be of the form
where \(a\) is to be selected for the phase-lag portion of the compensator, and \(b\) is to be selected for the phase-lead portion of the compensator. The ratio \(\alpha\) is chosen to be 10 for both the lead and lag portions.
P10.23 A system the loop transfer function with unity feedback has
We desire the steady-state error to a step input to be approximately \(4\%\) and the phase margin to be \(P.M. = 60^{\circ}\). Design a phase-lag compensator to meet these specifications.
P10.24 The stability and performance of the rotation of a robot (similar to waist rotation) presents a challenging control problem. The system requires high gains in order to achieve high resolution; yet a large percent overshoot of the transient response cannot be tolerated. The block diagram of an electrohydraulic system for rotation control is shown in Figure P10.24 [15].
We want to have \(K_{v} = 20\) for the compensated system. Design a compensator that results in a percent overshoot to a step input of P.O. \(\leq 10\%\).
P10.25 The possibility of overcoming wheel friction, wear, and vibration by contactless suspension for passenger-carrying mass-transit vehicles is being investigated throughout the world. One design uses a magnetic suspension with an attraction force between the vehicle and the guideway with an accurately controlled airgap. A system is shown in Figure P10.25, which incorporates feedback compensation.
FIGURE P10.24 Robot position control.
FIGURE P10.25
Airgap control of train.
Using root locus methods, select a suitable value for \(K_{1}\) and \(b\) so the system has a damping ratio for the underdamped roots of \(\zeta = 0.50\).
P10.26 A computer uses a printer as a fast output device. We desire to maintain accurate position control while moving the paper rapidly through the printer. Consider a system with unity feedback and a transfer function for the motor and amplifier of
Design a phase-lead compensator so that the system bandwidth is \(\omega_{B} = 0.8rad/s\) and the phase margin is \(P.M. \geq 30^{\circ}\).
P10.27 An engineering design team is attempting to control a process shown in Figure P10.27. It is agreed that a system with a phase margin of \(P.M. = 50^{\circ}\) is acceptable. Determine \(G_{c}(s)\).
First, let \(G_{c}(s) = K\) and find (a) a value of \(K\) that yields a phase margin of \(P.M. = 50^{\circ}\) and the system step response for this value of \(K\). (b) Determine the settling time, percent overshoot, and the peak time. (c) Obtain the system closed-loop frequency response, and determine \(M_{p\omega}\) and the bandwidth.
The team has decided to let
and to repeat parts (a), (b), and (c). Determine the gain \(K\) that results in a phase margin of P.M. \(= 50^{\circ}\) and then proceed to evaluate the time response and the closed-loop frequency response. Prepare a table contrasting the results of the two selected controllers for \(G_{c}(s)\) by comparing settling time (with a \(2\%\) criterion), percent overshoot, peak time, \(M_{p\omega}\), and bandwidth.
P10.28 An adaptive suspension vehicle uses a legged locomotion principle. The control of the leg can be represented by a unity feedback system with [12]
We desire to achieve a steady-state error for a ramp input of \(10\%\) and a damping ratio of the dominant roots of \(\zeta = 0.707\). Determine a suitable lag compensator, and determine the actual overshoot and the time to settle (to within \(2\%\) of the final value).
P10.29 A liquid-level control system has a loop transfer function
where \(G_{c}(s)\) is a compensator, and the plant is
where \(T = 50\text{ }ms\). Design a compensator so that \(M_{p\omega}\) does not exceed \(3.5\text{ }dB\) and \(\omega_{r}\) is approximately \(1.4rad/s\). Predict the percent overshoot and settling time (with a \(2\%\) criterion) of the compensated system when the input is a step. Plot the actual response.
P10.30 An automated guided vehicle (AGV) can be considered as an automated mobile conveyor designed
FIGURE P10.30 Steering control for vehicle.
to transport materials. Most AGVs require some type of guide path. The steering stability of the guidance control system has not been fully solved. The slight "snaking" of the AGV about the track generally has been acceptable, although it indicates instability of the steering guidance control system [9].
Most AGVs have a specification of maximum speed of about \(1\text{ }m/s\), although in practice they are usually operated at half that speed. In a fully automated manufacturing environment, there should be few personnel in the production area; therefore, the AGV should be able to be run at full speed. As the speed of the AGV increases, so does the difficulty in designing stable and smooth tracking controls.
A steering system for an \(AGV\) is shown in Figure P10.30, where \(\tau_{1} = 40\text{ }ms\) and \(\tau_{2} = 1\text{ }ms\). We require that the velocity constant \(K_{v}\) be 100 so that the steady-state error for a ramp input will be \(1\%\) of the slope of the ramp. Neglect \(\tau_{2}\) and design a lead compensator so that the phase margin is
Attempt to obtain the two limiting cases for phase margin, and compare your results for the two designs by determining the actual percent overshoot and settling time for a step input.
P10.31 For the system of Problem P10.30, use a phase-lag compensator and attempt to achieve a phase margin of P.M. \(= 50^{\circ}\). Determine the actual percent overshoot and peak time for the compensated system.
P10.32 When a motor drives a flexible structure, the structure's natural frequencies, as compared to the bandwidth of the servodrive, determine the contribution of the structural flexibility to the errors of the resulting motion. In current industrial robots, the drives are often relatively slow, and the structures are relatively rigid, so that overshoots and other errors are caused mainly by the servodrive. However, depending on the accuracy required, the structural deflections of the driven members can become significant. Structural flexibility must be considered the major source of motion errors in space structures and manipulators. Because of weight restrictions in space, large arm lengths result in flexible structures. Furthermore, future industrial robots should require lighter and more flexible manipulators.
To investigate the effects of structural flexibility and how different control schemes can reduce unwanted oscillations, an experimental apparatus was constructed consisting of a DC motor driving a slender aluminum beam. The purpose of the experiments was to identify simple and effective control strategies to deal with the motion errors that occur when a servomotor is driving a very flexible structure [13].
The experimental apparatus is shown in Figure P10.32(a), and the control system is shown in Figure \(P10.32(\text{ }b)\). The goal is that the system will have a \(K_{v}\) of 100. (a) When \(G_{c}(s) = K\), determine \(K\) and obtain the Bode plot. Find the phase margin and gain margin. (b) Using the Nichols chart, find \(\omega_{r},M_{p\omega}\), and \(\omega_{B}\). (c) Select a compensator so that the phase margin is P.M. \(\geq 35^{\circ}\) and find \(\omega_{r},M_{p\omega}\), and \(\omega_{B}\) for the compensated system.
P10.33 Consider the block diagram of the extender robot system shown in Figure P10.33 [14]. The goal is that the compensated system will have a velocity constant \(K_{v}\) equal to 80 , so that the settling time (with a \(2\%\) criterion) will be \(T_{s} = 1.6\text{ }s\), and that the percent overshoot will be P.O. \(= 16\%\), so that the dominant roots have a \(\zeta = 0.5\). Determine a lead-lag compensator using root locus methods.
P10.34 A magnetically levitated train operated in Berlin, Germany from 1989-1991. Fully automated trains can run at short intervals and operate with excellent energy efficiency. The control system for the levitation of the car is shown in Figure P10.34. Select a compensator so that the phase margin of the system is \(45^{\circ} \leq P.M. \leq 55^{\circ}\). Predict the response of the system to a step command, and determine the actual step response for comparison.
P10.35 A unity feedback system has the loop transfer function
where \(T\) is a time delay and \(K\) is the controller proportional gain. The block diagram is illustrated in Figure P10.35. The nominal value of \(K = 2\). Plot the phase margin of the system for \(0 \leq T \leq 2\) s when \(K = 2\). What happens to the phase margin as the time delay
(a)
FIGURE P10.32
Flexible arm control.
FIGURE P10.33 Extender robot control.
FIGURE P10.34 Magnetically levitated train control.
FIGURE P10.35 Unity feedback system with a time delay and PI controller.
(b)
increases? What is the maximum time delay allowed before the system becomes unstable? the compensated system and plot the step response. Assume a unity feedback system.
P10.36 A system transfer function is a pure time delay of P10.37 A unity feedback system has the loop transfer \(0.5\text{ }s\), so that \(G(s) = e^{- s/2}\). Select a compensator \(G_{c}(s)\) so that the steady-state error for a step input is less than \(2\%\) of the magnitude of the step and the phase margin is P.M. \(\geq 30^{\circ}\). Determine the bandwidth of function
Design a compensator \(G_{c}(s)\) so that the percent overshoot for a step input \(R(s)\) is \(P.O. \leq 5\%\) and the steady-state error is less than \(1\%\). Determine the bandwidth of the system.
P10.38 A unity feedback system has a plant
We desire to have a phase margin of \(P.M. = 35^{\circ}\) and a relatively large bandwidth. Select the crossover frequency \(\omega_{c} = 10rad/s\), and design a phase-lead compensator. Verify the results.
P10.39 A unity feedback system has a plant
We desire that the phase margin be P.M. \(= 30^{\circ}\). For a ramp input \(r(t) = t\), we want the steady-state error to be equal to 0.05 . Design a phase-lag compensator to satisfy the requirements. Verify the results.
P10.40 For the system and requirements of Problem P10.39, determine the required compensator when the steadystate error for the ramp input must be equal to 0.02 .
P10.41 Repeat Example 10.12 when we want the rise time to be \(T_{r} = 1\text{ }s\).
P10.42 Consider the system shown in Figure P10.42 and let \(R(s) = 0\) and \(T_{d}(s) = 0\). Design the compensator \(G_{c}(s) = K\) such that, in the steady-state, the response of the system is less than \(- 40\text{ }dB\) when the noise \(N(s)\) is a sinusoidal input at a frequency of \(\omega \geq 100rad/s\).
P10.43 A unity feedback system has a loop transfer function
Plot the percent overshoot of the closed-loop system response to a unit step input for \(K\) in the range \(0 < K \leq 100\). Explain the behavior of the system response for \(K\) in the range \(0.129 < K \leq 69.872\).
FIGURE P10.42 Unity feedback system with proportional controller and measurement noise.
364. ADVANCED PROBLEMS
AP10.1 A three-axis pick-and-place application requires the precise movement of a robotic arm in three-dimensional space, as shown in Figure AP10.1 for joint 2. The arm has specific linear paths it must follow to avoid other pieces of machinery. The overshoot for a step input should be less than \(13\%\).
(a) Let \(G_{c}(s) = K\), and determine the gain \(K\) that satisfies the requirement. Determine the resulting settling time (with a \(2\%\) criterion). (b) Use pole-zero cancellation to reduce the settling time to \(T_{s} \leq 5\text{ }s\).
AP10.2 The system of Advanced Problem AP10.1 is to have a percent overshoot of \(P.O. \leq 13\%\). In addition, we desire that the steady-state error for a unit ramp input will be less than \(0.125\left( K_{v} = 8 \right)\) [24]. Design a lag compensator to meet the specifications. Check the resulting percent overshoot and settling time (with a \(2\%\) criterion) for the design.
AP10.3 The system of Advanced Problem AP10.1 is required to have a percent overshoot of P.O. \(\leq 13\%\) with a steady-state error for a unit ramp input less than \(0.125\left( K_{v} = 8 \right)\). Design a proportional plus integral (PI) controller to meet the specifications.
AP10.4 A DC motor control system with unity feedback has the form shown in Figure AP10.4. Select \(K_{1}\) and \(K_{2}\) so that the system response has a settling time (with a \(2\%\) criterion) \(T_{s} \leq 0.5\text{ }s\) and a percent overshoot of P.O. \(\leq 10\%\) for a step input.
AP10.5 A unity feedback system is shown in Figure AP10.5. We want the step response of the system to have a percent overshoot of P.O. \(\leq 10\%\) and a settling time (with a \(2\%\) criterion) of \(T_{s} \leq 4\text{ }s\).
(a) Design a phase-lead compensator \(G_{c}(s)\) to achieve the dominant roots desired. (b) Determine the step response of the system when \(G_{p}(s) = 1\). (c) Select FIGURE AP10.1
Pick-and-place robot.
FIGURE AP10.4 Motor control system.
FIGURE AP10.5 Unity feedback with a prefilter.
(a)
(b)
a prefilter \(G_{p}(s)\), and determine the step response of the system with the prefilter.
AP10.6 Consider a unity feedback system with loop transfer function
We wish to minimize the settling time of the system while requiring that \(K < 52\). Determine the appropriate compensator parameters \(p\) and \(z\) that will minimize the settling time. Plot the system response.
AP10.7 A unity feedback system has
with a phase-lead compensator
Determine \(K\) so that the complex roots have \(\zeta = 1/\sqrt{2}\). Consider the prefilter
(a) Determine the percent overshoot and rise time for \(G_{p}(s) = 1\) and for \(p = 1\). (b) Select an appropriate value for \(p\) that will give an overshoot of P.O. \(\leq 1\%\), and compare the results.
AP10.8 The Manutec robot has large inertia and arm length resulting in a challenging control problem, as shown in Figure AP10.8(a). The block diagram model of the system is shown in Figure AP10.8(b).
The percentage overshoot for a step input should be \(P.O. \leq 20\%\) with a rise time of \(T_{r} \leq 0.5\text{ }s\) and a settling time (with a \(2\%\) criterion) of \(T_{s} \leq 1.2\text{ }s\). Also, we desire that for a ramp input \(K_{v} \geq 10\). Determine a suitable phase-lead compensator.
AP10.9 The plant dynamics of a chemical process are represented by
We desire that the unity feedback system have a small steady-state error for a ramp input so that \(K_{v} = 100\). For stability purposes, we desire a gain margin of G.M. \(\geq 10\text{ }dB\) and a phase margin of P.M. \(\geq 40^{\circ}\). Determine a lead-lag compensator that meets these specifications.
FIGURE AP10.8
(a) Manutec robot.
(b) Block diagram.
(a)
(b)
365. DESIGN PROBLEMS
CDP10.1 The capstan-slide system of Figure CDP4.1 uses a PD controller. Determine the necessary values of the gain constants of the PD controller so that the deadbeat response is achieved. Also, we want the settling time (with a \(2\%\) criterion) to be \(T_{s} \leq 250\text{ }ms\). Verify the results.
DP10.1 In Figure DP10.1, two robots are shown cooperating with each other to manipulate a long shaft to insert it into the hole in the block resting on the table. Long part insertion is a good example of a task that can benefit from cooperative control. The unity feedback control system of one robot joint has the process transfer function
The specifications require a steady-state error for a unit ramp input of 0.01 , and the step response has an overshoot of P.O. \(\leq 5\%\) with a settling time (with a \(2\%\) criterion) of \(T_{s} \leq 1\text{ }s\). Determine a lead-lag compensator that will meet the specifications, and plot the compensated responses for the ramp and step inputs.
FIGURE DP10.1 Two robots cooperate to insert a shaft.
DP10.2 The heading control of the traditional bi-wing aircraft, shown in Figure DP10.2(a), is represented by the block diagram of Figure DP10.2(b).
(a) Determine the minimum value of the gain \(K\) when \(G_{c}(s) = K\), so that the steady-state effect of a unit step disturbance \(T_{d}(s) = 1/s\) is less than or equal to \(5\%\) of the unit step \((y(\infty) = 0.05)\).
(b) Determine whether the system using the gain of part (a) is stable.
(c) Design a compensator using one stage of lead compensation, so that the phase margin is P.M. \(= 30^{\circ}\).
(d) Design a two-stage lead compensator so that the phase margin is P.M. \(= 55^{\circ}\).
(e) Compare the bandwidth of the systems of parts (c) and (d).
(f) Plot the step response \(y(t)\) for the systems of parts (c) and (d) and compare percent overshoot settling time (with a \(2\%\) criterion), and peak time.
DP10.3 NASA has identified the need for large deployable space structures, which will be constructed of lightweight materials and will contain large numbers of joints or structural connections. This need is evident for programs such as the space station. These deployable space structures may have precision shape requirements and a need for vibration suppression during in-orbit operations [16].
One such structure is the mast flight system, which is shown in Figure DP10.3(a). The intent of the system is to provide an experimental test bed for controls and dynamics. The basic element in the mast flight system is a \(60.7 - m\)-long truss beam structure, which is attached to the shuttle orbiter. Included at the tip of the truss structure are the primary actuators and collocated sensors. A deployment/retraction subsystem, which also secures the stowed beam package during launch and landing, is provided.
The system uses a large motor to move the structure and has the block diagram shown in Figure DP10.3(b). The goal is a percent overshoot to a step response of P.O. \(\leq 20\%\); thus, we estimate the system \(\zeta = 0.5\) and the required phase margin as P.M. \(= 50^{\circ}\). Design for \(0.75 < K < 2.0\) and record the percent overshoot, rise time, and phase margin for selected gains.
DP10.4 A high-speed train is under development in Texas [21] with a design based on the French Train à Grande Vitesse (TGV). Train speeds of 186 miles per hour are foreseen. To achieve these speeds on tight curves, the train may use independent axles combined with the ability to tilt the train. Hydraulic cylinders connecting the passenger compartments to their wheeled bogies allow the train to lean into curves like a motorcycle. A pendulum like device on the leading bogie of each car senses when it is entering a curve and feeds this information to the hydraulic system. Tilting does not make the train safer, but it does make passengers more comfortable.
(a)
FIGURE DP10.2 (a) Bi-wing aircraft. (Source: The Illustrated London News, October 9, 1920.) (b) Control system.
Wind
(b)
(a)
(b) FIGURE DP10.4
High-speed train
feedback control
system.
Consider the tilt control shown in Figure DP10.4. Design a compensator \(G_{c}(s)\) for a step-input command so that the percent overshoot is P.O. \(\leq 10\%\) and the settling time (with a \(2\%\) criterion) \(T_{s} \leq 1\text{ }s\). We also desire that the steady-state error for a velocity (ramp) input be less than \(0.2A\), where \(r(t) = At,t > 0\). Verify the results for the design.
DP10.5 High-performance tape transport systems are designed with a small capstan to pull the tape past the \(read/\) write heads and with take-up reels turned by DC motors. The tape is to be controlled at speeds up to 200 inches per second, with start-up as fast as possible, while preventing permanent distortion of the tape. Since we wish to control the speed and the tension of the tape, we will use a DC tachometer for the speed sensor and a potentiometer for the position sensor. We will use a DC motor for the actuator. Then the linear model for the system is a unity feedback system with
where \(Y(s)\) is position.
The specifications for the system are (1) settling time of \(T_{s} \leq 12\text{ }ms\), (2) a percent overshoot to a step position command of P.O. \(\leq 10\%\), and (3) a steadystate velocity error of less than \(0.5\%\). Determine a compensator to achieve these specifications.
DP10.6 The past years have witnessed a significant engine model-building activity in the automotive industry in a category referred to as "control-oriented" or "control design" models. These models contain representations of the throttle body, engine pumping phenomena, induction process dynamics, fuel system, engine torque generation, and rotating inertia.
The control of the fuel-to-air ratio in an automobile carburetor became of prime importance as automakers worked to reduce exhaust-pollution emissions. Thus, auto engine designers turned to the feedback control of the fuel-to-air ratio. Operation of an engine at or near a particular air-to-fuel ratio requires management of both air and fuel flow into the manifold system. The fuel command is considered the input and the engine speed is considered the output \(\lbrack 9,10\rbrack\).
The block diagram of the system is shown in Figure DP10.6, where \(T = 0.066\text{ }s\). A compensator is required to yield zero steady-state error for a step input and a percent overshoot of P.O. \(\leq 10\%\). We also desire that the settling time (with a \(2\%\) criterion) of \(T_{s} \leq 10\text{ }s\).
DP10.7 A high-performance jet airplane is shown in Figure DP10.7(a), and the roll-angle control system is shown in Figure DP10.7(b). Design a controller \(G_{c}(s)\) so that the step response is well behaved and the steady-state error is zero. That is, \(P.O. \leq 10\%\) and \(T_{s} \leq 2\text{ }s\).
DP10.8 A simple closed-loop control system has been proposed to demonstrate proportional-integral (PI) control of a windmill radiometer [27]. The windmill radiometer is shown in Figure DP10.8(a) and the control system is shown in Figure DP10.8(b). The variable to be controlled is the angular velocity \(\omega\) of the windmill radiometer whose vanes turn when exposed to infrared radiation. An experimental setup using a reflexive photoelectric sensor and basic electronic circuitry makes possible the design and implementation of a high performance control system.
Assume \(\tau = 20\text{ }s\). Design a PI controller so that the system achieves a deadbeat response with a settling time of \(T_{s} \leq 25\text{ }s\).
DP10.9 Consider the feedback control system shown in Figure DP10.9. Design a PID compensator \(G_{c1}(s)\) and a lead-lag compensator \(G_{c2}(s)\) such that, in each
FIGURE DP10.6
Engine control system.
FIGURE DP10.7
Roll-angle control of a jet airplane.
FIGURE DP10.8
(a) Radiometric windmill. (b) Control system.
FIGURE DP10.9
Feedback control system with a time-delay.
(a)
(b)
(a)
(b)
case, the closed-loop system is stable in the presence of a time-delay \(T = 0.1\text{ }s\). Discuss the capability of each compensator to insure stability in the presence of an increase in the time-delay uncertainty of up to 0.2 second.
DP10.10 A unity feedback system has the process transfer function
Design the controller \(G_{c}(s)\) such that the Bode magnitude plot of the loop transfer function \(L(s) =\) \(G_{c}(s)G(s)\) is greater than \(20\text{ }dB\) for \(\omega \leq 0.01rad/s\) and less than \(- 20\text{ }dB\) for \(\omega \leq 10rad/s\). The desired shape of the loop transfer function Bode plot magnitude is illustrated in Figure DP10.10. Explain why we would want the gain to be high at low-frequency and the gain to be low at high-frequency.
DP10.11 Modern microanalytical systems used for polymerase chain reaction (PCR) requires fast, damped tracking response [30]. The control of the temperature of the PCR reactor can be represented as shown in Figure DP10.11. The controller is chosen to be PID controller, denoted by \(G_{c}(s)\), with a prefilter, denoted by \(G_{p}(s)\).
It is required that the percent overshoot P.O. \(< 1\%\) and the settling time \(T_{s} < 3\) s to a unit step input. Design a controller \(G_{c}(s)\) and prefilter \(G_{p}(s)\) to achieve the control specifications.
FIGURE DP10.10
Bode plot loop shaping requirements.
FIGURE DP10.11 Polymerase chain reaction control system.
366. COMPUTER PROBLEMS
CP10.1 Consider the control system in Figure CP10.1, where
Develop an \(m\)-file to show that the phase margin is approximately P.M. \(= 50^{\circ}\) and that the percent overshoot to a unit step input is approximately \(P.O. = 18\%\).
FIGURE CP10.1 A feedback control system with compensation. FIGURE CP10.2
Single-loop feedback system
with proportional controller.
FIGURE
CP10.4 An aircraft pitch rate feedback control system.
CP10.2 A unity feedback control system is shown in Figure CP10.2. Design the proportional controller \(G_{c}(s) = K\) so that the system has a phase margin of \(P.M. = 55^{\circ}\). Develop an m-file to obtain a Bode plot, and verify that the design specification is satisfied.
CP10.3 Consider the system in Figure CP10.1, where
Design a compensator \(G_{c}(s)\) so that the steady-state tracking error to a ramp input is zero and the settling time (with a \(2\%\) criterion) is \(T_{S} \leq 5\text{ }s\). Obtain the response of the closed-loop system to the input \(R(s) = 1/s^{2}\), and verify that the settling time requirement has been satisfied and that the steady-state error is zero.
CP10.4 Consider the aircraft unity feedback control system in Figure CP10.4, where \(\overset{˙}{\theta}(t)\) is the pitch rate \((rad/s)\) and \(\delta(t)\) is the elevator deflection (rad). The four poles represent the phugoid and short-period modes. The phugoid mode has a natural frequency of \(0.1rad/s\), and the short period mode is \(1.4rad/s\).
(a) Using Bode plot methods, design a phaselead compensator to meet the following specifications: (1) settling time (with a \(2\%\) criterion) to a unit step of \(T_{s} \leq 2\text{ }s\), and (2) percent overshoot of P.O. \(\leq 10\%\).
(b) Simulate the closed-loop system with a step input of \(10\%\text{ }s\), and show the time history of \(\overset{˙}{\theta}(t)\).
CP10.5 The pitch attitude motion of a rigid spacecraft is described by
where \(J\) is the principal moment of inertia, and is the input torque on the vehicle [7]. Consider the PD controller
(a) Design a unity feedback control system to meet the following specifications: (1) closed-loop system bandwidth about \(\omega_{B} = 10rad/s\), and (2) percent overshoot of P.O. \(\leq 20\%\) to a \(10^{\circ}\) step input. Complete the design by developing and using an interactive m-file script. (b) Verify the design by simulating the response to a \(10^{\circ}\) step input. (c) Include a closed-loop transfer function Bode plot to verify that the bandwidth requirement is satisfied.
CP10.6 Consider the control system shown in Figure CP10.6. Design a lag compensator using root locus methods to meet the following specifications:(1) steadystate error less than \(10\%\) for a step input, (2) phase margin of P.M. \(\geq 45^{\circ}\), and (3) settling time (with a \(2\%\) criterion) of \(T_{s} \leq 5\text{ }s\) for a unit step input.
(a) Design a phase-lag compensator utilizing root locus methods to meet the design specifications. Develop a set of m-file scripts to assist in the design process. (b) Test the controller developed in part (a) by simulating the closed-loop system response to unit step input. Provide the time histories of the output \(y(t)\). (c) Compute the phase margin using the margin function.
CP10.7 A lateral beam guidance system has an inner loop as shown in Figure CP10.7 [26].
(a) Design a control system to meet the following specifications: (1) settling time (with a \(2\%\) criterion) to a unit step input of \(T_{s} \leq 1\text{ }s\), and (2) steady-state tracking error for a unit ramp input of less than 0.1 . (b) Verify the design by simulation. FIGURE CP10.6
A unity feedback control system.
FIGURE CP10.7
A lateral beam guidance system inner loop.
CP10.8 Consider a unity feedback system with the loop CP10.9 Consider a circuit with the transfer function transfer function
where $\ C_{1} = 1mF,C_{2} = 0.1\mu F,R_{1} = 10\Omega,\ $ and \(R_{2} = 10k\Omega\). Plot the frequency response of the circuit.
where \(z = 2\) and \(p = 5\). The actual percent overshoot CP10.10 Consider the feedback control system shown in of the compensated system will be P.O. \(= 59.1\%\). We want to reduce the percent overshoot to P.O. \(= 35\%\). Using an m-file script, determine an appropriate value for the zero of \(G_{c}(s)\).
Figure CP10.10. The time delay is \(T = 0.7\text{ }s\). Plot the phase margin for the system versus the gain in the range \(0.1 \leq K \leq 10\). Determine the gain \(K\) that maximizes the phase margin. What is the stability limit for \(K\) ?
FIGURE CP10.10 Feedback control system with a time delay.
367. ANSWERS TO SKILLS CHECK
True or False: (1) False; (2) False; (3) True; (4) True; (5) True
Multiple Choice: (6) d; (7) b; (8) d; (9) a; (10) b; (11) c; (12) a; (13) a; (14) b; (15) b
Word Match (in order, top to bottom): a, l, g, d, j, h, k, c, m, e, i, f, b
368. TERMS AND CONCEPTS
Cascade compensator A compensator placed in cascade or series with the system process.
Compensation The alteration or adjustment of a control Phase lead compensation A widely-used compensator system in order to provide a suitable performance.
Compensator An additional component or circuit that is inserted into the system to compensate for a performance deficiency.
Deadbeat response A system with a rapid response, minimal overshoot, and zero steady-state error for a step input.
Design of a control system The arrangement or the plan of the system structure and the selection of suitable components and parameters.
Integration compensator A compensator that acts, in part, like an integrator.
Lag network See Phase-lag compensator.
Lead-lag compensator A compensator with the characteristics of both a phase-lead compensator and a phase-lag compensator.
Lead compensator See Phase-lead compensator.
Phase lag compensation A widely-used compensator that possesses one zero and one pole with the pole closer to the origin of the \(s\)-plane. This compensator reduces the steady-state tracking errors. that possesses one zero and one pole with the zero closer to the origin of the \(s\)-plane. This compensator increases the system bandwidth and improves the dynamic response.
Phase-lag compensator A compensator that provides a negative phase angle and a significant attenuation over the frequency range of interest.
Phase-lead compensator A widely-used compensator that provides a positive phase angle over the frequency range of interest. Thus, phase lead can be used to cause a system to have an adequate phase margin.
PD controller Controller with a proportional term and a derivative term (Proportional-Derivation).
PI controller Controller with a proportional term and an integral term (Proportional-Integral).
Prefilter A transfer function \(G_{p}(s)\) that filters the input signal \(R(s)\) prior to calculating the error signal.
11.1 Introduction 813
11.2 Controllability and Observability 813
11.3 Full-State Feedback Control Design 819
11.4 Observer Design 825
11.5 Integrated Full-State Feedback and Observer 829
11.6 Reference Inputs 835
11.7 Optimal Control Systems 837
11.8 Internal Model Design 845
11.9 Design Examples 848
11.10 State Variable Design Using Control Design Software 855
11.11 Sequential Design Example: Disk Drive Read System 860
11.12 Summary 862
369. PREVIEW
The design of controllers utilizing state feedback is the subject of this chapter. We first present a system test for controllability and observability. Using the powerful notion of state variable feedback, we introduce the pole placement design technique. Ackermann's formula can be used to determine the state variable feedback gain matrix to place the system poles at the desired locations. The closed-loop system pole locations can be arbitrarily placed if and only if the system is controllable. When the full state is not available for feedback, we introduce an observer. The observer design process is described and the applicability of Ackermann's formula is established. The state variable compensator is obtained by connecting the fullstate feedback law to the observer. We consider optimal control system design and then describe the use of internal model design to achieve prescribed steady-state response to selected input commands. The chapter concludes by revisiting the Sequential Design Example: Disk Drive Read System.
370. DESIRED OUTCOMES
Upon completion of Chapter 11, students should be able to:
$\square\ $ Describe the concepts of controllability and observability.
$\square\ $ Design full-state feedback controllers and observers.
\(\square\) Explain pole-placement methods and Ackermann's formula.
\(\square\) Explain the separation principle and construct state variable compensators.
$\square\ $ Identify reference inputs, optimal control, and describe internal model design.
370.1. INTRODUCTION
The time-domain method, employing state variables, can be used to design a suitable compensation scheme for a control system. Typically, we are interested in controlling the system with a control signal \(\mathbf{u}(t)\) that is a function of several measurable state variables. Then we develop a state variable controller that operates on the information available in measured form. This type of system compensation is quite useful for system optimization and will be considered in this chapter.
State variable design typically comprises three steps. In the first step, we assume that all the state variables are measurable and utilize them in a full-state feedback control law. Full-state feedback is usually not practical because it is not possible (in general) to measure all the states. In practice, only certain states (or linear combinations thereof) are measured and provided as system outputs. The second step in state variable design is to construct an observer to estimate the states that are not directly measured and available as outputs. Observers can either be full-state observers or reduced-order observers. Reduced-order observers account for the fact that certain states are already available as system outputs; hence they do not need to be estimated [26]. In this chapter, we consider only full-state observers. The final step in the design process is to appropriately connect the observer to the full-state feedback control law. It is common to refer to the state-variable controller (full-state control law plus observer) as a compensator. The state variable design yields a compensator of the form depicted in Figure 11.1. Additionally, it is possible to consider non-zero reference inputs to the state variable compensator to complete the design. All three steps in the design process are discussed in the subsequent sections, as well as how to incorporate the reference inputs.
370.2. CONTROLLABILITY AND OBSERVABILITY
A key question that arises in the design of state variable compensators is whether or not all the poles of the closed-loop system can be arbitrarily placed in the complex plane. Recall that the poles of the closed-loop system are equivalent to the
FIGURE 11.1
State variable compensator employing full-state feedback in series with a full-state observer.
eigenvalues of the system matrix in state variable format. As we shall see, if the system is controllable and observable, then we can accomplish the design objective of placing the poles precisely at the desired locations to meet the performance specifications. Full-state feedback design commonly relies on pole-placement techniques \(\lbrack 2,27\rbrack\). Pole placement is discussed more fully in Section 11.3. It is important to note that a system must be completely controllable and completely observable to allow the flexibility to place all the closed-loop system poles arbitrarily. The concepts of controllability and observability (discussed in this section) were introduced by Kalman in the 1960s [28-30]. Rudolph Kalman was a central figure in the development of mathematical systems theory upon which much of the subject of state variable methods rests. Kalman is well known for his role in the development of the so-called Kalman filter, which was instrumental in the successful Apollo moon landings [31, 32].
371. A system is completely controllable if there exists an unconstrained control \(u(t)\) that can transfer any initial state \(\mathbf{x}\left( t_{0} \right)\) to any other desired location \(\mathbf{x}(\mathbf{t})\) in a finite time, \(t_{0} \leq t \leq T\).
For the system
we can determine whether the system is controllable by examining the algebraic condition
The matrix \(\mathbf{A}\) is an \(n \times n\) matrix and \(\mathbf{B}\) is an \(n \times 1\) matrix. For multi-input systems, \(\mathbf{B}\) can be \(n \times m\), where \(m\) is the number of inputs.
For a single-input, single-output system, the controllability matrix \(\mathbf{P}_{c}\) is described in terms of \(\mathbf{A}\) and \(\mathbf{B}\) as
which is an \(n \times n\) matrix. Therefore, if the determinant of \(\mathbf{P}_{c}\) is nonzero, the system is controllable [11].
Advanced state variable design techniques can handle situations wherein the system is not completely controllable, but where the states (or linear combinations thereof) that cannot be controlled are inherently stable. These systems are classified as stabilizable. If a system is completely controllable, it is also stabilizable. The Kalman state-space decomposition provides a mechanism for partitioning the state-space so that it becomes apparent which states (or state combinations) are controllable and which are not \(\lbrack 12,18\rbrack\). The controllable subspace is thus exposed, and if the system is stabilizable, the control system design can, in theory, proceed. In this chapter, we consider only completely controllable systems.
372. EXAMPLE 11.1 Controllability of a system
Let us consider the system
The signal-flow graph and block diagram model are illustrated in Figure 11.2. Then we have
Therefore, we obtain
The determinant of \(\mathbf{P}_{c} = - 1 \neq 0\), hence this system is controllable.
(a)
FIGURE 11.2
Third-order system.
(a) Signal-flow
graph model.
(b) Block diagram model.
(b)
373. EXAMPLE 11.2 Controllability of a two-state system
Let us consider a system represented by the two state equations
where \(d\) is a constant and determine the condition for controllability. Also, we have \(y(t) = x_{2}(t)\), as shown in Figure 11.3. The system state variable model is
We can determine the requirement on the parameter \(d\) by generating the matrix \(P_{c}\). So, with
we have
The determinant of \(\mathbf{P}_{c}\) is equal to \(d\), which is nonzero whenever \(d\) is nonzero.
All the poles of the closed-loop system can be placed arbitrarily in the complex plane if and only if the system is observable and controllable. Observability refers to the ability to estimate a state variable.
A system is completely observable if and only if there exists a finite time \(\mathbf{T}\) such that the initial state \(\mathbf{x}(0)\) can be determined from the observation history given \(\mathbf{y}(\mathbf{t})\) the control \(\mathbf{u}(\mathbf{t}),\mathbf{0} \leq \mathbf{t} \leq \mathbf{T}\).
FIGURE 11.3
(a) Flow graph model for Example 11.2. (b) Block diagram model. (a)
(b) Consider the single-input, single-output system
where \(\mathbf{C}\) is a \(1 \times n\) row vector, and \(\mathbf{x}\) is an \(n \times 1\) column vector. This system is completely observable when the determinant of the observability matrix \(\mathbf{P}_{o}\) is nonzero, where
which is an \(n \times n\) matrix.
As discussed in this section, advanced state variable design techniques can handle situations wherein the system is not completely controllable, as long as the system is stabilizable. These same techniques can handle cases wherein the system is not completely observable, but where the states (or linear combinations thereof) that cannot be observed are inherently stable. These systems are classified as detectable. If a system is completely observable, it is also detectable. The Kalman state-space decomposition provides a mechanism for partitioning the state-space so that it becomes apparent which states (or state combinations) are observable and which are not \(\lbrack 12,18\rbrack\). The unobservable subspace is thus exposed, and if the system is detectable, the control system design can, in theory, proceed. In this chapter, we consider only completely observable systems. The approach to state-variable design involves first verifying that the system under consideration is completely controllable and completely observable. If so, the pole placement design technique considered here can provide acceptable closed-loop system performance.
374. EXAMPLE 11.3 Observability of a system
Consider again the system of Example 11.1. The model is shown in Figure 11.2. To construct \(\mathbf{P}_{o}\), we use
Therefore,
Thus, we obtain
The det \(\mathbf{P}_{o} = 1\), and the system is completely observable.
375. EXAMPLE 11.4 Observability of a two-state system
Consider the system given by
The system is illustrated in Figure 11.4. We can check the system controllability and observability using the \(\mathbf{P}_{c}\) and \(\mathbf{P}_{o}\) matrices.
From the system definition, we obtain
Therefore, the controllability matrix is determined to be
and det \(\mathbf{P}_{c} = 0\). Thus, the system is not controllable.
(a)
FIGURE 11.4
Two state system model for Example 11.4. (a) Signal-flow graph model.
(b) Block diagram model.
(b) From the system definition, we have
Therefore, computing the observability matrix yields
and det \(\mathbf{P}_{o} = 0\). Hence, the system is not observable.
If we look again at the state model, we note that
However,
Thus, the system state variables do not depend on \(u\), and the system is not controllable. Similarly, the output \(x_{1}(t) + x_{2}(t)\) depends on \(x_{1}(0)\) plus \(x_{2}(0)\) and does not allow us to determine \(x_{1}(0)\) and \(x_{2}(0)\) independently. Consequently, the system is not observable.
375.1. FULL-STATE FEEDBACK CONTROL DESIGN
In this section, we consider full-state variable feedback to achieve the desired pole locations of the closed-loop system. The first step in the state variable design process requires us to assume that all the states are available for feedback - that is, we have access to the complete state \(\mathbf{x}(t)\) for all \(t\). Suppose the system input \(u(t)\) is given by
Determining the gain matrix \(\mathbf{K}\) is the objective of the full-state feedback design procedure. The beauty of the state variable design process is that the problem naturally separates into a full-state feedback component and an observer design component. These two design procedures can occur independently, and in fact, the separation principle provides the proof that this approach is optimal. We will show later that the stability of the closed-loop system is guaranteed if the full-state feedback control law stabilizes the system (under the assumption of access to the complete state) and the observer is stable (the tracking error is asymptotically stable). Observers are discussed in Section 11.4. The full-state feedback block diagram is illustrated in Figure 11.5. With the system defined by the state variable model
and the control feedback given by
FIGURE 11.5
Full-state feedback block diagram (with no reference input).
we find the closed-loop system to be
The characteristic equation associated with Equation (11.5) is
If all the roots of the characteristic equation lie in the left half-plane, then the closedloop system is stable. In other words, for any initial condition \(\mathbf{x}\left( t_{0} \right)\), it follows that
Given the pair (A, B), we can always determine \(\mathbf{K}\) to place all the system closed-loop poles in the left half-plane if and only if the system is completely controllable-that is, if and only if the controllability matrix \(\mathbf{P}_{c}\) is full rank (for a single-input, singleoutput system, full rank implies that \(\mathbf{P}_{c}\) is invertible).
The addition of a reference input can be written as
where \(r(t)\) is the reference input. The question of reference inputs is addressed in Section 11.6. When \(r(t) = 0\) for all \(t > t_{0}\), the control design problem is known as the regulator problem. That is, we want to compute \(\mathbf{K}\) so that all initial conditions are driven to zero in a specified fashion (as determined by the design specifications).
When using this state variable feedback, the roots of the characteristic equation are placed where the transient performance meets the desired response.
376. EXAMPLE 11.5 Design of a third-order system
Let us consider the third-order system with the differential equation
We can select the state variables as the phase variables so that \(x_{1}(t) = y(t)\), \(x_{2}(t) = dy(t)/dt,x_{3}(t) = d^{2}y(t)/dt^{2}\), and then
and
If the state variable feedback matrix is
and
then the closed-loop system is
The state feedback matrix is
and the characteristic equation is
If we seek a rapid response with a low overshoot, we choose a desired characteristic equation such as
We choose \(\zeta = 0.8\) for minimal overshoot and \(\omega_{n}\) to meet the settling time requirement. If we want a settling time (with a \(2\%\) criterion) equal to 1 second, then
If we choose \(\omega_{n} = 6\), the desired characteristic equation is
Comparing Equations (11.6) and (11.7) yields the three equations
Therefore, we require that \(k_{3} = 9.4,k_{2} = 79.1\), and \(k_{1} = 170.8\). The step response has no overshoot and a settling time of 1 second, as desired.
377. EXAMPLE 11.6 Inverted pendulum control
Consider the control of an unstable inverted pendulum balanced on a moving cart. We measure and utilize the state variables of the system in order to control the pendulum. Thus, if we want to measure the angle from vertical, \(\theta(t)\), we could use a potentiometer connected to the shaft of the pendulum hinge. Similarly, we could measure the rate of change of the angle \(\overset{˙}{\theta}(t)\) by using a tachometer generator. If the state variables are all measured, then they can be used in a feedback controller so that \(u(t) = - \mathbf{Kx}(t)\), where \(\mathbf{K}\) is the feedback matrix. The state vector \(\mathbf{x}(t)\) represents the state of the system; therefore, knowledge of \(x(t)\) and the equations describing the system dynamics provide sufficient information for control and stabilization of a system \(\lbrack 4,5,7\rbrack\).
To illustrate the use of state variable feedback, consider the unstable inverted pendulum and design a suitable state variable feedback control system. If we assume that the control input, \(u(t)\), is an acceleration signal, we can focus on the unstable dynamics of the pendulum. The equation of motion describing the angle, \(\overset{˙}{\theta}(t)\), from the vertical, is
Let the state vector be \(\left( x_{1}(t),x_{2}(t) \right) = (\theta(t),\overset{˙}{\theta}(t))\). The state vector differential equation is
The A matrix of Equation (11.8) has the characteristic equation \(\lambda^{2} - g/l = 0\) with one root in the right-hand \(s\)-plane. To stabilize the system, we generate a control signal that is a function of the two state variables, \(x_{1}(t)\) and \(x_{2}(t)\). Then we have
Substituting this control signal relationship into Equation (11.8), we have
Combining the two additive terms on the right side of the equation, we find that
Obtaining the characteristic equation, we have
Thus, for the system to be stable, we require that \(k_{2}/l < 0\) and \(k_{1} > - g\). Hence, we have stabilized an unstable system by measuring the state variables \(x_{1}\) and \(x_{2}\) and using the control function \(u(t) = - \mathbf{Kx}(t)\) to obtain a stable system. If we wish to achieve a rapid response with modest overshoot, we select \(\omega_{n} = 10\) and \(\zeta = 0.8\). Then we require
The step response would have a percent overshoot of \(P.O. = 1.5\%\) and a settling time of \(T_{s} = 0.5\text{ }s\).
Thus far, we have established an approach for the design of a feedback control system by using the state variables as the feedback variables in order to increase the stability of the system and obtain the desired system response. Now we face the task of computing the gain matrix \(\mathbf{K}\) to place the poles at desired locations. For a single-input, single-output system, Ackermann's formula is useful for determining the state variable feedback matrix
where
Given the desired characteristic equation
the state feedback gain matrix is
where
and \(\mathbf{P}_{c}\) is the controllability matrix of Equation (11.2).
378. EXAMPLE 11.7 Second-order system
Consider the system
and determine the feedback gain to place the closed-loop poles at \(s = - 1 \pm j\). Therefore, we require that
and \(\alpha_{1} = \alpha_{2} = 2\). With \(x_{1}(t) = y(t)\) and \(x_{2}(t) = \overset{˙}{y}(t)\), the matrix equation for the system \(G(s)\) is
The controllability matrix is
Thus, we obtain
where
and
Then we have
Note that computing the gain matrix K using Ackermann's formula requires the use of \(\mathbf{P}_{c}^{- 1}\). We see that complete controllability is essential because only then can we guarantee that the controllability matrix \(\mathbf{P}_{c}\) has full rank and hence that \(\mathbf{P}_{c}^{- 1}\) exists.
378.1. OBSERVER DESIGN
In the full-state feedback design procedure discussed in Section 11.3, it was assumed that all the states were available for feedback at all times. This is a good assumption for the control law design process. Only a subset of the states are typically measurable and available for feedback. Having all the states available for feedback implies that these states are measured with a sensor or sensor combinations. The cost and complexity of the control system increases as the number of required sensors increases. So, even in situations where extra sensors are available, it may not be cost effective to employ these extra sensors, if indeed, the control system design goals can be accomplished without them. Fortunately, if the system is completely observable with a given set of outputs, then it is possible to determine (or to estimate) the states that are not directly measured (or observed).
According to Luenberger [26], the full-state observer for the system
is given by
where \(\widehat{\mathbf{x}}(t)\) denotes the estimate of the state \(\mathbf{x}(t)\). The matrix \(\mathbf{L}\) is the observer gain matrix and is to be determined as part of the observer design procedure. The observer is depicted in Figure 11.6. The observer has two inputs, \(u(t)\) and \(y(t)\), and one output, \(\widehat{\mathbf{x}}(t)\).
The goal of the observer is to provide an estimate \(\widehat{\mathbf{x}}(t)\) so that \(\widehat{\mathbf{x}}(t) \rightarrow \mathbf{x}(t)\) as \(t \rightarrow \infty\). Remember that we do not know \(\mathbf{x}\left( t_{0} \right)\) precisely; therefore we must provide an initial estimate \(\widehat{\mathbf{x}}\left( t_{0} \right)\) to the observer. Define the observer estimation error as
The observer design should produce an observer with the property that \(\mathbf{e}(t) \rightarrow 0\) as \(t \rightarrow \infty\). One of the main results of systems theory is that if the system is completely observable, we can always find \(\mathbf{L}\) so that the tracking error is asymptotically stable, as desired.
Taking the time-derivative of the estimation error in Equation (11.12) yields
FIGURE 11.6
The full-state observer.
and using the system model and the observer in Equation (11.11), we obtain
or
We can guarantee that \(\mathbf{e}(t) \rightarrow 0\) as \(t \rightarrow \infty\) for any initial tracking error \(\mathbf{e}\left( t_{0} \right)\) if the characteristic equation
has all its roots in the left half-plane. Therefore, the observer design process reduces to finding the matrix \(\mathbf{L}\) such that the roots of the characteristic equation in Equation (11.14) lie in the left half-plane. This can always be accomplished if the system is completely observable; that is, if the observability matrix \(\mathbf{P}_{o}\) has full rank (for a single-input, single-output system, full rank implies that \(\mathbf{P}_{o}\) is invertible).
379. EXAMPLE 11.8 Second-order system observer design
Consider the second-order system
In this example, we can only directly observe the state \(y(t) = x_{1}(t)\). The observer will provide estimates of the second state \(x_{2}(t)\).
We only consider full-state observers, which implies that the observer will provide estimates of all the states. We might be inclined to suppose that since some states are directly measured, it may be possible to design an observer that provides just the estimates of the states not directly measured. This is, in fact, possible, and the resulting observers are known as reduced-order observers [12, 18]. However, since sensors are not noise free, even states that are directly measured are generally estimated in an effort to reduce the effect of sensor noise on the state estimate. The Kalman filter (which is a time-varying optimal observer) solves the observer problem in the presence of measurement noise (and process noise as well) [33, 34].
The observer design begins by checking the system observability to verify that an observer can be constructed to guarantee the stability of the estimation error. From the system model, we find that
The corresponding observability matrix is
FIGURE 11.7
Second-order observer response to initial estimation errors.
Since det \(\mathbf{P}_{o} = 3 \neq 0\), the system is completely observable. Suppose that the desired characteristic equation is given by
We can select \(\zeta = 0.8\) and \(\omega_{n} = 10\), resulting in an expected settling time of less than 0.5 second. Computing the actual characteristic equation yields
where \(\mathbf{L} = \begin{bmatrix} L_{1} & L_{2} \end{bmatrix}^{T}\). Equating the coefficients in Equation (11.15) to those in Equation (11.16) yields the two equations
which, when solved, produces
The observer is thus given by
The response of the estimation error to an initial error of
is shown in Figure 11.7. Ackermann's formula can also be employed to place the roots of the observer characteristic equation at the desired locations. Consider the observer gain matrix
and the desired observer characteristic equation
The \(\beta^{'}\) s are selected to meet given performance specifications for the observer. The observer gain matrix is then computed via
where \(\mathbf{P}_{O}\) is the observability matrix given in Equation (11.3) and
380. EXAMPLE 11.9 Second-order system observer design using Ackermann's formula
Consider the second-order system in Example 11.8. The desired characteristic equation was given as
where \(\zeta = 0.8\) and \(\omega_{n} = 10\); hence, \(\beta_{1} = 16\) and \(\beta_{2} = 100\). Computing \(p(\mathbf{A})\) yields
and from Example 11.8, we have the observability matrix
which implies that
Using Ackermann's formula in Equation (11.17) yields the observer gain matrix
This is the identical result obtained in Example 11.8 using other methods.
380.1. INTEGRATED FULL-STATE FEEDBACK AND OBSERVER
The state variable compensator is constructed by appropriately connecting the fullstate feedback control law (see Section 11.3) to the observer (see Section 11.4). The compensator is shown in Figure 11.1 (as discussed in Section 11.1). Our strategy was to design the state feedback control law as \(u(t) = - \mathbf{Kx}(t)\), where we assumed that we had access to the complete state \(\mathbf{x}(t)\). Then we designed an observer to provide an estimate of the state \(\widehat{\mathbf{x}}(t)\). It seems reasonable that we can employ the state estimate in the feedback control law in place of \(\mathbf{x}(t)\). In other words, we can consider the feedback law
But is this a good strategy? The feedback gain matrix \(\mathbf{K}\) was designed to guarantee stability of the closed-loop system; that is, the roots of the characteristic equation
are in the left half-plane. Under the assumption that the complete state \(\mathbf{x}(t)\) is available for feedback, the feedback control law (with properly designed gain matrix \(\mathbf{K}\) ) leads to the desired result that \(\mathbf{x}(t) \rightarrow 0\) as \(t \rightarrow \infty\) for any initial condition \(\mathbf{x}\left( t_{0} \right)\). We need to verify that, when using the feedback control law in Equation (11.18), we retain the stability of the closed-loop system.
Consider the observer (from Section 11.4)
Substituting the feedback law in Equation (11.18) and rearranging terms in the observer yields the compensator system
Notice that the system in Equation (11.19) has the form of a state variable model with input \(y(t)\) and output \(u(t)\), as illustrated in Figure 11.8.
FIGURE 11.8
State variable compensator with integrated full-state feedback and observer.
Computing the estimation error using the compensator in Equation (11.19) yields
or
This is the same result as we obtained for the estimation error in Section 11.4. The estimation error does not depend on the input as seen in Equation (11.20), where the input terms cancel. Recall that the underlying system model is given by
Substituting the feedback law \(u(t) = - \mathbf{K}\widehat{\mathbf{x}}(t)\) into the system model yields
and with \(\widehat{\mathbf{x}}(t) = \mathbf{x}(t) - \mathbf{e}(t)\), we obtain
Writing Equations (11.20) and (11.21) in matrix form, we have
Recall that our goal is to verify that, with \(u(t) = - \mathbf{K}\widehat{\mathbf{x}}(t)\), we retain stability of the closed-loop system and the observer. The characteristic equation associated with Equation (11.22) is
So if the roots of \(det(\lambda\mathbf{I} - (\mathbf{A} - \mathbf{BK})) = 0\) lie in the left half-plane (which they do by design of the full-state feedback law), and if the roots of \(det(\lambda\mathbf{I} - (\mathbf{A} - \mathbf{LC})) = 0\) lie in the left half-plane (which they do by design of the observer), then the overall system is stable. Therefore, employing the strategy of using the state estimates for the feedback is in fact a good strategy.
In other words, when we use \(u(t) = - \mathbf{K}\widehat{\mathbf{x}}(t)\) where \(\mathbf{K}\) is designed using the methods proposed in Section 11.3 and \(\widehat{\mathbf{x}}(t)\) is derived from the observer discussed in Section 11.4, then \(\mathbf{x}(t) \rightarrow 0\) as \(t \rightarrow \infty\) for any initial condition \(\mathbf{x}\left( t_{0} \right)\) and \(\mathbf{e}(t) \rightarrow 0\) as \(t \rightarrow \infty\) for any initial estimation error \(\mathbf{e}\left( t_{0} \right)\). The fact that the full-state feedback law and the observer can be designed independently is an illustration of the separation principle.
The design procedure is summarized as follows:
-
Determine \(\mathbf{K}\) such that \(det(\lambda\mathbf{I} - (\mathbf{A} - \mathbf{BK})) = 0\) has roots in the left half-plane and place the poles appropriately to meet the control system design specifications. The ability to place the poles arbitrarily in the complex plane is guaranteed if the system is completely controllable. 2. Determine \(\mathbf{L}\) such that \(det(\lambda\mathbf{I} - (\mathbf{A} - \mathbf{LC})) = 0\) has roots in the left half-plane and place the poles to achieve acceptable observer performance. The ability to place the observer poles arbitrarily in the complex plane is guaranteed if the system is completely observable.
-
Connect the observer to the full-state feedback law using
Compensator Transfer Function. The compensator given in Equation (11.19) can be given equivalently in transfer function form with input \(Y(s)\) and output \(U(s)\). Taking the Laplace transform (with zero initial conditions) of the compensator yields
and rearranging and solving for \(U(s)\), we obtain the transfer function
Note that the compensator transfer function itself (when viewed as a system) may or may not be stable. Even though \(\mathbf{A} - \mathbf{BK}\) is stable and \(\mathbf{A} - \mathbf{LC}\) is stable, it does not necessarily follow that \(\mathbf{A} - \mathbf{BK} - \mathbf{LC}\) is stable. However, the overall closedloop system is stable (as we proved in the previous discussions). The controller in Equation (11.23) is commonly referred to as a stabilizing controller.
381. EXAMPLE 11.10 Compensator design for the inverted pendulum
The state variable model representing the inverted pendulum atop a moving cart is
where \(\mathbf{x}(t) = \left( x_{1}(t),x_{2}(t),x_{3}(t),x_{4}(t) \right)^{T},x_{1}(t)\) is the cart position, \(x_{2}(t)\) is the cart velocity, \(x_{3}(t)\) is the pendulum angular position (measured from the vertical), \(x_{4}(t)\) is the pendulum angular rate, and \(u(t)\) is the input applied to the cart. Typically, we can measure the state variable \(x_{3}(t) = \theta\) using a potentiometer attached to the shaft, or measure \(x_{4}(t) = \overset{˙}{\theta}(t)\) using a tachometer generator. However, suppose that we have a sensor available to measure the position of the cart. Is it possible to hold the angular position of the pendulum at the desired value \(\left( \theta(t) = 0^{\circ} \right)\) when only the output \(y(t) = x_{1}(t)\) (the cart position) is available? In this case, we have the output equation
Let the system parameters be \(l = 0.098\text{ }m,g = 9.8\text{ }m/s^{2},m = 0.825\text{ }kg\) and \(M = 8.085\text{ }kg\). Therefore, using the parameter values, the system state and input matrices are
Checking controllability yields the controllability matrix
Computing det \(\mathbf{P}_{c} = 196.49 \neq 0\); hence, the system is completely controllable. Likewise, computing the observability matrix yields
and det \(\mathbf{P}_{o} = 1 \neq 0\); hence, the system is completely observable. We can now proceed with the three-step design procedure knowing that we can determine a control gain matrix \(\mathbf{K}\) and observer gain matrix \(\mathbf{L}\) to place all the closed-loop system poles at desired locations.
382. STEP 1: Design the Full-State Feedback Control Law
The open-loop system poles are located at \(\lambda = 0,0, - 10\), and 10 , hence the openloop system is unstable (there is a pole in the right half-plane). Suppose that the desired closed-loop system characteristic equation is given by
where we choose (1) the pair \(\left( \zeta,\omega_{n} \right)\) so that these poles are the dominant poles, and (2) the pair \((a,b)\) farther in the left half-plane so as not to dominate the response. To obtain a settling time less than 10 seconds with low overshoot, we can select \(\left( \zeta,\omega_{n} \right) = (0.8,0.5)\). Then, we choose a separation factor of 20 between the dominant poles and the remaining poles, from which it follows that \((a,b) = (16,100)\). Figure 11.9 shows the pole zero map for the system design. The separation factor between the dominant and nondominant poles is a FIGURE 11.9
System pole map: open-loop poles, desired closed-loop poles, and observer poles.
parameter that can be varied as part of the design process. The larger the separation selected, the further left in the left half-plane the nondominant poles will be placed, and hence the larger the required control law gains. The desired roots are then specified to be
The poles at \(\lambda = - 0.4 \pm 0.3j\) are the dominant poles. Using Ackermann's formula yields the feedback gain matrix
383. STEP 2: Observer Design
The observer needs to provide an estimate of the states that cannot be directly observed. The goal is to achieve an accurate estimate as fast as possible without resulting in too large a gain matrix \(\mathbf{L}\). How large is too large depends on the problem under consideration. In particular, if there are significant levels of measurement noise (this is sensor dependent), then the magnitude of the observer matrix should be kept correspondingly low to avoid amplifying the measurement noise. The trade-off between the time required to obtain accurate observer performance and the amount of noise amplification is a primary design issue. For design purposes, we will attempt to insure a separation of the desired closedloop system poles and the observer poles on the order of 2 to 10 (as illustrated in Figure 11.9). The desired observer characteristic equation is selected to be of the form
(a)
(b)
FIGURE 11.10 (a) Pendulum performance under full-state feedback control with the observer in the loop, (b) Observer performance.
where the constants \(c_{1}\) and \(c_{2}\) are appropriately chosen. As a first attempt, we select \(c_{1} = 32\) and \(c_{2} = 711.11\). These values should produce a response to an initial state estimation error that settles in less than 0.5 second with minimal percent overshoot. Using Ackermann's formula from Section 11.3, we determine that the observer gain that achieves the desired observer pole locations \(det(\lambda\mathbf{I} - (\mathbf{A} - \mathbf{LC})) = ((\lambda + 16 + j21.3)(\lambda + 16 - j21.3))^{2}\) is
384. STEP 3: Compensator Design
The final step in the design is to connect the observer to the full-state feedback control law via \(u(t) = - \mathbf{K}\widehat{\mathbf{x}}(t)\). As proved earlier, the closed-loop system will remain stable; however, we should not expect the closed-loop performance to be as good when using the state estimate from the observer. This makes sense, since it takes a finite amount of time for the observer to provide accurate state estimates. The response of the inverted pendulum design is shown in Figure 11.10. The pendulum is initially stationary at \(\theta\left( t_{0} \right) = {5.72}^{\circ}\), and the cart is initially not moving. The initial state estimate in the observer is set to zero.
In Figure 11.10(a), we see that, indeed, the pendulum is balanced to the vertical in under 4 seconds. The response of the compensator (with the observer) is more oscillatory than without the observer in the loop-but this difference in performance is expected, since it takes about 0.4 second for the observer to converge to a minimal state tracking error, as seen in Figure 11.10(b).
384.1. REFERENCE INPUTS
The feedback strategies discussed in the previous sections (and illustrated in Figure 11.1) were constructed without consideration of reference inputs. We referred to the design of state variable feedback compensators without reference inputs (i.e., \(r(t) = 0\) ) as regulators. Since command following is also an important aspect of feedback design, it is important to consider how we can introduce a reference signal into the state variable feedback compensator. There are, in fact, many different techniques that can be employed to permit the tracking of a reference input. Two of the more common methods are discussed in this section.
The general form of the state variable feedback compensator is
where \(\widetilde{y} = y(t) - \mathbf{C}\widehat{\mathbf{x}}(t)\) and \(\widetilde{u}(t) = - \mathbf{K}\widehat{\mathbf{x}}(t)\). The state variable compensator with the reference input is illustrated in Figure 11.11. Notice that when \(\mathbf{M} = 0\) and \(N = 0\), the compensator in Equation (11.24) reduces to the regulator described in Section 11.5 and illustrated in Figure 11.1.
The compensator key design parameters required to implement the command tracking of the reference input are \(\mathbf{M}\) and \(N\). When the reference input is a scalar signal (i.e., a single input), the parameter \(\mathbf{M}\) is a column vector of length \(n\), where \(n\) is the length of the state vector, \(\mathbf{x}(t)\) and \(N\) is a scalar. Here, we consider two possibilities for selecting \(\mathbf{M}\) and \(N\). In the first case, we select \(\mathbf{M}\) and \(N\) so that the estimation error \(\mathbf{e}(t)\) is independent of the reference input \(r(t)\). In the second case, we select \(\mathbf{M}\) and \(N\) so that the tracking error \(y(t) - r(t)\) is used as an input to the compensator. These two cases will result in implementations wherein the compensator is in the feedback loop in the first case and in the forward loop in the second case.
Employing the generalized compensator in Equation (11.24), the estimation error is found to be described by the differential equation
FIGURE
11.11
State variable compensator with a reference input.
FIGURE 11.12
State variable compensator with reference input and \(\mathbf{M} = \mathbf{B}N\).
or
Suppose that we select
Then the corresponding estimation error is given by
In this case, the estimation error is independent of the reference input \(r(t)\). This is the identical result found in Section 11.4, where we considered the observer design assuming no reference inputs. The remaining task is to determine a suitable value of \(N\), since the value of \(\mathbf{M}\) follows from Equation (11.25). For example, we might choose \(N\) to obtain a zero steady-state tracking error to a step input \(r(t)\).
With \(\mathbf{M} = \mathbf{B}N\), we find that the compensator is given by
This implementation of the state variable compensator is illustrated in Figure 11.12.
As an alternative approach, suppose that we select \(N = 0\) and \(\mathbf{M} = - L\). Then, the compensator in Equation (11.24) is given by
which can be rewritten as
In this formulation, the observer is driven by the tracking error \(y(t) - r(t)\). The reference input tracking implementation is illustrated in Figure 11.13.
Notice that in the first implementation (with \(\mathbf{M} = \mathbf{B}N\) ) the compensator is in the feedback loop, whereas in the second implementation \((N = 0\) and \(\mathbf{M} = - \mathbf{L})\) FIGURE 11.13
State variable compensator with reference
input and \(N = 0\) and \(\mathbf{M} = - \mathbf{L}\).
the compensator is in the forward path. These two implementations are representative of the possibilities open to control system designers when considering reference inputs.
Depending on the choice of \(N\) and \(\mathbf{M}\), other implementations are possible. For example, Section 11.8 presents a method of tracking reference inputs with guaranteed steady-state tracking errors using internal model design techniques.
384.2. OPTIMAL CONTROL SYSTEMS
The design of optimal control systems is an important function of control engineering. The purpose of design is to realize a system with practical components that will provide the desired performance. The desired performance can be readily stated in terms of time-domain performance indices, such as the integral performance measures. The design of a system can be based on minimizing a performance index, such as the integral of the squared error (ISE). Systems that are adjusted to provide a minimum performance index are called optimal control systems. In this section, we consider the design of an optimal control system that is described by a state variable formulation.
The performance of a control system, written in terms of the state variables of a system, can be expressed as
where \(\mathbf{x}(t)\) is the state vector, and \(\mathbf{u}(t)\) is the control vector. In this section, we consider the design of optimal control systems using state variable feedback and error-squared performance indices [1-3].
Consider the system
We will select a feedback controller as
where \(\mathbf{K}\) is an \(1 \times n\) matrix.
Substituting Equation (11.28) into Equation (11.27) yields
where \(\mathbf{H}\) is the \(n \times n\) matrix resulting from the addition of the elements of \(\mathbf{A}\) and - BK. The error-squared performance index for a single state variable, \(x_{1}(t)\), is written as
A performance index written in terms of two state variables would then be
Since we wish to define the performance index in terms of an integral of the sum of the state variables squared, we will use the matrix operation
where \(\mathbf{x}^{T}(t)\) indicates the transpose of \(\mathbf{x}(t).^{\dagger}\) Then the specific form of the performance index, in terms of the state vector, is
The general form of the performance index (Equation 11.26) incorporates a term with \(u(t)\) that we have not included at this point, but we will do so later in this section.
To obtain the minimum value of \(J\), we postulate the existence of an exact differential so that
where \(\mathbf{P}\) is to be determined. A symmetric \(\mathbf{P}\) matrix will be used to simplify the algebra without any loss of generality. Then, for a symmetric \(\mathbf{P}\) matrix, \(p_{ij} = p_{ji}\). Completing the differentiation indicated on the left-hand side of Equation (11.34), we have
Substituting Equation (11.29) into Equation (11.35), we obtain
If we let
then Equation (11.36) becomes
which is the exact differential we are seeking. Substituting Equation (11.38) into Equation (11.33), we obtain
In the evaluation of the limit at \(t = \infty\), we have assumed that the system is stable, and hence \(\mathbf{x}(\infty) = 0\), as desired. Therefore, to minimize the performance index \(J\), we consider the two equations
and
The design steps are then as follows:
-
Determine the matrix \(\mathbf{P}\) that satisfies Equation (11.41), where \(\mathbf{H}\) is known.
-
Minimize \(J\) by determining the minimum of Equation (11.40) by adjusting one or more unspecified system parameters.
385. EXAMPLE 11.11 State variable feedback
Consider the open-loop control system shown in Figure 11.14. The state variables are identified as \(x_{1}(t)\) and \(x_{2}(t)\). The performance of this system is quite unsatisfactory because an undamped response results for a step input. The vector differential equation of this system is
We choose a feedback control system so that
and therefore the control signal is a linear function of the two state variables. Then Equation (11.42) becomes
In matrix form, we have
Let \(k_{1} = 1\) and determine a suitable value for \(k_{2}\) so that the performance index is minimized. From Equation (11.41), it follows that
Completing the matrix multiplication and addition yields
Solving these simultaneous equations, we obtain
The integral performance index is then
and we will consider the case where each state is initially displaced one unit from equilibrium so that \(\mathbf{x}^{T}(0) = (1,1)\). Therefore Equation (11.48) becomes
Substituting the values of the elements of \(\mathbf{P}\), we have
To minimize as a function of \(k_{2}\), we take the derivative with respect to \(k_{2}\) and set it equal to zero yielding
Therefore, \(k_{2}^{2} = 4\), and \(k_{2} = 2\) when \(J\) is a minimum. The minimum value of \(J\) is obtained by substituting \(k_{2} = 2\) into Equation (11.50). Thus, we obtain
The system matrix \(\mathbf{H}\), obtained for the compensated system, is then
The characteristic equation of the compensated system is therefore
Because this is a second-order system, we note that the characteristic equation is of the form \(s^{2} + 2\zeta\omega_{n}s + \omega_{n}^{2} = 0\), and therefore the damping ratio of the compensated system is \(\zeta = 1.0\). This compensated system is considered to be an optimal system in that the compensated system results in a minimum value for the performance index when \(k_{1} = 1\) is fixed. Of course, we recognize that this system is optimal only for the specific set of initial conditions that were assumed. The compensated system is shown in Figure 11.15. A curve of the performance index as a function of \(k_{2}\) is shown in Figure 11.16. It is clear that this system is not very sensitive to changes in \(k_{2}\) and will maintain a near-minimum performance index if the \(k_{2}\) is altered by some percentage. We define the sensitivity of an optimal system as
FIGURE 11.15 Compensated control system of Example 11.11.
FIGURE 11.16 Performance index versus the parameter \(k_{2}\). where \(k\) is the design parameter. Then, for this example, we have \(k = k_{2}\), and considering \(k_{2} = 2.5\), for which \(J = 3.05\), we obtain
386. EXAMPLE 11.12 Determination of an optimal system
Now let us consider again the system of Example 11.11, where both the feedback gains, \(k_{1}\) and \(k_{2}\), are unspecified. To simplify the algebra without any loss in insight into the problem, let us set \(k_{1} = k_{2} = k\). We can prove that if \(k_{1}\) and \(k_{2}\) are unspecified, then \(k_{1} = k_{2}\) when the minimum of the performance index (Equation 11.40) is obtained. Then, for the system of Example 11.11, Equation (11.45) becomes
To determine the \(\mathbf{P}\) matrix, we use Equation (11.41), yielding
Let us consider the case where the system is initially displaced one unit from equilibrium so that \(\mathbf{x}^{T}(0) = \begin{pmatrix} 1 & 0 \end{pmatrix}\). Then the performance index becomes
Thus, the performance index to be minimized is
The minimum value of \(J\) is obtained when \(k\) approaches infinity; the result is \(J_{\min} = 1\). A plot of \(J\) versus \(k\), shown in Figure 11.17, illustrates that the performance index approaches a minimum asymptotically as \(k\) approaches an infinite value. Now, we recognize that, in providing a very large gain \(k\), we can cause the feedback signal
to be very large. However, we are restricted to realizable magnitudes of the control signal \(u(t)\). Therefore, we should introduce a constraint on \(u(t)\) so that the gain \(k\) is not too large. Then, for example, if we establish a constraint on \(u(t)\) so that
FIGURE 11.17
Performance index versus the feedback gain \(k\) for Example 11.12.
we require that the maximum acceptable value of \(k\) in this case be
Then the minimum value of \(J\) is
which is sufficiently close to the absolute minimum of \(J\) to satisfy our requirements.
Upon examining the performance index, we recognize that the reason the magnitude of the control signal is not accounted for in the original calculations is that \(u(t)\) is not included within the expression for the performance index. However, in many cases, we have physical limits on the control magnitude. To account for the control magnitude, we can consider the performance index
where \(\lambda\) is a scalar weighting factor and \(\mathbf{I} =\) identity matrix. The weighting factor \(\lambda\) will be chosen so that the relative importance of the state variable performance is contrasted with the importance of the control energy represented by \(\mathbf{u}^{T}(t)\mathbf{u}(t)\). We represent the state variable feedback via
and the system with this state variable feedback as
Substituting Equation (11.64) into Equation (11.63) yields
where \(\mathbf{Q} = \mathbf{I} + \lambda\mathbf{K}^{T}\mathbf{K}\) is an \(n \times n\) matrix. Following the development of Equations (11.33) through (11.39), we postulate the existence of an exact differential so that
Then, in this case, we require that
and thus, as before, we have
If \(\lambda = 0\), Equation (11.68) reduces to Equation (11.41). Now, let us consider again Example 11.11 when \(\lambda\) is other than zero and account for the expenditure of control signal energy.
387. EXAMPLE 11.13 Optimal system with control energy considerations
Consider again the system of Example 11.11, which is shown in Figure 11.14. For this system, we use a state variable feedback so that
Therefore, the matrix
As in Example 11.12, we will let \(\mathbf{x}^{T}(0) = (1,0)\) so that \(J = p_{11}\). We evaluate \(p_{11}\) from Equation (11.68), yielding
The minimum of \(J\) is found by taking the derivative of \(J\), setting the result to zero, and solving for \(k\), yielding
Therefore, the minimum of the performance index occurs when \(k = k_{\min} = 1/\sqrt{\lambda}\), where \(k_{\min}\) is the solution of Equation (11.73).
Let us complete this example for the case where the control energy and the state variables squared are equally important, so that \(\lambda = 1\). Then Equation (11.73) is satisfied when \(k^{2} - 1 = 0\), and we find that \(k_{\min} = 1.0\). The plot of \(J\) versus \(k\) for FIGURE 11.18
Performance index versus the feedback gain \(k\) for Example 11.13.
this case is shown in Figure 11.18. The plot of \(J\) versus \(k\) for Example 11.12 is also shown for comparison in Figure 11.18.
The design of several parameters can be accomplished in a manner similar to that illustrated in the examples. Also, the design procedure can be carried out for higher-order systems. Consider the single-input, single-output system with
and feedback
We can consider the performance index
where \(R > 0\) is a scalar weighting factor. This index is minimized when
The \(n \times n\) matrix \(\mathbf{P}\) is determined from the solution of the equation
Equation (11.76) is often called the algebraic Riccati equation. This optimal control problem is called the linear quadratic regulator (LQR) \(\lbrack 12,19\rbrack\).
387.1. INTERNAL MODEL DESIGN
In this section, we consider the problem of designing a compensator that provides asymptotic tracking of a reference input with zero steady-state error. The reference inputs considered can include steps, ramps, and other persistent signals, such as sinusoids. For a step input, we know that zero steady-state tracking errors can be achieved with a type-one system. This idea is formalized here by introducing an internal model of the reference input in the compensator \(\lbrack 5,18\rbrack\). Consider a state variable model given by
We consider a reference input to be generated by a linear system of the form
with unknown initial conditions.
We begin by considering a familiar design problem, namely, the design of a controller to enable the tracking of a step reference input with zero steady-state error. In this case, the reference input is generated by
or equivalently
and the tracking error \(e(t)\) is defined as
Taking the time derivative yields
If we define the two intermediate variables
we have
If the system in Equation (11.81) is controllable, we can find a feedback of the form
such that Equation (11.81) is stable. This implies that the tracking error \(e\) is stable; thus, we will have achieved the objective of asymptotic tracking with zero steadystate error. The control input, found by integrating Equation (11.82), is
The corresponding block diagram is shown in Figure 11.19. We see that the compensator includes an internal model (that is, an integrator) of the reference step input.
FIGURE 11.19 Internal model design for a step input.
388. EXAMPLE 11.14 Internal model design for a unit step input
Consider a process given by
We want to design a controller for this system to track a reference step input with zero steady-state error. From Equation (11.81), we have
A check of controllability shows that the system described by Equation (11.84) is completely controllable. We use
in order to locate the roots of the characteristic equation of Equation (11.84) at \(s = - 1 \pm j, - 10\). With \(w(t)\) given in Equation (11.82), Equation (11.84) is asymptotically stable. So, for any initial tracking error \(e(0)\) we are guaranteed that \(e(t) \rightarrow 0\) as \(t \rightarrow \infty\).
Consider a block diagram where the process is represented by \(G(s)\) and the cascade controller is \(G_{c}(s) = K_{1}/s\). The internal model principle states that if \(G(s)G_{c}(s)\) contains \(R(s)\), then \(y(t)\) will track \(r(t)\) asymptotically. In this case \(R(s) = 1/s\), which is contained in \(G(s)G_{c}(s)\), as we expect.
Consider the problem of designing a controller to provide asymptotic tracking of a ramp input with zero steady-state error \(r(t) = Mt,t \geq 0\), where \(M\) is the ramp magnitude. In this case, the reference input model is
In input-output form, the reference model in Equation (11.85) is given by
Proceeding as before, we take the time-derivative of the tracking error twice yielding
With the definitions
we have
FIGURE 11.20 Internal model design for a ramp input. Note that \(G(s)G_{c}(s)\) contains \(1/s^{2}\), the reference input \(R(s)\).
So if the system of Equation (11.86) is controllable, then we can compute the gains \(K_{1},K_{2}\), and \(\mathbf{K}_{3}\) such that with
the system represented by Equation (11.86) is asymptotically stable; hence, the tracking error \(e(t) \rightarrow 0\) as \(t \rightarrow \infty\), as desired. The control, \(u(t)\), is found by integrating Equation (11.87) twice. In Figure 11.20, we see that the resulting controller has a double integrator, which is the internal model of the reference ramp input.
The internal model approach can be extended to other reference inputs by following the same general procedure outlined for the step and ramp inputs. In addition, the internal model design can be used to reject persistent disturbances by including models of the disturbances in the compensator.
388.1. DESIGN EXAMPLES
In this section we present a control system designed to manage the speed of the electric motor shaft of a diesel electric locomotive. The design process focuses on the design of a full-state feedback control system using pole-placement methods.
389. EXAMPLE 11.15 Diesel electric locomotive control
The diesel electric locomotive is depicted in Figure 11.21. The efficiency of the diesel engine is very sensitive to the speed of rotation of the motors. We want to design a control system that drives the electric motors of a diesel electric locomotive for use on railroad trains. The locomotive is driven by DC motors located on each of the axles. The throttle position is set by moving the input potentiometers. The elements of the design process emphasized in this example are highlighted in Figure 11.22.
The control objective is to regulate the shaft rotation speed \(\omega_{o}(t)\) to the desired value \(\omega_{r}(t)\).
390. Control Goal
Regulate the shaft rotation speed to the desired value in the presence of external load torque disturbances.
The corresponding variable to be controlled is the shaft rotation speed \(\omega_{o}(t)\). FIGURE 11.21
Diesel electric locomotive system.
Topics emphasized in this example
If the performance does not meet the specifications, then iterate the configuration.
If the performance meets the specifications, then finalize the design.
FIGURE 11.22
Elements of the control system design process emphasized in this diesel electric locomotive example.
391. Variable to Be Controlled
Shaft rotation speed \(\omega_{o}(t)\).
The controlled speed \(\omega_{o}(t)\) is sensed by a tachometer, which supplies a feedback voltage \(v_{o}(t)\). The electronic amplifier amplifies the error signal, \(v_{r}(t) - v_{o}(t)\), between the reference and feedback voltage signals and provides a voltage \(v_{f}(t)\) that is supplied to the field winding of a DC generator.
The generator is run at a constant speed \(\omega_{d}\) by the diesel engine and generates a voltage \(v_{g}\) that is supplied to the armature of a DC motor. The motor is armature controlled, with a fixed current supplied to its field. As a result, the motor produces a torque \(T\) and drives the load connected to its shaft so that the controlled speed \(\omega_{o}(t)\) tends to equal the command speed \(\omega_{r}(t)\).
A block diagram and signal flow graph of the system are shown in Figure 11.23. In Figure 11.23 we use \(L_{t}\) and \(R_{t}\), which are defined as
Values for the parameters of the diesel electric locomotive are given in Table 11.1.
(a)
(b)
FIGURE 11.23 Signal flow graph of the diesel electric locomotive. (a) Signal flow graph. (b) Block diagram controller feedback loops are shown in green.
392. Table 11.1 Parameter Values for the Diesel Electric Locomotive
| $$K_{m}$$ | $$K_{g}$$ | $$K_{b}$$ | $$J$$ | $$b$$ | $$L_{a}$$ | $$R_{a}$$ | $$R_{f}$$ | $$L_{f}$$ | $$K_{t}$$ | $$K_{pot}$$ | $$L_{g}$$ | $$R_{g}$$ |
| 10 | 100 | 0.62 | 1 | 1 | 0.2 | 1 | 1 | 0.1 | 1 | 1 | 0.1 | 1 |
Notice that the system has a feedback loop; we use the tachometer voltage \(v_{o}(t)\) as a feedback signal to form an error signal \(v_{r}(t) - v_{o}(t)\). Without additional state feedback, the only tuning parameter is the amplifier gain \(K\). As a first step, we can investigate the system performance with tachometer voltage feedback only.
The key tuning parameters are given by
393. Select Key Tuning Parameters
\(K\) and \(\mathbf{K}\)
The matrix \(\mathbf{K}\) is the state feedback gain matrix. The design specifications are
394. Design Specifications
DS1 Steady-state tracking \(e_{ss} \leq 2\%\) to a unit step input.
DS2 Percent overshoot of \(\omega_{O}(t)\) of P.O. \(\leq 10\%\) to a unit step input \(\omega_{r}(s) = 1/s\).
DS3 Settling time of \(T_{s} \leq 1\text{ }s\) to a unit step input.
The first step in the development of the vector differential equation that accurately describes the system is to choose a set of state variables. In practice the selection of state variables can be a difficult process, especially for complex systems. The state variables must be sufficient in number to determine the future behavior of the system when the present state and all future inputs are known. The selection of state variables is related to the issue of complexity.
The diesel electric locomotive system has three major components: two electrical circuits and one mechanical system. It seems logical that the state vector will include state variables from both electrical circuits and from the mechanical system. One reasonable choice of state variables is \(x_{1}(t) = \omega_{o}(t),x_{2}(t) = i_{a}(t)\), and \(x_{3}(t) = i_{f}(t)\). This state variable selection is not unique. With the state variables defined above, the state variable model is
where
In matrix form (with \(\left. \ T_{d}(t) = 0 \right)\), we have
where
The corresponding transfer function is
Assume the tachometer feedback is available, that is, that \(K_{t}\) is in the loop. If we take advantage of the fact that
then (from an input-output perspective) the system has the feedback configuration shown in Figure 11.24.
Using the parameter values given in Table 11.1 and computing the steady-state tracking error for a unit step input yields
Using the Routh-Hurwitz method, we also find that the closed-loop system is stable for
The smallest steady-state tracking error is achieved for the largest value of \(K\). At best we can obtain a \(15\%\) tracking error, which does not meet the design specification DS1. Also, as \(K\) gets larger, the response becomes unacceptably oscillatory.
We now consider a full state feedback controller design. The feedback loops are shown in Figure 11.23, which shows that \(\omega_{0}(t),i_{a}(t)\), and \(i_{f}(t)\) are available for feedback. Without any loss of generality, we set \(K = 1\). Any value of \(K > 0\) would work as well.
FIGURE 11.24 Block diagram representation of the diesel electric locomotive.
The control input is
The feedback gains to be determined are \(K_{t},K_{2}\), and \(K_{3}\). The tachometer gain, \(K_{t}\), is now a key parameter of the design process. Also \(K_{\text{pot}\text{~}}\) is a key variable for tuning. By adjusting the parameter \(K_{\text{pot}\text{~}}\), we have the freedom to scale the input \(\omega_{r}(t)\). When we define
then
The closed-loop system with state feedback is
where
We will use pole-placement methods to determine \(\mathbf{K}\) such that the eigenvalues of \(\mathbf{A} - \mathbf{BK}\) are in the desired locations. First we make sure the system is controllable. When \(n = 3\) the controllability matrix is
Computing the determinant of \(\mathbf{P}_{c}\) yields
Since \(K_{g} \neq 0\) and \(K_{m} \neq 0\) and \(JL_{f}^{3}L_{t}^{2}\) is nonzero, we determine that
Thus the system is controllable. We can place all the poles of the system appropriately to satisfy DS2 and DS3.
The desired region to place the eigenvalues of \(\mathbf{A} - \mathbf{BK}\) is illustrated in Figure 11.25. The specific pole locations are selected to be
Selecting \(p_{1} = - 50\) allows for a good second-order response governed by \(p_{2}\) and \(p_{3}\). The gain matrix \(\mathbf{K}\) that achieves the desired closed-loop poles is
To select the gain \(K_{\text{pot}\text{~}}\), we first compute the DC gain of the closed-loop transfer function. With the state feedback in place, the closed-loop transfer function is
FIGURE 11.25
Desired location of the closed-loop poles (that is, the eigenvalues of A - BK).
FIGURE 11.26
Closed-loop step response of the diesel electric locomotive.
Then
Using the gain \(K_{\text{pot}\text{~}}\) in this manner effectively scales the closed-loop transfer function so that the DC gain is equal to 1 . We then expect that a unit step input representing a \(1\%\) s step command results in a \(1\%\) s steady-state output at \(\omega_{o}\).
The step response of the system is shown in Figure 11.26. We can see that all the design specifications are satisfied.
394.1. STATE VARIABLE DESIGN USING CONTROL DESIGN SOFTWARE
Controllability and observability of a system in state variable feedback form can be checked using the functions ctrb and obsv, respectively. The inputs to the ctrb function, shown in Figure 11.27, are the system matrix \(\mathbf{A}\) and the input matrix \(\mathbf{B}\); the output of ctrb is the controllability matrix \(\mathbf{P}_{c}\). Similarly, the input to the obsv function, shown in Figure 11.27, is the system matrix \(\mathbf{A}\) and the output matrix \(\mathbf{C}\); the output of obsv is the observability matrix \(\mathbf{P}_{o}\).
Notice that the controllability matrix \(\mathbf{P}_{c}\) is a function only of \(\mathbf{A}\) and \(\mathbf{B}\), while the observability matrix \(P_{o}\) is a function only of \(\mathbf{A}\) and \(\mathbf{C}\).
395. EXAMPLE 11.16 Satellite trajectory control
Let us consider a satellite in a circular, equatorial orbit at an altitude of 250 nautical miles above the Earth, as illustrated in Figure 11.28 [14, 24]. The satellite motion (in the orbit plane) is described by the normalized state variable model
where the state vector \(\mathbf{x}(t)\) represents normalized perturbations from the circular, equatorial orbit; \(u_{r}(t)\) is the input from a radial thruster; \(u_{t}(t)\) is the input from a tangential thruster; and \(\omega = 0.0011rad/s\) (approximately one orbit of 90 minutes) is the orbital rate for the satellite at the specific altitude. In the absence of perturbations, the satellite will remain in the nominal circular equatorial orbit. However, disturbances such as aerodynamic drag can cause the satellite to deviate from its nominal path. The problem is to design a controller that commands the satellite thrusters in
Controllability \(\overset{˙}{\mathbf{x}} = \mathbf{Ax} + \mathbf{B}u\) matrix \(\ y = \mathbf{Cx} + \mathbf{D}u\)
FIGURE 11.27 The ctrb and obsv functions.
FIGURE 11.28 The satellite in an equatorial circular orbit. FIGURE 11.29
Controllability with radial thrusters only: (a) \(m\)-file script, (b) output. radial.m
(a)
(b)
such a manner that the actual orbit remains near the desired circular orbit. Before commencing with the design, we check controllability. In this case, we investigate controllability using the radial and tangential thrusters independently.
Suppose the tangential thruster fails (i.e., \(u_{t}(t) = 0\) ), and only the radial thruster is operational. Is the satellite controllable from \(u_{r}(t)\) only? We answer this question by using an m-file script to determine the controllability. Using the script shown in Figure 11.29, we find that the determinant \(\mathbf{P}_{c}\) is zero; thus, the satellite is not completely controllable when the tangential thruster fails.
Suppose now that the radial thruster fails (i.e., \(\left. \ u_{r}(t) = 0 \right)\) and that the tangential thruster is functioning properly. Is the satellite controllable from \(u_{t}\) only? Using the script in Figure 11.30, we find that the satellite is completely controllable using the tangential thruster only.
We conclude this section with a controller design for a third-order system using state variable models. The design approach utilizes root locus methods and incorporates \(m\)-file scripts to assist in the procedure.
396. EXAMPLE 11.17 Third-order system
Consider a system with the state-space representation
where
FIGURE 11.30
Controllability with tangential thrusters only: (a) m-file script, (b) output. tangent.m
\(\%\) This script computes the satellite controllability
\(\%\) with a tangential thruster only (i.e. failed radial thruster)
\(W = 0.0011\);
b2 \(= \lbrack 0;0;0;1\rbrack\);
\(Pc = ctrb(A,b2)\);
\(n = det(Pc)\);
if \(abs(n) < eps\)
disp('Satellite is uncontrollable with tangential thruster only!')
else
disp('Satellite is controllable with tangential thruster only!')
end
(a)
tangent
Satellite is controllable with tangential thruster only!
Input matrix associated with tangential thruster
Compute controllability matrix \(n =\) determinant of controllability matrix FIGURE 11.31
(a) Root locus.
(b) m-file script.
(a)
\(\%\) Root locus script
\(\%\) Including performance specs regions
num=[1 8 20]; den=[1 \(\left. \ 6\begin{matrix} 1 & 5 & 0 \end{matrix} \right\rbrack\); sys=tf(num,den);
clf; rlocus(sys); hold on $4 - \ $ Hold plot to add \(\%\)
zeta \(= 0.72;wn = 2.8\);
stability regions
(b)
where \(\mathbf{H} = \mathbf{A} - \mathbf{BK}\). The characteristic equation associated with Equation (11.93) can be obtained by evaluating \(det(s\mathbf{I} - \mathbf{H}) = 0\), resulting in
If we view \(KK_{3}\) as a parameter and let \(K_{1} = 1\), then we can write Equation (11.94) as
We place the zeros at \(s = - 4 \pm 2j\) in order to pull the locus to the left in the \(s\)-plane. Thus, our desired numerator polynomial is \(s^{2} + 8s + 20\). Comparing corresponding coefficients leads to
FIGURE 11.32 Step response.
FIGURE 11.33 The acker function.
Therefore, \(K_{2} = 0.35\) and \(K_{3} = 0.05\). We can now plot a root locus with \(KK_{3}\) as the parameter, as shown in Figure 11.31.
The characteristic equation, Equation (11.94), is
The roots for the selected gain, \(KK_{3} = 12\), lie in the performance region, as shown in Figure 11.31. The rlocfind function is used to determine the value of \(KK_{3}\) at the selected point. The final gains are \(K = 240.00,K_{1} = 1.00,K_{2} = 0.35\), and \(K_{3} = 0.05\). The controller design results in a settling time of about 1.8 seconds and an overshoot of \(3\%\), as shown in Figure 11.32.
In Section 11.4, we discussed Ackermann's formula to place the poles of the system at desired locations. The function acker calculates the gain matrix \(\mathbf{K}\) to place the closed-loop poles at the desired locations. The acker function is illustrated in Figure 11.33.
397. EXAMPLE 11.18 Second-order system design using the acker function
Consider again the second-order system in Example 11.7. The system model is
FIGURE 11.34
Using acker to compute \(K\) to place the poles at $\mathbf{P} = \begin{bmatrix}
- 1 + j & - 1 - j
\end{bmatrix}^{T}$.
The desired closed-loop pole locations are \(s_{1,2} = - 1 \pm j\). To apply Ackermann's formula using the acker function, form the vector
Then, with
the acker function, illustrated in Figure 11.34, determines that the gain matrix that achieves the desired pole locations is
This confirms the result in Example 11.7.
397.1. SEQUENTIAL DESIGN EXAMPLE: DISK DRIVE READ SYSTEM
In this chapter, we will design a state variable feedback system that will achieve the desired system response. The specifications for the system are given in Table 11.2. The second-order open-loop model is shown in Figure 11.35. We will design the system for this second-order model and then test the system response for both the second-order and third-order models.
First, we select the two state variables as \(x_{1}(t) = y(t)\) and \(x_{2}(t) = dy(t)/dt =\) \(dx_{1}(t)/dt\), as shown in Figure 11.36. It is practical to measure these variables as the position and velocity of the reader head. We then add the state variable feedback, as shown in Figure 11.36. We choose \(K_{1} = 1\), since our goal is for \(y(t)\) to closely and accurately follow the command \(r(t)\). The state variable differential equation for the open-loop system is
The closed-loop state variable differential equation obtained from Figure 11.36 is
FIGURE 11.35
Open-loop model of head control system.
FIGURE 11.36
Closed-loop system with feedback of the two state variables.
Table 11.2 Disk Drive Control System Specifications and Actual Performance
| $$\begin | |||
| \text{}\text{Performance}\text{} \ | |||
| \text{}\text{Measure}\text{} | |||
| \end{matrix}$$ | $$\begin |
\text{~}\text{Desired}\text{~} \\
\text{~}\text{Value}\text{~}
\end{matrix}$$ | $$\begin{matrix}
\text{~}\text{Response for}\text{~} \\
\text{~}\text{Second-Order}\text{~} \\
\text{~}\text{Model}\text{~}
\end{matrix}$$ | $$\begin{matrix}
\text{~}\text{Response for}\text{~} \\
\text{~}\text{Third-Order}\text{~} \\
\text{~}\text{Model}\text{~}
\end{matrix}$$ |
| Percent overshoot | $$< 5%$$ | $$< 1%$$ | $$0%$$ |
| $$\begin{matrix}
\text{~}\text{Settling time}\text{~} \
\begin{matrix}
\text{~}\text{Maximum response for a}\text{~} \
\text{~}\text{unit step disturbance}\text{~}
\end{matrix}
\end{matrix}$$ | $$< 50\text{ }ms$$ | $$34.3\text{ }ms$$ | $$34.2\text{ }ms$$ |
The characteristic equation of the closed-loop system is
since \(K_{1} = 1\). In order to achieve the specifications, we select \(\zeta = 0.90\) and \(\zeta\omega_{n} = 125\). Then the desired closed-loop characteristic equation is
Therefore, we require that \(5K_{a} = 19290\) or \(K_{a} = 3858\). Furthermore, we require that
or \(K_{2} = 0.012\).
The system with the second-order model has the desired response and meets all the specifications, as shown in Table 11.2. If we add the field inductance \(L = 1mH\), we have a third-order model with
Using this model, which incorporates the field inductance, we test the response of the system with the feedback gains selected for the second-order model. The results are provided in Table 11.2, illustrating that the second-order model is a very good model of the system. The actual results of the third-order system meet the specifications.
397.2. SUMMARY
In this chapter, the design of control systems in the time domain was examined. The three-step design procedure for constructing state variable compensators was presented. The optimal design of a system using state variable feedback and an integral performance index was considered. Also, the \(s\)-plane design of systems utilizing state variable feedback was examined. Finally, internal model design was discussed.
398. 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 11.37 as specified in the various problem statements.
FIGURE 11.37 Block diagram for the Skills Check.
In the following True or False and Multiple Choice problems, circle the correct answer.
-
A system is said to be controllable on the interval \(\left\lbrack t_{0},t_{f} \right\rbrack\) if there exists a continuous input \(u(t)\) such that any initial state \(\mathbf{x}\left( t_{0} \right)\) can be transformed to any arbitrary state \(\mathbf{x}\left( t_{f} \right)\) in a finite interval \(t_{f} - t_{0} > 0\). True or False
-
The poles of a system can be arbitrarily assigned through full-state feedback if and only if the system is completely controllable and observable.
True or False
- The problem of designing a compensator that provides asymptotic tracking of a reference input with zero steady-state error is called state-variable feedback.
True or False
- Optimal control systems are systems whose parameters are adjusted so that the performance index reaches an extremum value.
True or False
- Ackerman's formula is used to check observability of a system.
True or False
- Consider the system
The system is:
a. Controllable, observable
b. Not controllable, not observable
c. Controllable, not observable
d. Not controllable, observable
- Consider the system
This system is:
a. Controllable, observable
b. Not controllable, not observable
c. Controllable, not observable
d. Not controllable, observable
- A system has the state variable representation
Determine the associated transfer function model \(G(s) = \frac{Y(s)}{U(s)}\).
a. \(G(s) = \frac{5s^{2} + 32s + 35}{s^{3} + 9s^{2} + 23s + 15}\)
b. \(G(s) = \frac{5s^{2} + 32s + 35}{s^{4} + 9s^{3} + 23s + 15}\)
c. \(G(s) = \frac{2s^{2} + 16s + 22}{s^{3} + 9s^{2} + 23s + 15}\)
d. \(G(s) = \frac{5s + 32}{s^{2} + 32s + 9}\)
- Consider the closed-loop system in Figure 11.37, where
Determine the state-variable feedback control gain matrix \(\mathbf{K}\) so that the closed-loop system poles are \(s = - 3, - 4\), and -6 .
a. \(\mathbf{K} = \begin{bmatrix}
1 & 44 & 67
\end{bmatrix}\)
b. \(\mathbf{K} = \begin{bmatrix}
10 & 44 & 67
\end{bmatrix}\)
c. \(\mathbf{K} = \begin{bmatrix}
44 & 1 & 1
\end{bmatrix}\)
d. \(\mathbf{K} = \begin{bmatrix}
1 & 67 & 44
\end{bmatrix}\)
- Consider the system depicted in the block diagram in Figure 11.38.
This system is:
a. Controllable, observable
b. Not controllable, not observable c. Controllable, not observable
d. Not controllable, observable
FIGURE 11.38 Two-loop feedback control system.
- A system has the transfer function
Determine the values of \(a\) that render the system unobservable.
a. \(a = 1.30\) or \(a = - 1.44\)
b. \(a = 3.30\) or \(a = 1.44\)
c. \(a = - 3.30\) or \(a = - 1.44\)
d. \(a = - 5.7\) or \(a = - 2.04\)
- Consider the closed-loop system in Figure 11.37, where
Determine the state variable feedback control gain matrix \(K\) for a zero steady-state tracking error to a step input.
a. \(\mathbf{K} = \begin{bmatrix}
3 & - 9
\end{bmatrix}\)
b. \(\mathbf{K} = \begin{bmatrix}
3 & - 6
\end{bmatrix}\)
c. $\mathbf{K} = \begin{bmatrix}
- 3 & 2
\end{bmatrix}$
d. $\mathbf{K} = \begin - 1 & 4
\end{bmatrix}$
- Consider the system where
It is desired to place the observer poles at \(s_{1,2} = - 3 \pm j3\). Determine the appropriate state-variable feedback control gain matrix \(\mathbf{L}\).
a. $\mathbf{L} = \begin{bmatrix}
- 9 \
3
\end{bmatrix}$
b. \(\mathbf{L} = \begin{bmatrix} 9 \\ 3 \end{bmatrix}\)
c. \(\mathbf{L} = \begin{bmatrix} 3 \\ 9 \end{bmatrix}\)
d. None of the above 14. A feedback system has a state-space representation
where the feedback is \(u(t) = - \mathbf{Kx} + r(t)\). The control system design specifications are: (i) the overshoot to a step input approximately P.O. \(\approx 6\%\), and (ii) the settling time \(T_{s} \approx 0.1\text{ }s\). A state variable feedback gain matrix which satisfies the specifications is:
a. \(\mathbf{K} = \begin{bmatrix}
10 & 200
\end{bmatrix}\)
b. \(\mathbf{K} = \begin{bmatrix}
6 & 3600
\end{bmatrix}\)
c. \(\mathbf{K} = \begin{bmatrix}
3600 & 10
\end{bmatrix}\)
d. \(\mathbf{K} = \begin{bmatrix}
100 & 40
\end{bmatrix}\)
- Consider the system
Determine the eigenvalues of the closed-loop system when utilizing state variable feedback, where \(u(t) = - 2x_{2}(t) - 2x_{1}(t) + r(t)\). We define \(x_{1}(t) = y(t),x_{2}(t) = {\overset{˙}{x}}_{1}(t)\), and \(r(t)\) is a reference input.
a. \(s_{1} = - 1 + j1\ s_{2} = - 1 - j1\)
b. \(s_{1} = - 2 + j2\ s_{2} = - 2 - j2\)
c. \(s_{1} = - 1 + j2\ s_{2} = - 1 - j2\)
d. \(s_{1} = - 1\ s_{2} = - 1\)
In the following Word Match problems, match the term with the definition by writing the correct letter in the space provided.
a. Stabilizing controller Occurs when the control signal for the process is a direct function of all the state variables.
b. Controllability matrix
c. Stabilizable
d. Command following
e. State variable feedback
f. Full-state feedback control law
g. Observer
h. Linear quadratic regulator
i. Optimal control system
A system in which any initial state \(\mathbf{x}\left( t_{0} \right)\) is uniquely determined by observing the output \(y(t)\) on the interval \(\left\lbrack t_{0},t_{f} \right\rbrack\).
A system in which there exists a continuous input \(u(t)\) such that any initial state \(\mathbf{x}\left( t_{0} \right)\) can be driven to any arbitrary trial state \(\mathbf{x}\left( t_{f} \right)\) in a finite time interval \(t_{f} - t_{0} > 0\).
A system whose parameters are adjusted so that the performance index reaches an extremum value.
An important aspect of control system design wherein a nonzero reference input is tracked.
A linear system is (completely) controllable if and only if this matrix has full rank.
A system in which the states that are unobservable are naturally stable.
The difference between the actual state and the estimated state.
A control law of the form \(u(t) = - \mathbf{Kx}(t)\) where \(\mathbf{x}(t)\) is the state of the system assumed known at all times. j. Detectable
k. Controllable system
I. Pole placement
m. Estimation error
n. Kalman state-space decomposition
o. Observable system
p. Separation principle
q. Observability matrix
A partition of the state space that illuminates the states that are controllable and unobservable, uncontrollable and unobservable, controllable and observable, and uncontrollable and observable.
An optimal controller designed to minimize a quadratic performance index.
A linear system is (completely) observable if and only if this matrix has full rank.
A dynamic system used to estimate the state of another dynamic system given knowledge of the system inputs and measurements of the system outputs.
A design methodology wherein the objective is to place the eigenvalues of the closed-loop system in desired regions of the complex plane.
The principle that states that the full-state feedback law and the observer can be designed independently and when connected will function as an integrated control system in the desired manner (that is, stable).
A system in which the states that are not controllable are naturally stable.
A controller that stabilizes the closed-loop system.
399. EXERCISES
E11.1 The ability to balance actively is a key ingredient in the mobility of a device that hops and runs on one springy leg, as shown in Figure E11.1 [8]. The control of the attitude of the device uses a gyroscope and a feedback such that \(u(t) = \mathbf{Kx}(t)\), where
and
where
Determine a value for \(k\) so that the response of each hop is critically damped.
E11.2 A magnetically suspended steel ball can be described by the linear equation
FIGURE E11.1 Single-leg control. The state variables are \(x_{1}(t) =\) position and \(x_{2}(t) =\) velocity. Select a feedback so that the settling time (with a \(2\%\) criterion) is \(T_{s} = 4\text{ }s\) and \(P.O. \leq 10\%\) to a unit step input. Choose the feedback in the form
where \(r(t)\) is the reference input and the gains \(k_{1}\) and \(k_{2}\) are to be determined.
E11.3 A system is described by the matrix equations
Determine whether the system is controllable and observable.
Answer: controllable and observable
E11.4 A system is described by the matrix equations
Determine whether the system is controllable and observable.
E11.5 A system is described by the matrix equations
Determine whether the system is controllable and observable.
E11.6 A system is described by the matrix equations
Determine whether the system is controllable and observable.
Answer: uncontrollable and unobservable
E11.7 Consider the system represented in state variable form
where
Sketch a block diagram model of the system.
E11.8 Consider the third-order system
Sketch a block diagram model of the system.
E11.9 Consider the second-order system
For what values of \(k_{1}\) and \(k_{2}\) is the system completely controllable?
E11.10 Consider the block diagram model in Figure E11.10. Write the corresponding state variable model in the form
FIGURE E11.10
State variable block diagram.
E11.11 Consider the system shown in block diagram form in Figure E11.11. Obtain a state variable representation of the system. Determine if the system is controllable and observable.
E11.12 Consider a single-input, single-output system that is described by
where
Compute the corresponding transfer function representation of the system. If the initial conditions are zero (i.e., \(x_{1}(0) = 0\) and \(\left. \ x_{2}(0) = 0 \right)\), determine the response when \(u(t)\) is a unit step input for \(t \geq 0\).
FIGURE E11.11
State variable block diagram with a
feedforward term.
400. PROBLEMS
P11.1 A first-order system is represented by the time-domain differential equation
A feedback controller is to be designed such that
and the desired equilibrium condition is \(x(t) = 0\) as \(t \rightarrow \infty\). The performance integral is defined as
and the initial value of the state variable is \(x(0) = 3\). Obtain the value of \(k\) in order to make \(J\) a minimum. Is this \(k\) physically realizable? Select a practical value for the gain \(k\), and evaluate the performance index with that gain. Is the system stable without the feedback due to \(u(t)\) ?
P11.2 To account for the expenditure of energy and resources, the control signal is often included in the performance integral. Then the operation will not involve an unlimited control signal \(u(t)\). One suitable performance index, which includes the effect of the magnitude of the control signal, is
a. Repeat Problem P11.1 for the performance index.
b. If \(\lambda = 0.5\), obtain the value of \(k\) that minimizes the performance index. Calculate the resulting minimum value of \(J\).
P11.3 An unstable robot system is described by the vector differential equation [9]
where \(\mathbf{x}(t) = \left( x_{1}(t) = x_{2}(t) \right)^{T}\). Both state variables are measurable, and so the control signal is set as \(u(t) =\) \(- k\left( x_{1}(t) + x_{2}(t) \right)\). Design the gain \(k\) so that the performance index
is minimized. Evaluate the minimum value of the performance index. Determine the sensitivity of the performance to a change in \(k\). Assume that the initial conditions are
Is the system stable without the feedback signals due to \(u(t)\) ? P11.4 Consider the system
where $\mathbf{H} = \begin{bmatrix}
0 & 1 \
- k & - k
\end{bmatrix}$. Determine the feedback gain \(k\) that minimizes the performance index
when \(\mathbf{x}^{T}(0) = \begin{bmatrix} 2, & - 1 \end{bmatrix}\). Plot the performance index \(J\) versus the gain \(k\).
P11.5 Consider the system described by
where \(\mathbf{x}(t) = \left( x_{1}(t),x_{2}(t) \right)^{T}\),
The state feedback of the system is \(u(t) = - k_{1}x_{1}(t)\) \(- k_{2}x_{2}(t)\), where \(k_{1} = 3k\) and \(k_{2} = k,k > 0\). Given the initial condition \(\mathbf{x}(0) = (0,1)^{T}\), and determine the gain \(k\) to minimize the performance index
Plot the function \(J(k)\).
P11.6 For the solutions of Problems P11.3, P11.4, and P11.5, determine the roots of the closed-loop optimal control system. Note that the resulting closed-loop roots depend on the performance index selected.
P11.7 A system has the vector differential equation
where
We want both state variables to be used in the feedback so that \(u(t) = - k_{1}x_{1}(t) + k_{2}x_{2}(t)\). Also, we desire to have a natural frequency \(\omega_{n} = 3\). Find a set of gains \(k_{1}\) and \(k_{2}\) in order to achieve an optimal system when \(J\) is given by
Assume \(\mathbf{x}^{T}(0) = \begin{bmatrix} 1, & 0 \end{bmatrix}\).
P11.8 For the system of P11.7, determine the optimum value for \(k_{2}\) when \(k_{1} = 0.25\), and \(\mathbf{x}^{T}(0) = \begin{bmatrix} 1, & 0 \end{bmatrix}\).
P11.9 An interesting mechanical system with a challenging control problem is the ball and beam, shown in Figure P11.9(a) [10]. It consists of a rigid beam that is free to rotate in the plane of the paper around a center pivot, with a solid ball rolling along a groove in the top of the beam. The control problem is to position the ball at a desired point on the beam using a torque applied to the beam as a control input at the pivot.
A linear model of the system with a measured value of the angle \(\phi(t)\) and its angular velocity \(\phi = \omega(t)\) is available. Select a feedback scheme so that the response of the closed-loop system has a percent overshoot of P.O. \(= 5\%\) and a settling time (with a \(2\%\) criterion) of \(T_{s} = 1.5\text{ }s\) for a step input.
P11.10 The dynamics of a rocket are represented by
and state variable feedback is used, where \(u(t) = - 7x_{1}(t) - 12x_{2}(t) + r(t)\). Determine the roots of the characteristic equation of this system and the response of the system when the initial conditions are \(x_{1}(0) = 1\) and \(x_{2}(0) = - 1\). Assume the reference input \(r(t) = 0\).
(a)
FIGURE P11.9
(a) Ball and beam.
(b) Model of the ball and beam.
(b) P11.11 The state variable model of a plant to be controlled is
Use state variable feedback, and incorporate a command input \(u(t) = - \mathbf{Kx}(t) + \alpha r(t)\). Select the gains \(\mathbf{K}\) and \(\alpha\) so that the system has a rapid response with a percent overshoot of P.O. \(= 1\%\), a settling time (with a \(2\%\) criterion) of \(T_{s} \leq 1\text{ }s\), and a zero steadystate error to a unit step input.
P11.12 A voice-coil actuator-driven electromechanical system has the following state-space model:
where
Compute the transfer function and the poles of this system.
Determine whether it is controllable and observable.
P11.13 A feedback system has a plant transfer function
We want the percent overshoot to a step to be P.O. \(\leq 10\%\) and the settling time (with a \(2\%\) criterion) \(T_{s} \leq 1\text{ }s\). Design an appropriate state variable feedback system for \(r(t) = - k_{1}x_{1}(t) - k_{2}x_{2}(t)\).
P11.14 A process has the transfer function
Determine the state variable feedback gains to achieve a settling time (with a \(2\%\) criterion) of \(T_{S} = 1.5\text{ }s\) and a percent overshoot of P.O. \(= 8\%\). Assume the complete state vector is available for feedback.
P11.15 A telerobot system has the matrix equations [16]
(a) Determine the transfer function, \(G(s) = Y(s)/\) \(U(s)\). (b) Draw the block diagram indicating the state variables. (c) Determine whether the system is controllable. (d) Determine whether the system is observable.
P11.16 Hydraulic power actuators were used to drive the dinosaurs of the movie Jurassic Park [20]. The motions of the large monsters required high-power actuators requiring 1200 watts.
One specific limb motion has dynamics represented by
We want to place the closed-loop poles at \(s = - 5 \pm j\). Determine the required state variable feedback using Ackermann's formula. Assume that the complete state vector is available for feedback.
P11.17 A system has a transfer function
Determine real values of \(a\) and \(b\) so that the system is either uncontrollable or unobservable.
P11.18 A system has a plant
(a) Find the matrix differential equation to represent this system. Identify the state variables on a block diagram model. (b) Select a state variable feedback structure using \(u(t)\), and select the feedback gains so that the response \(y(t)\) of the unforced system is critically damped when the initial condition is \(x_{1}(0) = 1\) and \(x_{2}(0) = 0\), where \(x_{1} = y(t)\). The repeated roots are at \(s = - 3\).
P11.19 The block diagram of a system is shown in Figure P11.19. Determine whether the system is controllable and observable.
P11.20 Consider the automatic ship-steering system. The state variable form of the system differential equation is
FIGURE P11.19
Multiloop feedback control system.
where \(\mathbf{x}^{T}(t) = \begin{bmatrix} v(t) & \omega_{s}(t) & y(t) & \theta(t) \end{bmatrix}\). The state variables are \(x_{1}(t) = v(t) =\) the transverse velocity; \(x_{2}(t) = \omega_{s}(t) =\) angular rate of ship's coordinate frame relative to response frame; \(x_{3}(t) = y(t) =\) deviation distance on an axis perpendicular to the track; \(x_{4}(t) = \theta(t) =\) deviation angle. (a) Determine whether the system is stable. (b) Feedback can be added so that
Determine whether this system is stable for suitable values of \(k_{1}\) and \(k_{3}\). If so, determine \(k_{1}\) and \(k_{3}\) such that the percent overshoot P.O. \(\leq 25\%\) to a unit step, \(R(s) = 1/s\) and \(T_{S} \leq 20\text{ }s\).
P11.21 An op-amp circuit is shown in Figure P11.21 with input \(u(t) = v_{i}(t)\), and output \(y(t) = v_{0}(t)\).
a. Determine the system transfer function.
b. Select the state variables, and write the state equations for this system.
c. Determine its observability, and find the condition when the system has a double root.
FIGURE P11.21 An op-amp circuit.
P11.22 The speed control system of an electric car has a plant transfer function of
in the open loop and a negative feedback transfer function of \(H(s) = k\) with a reference input of \(r(t)\) and an output \(y(t)\).
a. Represent the system in the block diagram for unity feedback systems by scaling the input, and write the corresponding vector differential equation for this system.
b. The system uses state variable feedback with a feedback signal from the output derivative \(\overset{˙}{y}(t)\) with gain \(b\). Design \(k\) and \(b\) so that the car speed has a settling time of \(0.5\text{ }s\) and a percent overshoot of \(10\%\).
c. Plot the response of the state variable feedback system to a step input.
P11.23 Let
and \(\mathbf{D} = \lbrack 0\rbrack\). Then design a controller using internal model methods so that the steady-state error for a step input is zero and the desired roots of the characteristic equation are \(s = - 2 \pm 2j\) and \(s = - 20\).
P11.24 Let \(\mathbf{A} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix},\ \mathbf{B} = \begin{bmatrix} 0 \\ 1 \end{bmatrix},\ \mathbf{C} = \begin{bmatrix} 1 & 0 \end{bmatrix}\) and \(\mathbf{D} = \lbrack 0\rbrack\). Then design a controller using internal model methods so that the steady-state error for a ramp input is zero and the roots of the characteristic equation are \(s = - 2 \pm j2,s = - 2\), and \(s = - 1\).
P11.25 Consider the system represented in state variable form
where
Verify that the system is observable. Then design a full-state observer by placing the observer poles at \(s_{1,2} = - 1\). Plot the response of the estimation error \(\mathbf{e}(t) = \mathbf{x}(t) - \widehat{\mathbf{x}}(t)\) with an initial estimation error of \(\mathbf{e}(0) = \lbrack 1,1\rbrack^{T}\).
P11.26 Consider the second-order system
Verify that the system is observable. If so, determine the observer gain matrix required to place the observer poles at \(s_{1,2} = - 2 \pm j3\) and \(s_{3} = - 6\).
P11.27 Consider the second-order system
Determine the observer gain matrix required to place the observer poles at \(s_{1,2} = - 1 \pm j2\).
P11.28 Consider the single-input, single-output system is described by
where
a. Determine the value of \(K\) resulting in a zero steady-state tracking error when \(u(t)\) is a unit step input for \(t \geq 0\). The tracking error is defined here as \(e(t) = u(t) - y(t)\).
b. Plot the response to a unit step input and verify that the tracking error is zero for the gain \(K\) determined in part (a).
P11.29 The block diagram shown in Figure P11.29 is an example of an interacting system. Determine a state variable representation of the system in the form
FIGURE P11.29 Interacting feedback system.
401. ADVANCED PROBLEMS
AP11.1 A DC motor control system has the form shown in Figure AP11.1 [6]. The three state variables are available for measurement; the output position is \(x_{1}(t)\). Select the feedback gains so that the system has a steady-state error equal to zero for a step input and a response with a percent overshoot of P.O. \(\leq 4\%\).
AP11.2 A system has the model
Add state variable feedback so that the closed-loop poles are \(s = - 2 \pm 2j\) and \(s = - 20\).
FIGURE AP11.1 Field-controlled DC motor.
AP11.3 A system has a matrix differential equation
What values for \(b_{1}\) and \(b_{2}\) are required so that the system is controllable?
AP11.4 The vector differential equation describing the inverted pendulum of Example 3.3 is
Assume that all state variables are available for measurement, and use state variable feedback. Place the system characteristic roots at \(s = - 5 \pm j2, - 3\), and -3 .
AP11.5 An automobile suspension system has three physical state variables, as shown in Figure AP11.5 [13]. The state variable feedback structure is shown in the figure, with \(K_{1} = 1\). Select \(K_{2}\) and \(K_{3}\) so that the roots of the characteristic equation are three real roots lying between \(s = - 3\) and \(s = - 6\). Also, select \(K_{p}\) so that the steady-state error for a step input is equal to zero.
AP11.6 A system is represented by the differential equation
where \(y(t) =\) output and \(u(t) =\) input.
(a) Define the state variables as \(x_{1}(t) = y(t)\) and \(x_{2}(t)\) \(= \overset{˙}{y}(t)\). Develop a state variable representation and show that it is a controllable system. (b) Define the state variables as \(x_{1}(t) = y(t)\) and \(x_{2}(t) = \overset{˙}{y}(t) - u(t)\). Develop a state variable represetation and determine whether the system is controllable. (c) Explain why the system controllability differs in these two cases.
AP11.7 The Radisson Diamond uses pontoons and stabilizers to damp out the effect of waves hitting the ship, as shown in Figure AP11.7(a). The block diagram of the ship roll control system is shown in Figure AP11.7(b).
Determine the feedback gains \(K_{2}\) and \(K_{3}\) so that the characteristic roots are \(s = - 15\) and \(s = - 2 \pm j2\). Plot the roll angle for a unit step disturbance.
AP11.8 Consider the system
where
(a) Design a state variable controller using only \(x_{1}(t)\) as the feedback variable, so that the step response has a percent overshoot of P.O. \(\leq 10\%\) and a settling time (with a \(2\%\) criterion) of \(T_{s} \leq 5\text{ }s\). (b) Design a state variable controller feedback using two state variables, level \(x_{1}(t)\) and shaft position \(x_{2}(t)\), to satisfy the specifications of part (a). (c) Compare the results of parts (a) and (b).
AP11.9 The motion control of a lightweight hospital transport vehicle can be represented by a system of two masses, as shown in Figure AP11.9, where \(m_{1} = m_{2} = 1\) and \(k_{1} = k_{2} = 1\) [21]. (a) Determine the state vector differential equation. (b) Find the roots of the characteristic equation. (c) We wish to stabilize the system by letting \(u(t) = - kx_{i}(t)\), where \(u\) is the force on the lower mass, and \(x_{i}(t)\) is one of the state variables. Select an appropriate state variable \(x_{i}(t)\). (d) Choose a value for the gain \(k\) and sketch the root locus as \(k\) varies.
AP11.10 Consider the inverted pendulum mounted to a motor, as shown in Figure AP11.10. The motor and load are assumed to have no friction damping. The pendulum to be balanced is attached to the horizontal shaft of a servomotor. The servomotor carries a tachogenerator, so that a velocity signal is available, but there is no position signal. When the motor is unpowered, the pendulum will hang vertically downward and, if slightly disturbed, will perform oscillations. If lifted to the top of its arc,
FIGURE AP11.5
Automobile suspension system.
(a)
FIGURE AP11.7
(a) Radisson
Diamond.
(b) Control system to reduce the effect of the disturbance.
(b)
the pendulum is unstable in that position. Devise a feedback compensator using only the velocity signal from the tachometer.
AP11.11 Determine an internal model controller \(G_{c}(s)\) for the system shown in Figure AP11.11. We want the steady-state error to a step input to be zero. We also want the settling time (with a \(2\%\) criterion) to be \(T_{s} \leq 5\text{ }s\).
FIGURE AP11.10 Motor and inverted pendulum.
AP11.12 A fourth-order system has the model.
where
FIGURE AP11.9 Model of hospital vehicle. What are the poles of the open-loop system? By using the state variable feedback control \(u(t) = - \mathbf{Kx}(t)\), determine matrix \(\mathbf{K}\) to place the closed-loop poles at \(- 1, - 2, - 4\), and -8 .
AP11.13 Consider the system represented in state variable form
where
Verify that the system is observable and controllable. If so, design a full-state feedback law and an observer by placing the closed-loop system poles at \(s_{1,2} = - 2 \pm j2\) and the observer poles at \(s_{1,2} = - 13\).
AP11.14 Consider the third-order system
Verify that the system is observable and controllable. Then, design a full-state feedback law and an observer by placing the closed-loop system poles at \(s_{1,2} = - 4 \pm j1,s_{3} = - 5\) and the observer poles at \(s_{1,2} = - 10 \pm j5,s_{3} = - 25\).
AP11.15 Consider the system depicted in Figure AP11.15. Design a full-state observer for the system. Determine the observer gain matrix \(\mathbf{L}\) to place the observer poles at \(s_{1,2} = - 10 \pm j10\).
FIGURE AP11.15 A second-order system block diagram.
FIGURE AP11.11
Internal model control.
402. DESIGN PROBLEMS
CDP11.1 We wish to obtain a state variable feedback system for the capstan-slide the state variable model developed in CDP3.1 and determine the feedback system. The step response should have a percent overshoot of P.O. \(\leq 2\%\) and a settling time of \(T_{s} \leq 250\text{ }ms\).
DP11.1 Consider the device for the magnetic levitation of a steel ball, as shown in Figures DP11.1(a) and (b). Design a feedback controller \(i = - k_{1}x_{1} - k_{2}x_{2} + \beta r\) where \(x_{1}(t) = y(t),x_{2}(t) = \overset{˙}{y}(t)\), and \(\beta\) is selected to produce a zero steady-state error to a unit step. The goal for \(y(t)\) is \(P.O. \leq 10\%\) for a unit step. Assume that \(y(t)\) and \(\overset{˙}{y}(t)\) are measurable.
DP11.2 The control of the fuel-to-air ratio in an automobile of prime importance as automakers work to reduce exhaust-pollution emissions. Thus, auto engine designers turned to the feedback control of the fuel-to-air ratio. A sensor was placed in the exhaust stream and used as an input to a controller. The
(a)
(b)
FIGURE DP11.1 (a) The levitation of a ball using an electromagnet. (b) The model of the electromagnet and the ball.
controller actually adjusts the orifice that controls the flow of fuel into the engine [3].
Select the devices and develop a linear model for the entire system. Assume that the sensor measures the actual fuel-to-air ratio with a negligible delay. With this model, determine the optimum controller when we desire a system with a zero steadystate error to a step input and a percent overshoot for a step command of P.O. \(\leq 10\%\).
DP11.3 Consider the feedback system depicted in Figure DP11.3. The system model is given by
where
Design the compensator to meet the following specifications:
-
The steady-state error to a unit step input is zero.
-
The settling time \(T_{s} < 1\text{ }s\) and the percent overshoot is P.O. \(< 5\%\).
-
Select initial conditions \(x(0)\) and different initial conditions \(\widehat{x}(0)\) and simulate the response of the closed-loop system to a unit step input.
DP11.4 A high-performance helicopter has a model shown in Figure DP11.4. The goal is to control the pitch angle \(\theta(t)\) of the helicopter by adjusting the rotor thrust angle \(\delta(t)\). The equations of motion of the helicopter are
FIGURE DP11.4 Helicopter pitch angle, \(\theta\), control.
FIGURE DP11.3
Feedback system constructed to track a desired input \(r(t)\).
where \(x(t)\) is the translation in the horizontal direction. For a high-performance helicopter, we find that
all in appropriate SI units.
Find (a) a state variable representation of this system and (b) the transfer function representation for \(\theta(s)/\delta(s)\). (c) Use state variable feedback to achieve adequate performances for the controlled system.
Desired specifications include (1) a steady-state for an input step command for \(\theta_{d}(s)\), the desired pitch angle, less than \(20\%\) of the input step magnitude; (2) an overshoot for a step input command less than \(20\%\); and (3) a settling (with a \(2\%\) criterion) time for a step command of less than 1.5 seconds.
DP11.5 The headbox process is used in the manufacture of paper to transform the pulp slurry flow into a jet of \(2\text{ }cm\) and then spread it onto a mesh belt [22]. To achieve desirable paper quality, the pulp slurry must be distributed as evenly as possible on the belt, and the relationship between the velocity of the jet and that of the belt, called the jet/belt ratio, must be maintained. One of the main control variables is the pressure in the headbox, which in turn controls the velocity of the slurry at the jet. The total pressure in the headbox is the sum of the liquid-level pressure and the air pressure that is pumped into the headbox. Because the pressurized headbox is a highly dynamic and coupled system, manual control would be difficult to maintain and could result in degradation in the sheet properties.
The state-space model of a typical headbox, linearized about a particular stationary point, is given by
and \(y(t) = \begin{bmatrix} 1, & 0 \end{bmatrix}\mathbf{x}(t)\).
The state variables are \(x_{1}(t) =\) liquid level and \(x_{2}(t) =\) pressure. The control variable is \(u_{1}(t) =\) pump current. (a) Design a state variable feedback system that has a characteristic equation with real roots with a magnitude greater than five. (b) Design an observer with observer poles located at least ten times farther in the left half-plane than the state variable feedback system. (c) Connect the observer and full-state feedback system and sketch the block diagram of the integrated system.
DP11.6 A coupled-drive apparatus is shown in Figure DP11.6. The coupled drives consist of two pulleys connected via an elastic belt, which is tensioned by a third pulley mounted on springs providing an underdamped dynamic mode. One of the main pulleys, pulley \(A\), is driven by an electric DC motor. Both pulleys \(A\) and \(B\) are fitted with tachometers that generate measurable voltages proportional to the rate of rotation of the pulley. When a voltage is applied to the DC motor, pulley \(A\) will accelerate at a rate governed by the total inertia experienced by the system. Pulley \(B\), at the other end of the elastic belt, will also accelerate owing to the applied voltage or torque, but with a lagging effect caused by the elasticity of the belt. Integration of the velocity signals measured at each pulley will provide an angular position estimate for the pulley [23].
The second-order model of a coupled-drive is
and \(y(t) = x_{1}(t)\).
(a) Design a state variable feedback controller that will yield a step response with deadbeat response and a settling time (with a \(2\%\) criterion) of \(T_{s} \leq 0.5\text{ }s\). (b) Design an observer for the system by placing the observer poles appropriately in the left half-plane. (c) Draw the block diagram of the system including the compensator with the observer and state feedback. (d) Simulate the response to an initial state at \(\mathbf{x}(0) = \begin{bmatrix} 1 & 0 \end{bmatrix}^{T}\) and \(\widehat{\mathbf{x}}(0) = \begin{bmatrix} 0 & 0 \end{bmatrix}^{T}\).
DP11.7 A closed-loop feedback system is to be designed to track a reference input. The desired feedback block diagram is shown in Figure DP11.3. The system
FIGURE DP11.6 FIGURE DP11.7
Feedback system constructed to track a desired input \(r(t)\).
model is given by
where
Design the observer and the control law to meet the following specifications:
-
The steady-state error of the closed-loop system to a unit step input is zero.
-
P.O. \(\leq 20\%\) to a unit step.
-
\(T_{s} \leq 1\) s to a unit step.
-
Select initial conditions \(x(0)\) and different initial conditions \(\widehat{x}(0)\) and simulate the response of the closed-loop system to a unit step input. Verify that the tracking error is zero in the steady-state.
403. COMPUTER PROBLEMS
CP11.1 Consider the system
Using the ctrb and obsv functions, show that the system is controllable and observable.
CP11.2 Consider the system
Determine if the system is controllable and observable. Compute the transfer function from \(u(t)\) to \(y(t)\).
CP11.3 Find a gain matrix \(\mathbf{K}\) so that the closed-loop poles of the system,
are \(s_{1} = - 2\) and \(s_{2} = - 5\). Use state feedback \(u(t) = - \mathbf{Kx}(t)\).
CP11.4 The following model has been proposed to describe the motion of a constant-velocity rocket:
a. Verify that the system is not controllable by analyzing the controllability matrix using the ctrb function.
b. Develop a controllable state variable model by first computing the transfer function from \(u(t)\) to \(y(t)\), then cancel any common factors in the numerator and denominator polynomials of the transfer function. With the modified transfer function just obtained, use the ss function to determine a modified state variable model for the system.
c. Verify that the modified state variable model in part (b) is controllable.
d. Is the constant velocity rocket stable?
e. Comment on the relationship between the controllability and the complexity of the state variable model (where complexity is measured by the number of state variables). CP11.5 A linearized model of a vertical takeoff and landing (VTOL) aircraft is [24]
where
The state vector components are: (i) \(x_{1}(t)\) is the horizontal velocity (knots), (ii) \(x_{2}(t)\) is the vertical velocity (knots), (iii) \(x_{3}(t)\) is the pitch rate (degrees/second), and (iv) \(x_{4}(t)\) is the pitch angle (degrees). The input \(u_{1}(t)\) is used mainly to control the vertical motion, and \(u_{2}(t)\) is used for the horizontal motion.
(a) Compute the eigenvalues of the system matrix A. Is the system stable? (b) Determine the characteristic polynomial associated with \(\mathbf{A}\) using the poly function. Compute the roots of the characteristic equation, and compare them with the eigenvalues in part (a). (c) Is the system controllable from \(u_{1}(t)\) alone? What about from \(u_{2}(t)\) alone? Comment on the results.
CP11.6 In an effort to open up the far side of the Moon to exploration, studies have been conducted to determine the feasibility of operating a communication satellite around the translunar equilibrium point in the Earth-Sun-Moon system. The desired satellite orbit, known as a halo orbit, is shown in Figure CP11.6. The objective of the controller is to keep the satellite on a halo orbit trajectory that can be seen from the Earth so that the lines of communication are accessible at all times. The communication link is from the Earth to the satellite and then to the far side of the Moon.
The linearized (and normalized) equations of motion of the satellite around the translunar equilibrium point are [25]
The state vector \(\mathbf{x}(t)\) is the satellite position and velocity, and the inputs \(u_{i}(t),i = 1,2,3\), are the engine thrust accelerations in the \(\xi,\eta\), and \(\zeta\) directions, respectively.
(a) Is the translunar equilibrium point a stable location? (b) Is the system controllable from \(u_{1}(t)\) alone? (c) Repeat part (b) for \(u_{2}(t)\). (d) Repeat part (b) for \(u_{3}(t)\). (e) Suppose that we can observe the position in the \(\eta\) direction. Determine the transfer function from \(u_{2}(t)\) to \(x_{2}(t)\). (Hint: Let \(\left. \ y(t) = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}x(t). \right)\ (f)\) Compute a state-space representation of the transfer function in part (e) using the ss function. Verify that the system is controllable. (g) Using state feedback
design a controller (i.e., find \(\mathbf{K}\) ) for the system in part (f) such that the closed-loop system poles are at \(s_{1,2} = - 1 \pm j\) and \(s_{3,4} = - 10\).
CP11.7 Consider the system
Suppose that we are given three observations \(y\left( t_{i} \right),i = 1,2,3\), as follows:
(a) Using the three observations, develop a method to determine the initial value of the state vector \(\mathbf{x}\left( t_{0} \right)\) for the system in Equation CP11.1 that will reproduce the three observations when simulated using the Isim function. (b) With the observations given, compute \(\mathbf{x}\left( t_{0} \right)\) and discuss the condition under which this problem can be solved in general. (c) Verify the result by simulating the system response to the computed initial condition. (Hint: Recall that \(\mathbf{x}(t) = e^{A\left( t - t_{0} \right)}\mathbf{x}\left( t_{0} \right)\) for the system in Equation CP11.1.)
CP11.8 Consider the system
FIGURE CP11.6 The translunar satellite halo orbit.
where
Let \(u(t) = - \mathbf{Kx}(t)\) and consider the performance index
Determine the optimal system when \(\mathbf{x}^{T}(0) = (1,0)\).
CP11.9 A first-order system is given by
with the initial condition \(x(0) = x_{0}\). We want to design a feedback controller
such that the performance index
is minimized.
(a) Let \(\lambda = 1\). Develop a formula for \(J\) in terms of \(k\), valid for any \(x_{0}\), and use an m-file to plot \(J/x_{0}^{2}\) versus \(k\). From the plot, determine the approximate value of \(k = k_{\min}\) that minimizes \(J/x_{0}^{2}\). (b) Verify the result in part (a) analytically. (c) Using the procedure developed in part (a), obtain a plot of \(k_{\min}\) versus \(\lambda\), where \(k_{\min}\) is the gain that minimizes the performance index.
CP11.10 Consider the system represented in state variable form
where
Using the acker function, determine a full-state feedback gain matrix and an observer gain matrix to place the closed-loop system poles at \(s_{1,2} = - 3\) and the observer poles at \(s_{1,2} = - 18 \pm j5\).
CP11.11 Consider the third-order system
(a) Using the acker function, determine a fullstate feedback gain matrix and an observer gain matrix to place the closed-loop system poles at \(s_{1,2} = - 1.4 \pm j1.4,s_{3} = - 2\) and the observer poles at \(s_{1,2} = - 18 \pm j5,s_{3} = - 20\). (b) Construct the state variable compensator. (c) Simulate the closed-loop system with the state initial conditions \(\mathbf{x}(0) = \begin{pmatrix} 1 & 0 & 0 \end{pmatrix}^{T}\) and initial state estimate of \(\widehat{\mathbf{x}}(0) = \begin{pmatrix} 0.5 & 0.1 & 0.1 \end{pmatrix}^{T}\).
CP11.12 Implement the system shown in Figure CP11.12 in an m-file. Obtain the step response of the system.
CP11.13 Consider the system in state variable form
FIGURE CP11.12 Control system for m-file implementation. Design a full-state feedback gain matrix and an observer gain matrix to place the closed-loop system poles at \(s_{1,2} = - 2 \pm j2,s_{3,4} = - 5 \pm j\) and the observer poles \(s_{1,2} = - 9 \pm j2,s_{3,4} = - 15\).
Construct the state variable compensator, and simulate the closed-loop system. Select several values of initial states and initial state estimates in the observer, and display the tracking results.
404. ANSWERS TO SKILLS CHECK
True or False: (1) True; (2) True; (3) False; (4) True; (5) False
Multiple Choice: (6) c; (7) a; (8) c; (9) a; (10) a; (11) b; (12) a; (13) b; (14) b; (15) a
Word Match (in order, top to bottom): e, o, k, i, d, b, j, m, f, n, h, q, g, l, p, c, a
405. TERMS AND CONCEPTS
Command following An important aspect of control system design wherein a nonzero reference input is tracked.
Controllability matrix A linear system is (completely) controllable if and only if the controllability matrix \(\mathbf{P}_{c} = \begin{bmatrix} \mathbf{B} & \mathbf{AB} & \mathbf{A}^{2}\mathbf{B}\ldots & \mathbf{A}^{n - 1}\mathbf{B} \end{bmatrix}\) has full rank, where \(\mathbf{A}\) is an \(n \times n\) matrix. For single-input, single-output linear systems, the system is controllable if and only if the determinant of the \(n \times n\) controllability matrix \(\mathbf{P}_{c}\) is nonzero.
Controllable system A system is controllable on the interval \(\left\lbrack t_{0},t_{f} \right\rbrack\) if there exists a continuous input \(u(t)\) such that any initial state \(\mathbf{x}\left( t_{0} \right)\) can be driven to any arbitrary trial state \(\mathbf{x}\left( t_{f} \right)\) in a finite time interval \(t_{f} - t_{0} > 0\).
Detectable A system in which the states that are unobservable are naturally stable.
Estimation error The difference between the actual state and the estimated state \(\mathbf{e}(t) = \mathbf{x}(t) - \widehat{\mathbf{x}}(t)\).
Full-state feedback control law A control law of the form \(\mathbf{u} = - \mathbf{Kx}\) where \(\mathbf{x}\) is the state of the system assumed known at all times.
Internal model design A method of tracking reference inputs with guaranteed steady-state tracking errors.
Kalman state-space decomposition A partition of the state space that illuminates the states that are controllable and unobservable, uncontrollable and unobservable, controllable and observable, and uncontrollable and observable.
Linear quadratic regulator An optimal controller designed to minimize the quadratic performance index \(J = \int_{0}^{\infty}\mspace{2mu}\left( \mathbf{x}^{T}\mathbf{Qx} + \mathbf{u}^{T}\mathbf{Ru} \right)dt\), where \(\mathbf{Q}\) and \(\mathbf{R}\) are design parameters.
Observability matrix A linear system is (completely) observable if and only if the observability matrix
\(\mathbf{P}_{o} = \begin{bmatrix}
\mathbf{C}^{T} & (\mathbf{CA})^{T} & \left( \mathbf{CA}^{2} \right)^{T}\ldots\left( \mathbf{CA}^{n - 1} \right)^{T}
\end{bmatrix}^{T}\) has full rank, where \(\mathbf{A}\) is an \(n \times n\) matrix. For single-input, single-output linear systems, the system is observable if and only if the determinant of the \(n \times n\) observability matrix \(\mathbf{P}_{O}\) is nonzero.
Observable system A system is observable on the interval \(\left\lbrack t_{0},t_{f} \right\rbrack\) if any initial state \(\mathbf{x}\left( t_{0} \right)\) is uniquely determined by observing the output \(y(t)\) on the interval \(\left\lbrack t_{0},t_{f} \right\rbrack\).
Observer A dynamic system used to estimate the state of another dynamic system given knowledge of the system inputs and measurements of the system outputs.
Optimal control system A system whose parameters are adjusted so that the performance index reaches an extremum value.
Pole placement A design methodology wherein the objective is to place the eigenvalues of the closed-loop system in desired regions of the complex plane.
Regulator problem The control design problem when the reference input \(r(t) = 0\) for all \(t \geq t_{0}\).
Separation principle The principle that states that the full-state feedback law and the observer can be designed independently and when connected will function as an integrated control system in the desired manner (i.e., stable).
Stabilizable A system in which the states that are not controllable are naturally stable.
Stabilizing controller A controller that stabilizes the closed-loop system.
State variable feedback Occurs when the control signal \(u\) for the process is a direct function of all the state variables.
\section{CHAPTER
406. Robust Control Systems
12.1 Introduction 883
12.2 Robust Control Systems and System Sensitivity 884
12.3 Analysis of Robustness 888
12.4 Systems with Uncertain Parameters 890
12.5 The Design of Robust Control Systems 892
12.6 The Design of Robust PID-Controlled Systems 896
12.7 The Robust Internal Model Control System 900
12.8 Design Examples 903
12.9 The Pseudo-Quantitative Feedback System 914
12.10 Robust Control Systems Using Control Design Software 916
12.11 Sequential Design Example: Disk Drive Read System 919
12.12 Summary 921
407. PREVIEW
Physical systems and the external environment in which they operate cannot be modeled precisely, may change in an unpredictable manner, and may be subject to significant disturbances. The design of control systems in the presence of significant uncertainty motivates the concept of robust control system design. Advances in robust control design methodologies can address stability robustness and performance robustness in the presence of uncertainty. In this chapter, we describe five methods for robust design, including root locus, frequency response, ITAE methods for a robust PID systems, internal model control, and pseudo-quantitative feedback methods. However, we should also realize that classical design techniques may also produce robust control systems. Control engineers who are aware of these issues can design robust PID controllers, robust lead-lag controllers, and so forth. The chapter concludes with a PID controller design for the Sequential Design Example: Disk Drive Read System.
408. DESIRED OUTCOMES
Upon completion of Chapter 12, students should be able to:
\(\square\) Describe the role of robustness in control system design.
\(\square\) Identify uncertainty models, including additive uncertainty, multiplicative uncertainty, and parameter uncertainty.
\(\square\) Explain the various methods of tackling the robust control design problem using root locus, frequency response, ITAE methods for PID control, internal model, and pseudoquantitative feedback methods.
408.1. INTRODUCTION
A control system designed using the methods and concepts of the preceding chapters assumes knowledge of the model of the process and controller and constant parameters. The process model will be an inaccurate representation of the actual physical system due to
parameter changes
unmodeled dynamics
\(◯\) unmodeled time delays
changes in equilibrium point (operating point)
○ sensor noise
unpredicted disturbance inputs.
The goal of robust control system design is to maintain acceptable closed-loop system performance in the presence of model inaccuracies and changes.
A robust control system maintains acceptable performance in the presence of significant model uncertainty, disturbances, and noise.
A system structure that incorporates system uncertainties is shown in Figure 12.1. This model includes the sensor noise \(N(s)\), the disturbance input \(T_{d}(s)\), and a process \(G(s)\) with unmodeled dynamics or parameter changes. The
(a)
FIGURE 12.1
Closed-loop control system. (a) Signal flow graph.
(b) Block diagram.
(b) unmodeled dynamics and parameter changes may be significant, and for these systems the challenge is to create a design that retains the desired performance.
408.2. ROBUST CONTROL SYSTEMS AND SYSTEM SENSITIVITY
Designing highly accurate systems in the presence of significant plant uncertainty is a classical feedback design problem. The theoretical bases for the solution of this problem date back to the works of H. S. Black and H. W. Bode in the early 1930s, when this problem was referred to as the sensitivities design problem. A significant amount of literature has been published since then regarding the design of systems subject to large process uncertainty. The designer seeks to obtain a system that performs adequately over a large range of uncertain parameters. A system is said to be robust when it is durable, hardy, and resilient.
A control system is robust when (1) it has low sensitivities, (2) it is stable over the expected range of parameter variations, and (3) the performance continues to meet the specifications in the presence of a set of changes in the system parameters \(\lbrack 3,4\rbrack\). Robustness is the low sensitivity to effects that are not considered in the analysis and design phase - for example, disturbances, measurement noise, and unmodeled dynamics. The system should be able to withstand these neglected effects when performing the tasks for which it was designed.
For small-parameter perturbations, we may use, as a measure of robustness, the differential sensitivities discussed in Sections 4.3 (system sensitivity) and Section 7.5 (root sensitivity) [6]. The system sensitivity is defined as
where \(\alpha\) is the parameter and \(T\) the transfer function. The root sensitivity is defined as
When the zeros of \(T(s)\) are independent of the parameter \(\alpha\), we showed that
for an \(n\) th-order system. For example, if we have a closed-loop system, as shown in Figure 12.2, where the variable parameter is \(\alpha\), then \(T(s) = 1/\lbrack s + (\alpha + 1)\rbrack\), and
This follows because \(r_{1} = + (\alpha + 1)\), and
FIGURE 12.2
A first-order system.
FIGURE 12.3
A second-order system.
Therefore,
Let us examine the sensitivity of the second-order system shown in Figure 12.3. The transfer function of the closed-loop system is
The system sensitivity for \(K\) is
A Bode plot of the asymptotes of \(20log|T(j\omega)|\) and \(20log|S(j\omega)|\) is shown in Figure 12.4 for \(K = 1/4\) (critical damping). Note that the sensitivity is small for lower frequencies, while the transfer function primarily passes low frequencies.
Of course, the sensitivity \(S(s)\) only represents robustness for small changes in gain. If \(K\) changes from \(K = 1/4\) within the range \(K = 1/16\) to \(K = 1\), the resulting range of step response is shown in Figure 12.5. This system, with an expected wide
FIGURE 12.4 Sensitivity and \(20log|T(j\omega)|\) for the second-order system in Figure 12.3. The asymptotic approximations are shown for \(K = 1/4\).
FIGURE 12.5 The step response for selected gain \(K\). FIGURE 12.6
A system with a PD controller.
range of \(K\), may not be considered adequately robust. A robust system would be expected to yield essentially the same (within an agreed-upon variation) response to a selected input.
409. EXAMPLE 12.1 Sensitivity of a controlled system
Consider the system shown in Figure 12.6, where \(G(s) = 1/s^{2}\) and a PD controller \(G_{c}(s) = K_{p} + K_{D}s\). Then the sensitivity with respect to changes in \(G(s)\) is
and
Consider the nominal condition \(\zeta = 1\) and \(\omega_{n} = \sqrt{K_{p}}\). Then, \(K_{D} = 2\omega_{n}\) to achieve \(\zeta = 1\). Therefore, we may plot \(20log|S|\) and \(20log|T|\) on a Bode plot, as shown in Figure 12.7. Note that the frequency \(\omega_{n}\) is an indicator on the boundary between the frequency region in which the sensitivity is the important design criterion and the region in which the stability margin is important. Thus, if we specify \(\omega_{n}\) properly to take into consideration the extent of modeling error and the frequency of external disturbance, we can expect the system to have an acceptable amount of robustness.
410. EXAMPLE 12.2 System with a right-hand-plane zero
Consider the system shown in Figure 12.8, where the plant has a zero in the righthand plane. The closed-loop transfer function is
The system is stable for a gain \(- 2 < K < 1\). The steady-state error due to a negative unit step input \(R(s) = - 1/s\) is
FIGURE 12.7 Sensitivity and \(T(s)\) for the secondorder system in Figure 12.6.
FIGURE 12.8 A second-order system.
FIGURE 12.9
Step response of the system in Figure 12.8 with \(K = 1/2\).
and \(e_{ss} = 0\) when \(K = 1/2\). The response is shown in Figure 12.9. Note the initial undershoot at \(t = 1\text{ }s\). This system is sensitive to changes in \(K\), as recorded in Table 12.1. The performance of this system might not be acceptable for a change of gain of only \(\pm 10\%\). Thus, this system would not be considered robust. The steadystate error of this system changes greatly as \(K\) changes.
411. Table 12.1 Results for Example 12.2
| $$\mathbf{K}$$ | 0.25 | 0.45 | 0.50 | 0.55 | 0.75 |
| $$\left | e_{SS} \right | $$ | 0.67 | 0.18 | 0 |
| Undershoot | $$5%$$ | $$9%$$ | $$10%$$ | $$11%$$ | $$15%$$ |
| Settling time (seconds) | 15 | 24 | 27 | 30 | 45 |
411.1. ANALYSIS OF ROBUSTNESS
System goals include maintaining a small tracking error for an input \(R(s)\) and keeping the output \(Y(s)\) small for a disturbance \(T_{d}(s)\). The sensitivity function is
and the complementary sensitivity function is
We also have the relationship
For physically realizable systems, the loop gain \(L(s) = G_{c}(s)G(s)\) is small for high frequencies. This means that \(S(j\omega)\) approaches 1 at high frequencies.
Consider the closed-loop system shown in Figure 12.1. An additive perturbation characterizes the set of possible processes as follows:
where \(G(s)\) is the nominal process, and \(A(s)\) is the perturbation that is bounded in magnitude. We assume that \(G_{a}(s)\) and \(G(s)\) have the same number of poles in the right-hand \(s\)-plane (if any) [32]. Then the system stability will not change if
This assures stability but not dynamic performance.
A multiplicative perturbation is modeled as
The perturbation is bounded in magnitude, and it is again assumed that \(G_{m}(s)\) and \(G(s)\) have the same number of poles in the right-hand \(s\)-plane. Then the system stability will not change if
Equation (12.15) is called the robust stability criterion. This is a test for robustness with respect to a multiplicative perturbation. This form of perturbation is often used because it satisfies the intuitive properties of (1) being small at low frequencies, where the nominal process model is usually well known, and (2) being large at high frequencies, where the nominal model is always inexact.
412. EXAMPLE 12.3 System with multiplicative perturbation
Consider the system of Figure 12.1 with \(G_{c} = K\), and
The system is unstable with \(K = 1\), but a reduction in gain to \(K = 0.5\) will stabilize it. Now, consider the effect of an unmodeled pole at \(50rad/s\). In this case, the multiplicative perturbation is determined from
or \(M(s) = - s/(s + 50)\). The magnitude bound is then
\(|M(j\omega)|\) and \(|1 + 1/(KG(j\omega))|\) are shown in Figure 12.10(a), where it is seen that the criterion of Equation (12.15) is not satisfied. Thus, the system may not be stable.
If we use a lag compensator
the loop transfer function is \(L(s) = 1 + G_{c}(s)G(s)\). We reshape the function \(G_{c}(j\omega)G(j\omega)\) in the frequency range \(2 < \omega < 25\) and check the condition
as shown in Figure 12.10(b). Here the robustness inequality is satisfied, and the system is robustly stable.
The control objective is to design a compensator \(G_{c}(s)\) so that the transient, steady-state, and frequency-domain specifications are achieved and the cost of feedback measured by the bandwidth of the compensator \(G_{c}(j\omega)\) is sufficiently small. This bandwidth constraint is needed mainly because of measurement noise. In subsequent sections, we discuss including a pre-filter in a two-degree-of-freedom configuration to help achieve the design goals. FIGURE 12.10
The robust stability criterion for Example 12.3.
(a)
(b)
412.1. SYSTEMS WITH UNCERTAIN PARAMETERS
Many systems have several parameters that are constants but uncertain within a range. For example, consider a system with a characteristic equation
with known coefficients within bounds
where \(a_{n} = 1\).
To ascertain the stability of the system, we might have to investigate all possible combinations of parameters. Fortunately, it is possible to investigate a limited number of worst-case polynomials [20]. The analysis of only four polynomials is sufficient, and they are readily defined for a third-order system with a characteristic equation
The four polynomials are
One of the four polynomials represents the worst case and may indicate either unstable performance or at least the worst performance for the system in that case.
413. EXAMPLE 12.4 Third-order system with uncertain parameters
Consider a third-order system with uncertain coefficients such that
The four polynomials are
We then proceed to check these four polynomials by means of the Routh-Hurwitz criterion, and determine that the system is stable for all the range of uncertain parameters.
414. EXAMPLE 12.5 Stability of uncertain system
Consider a unity feedback system with a process transfer function (under nominal conditions)
The nominal characteristic equation is then
where \(a_{0} = 4.5,a_{1} = 2\), and \(a_{2} = 3\). Using the Routh-Hurwitz criterion, we find that this system is nominally stable. However, if the system has uncertain coefficients such that
then we must examine the four polynomials:
Using the Routh-Hurwitz criterion, \(q_{1}(s)\) and \(q_{3}(s)\) are stable and \(q_{2}(s)\) is marginally stable. For \(q_{4}(s)\), we have
| $$s^{3}$$ | 1 | 1 |
| $$s^{2}$$ | 2 | 5 |
| $$s^{1}$$ | $$- 3/2$$ | |
| $$s^{0}$$ | 5 |
.
Therefore, the system is unstable for the worst case, where \(\alpha_{2} =\) minimum, \(\alpha_{1} =\) minimum, and \(\beta_{0} =\) maximum. This occurs when the process has changed to
Note that the third pole has moved toward the \(j\omega\)-axis to its limit at \(s = - 1\) and that the gain has increased to its limit at \(K = 5\).
414.1. THE DESIGN OF ROBUST CONTROL SYSTEMS
The design of robust control systems involves determining the structure of the controller and adjusting the controller parameters to achieve acceptable performance in the presence of uncertainty. The structure of the controller is chosen such that the system response can meet certain performance criteria.
One possible objective in the design of a control system is that the controlled system output should very accurately track the input. That is, we want to minimize the tracking error. In an ideal setting, the Bode plot of the loop gain, \(L(s)\), would be \(0 - dB\) gain of infinite bandwidth and zero phase shift. In practice, this is not possible. One possible design objective is to maintain the magnitude response curve as flat and as close to unity for as large a bandwidth as possible for a given plant and controller combination [20].
Another important goal of a control system design is that the effect on the output of the system due to disturbances is minimized. Consider the control system shown in Figure 12.11, where \(G(s)\) is the plant and \(T_{d}(s)\) is the disturbance. We then have
FIGURE 12.11
A system with a disturbance.
and
Note that both the reference and disturbance transfer functions have the same denominator; in other words, they have the same characteristic equation-namely,
Recall that the sensitivity of \(T(s)\) with respect to \(G(s)\) is
Equation (12.21) shows that for low sensitivity, we desire a high value of loop gain \(L(j\omega)\). But it is known that a high gain can lead to instability and amplification of the measurement noise. Thus, we seek the following:
-
\(T(s)\) with wide bandwidth.
-
Large loop gain \(L(s)\) at low frequencies.
-
Small loop gain \(L(s)\) at high frequencies.
Setting the design of robust systems in frequency-domain terms, we scale a compensator \(G_{c}(s)\) such that the closed-loop sensitivity is less than some tolerance value. But sensitivity minimization involves finding a compensator such that the closed-loop sensitivity is minimized.
The gain and phase margin problem is to find a compensator to achieve prescribed gain and phase margins. The disturbance rejection problem and measurement noise attenuation problem seeks a solution with high loop gain at low frequencies and low loop gain at high frequencies, respectively. For the frequency-domain specifications, we seek the following conditions for the Bode plot of \(G_{c}(j\omega)G(j\omega)\), shown in Figure 12.12:
-
For relative stability, the loop gain must have not more than a \(- 20 - dB/\) decade slope at or near the crossover frequency \(\omega_{c}\).
-
Steady-state accuracy and measurement noise rejection achieved by the low gain at high frequency.
-
Disturbance rejection by a high gain over low frequencies.
-
Accuracy over a bandwidth \(\omega_{B}\), by maintaining the loop gain above a prescribed level. FIGURE 12.12
Bode plot for \(20log\) \(\left| G_{c}(j\omega)G(j\omega) \right|\).
Using the root sensitivity concept, we can state that \(S_{a}^{r}\) must be minimized while attaining \(T(s)\) with dominant roots that will provide the appropriate response and minimize the effect of \(T_{d}(s)\). As an example, let \(G_{c}(s) = K\) and \(G(s) = 1/(s(s + 1))\) for the system in Figure 12.11. This system has two roots, and we select a gain \(K\) so that \(Y(s)/T_{d}(s)\) is minimized, \(S_{K}^{r}\) is minimized, and \(T(s)\) has desirable dominant roots. The sensitivity is
and the characteristic equation is
Therefore, \(dK/ds = - (2s + 1)\), since \(K = - s(s + 1)\). We then obtain
When \(\zeta < 1\), the roots are complex and \(r = - 0.5 + j\frac{1}{2}\sqrt{4K - 1}\). Then,
The magnitude of the root sensitivity is shown in Figure 12.13 for \(K = 0.2\) to \(K = 5\). The percent overshoot to a step is also shown. As illustrated in Figure 12.13, select \(K \approx 1.25\) yields a near minimum sensitivity while maintaining good performance for the step response. To reduce the root sensitivity while simultaneously minimizing the effect of disturbances, we can use the design procedure as follows:
-
Sketch the root locus of the compensated system with \(G_{c}(s)\) chosen to attain the desired location for the dominant roots.
-
Maximize the gain of \(G_{c}(s)\) to reduce the effect of the disturbance.
-
Determine \(S_{K}^{r}\) and attain the minimum value of the root sensitivity consistent with the transient response required, as described in Step 1. FIGURE 12.13
Sensitivity and percent overshoot for a second-order system.
FIGURE 12.14
Bode plot for Example 12.6.
415. EXAMPLE 12.6 Sensitivity and compensation
Consider the system in Figure 12.11 when \(G(s) = 1/s^{2}\) and \(G_{c}(s)\) is to be selected by frequency response methods. Therefore, the compensator is to be selected to achieve an appropriate gain and phase margin while minimizing sensitivity and the effect of the disturbance. Thus, we choose
Choose \(K = 10\) to reduce the effect of the disturbance. To attain a phase margin of \(45^{\circ}\), select \(z = 2.0\) and \(p = 12.0\). The compensated diagram is shown in Figure 12.14. The closed-loop bandwidth is \(\omega_{B} = 1.6\omega_{c}\). Thus, we will increase the bandwidth by using the compensator.
The sensitivity at \(\omega_{c}\) is
To estimate \(\left| S_{G}^{T} \right|\), we recall that the Nichols chart enables us to obtain
We can plot points of \(G_{c}(j\omega)G(j\omega)\) on the Nichols chart and then read \(|T(j\omega)|\) from the chart. Then, we have
where \(\omega_{1}\) is chosen at a frequency below \(\omega_{c}\). The Nichols chart for the compensated system is shown in Figure 12.15. For \(\omega_{1} = \omega_{c}/2.5 = 2\), we have \(20log\left| T\left( j\omega_{1} \right) \right| = 2.5dB\) and \(20log\left| G_{c}\left( j\omega_{1} \right)G\left( j\omega_{1} \right) \right| = 9\text{ }dB\). Therefore,
415.1. THE DESIGN OF ROBUST PID-CONTROLLED SYSTEMS
The PID controller has the transfer function
The popularity of PID controllers can be attributed partly to their robust performance over a wide range of operating conditions and partly to their functional simplicity, which allows engineers to operate them in a straightforward manner. To implement the PID controller, three parameters must be determined for the given process: proportional gain, integral gain, and derivative gain [31].
Consider the PID controller
where \(a = K_{P}/K_{D}\) and \(b = K_{I}/K_{D}\) Therefore, a PID controller introduces a transfer function with one pole at the origin and two zeros.
Recall that a root locus begins at the poles and ends at the zeros. If we have a system as shown in Figure 12.16 with
and we use a PID controller with complex zeros, we can plot the root locus as shown in Figure 12.17. As the gain \(K_{D}\) of the controller is increased, the complex roots approach the zeros. The closed-loop transfer function is FIGURE 12.15
Nichols chart for Example 12.7.
FIGURE 12.16
Feedback control system with a desired input \(R(s)\) and an undesired input \(T_{d}(s)\).
FIGURE 12.17
Root locus with \(- z_{1} = - 6 + j2\).
because the zeros and the complex roots are approximately equal \(\left( r_{1} \approx z_{1} \right)\). Setting \(G_{p}(s) = 1\), we have
when \(K_{D} \gg 1\). The only limiting factor is the allowable magnitude of \(U(s)\) (Figure 12.16) when \(K_{D}\) is large. If \(K_{D}\) is 100 , the system has a fast response and zero steady-state error. Furthermore, the effect of the disturbance is reduced significantly.
In general, we note that PID controllers are particularly useful for reducing 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).
The main problem in the selection of the three PID controller parameters is that these coefficients do not readily translate into the desired performance and robustness characteristics that the control system designer has in mind. Several rules and methods have been proposed to solve this problem. In this section, we consider several design methods using root locus and performance indices.
The first design method uses the ITAE performance index. Hence, we select the three PID coefficients to minimize the ITAE performance index, which produces an excellent transient response to a step or a ramp. The design procedure consists of three steps:
-
Select the \(\omega_{n}\) of the closed-loop system by specifying the settling time.
-
Determine the three coefficients using the appropriate optimum equation (Table 5.3 and Table 5.4) and the \(\omega_{n}\) of step 1 to obtain \(G_{c}(s)\).
-
Determine a prefilter \(G_{p}(s)\) so that the closed-loop system transfer function, \(T(s)\), does not have any zeros.
416. EXAMPLE 12.7 Robust control of temperature
Consider a temperature controller with a control system as shown in Figure 12.16 and a process
If \(G_{c}(s) = 1\), the steady-state error is \(e_{ss} = 50\%\), and the settling time (with a \(2\%\) criterion) is \(T_{s} = 3.2\text{ }s\) for a step input. We want to obtain an optimum ITAE performance for a step input and a settling time of \(T_{s} \leq 0.5\text{ }s\). Using a PID controller, we have
where \(G_{p}(s) = 1\). The optimum coefficients of the characteristic equation for ITAE are
We need to select \(\omega_{n}\) in order to meet the settling time requirement. Since \(T_{s} = 4/\left( \zeta\omega_{n} \right)\) and \(\zeta\) is unknown but near 0.8 , we set \(\omega_{n} = 10\). Then, equating the denominator of Equation (12.34) to Equation (12.35), we obtain the three coefficients as \(K_{P} = 214,K_{D} = 15.5\), and \(K_{I} = 1000\).
Then Equation (12.34) becomes
The response of this system to a step input has a percent overshoot of P.O. \(= 33.9\%\), as recorded in Table 12.2.
We select a prefilter \(G_{p}(s)\) so that we achieve the desired ITAE response with
Table 12.2 Results for Example 12.7
| Controller | $$G_{c}(s) = \mathbf{1}$$ | $$\begin |
PID\text{~}\text{and}\text{~} \\
G_{p}(\mathbf{s}) = \mathbf{1}
\end{matrix}$$ | $$\begin{matrix}
\text{~}\text{PID with}\text{~} \\
G_{p}(\mathbf{s})\text{~}\text{Prefilter}\text{~}
\end{matrix}$$ |
| Percent overshoot | $$4.2%$$ | $$33.9%$$ | $$1.9%$$ |
| Settling time (seconds) | 4.2 | 0.6 | 0.75 |
| Steady-state error | $$50%$$ | $$0.0%$$ | $$0.0%$$ |
| Disturbance error | $$52%$$ | $$0.4%$$ | $$0.4%$$ |
Therefore, we require that
in order to eliminate the zeros in Equation (12.36) and bring the overall numerator to 1000 . The response of the system \(T(s)\) to a step input is indicated in Table 12.2. The system has a small percent overshoot, a settling time of \(T_{s} \leq 0.5\text{ }s\), and zero steady-state error. Furthermore, for a disturbance \(T_{d}(s) = 1/s\), the maximum value of \(y(t)\) due to the disturbance is \(0.4\%\) of the magnitude of the disturbance. This is a very favorable design.
Let us consider the system when the plant varies significantly, so that
where \(0.5 \leq \tau \leq 1\) and \(1 \leq K \leq 2\). We want to investigate the behavior using the ITAE optimum system with the prefilter. The objective is to have an overshoot of P.O. \(\leq 4\%\) and a settling time (with a \(2\%\) criterion) of \(T_{s} \leq 2\text{ }s\) while \(G(s)\) can attain any value in the range indicated.
We then obtain the step response for the four conditions: \(\tau = 1,K = 1;\tau = 0.5\), \(K = 1;\tau = 1,K = 2;\) and \(\tau = 0.5,K = 2\). The results are summarized in Figure 12.18. This is a very robust system.
The value of \(\omega_{n}\) that can be chosen will be limited by considering the maximum allowable \(u(t)\), where \(u(t)\) is the output of the controller, as shown in Figure 12.16. As an example, consider the system in Figure 12.16 with a PID controller, \(G(s) = 1/(s(s + 1))\), and the necessary prefilter \(G_{p}(s)\) to achieve ITAE performance. If we select \(\omega_{n} = 10,20\), and 40 , the maximum value of \(u(t)\) is as recorded in Table 12.3. If we wish to limit \(u(t)\), we need to limit \(\omega_{n}\). Thus, we are limited in the settling time we can achieve.
416.1. THE ROBUST INTERNAL MODEL CONTROL SYSTEM
The internal model control system is shown in Figure 12.19. We now consider the use of the internal model design with special attention to robust system performance. The internal model principle states that if \(G_{c}(s)G(s)\) contains \(R(s)\) then \(Y(s)\) will track \(R(s)\) asymptotically (in the steady state), and the tracking is robust. FIGURE 12.18
Response of the closed loop system in the presence of uncertainty in \(K\) and \(\tau\).
Table 12.3 Maximum Value of Plant Input
Consider a simple system with \(G(s) = 1/s\), for which we seek a ramp response with a steady-state error of zero. A PI controller is sufficient, and we let \(\mathbf{K} = \mathbf{0}\) (no state variable feedback). Then we have
Note that for a ramp, \(R(s) = 1/s^{2}\), which is contained as a factor of Equation (12.40), and the closed-loop transfer function is
FIGURE 12.19
The internal model control system.
Using the ITAE specifications for a ramp response, we require that
We select \(\omega_{n}\) to satisfy a specification for the settling time. For a settling time (with a \(2\%\) criterion) of \(T_{s} = 1\text{ }s\), we select \(\omega_{n} = 5\). Then we require \(K_{p} = 16\) and \(K_{I} = 25\). The response of this system settles in \(T_{s} = 1\text{ }s\) and then tracks the ramp with zero steady-state error. If this system (designed for a ramp input) receives a step input, the response has a percent overshoot of P.O. \(= 5\%\) and a settling time of \(T_{s} = 1.5\text{ }s\). This system is very robust to changes in the plant. For example, if \(G(s) = K/s\) changes gain so that \(K\) varies by \(\pm 50\%\), the change in the ramp response is insignificant.
417. EXAMPLE 12.8 Design of an internal model control system
Consider the system of Figure 12.20 with state variable feedback and a compensator \(G_{c}(s)\). We wish to track a step input with zero steady-state error. Here, we select a PID controller for \(G_{c}(s)\). We then have
and \(G(s)G_{c}(s)\) will contain \(R(s) = 1/s\), the input command. Note that we feed back both state variables and add these additional signals after \(G_{c}(s)\) in order to retain the integrator in \(G_{c}(s)\).
The goal is to achieve a settling time (to within \(2\%\) of the final value) of \(T_{s} \leq 1\) second and a deadbeat response while retaining a robust response. Here, we assume that the two poles of \(G(s)\) can change by \(\pm 50\%\). Then the worst-case condition is
One design approach is to design the control for this worst-case condition. Another approach, which we use here, is to design for the nominal \(G(s)\) and one-half the desired settling time. Then we expect to meet the settling time requirement and attain a very fast, highly robust system. Note that the prefilter \(G_{p}(s)\) is used to attain the desired form for \(T(s)\).
FIGURE 12.20
An internal model control with state variable feedback and \(G_{C}(s)\).
