Modern Control Systems_P2
FIGURE 3.18
An inverted pendulum balanced on a person's hand by moving the hand to reduce \(\theta(t)\).
Assume, for ease, that the pendulum rotates in the \(x - y\) plane.
FIGURE 3.19
A cart and an inverted pendulum. The pendulum is constrained to pivot in the vertical plane.
To obtain the necessary first-order differential equations, we solve for \(l{\overset{˙}{x}}_{4}(t)\) in Equation (3.66) and substitute into Equation (3.65) to obtain
since \(M \gg m\). Substituting \({\overset{˙}{x}}_{2}(t)\) from Equation (3.65) into Equation (3.66), we have
Therefore, the four first-order differential equations can be written as
Thus, the system matrices are
123.1. THE TRANSFER FUNCTION FROM THE STATE EQUATION
Given a transfer function \(G(s)\), we can obtain the state variable equations using the signal-flow graph model. Now we turn to the matter of determining the transfer function \(G(s)\) of a single-input, single-output (SISO) system. Recalling Equations (3.16) and (3.17), we have
and
where \(y(t)\) is the single output and \(u(s)\) is the single input. The Laplace transforms of Equations (3.71) and (3.72) are
and
where \(\mathbf{B}\) is an \(n \times 1\) matrix, since \(U(s)\) is a single input. Note that we do not include initial conditions, since we seek the transfer function. Rearranging Equation (3.73), we obtain
Since \(\lbrack s\mathbf{I} - \mathbf{A}\rbrack^{- 1} = \mathbf{\Phi}(s)\), we have
Substituting \(\mathbf{X}(s)\) into Equation (3.74), we obtain
Therefore, the transfer function \(G(s) = Y(s)/U(s)\) is
124. EXAMPLE 3.4 Transfer function of an \(\mathbf{RLC}\) circuit
Let us determine the transfer function \(G(s) = Y(s)/U(s)\) for the \(RLC\) circuit of Figure 3.3 as described by the differential equations (see Equations 3.18 and 3.19):
Then we have
Therefore, we obtain
where
Then the transfer function is
which agrees with the result Equation (3.40) obtained from the flow graph model using Mason's signal-flow gain formula.
124.1. THE TIME RESPONSE AND THE STATE TRANSITION MATRIX
It is often desirable to obtain the time response of the state variables of a control system and thus examine the performance of the system. The transient response of a system can be readily obtained by evaluating the solution to the state vector differential equation. In Section 3.3, we found that the solution for the state differential Equation (3.26) was
If the initial conditions \(\mathbf{x}(0)\), the input \(\mathbf{u}(\tau)\), and the state transition matrix \(\mathbf{\Phi}(t)\) are known, the time response of \(\mathbf{x}(t)\) can be obtained. Thus the problem focuses on the evaluation of \(\mathbf{\Phi}(t)\), the state transition matrix that represents the response of the system. Fortunately, the state transition matrix can be readily evaluated by using the signal-flow graph techniques.
Before proceeding to the evaluation of the state transition matrix using signal-flow graphs, we should note that several other methods exist for evaluating the transition matrix, such as the evaluation of the exponential series
in a truncated form \(\lbrack 2,8\rbrack\). Several efficient methods exist for the evaluation of \(\Phi(t)\) by means of a computer algorithm [21].
In Equation (3.25), we found that \(\mathbf{\Phi}(s) = \lbrack s\mathbf{I} - \mathbf{A}\rbrack^{- 1}\). Therefore, if \(\mathbf{\Phi}(s)\) is obtained by completing the matrix inversion, we can obtain \(\mathbf{\Phi}(t)\) by noting that \(\mathbf{\Phi}(t) = \mathcal{L}^{- 1}\{\mathbf{\Phi}(s)\}\). The matrix inversion process is generally unwieldy for higher-order systems.
The usefulness of the signal-flow graph state model for obtaining the state transition matrix becomes clear upon consideration of the Laplace transformation version of Equation (3.80) when the input is zero. Taking the Laplace transformation of Equation (3.80) when \(\mathbf{u}(\tau) = 0\), we have
Therefore, we can evaluate the Laplace transform of the transition matrix from the signal-flow graph by determining the relation between a state variable \(X_{i}(s)\) and the state initial conditions \(\left\lbrack x_{1}(0),x_{2}(0),\ldots,x_{n}(0) \right\rbrack\). Then the state transition matrix is the inverse transform of \(\mathbf{\Phi}(s)\); that is,
The relationship between a state variable \(X_{i}(s)\) and the initial conditions \(\mathbf{x}(0)\) is obtained by using Mason's signal-flow gain formula. Thus, for a second-order system, we would have
and the relation between \(X_{2}(s)\) as an output and \(x_{1}(0)\) as an input can be evaluated by Mason's signal-flow gain formula. All the elements of the state transition matrix, \(\phi_{ij}(s)\), can be obtained by evaluating the individual relationships between \(X_{i}(s)\) and \(x_{j}(0)\) from the state model flow graph. An example will illustrate this approach to determining the transition matrix.
125. EXAMPLE 3.5 Evaluation of the state transition matrix
We will consider the \(RLC\) network of Figure 3.3. We seek to evaluate \(\mathbf{\Phi}(s)\) by (1) determining the matrix inversion \(\mathbf{\Phi}(s) = \lbrack s\mathbf{I} - \mathbf{A}\rbrack^{- 1}\), and (2) using the signal-flow diagram and Mason's signal-flow gain formula. First, we determine \(\mathbf{\Phi}(s)\) by evaluating \(\mathbf{\Phi}(s) = \lbrack s\mathbf{I} - \mathbf{A}\rbrack^{- 1}\). We note from Equation (3.18) that
Then
The inverse matrix is
where \(\Delta(s) = s(s + 3) + 2 = s^{2} + 3s + 2 = (s + 1)(s + 2)\).
The signal-flow graph state model of the \(RLC\) network of Figure 3.3, is shown in Figure 3.7. This \(RLC\) network can be represented by the state variables \(x_{1}(t) = v_{c}(t)\) and \(x_{2}(t) = i_{L}(t)\). The initial conditions, \(x_{1}(0)\) and \(x_{2}(0)\), represent the initial capacitor voltage and inductor current, respectively. The flow graph, including the initial conditions of each state variable, is shown in Figure 3.20. The initial conditions appear as the initial value of the state variable at the output of each integrator.
To obtain \(\Phi(s)\), we set \(U(s) = 0\). When \(R = 3,L = 1\), and \(C = 1/2\), we obtain the signal-flow graph shown in Figure 3.21, where the output and input nodes are deleted because they are not involved in the evaluation of \(\mathbf{\Phi}(s)\). Then, using Mason's signal-flow gain formula, we obtain \(X_{1}(s)\) in terms of \(x_{1}(0)\) as
FIGURE 3.20
Flow graph of the \(RLC\) network.
FIGURE 3.21
Flow graph of the \(RLC\) network with \(U(s) = 0\).
where \(\Delta(s)\) is the graph determinant, and \(\Delta_{1}(s)\) is the path cofactor. The graph determinant is
The path cofactor is \(\Delta_{1} = 1 + 3s^{- 1}\) because the path between \(x_{1}(0)\) and \(X_{1}(s)\) does not touch the loop with the factor \(- 3s^{- 1}\). Therefore, the first element of the transition matrix is
The element \(\phi_{12}(s)\) is obtained by evaluating the relationship between \(X_{1}(s)\) and \(x_{2}(0)\) as
Therefore, we obtain
Similarly, for \(\phi_{21}(s)\) we have
Finally, for \(\phi_{22}(s)\), we obtain
Therefore, the state transition matrix in Laplace transformation form is
The factors of the characteristic equation are \((s + 1)\) and \((s + 2)\), so that
Then the state transition matrix is
The evaluation of the time response of the \(RLC\) network to various initial conditions and input signals can now be evaluated by using Equation (3.80). For example, when \(x_{1}(0) = x_{2}(0) = 1\) and \(u(t) = 0\), we have
FIGURE 3.22
Time response of the state variables of the \(RLC\) network for \(x_{1}(0) = x_{2}(0) = 1\).
FIGURE 3.23
Trajectory of the state vector in the \(\left( x_{1},x_{2} \right)\)-plane.
The response of the system for these initial conditions is shown in Figure 3.22. The trajectory of the state vector \(\left( x_{1}(t),x_{2}(t) \right)\) on the \(\left( x_{1},x_{2} \right)\)-plane is shown in Figure 3.23.
The evaluation of the time response is facilitated by the determination of the state transition matrix. Although this approach is limited to linear systems, it is a powerful method and utilizes the familiar signal-flow graph to evaluate the transition matrix.
125.1. DESIGN EXAMPLES
In this section we present two illustrative design examples. In the first example, we present a detailed look at modeling a large space vehicle (such as a space station) using a state variable model. The state variable model is then used to take a look at the stability of the orientation of the spacecraft in a low Earth orbit. The design process is highlighted in this example. The second example is a printer belt drive design. The relationship between the state variable model and the block diagram is illustrated and, using block diagram reduction methods, the transfer function equivalent of the state variable model is obtained.
126. EXAMPLE 3.6 Modeling the orientation of a space station
The International Space Station, shown in Figure 3.24, is a good example of a multipurpose spacecraft that can operate in many different configurations. An important step in the control system design process is to develop a mathematical model of the spacecraft motion. In general, this model describes the translation and attitude motion of the spacecraft under the influence of external forces and torques, and controller and actuator forces and torques. The resulting spacecraft dynamic model
FIGURE 3.24 The International Space Station. (Courtesy of NASA.)
is a set of highly coupled, nonlinear ordinary differential equations. Our objective is to simplify the model while retaining important system characteristics. This is not a trivial task, but an important, and often neglected component of control engineering. In this example, the rotational motion is considered. The translational motion, while critically important to orbit maintenance, can be decoupled from the rotational motion.
Many spacecraft (such as the International Space Station) will maintain an Earth-pointing attitude. This means that cameras and other scientific instruments pointing down will be able to sense the Earth. Conversely, scientific instruments pointing up will see deep space, as desired. To achieve Earth-pointing attitude, the spacecraft needs an attitude hold control system capable of applying the necessary torques. The torques are the inputs to the system, in this case, the space station. The attitude is the output of the system. The International Space Station employs control moment gyros and reaction control jets as actuators to control the attitude. The control moment gyros are momentum exchangers and are preferable to reaction control jets because they do not expend fuel. They are actuators that consist of a constant-rate flywheel mounted on a set of gimbals. The flywheel orientation is varied by rotating the gimbals, resulting in a change in direction of the flywheel angular momentum. In accord with the basic principle of conservation of angular momentum, changes in control moment gyro momentum must be transferred to the space station, thereby producing a reaction torque. The reaction torque can be employed to control the space station attitude. However, there is a maximum limit of control that can be provided by the control moment gyro. When that maximum is attained, the device is said to have reached saturation. So, while control moment gyros do not expend fuel, they can provide only a limited amount of control. In practice, it is possible to control the attitude of the space station while simultaneously desaturating the control moment gyros.
Several methods for desaturating the control moment gyros are available, but using existing natural environmental torques is the preferred method because it minimizes the use of the reaction control jets. A clever idea is to use gravity gradient torques (which occur naturally) to continuously desaturate the momentum exchange devices. Due to the variation of the Earth's gravitational field over the International Space Station, the total moment generated by the gravitational forces about the spacecraft's center of mass is nonzero. This nonzero moment is called the gravity gradient torque. A change in attitude changes the gravity gradient torque acting on the vehicle. Thus, combining attitude control and momentum management becomes a matter of compromise.
The elements of the design process emphasized in this example are illustrated in Figure 3.25. We can begin the modeling process by defining the attitude of the
FIGURE 3.25
Elements of the control system design process emphasized in the spacecraft control example.
Topics emphasized in this example
space station using the three angles, \(\theta_{2}(t)\) (the pitch angle), \(\theta_{3}(t)\) (the yaw angle), and \(\theta_{1}(t)\) (the roll angle). These three angles represent the attitude of the space station relative to the desired Earth-pointing attitude. When \(\theta_{1}(t) = \theta_{2}(t) = \theta_{3}(t) = 0\), the space station is oriented in the desired direction. The goal is to keep the space station oriented in the desired attitude while minimizing the amount of momentum exchange required by the control momentum gyros (keeping in mind that we want to avoid saturation). The control goal can be stated as
127. Control Goal
Minimize the roll, yaw, and pitch angles in the presence of persistent external disturbances while simultaneously minimizing the control moment gyro momentum.
The time rate of change of the angular momentum of a body about its center of mass is equal to the sum of the external torques acting on that body. Thus the attitude dynamics of a spacecraft are driven by externally acting torques. The main external torque acting on the space station is due to gravity. Since we treat the Earth as a point mass, the gravity gradient torque [30] acting on the spacecraft is given by
where \(n\) is the orbital angular velocity ( \(n = 0.0011rad/s\) for the space station), and \(\mathbf{c}(t)\) is
The notation ' \(X\) ' denotes vector cross-product. Matrix I is the spacecraft inertia matrix and is a function of the space station configuration. It also follows from Equation (3.95) that the gravity gradient torques are a function of the attitude \(\theta_{1}(t),\theta_{2}(t)\), and \(\theta_{3}(t)\). We want to maintain a prescribed attitude (that is Earth-pointing \(\left. \ \theta_{1} = \theta_{2} = \theta_{3} = 0 \right)\), but sometimes we must deviate from that attitude so that we can generate gravity gradient torques to assist in the control moment gyro momentum management. Therein lies the conflict; as engineers we often are required to develop control systems to manage conflicting goals.
Now we examine the effect of the aerodynamic torque acting on the space station. Even at the high altitude of the space station, the aerodynamic torque does affect the attitude motion. The aerodynamic torque acting on the space station is generated by the atmospheric drag force that acts through the center of pressure. In general, the center of pressure and the center of mass do not coincide, so aerodynamic torques develop. In low Earth orbit, the aerodynamic torque is a sinusoidal function that tends to oscillate around a small bias. The oscillation in the torque is primarily a result of the Earth's diurnal atmospheric bulge. Due to heating, the atmosphere closest to the Sun extends further into space than the atmosphere on the side of the Earth away from the Sun. As the space station travels around the Earth (once every 90 minutes or so), it moves through varying air densities, thus causing a cyclic aerodynamic torque. Also, the space station solar panels rotate as they track the Sun. This results in another cyclic component of aerodynamic torque. The aerodynamic torque is generally much smaller than the gravity gradient torque. Therefore, for design purposes we can ignore the atmospheric drag torque and view it as a disturbance torque. We would like the controller to minimize the effects of the aerodynamic disturbance on the spacecraft attitude.
Torques caused by the gravitation of other planetary bodies, magnetic fields, solar radiation and wind, and other less significant phenomena are much smaller than the Earth's gravity-induced torque and aerodynamic torque. We ignore these torques in the dynamic model and view them as disturbances.
Finally, we need to discuss the control moment gyros themselves. First, we will lump all the control moment gyros together and view them as a single source of torque. We represent the total control moment gyro momentum with the variable \(\mathbf{h}(t)\). We need to know and understand the dynamics in the design phase to manage the angular momentum. But since the time constants associated with these dynamics are much shorter than for attitude dynamics, we can ignore the dynamics and assume that the control moment gyros can produce precisely and without a time delay the torque demanded by the control system.
Based on the above discussion, a simplified nonlinear model that we can use as the basis for the control design is
where
where \(\mathbf{u}(t)\) is the control moment gyro input torque, \(\Omega(t)\) in the angular velocity, \(\mathbf{I}\) is the moment of inertia matrix, and \(\mathbf{n}\) is the orbital angular velocity. Two good references that describe the fundamentals of spacecraft dynamic modeling are [26] and [27]. There have been many papers dealing with space station control and momentum management. One of the first to present the nonlinear model in Equations (3.96-3.98) is Wie et al. [28]. Other related information about the model and the control problem in general appears in [29-33]. Articles related to advanced control topics on the space station can be found in [34-40]. Researchers are developing nonlinear control laws based on the nonlinear model in Equations (3.96)-(3.98). Several good articles on this topic appear in [41-50]. Equation (3.96) represents the kinematics - the relationship between the Euler angles, denoted by \(\Theta(t)\), and the angular velocity vector, \(\Omega(t)\). Equation (3.97) represents the space station attitude dynamics. The terms on the right side represent the sum of the external torques acting on the spacecraft. The first torque is due to inertia cross-coupling. The second term represents the gravity gradient torque, and the last term is the torque applied to the spacecraft from the actuators. The disturbance torques (due to such factors as the atmosphere) are not included in the model used in the design. Equation (3.98) represents the control moment gyro total momentum.
The conventional approach to spacecraft momentum management design is to develop a linear model, representing the spacecraft attitude and control moment gyro momentum by linearizing the nonlinear model. This linearization is accomplished by a standard Taylor series approximation. Linear control design methods can then be readily applied. For linearization purposes we assume that the spacecraft has zero products of inertia (that is, the inertia matrix is diagonal) and the aerodynamic disturbances are negligible. The equilibrium state that we linearize about is
and where we assume that
In reality, the inertia matrix, \(\mathbf{I}\), is not a diagonal matrix. Neglecting the off-diagonal terms is consistent with the linearization approximations and is a common assumption. Applying the Taylor series approximations yields the linear model, which as it turns out decouples the pitch axis from the roll/yaw axis.
The linearized equations for the pitch axis are
where
The subscript 2 refers to the pitch axis terms, the subscript 1 is for the roll axis terms, and 3 is for the yaw axis terms. The linearized equations for the roll/yaw axes are
where
Consider the analysis of the pitch axis. Define the state-vector as
and the output as
Here we are considering the spacecraft attitude, \(\theta_{2}(t)\), as the output of interest. We can just as easily consider both the angular velocity, \(\omega_{2}(t)\), and the control moment gyro momentum, \(h_{2}(t)\), as outputs. The state variable model is
where
and where \(u(t)\) is the control moment gyro torque in the pitch axis. The solution to the state differential equation, given in Equation (3.101), is
where
We can see that if \(\Delta_{2} > 0\), then some elements of the state transition matrix will have terms of the form \(e^{at}\), where \(a > 0\). This indicates that our system is unstable. Also, if we are interested in the output, \(y(t) = \theta_{2}(t)\), we have
With \(\mathbf{x}(t)\) given by
it follows that
The transfer function relating the output \(Y(s)\) to the input \(U(s)\) is
The characteristic equation is
If \(\Delta_{2} > 0\) (that is, if \(I_{3} > I_{1}\) ), then we have two real poles-one in the left halfplane and the other in the right half-plane. For spacecraft with \(I_{3} > I_{1}\), we can say that an Earth-pointing attitude is an unstable orientation. This means that active control is necessary.
Conversely, when \(\Delta_{2} < 0\) (that is, when \(I_{1} > I_{3}\) ), the characteristic equation has two imaginary roots at
This type of spacecraft is marginally stable. In the absence of any control moment gyro torques, the spacecraft will oscillate around the Earth-pointing orientation for any small initial deviation from the desired attitude.
128. EXAMPLE 3.7 Printer belt drive modeling
A commonly used low-cost printer for a computer uses a belt drive to move the printing device laterally across the printed page [11]. The printing device may be a laser printer, a print ball, or thermal printhead. An example of a belt drive printer with a DC motor actuator is shown in Figure 3.26. In this model, a light sensor is used to measure the position of the printing device, and the belt tension adjusts the spring flexibility of the belt. The goal of the design is to determine the effect of the belt spring constant \(k\) and select appropriate parameters for the motor, the belt pulley, and the controller. To achieve the analysis, we will determine a model of the belt-drive system and select many of its parameters. Using this model, we will obtain the signal-flow graph model and select the state variables. We then will determine an appropriate transfer function for the system and select its other parameters, except for the spring constant. Finally, we will examine the effect of varying the spring constant within a realistic range.
We propose the model of the belt-drive system shown in Figure 3.27. This model assumes that the spring constant of the belt is \(k\), the radius of the pulley is \(r\), the angular rotation of the motor shaft is \(\theta(t)\), and the angular rotation of the righthand pulley is \(\theta_{p}(t)\). The mass of the printing device is \(m\), and its position is \(y(t)\). A light sensor is used to measure \(y(t)\), and the output of the sensor is a voltage \(v_{1}(t)\),
FIGURE 3.26
Printer belt-drive system.
FIGURE 3.27
Printer belt-drive model.
where \(v_{1}(t) = k_{1}y(t)\). The controller provides an output voltage \(v_{2}(t)\), where \(v_{2}(t)\) is a function of \(v_{1}(t)\). The voltage \(v_{2}(t)\) is connected to the field of the motor. Let us assume that we can use the linear relationship
and elect to use \(k_{2} = 0.1\) and \(k_{3} = 0\) (velocity feedback).
The inertia of the motor and pulley is \(J = J_{\text{motor}\text{~}} + J_{\text{pulley. We plan to}\text{~}}\) use a moderate-DC motor. Selecting a typical 1/8-hp DC motor, we find that \(J = 0.01\text{ }kg{\text{ }m}^{2}\), the field inductance is negligible, the field resistance is \(R = 2\Omega\), the motor constant is \(K_{m} = 2Nm/A\), and the motor and pulley friction is \(b = 0.25Nms/rad\). The radius of the pulley is \(r = 0.15\text{ }m,m = 0.2\text{ }kg\), and \(k_{1} = 1\text{ }V/m\).
We now proceed to write the equations of the motion for the system; note that \(y(t) = r\theta_{p}(t)\). Then the tension from equilibrium \(T_{1}\) is
The tension from equilibrium \(T_{2}(t)\) is
The net tension at the mass \(m\) is
and
where the first state variable is \(x_{1}(t) = r\theta(t) - y(t)\). Let the second state variable be \(x_{2}(t) = dy(t)/dt\), and use Equations (3.102) and (3.103) to obtain
The first derivative of \(x_{1}(t)\) is
when we select the third state variable as \(x_{3}(t) = d\theta(t)/dt\). We now require a differential equation describing the motor rotation. When \(L = 0\), we have the field current \(i(t) = v_{2}(t)/R\) and the motor torque \(T_{m}(t) = K_{m}i(t)\). Therefore,
and the motor torque provides the torque to drive the belts plus the disturbance or undesired load torque, so that
The torque \(T(t)\) drives the shaft to the pulley, so that
Therefore,
Hence,
where
Thus, we obtain
Equations (3.104)-(3.106) are the three first-order differential equations required to describe this system. The matrix differential equation is
The signal-flow graph and block diagram models representing the matrix differential equation are shown in Figure 3.28, where we include the identification of the node for the disturbance torque \(T_{d}(t)\).
We can use the flow graph to determine the transfer function \(X_{1}(s)/T_{d}(s)\). The goal is to reduce the effect of the disturbance \(T_{d}(s)\), and the transfer function will show us how to accomplish this goal. Using Mason's signal-flow gain formula, we obtain
FIGURE 3.28
Printer belt drive. (a) Signal-flow graph. (b) Block diagram model.
(a)
(b)
where
We therefore have
We can also determine the closed-loop transfer function using block diagram reduction methods, as illustrated in Figure 3.29. Remember, there is no unique path to follow in reducing the block diagram; however, there is only one correct solution (a)
(b)
(c)
FIGURE 3.29
Printer belt drive block diagram reduction. (d)
in the end. The original block diagram is shown in Figure 3.28(b). The result of the first step is shown in Figure 3.29(a), where the upper feedback loop has been reduced to a single transfer function. The second step illustrated in Figure 3.29(b) then reduces the two lower feedback loops to a single transfer function. In the third step shown in Figure 3.29(c), the lower feedback loop is closed and then the remaining transfer functions in series in the lower loop are combined. The final step closed-loop transfer function is shown in Figure 3.29(d).
Substituting the parameter values, we obtain
We wish to select the spring constant \(k\) so that the state variable \(x_{1}(t)\) will quickly decline to a low value when a disturbance occurs. For test purposes, consider a step disturbance \(T_{d}(s) = a/s\). Recalling that \(x_{1}(t) = r\theta(t) - y(t)\), we thus seek a small magnitude for \(x_{1}\) so that \(y\) is nearly equal to the desired \(r\theta\). If we have a perfectly stiff belt with \(k \rightarrow \infty\), then \(y(t) = r\theta(t)\) exactly. With a step disturbance, \(T_{d}(s) = a/s\), we have
The final value theorem gives
and thus the steady-state value of \(x_{1}(t)\) is zero. We need to use a realistic value for \(k\) in the range \(1 \leq k \leq 40\). For an average value of \(k = 20\), we have
The characteristic equation has one real root and two complex roots. The partial fraction expansion yields
where we find \(A = - 0.0218,B = 0.0218\), and \(C = - 0.4381\). Clearly with these small residues, the response to the unit disturbance is relatively small. Because \(A\) and \(B\) are small compared to \(C\), we may approximate \(X_{1}(s)\) as
Taking the inverse Laplace transform, we obtain
The actual response of \(x_{1}\) is shown in Figure 3.30. This system will reduce the effect of the unwanted disturbance to a relatively small magnitude. Thus we have achieved our design objective. FIGURE 3.30
Response of \(x_{1}(t)\) to a step disturbance: peak value \(= - 0.0325\).
128.1. ANALYSIS OF STATE VARIABLE MODELS USING CONTROL DESIGN SOFTWARE
The time-domain method utilizes a state-space representation of the system model, given by
The vector \(\mathbf{x}(t)\) is the state of the system, \(\mathbf{A}\) is the constant \(n \times n\) system matrix, \(\mathbf{B}\) is the constant \(n \times m\) input matrix, \(\mathbf{C}\) is the constant \(p \times n\) output matrix, and \(\mathbf{D}\) is a constant \(p \times m\) matrix. The number of inputs, \(m\), and the number of outputs, \(p\), are taken to be one, since we are considering only single-input, single-output (SISO) problems. Therefore \(y(t)\) and \(u(t)\) are not bold (matrix) variables.
The main elements of the state-space representation in Equation (3.114) are the state vector \(\mathbf{x}(t)\) and the constant matrices \((\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D})\). Two new functions covered in this section are ss and Isim. We also consider the use of the expm function to calculate the state transition matrix.
Given a transfer function, we can obtain an equivalent state-space representation and vice versa. The function tf can be used to convert a state-space representation to a transfer function representation; the function ss can be used to convert a transfer function representation to a state-space representation. These functions are shown in Figure 3.31, where sys_tf represents a transfer function model and sys_ss is a state-space representation.
For instance, consider the third-order system
FIGURE 3.31
(a) The ss function.
(b) Linear system model conversion.
(a)
(b)
(a) m-file script.
(b) Output printout. (a)
| >>convert | |||
| $$a =$$ | |||
| $$x1$$ | $$x2$$ | x3 | |
| $$x1$$ | -8 | -4 | -1.5 |
| $$x2$$ | 4 | 0 | 0 |
| $$x3$$ | 0 | 1 | 0 |
| $$b =$$ | |||
| u1 | |||
| $$x1$$ | 2 | ||
| $$x2$$ | 0 | ||
| $$x3$$ | 0 | ||
| $$C =$$ | |||
| $$x1$$ | $$x2$$ | x3 | |
| $$y1$$ | 1 | 1 | 0.75 |
| $$d =$$ | |||
| u1 | |||
| $$y1$$ | 0 |
(b)
We can obtain a state-space representation using the ss function, as shown in Figure 3.32. A state-space representation of Equation (3.115) is given by Equation (3.114), where
The state-space representation of the transfer function in Equation (3.115) is depicted in Figure 3.33.
FIGURE 3.33 Block diagram with \(X_{1}(\text{ }s)\) defined as the leftmost state variable.
The state variable representation is not unique. For example, another equally valid state variable representation is given by
It is possible that when using the ss function, the state variable representation provided by your control design software will be different from the above two examples depending on the specific software and version.
The time response of the system in Equation (3.114) is given by the solution to the vector integral equation
The matrix exponential function in Equation (3.116) is the state transition matrix, \(\Phi(t)\), where
We can use the function expm to compute the transition matrix for a given time, as illustrated in Figure 3.34. The \(expm(A)\) function computes the matrix exponential. In contrast, the \(exp(A)\) function calculates \(e^{a_{ij}}\) for each of the elements \(a_{ij} \in \mathbf{A}\).
For example, let us consider the \(RLC\) network of Figure 3.3 described by the state-space representation of Equation (3.18) with
FIGURE 3.34
Computing the state transition matrix for a given time, \(\Delta t = dt\).
The initial conditions are \(x_{1}(0) = x_{2}(0) = 1\) and the input \(u(t) = 0\). At \(t = 0.2\), the state transition matrix is as given in Figure 3.34. The state at \(t = 0.2\) is predicted by the state transition methods to be
The time response of the system of Equation (3.115) can also be obtained by using the Isim function. The Isim function can accept as input nonzero initial conditions as well as an input function, as shown in Figure 3.35. Using the Isim function, we can calculate the response for the \(RLC\) network as shown in Figure 3.36.
The state at \(t = 0.2\) is predicted with the Isim function to be \(x_{1}(0.2) =\) \(x_{2}(0.2) = 0.6703\). If we can compare the results obtained by the Isim function and by multiplying the initial condition state vector by the state transition matrix, we find identical results.
(a)
(b)
FIGURE 3.35
The Isim function for calculating the output and state response. FIGURE 3.36
Computing the time response for nonzero initial conditions and zero input using Isim.
128.2. SEQUENTIAL DESIGN EXAMPLE: DISK DRIVE READ SYSTEM
Advanced disks have as many as 8000 tracks per \(cm\). These tracks are typically \(1\mu m\) wide. Thus, there are stringent requirements on the accuracy of the reader head position and of the movement from one track to another. In this chapter, we will develop a state variable model of the disk drive system that will include the effect of the flexure mount.
Since we want a lightweight arm and flexure for rapid movement, we must consider the effect of the flexure, which is a very thin mount made of spring steel. Again, we wish to accurately control the position of the head \(y(t)\) as shown in Figure 3.37(a). We will attempt to derive a model for the system shown in Figure 3.37(a). Here we identify the motor mass as \(M_{1}\) and the head mount mass as \(M_{2}\). The flexure spring is represented by the spring constant \(k\). The force \(u(t)\) to drive the mass \(M_{1}\) is generated by the DC motor. If the spring is absolutely rigid (nonspringy), then we obtain the simplified model shown in Figure 3.37(b). Typical parameters for the two-mass system are given in Table 3.1.
Let us obtain the transfer function model of the simplified system of Figure 3.37(b). Note that \(M = M_{1} + M_{2} = 20.5\text{ }g = 0.0205\text{ }kg\). Then we have
FIGURE 3.37
(a) Model of the two-mass system with a spring flexure. (b) Simplified model with a rigid spring.
FIGURE 3.38
Transfer function model of head reader device with a rigid spring.
Table 3.1 Typical Parameters of the Two-Mass Model
| Parameter | Symbol | Value |
| Motor mass | $$M_{1}$$ | $$20\text{ }g = 0.02\text{ }kg$$ |
| Flexure spring | $$k$$ | $$10 \leq k \leq \infty$$ |
| Head mounting mass | $$M_{2}$$ | $$0.5\text{ }g = 0.0005\text{ }kg$$ |
| Head position | $$x_{2}(t)$$ | variable in \(mm\) |
| Friction at mass 1 | $$b_{1}$$ | $$410 \times 10^{- 3}\text{ }N/(m/s)$$ |
| Field resistance | $$R$$ | $$1\Omega$$ |
| Field inductance | $$L$$ | $$1mH$$ |
| Motor constant | $$K_{m}$$ | $$0.1025\text{ }N\text{ }m/A$$ |
| Friction at mass 2 | $$b_{2}$$ | $$4.1 \times 10^{- 3}\text{ }N/(m/s)$$ |
(a)
(b)
Therefore, the transfer function model is
For the parameters of Table 3.1, we obtain
The transfer function model of the head reader, including the effect of the motor coil, is shown in Figure 3.38. When \(R = 1\Omega,L = 1mH\), and \(K_{m} = 0.1025\), we obtain
Now let us obtain the state variable model of the two-mass system shown in Figure 3.37(a). Write the differential equations as
To develop the state variable model, we choose the state variables as \(x_{1}(t) = q(t)\) and \(x_{2}(t) = y(t)\). Then we have
Then, in matrix form,
and we have
and
Note that the output is \(\overset{˙}{y}(t) = x_{4}(t)\). Also, for \(L = 0\) or negligible inductance, then \(u(t) = K_{m}v(t)\). For the typical parameters and for \(k = 10\), we have
The response of \(\overset{˙}{y}(t)\) for \(u(t) = 1,t > 0\) is shown in Figure 3.39. This response is quite oscillatory, and it is clear that we want a very rigid flexure with \(k > 100\). FIGURE 3.39 Response of \(y\) for a step input for the two-mass model with \(k = 10\).
128.3. SUMMARY
In this chapter, we have considered the description and analysis of systems in the time domain. The concept of the state of a system and the definition of the state variables of a system were discussed. The selection of a set of state variables of a system was examined, and the nonuniqueness of the state variables was noted. The state differential equation and the solution for \(\mathbf{x}(t)\) were discussed. Alternative signal-flow graph and block diagram model structures were considered for representing the transfer function (or differential equation) of a system. Using Mason's signal-flow gain formula, we noted the ease of obtaining the flow graph model. The state differential equation representing the flow graph and block diagram models was also examined. The time response of a linear system and its associated transition matrix was discussed, and the utility of Mason's signal-flow gain formula for obtaining the transition matrix was illustrated. A detailed analysis of a space station model development was presented for a realistic scenario where the attitude control is accomplished in conjunction with minimizing the actuator control. The relationship between modeling with state variable forms and control system design was established. The use of control design software to convert a transfer function to state variable form and calculate the state transition matrix was discussed and illustrated. The chapter concluded with the development of a state variable model for the Sequential Design Example: Disk Drive Read System.
129. 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.
In the following True or False and Multiple Choice problems, circle the correct answer.
- The state variables of a system comprise a set of variables that describe the future response of the system, when given the present state, all future excitation inputs, and the mathematical model describing the dynamics.
True or False
- The matrix exponential function describes the unforced response of the system and is called the state transition matrix.
True or False
- The outputs of a linear system can be related to the state variables and the input signals by the state differential equation.
True or False
- A time-invariant control system is a system for which one or more of the parameters of the system may vary as a function of time.
True or False
- A state variable representation of a system can always be written in diagonal form.
True or False
- Consider a system with the mathematical model given by the differential equation:
A state variable representation of the system is:
a.
b. $\overset{˙}{\mathbf{x}}(t) = \begin{bmatrix}
- 5 & - 1 & - 0.7 \
1 & 0 & 0 \
0 & - 1 & 0
\end{bmatrix}\mathbf{x}(t) + \begin - 1 \
0 \
0
\end{bmatrix}u(t)$
c. $\overset{˙}{\mathbf{x}}(t) = \begin{bmatrix}
- 2 & - 1 \
1 & - 0
\end{bmatrix}\mathbf{x}(t) + \begin{bmatrix}
1 \
0
\end{bmatrix}u(t)$
d.
For Problems 7 and 8, consider the system represented by
where
-
The associated state-transition matrix is:
a. \(\Phi(t) = \lbrack 5t\rbrack\)
b. \(\mathbf{\Phi}(t) = \begin{bmatrix} 1 & 5t \\ 0 & 1 \end{bmatrix}\)
c. $\mathbf{\Phi}(t) = \left\lbrack \begin{matrix}
1 & 5t \
1 & 1
\end{matrix} \right.\ $
d. \(\Phi(t) = \begin{bmatrix} 1 & 5t & t^{2} \\ 0 & 1 & t \\ 0 & 0 & 1 \end{bmatrix}\) -
For the initial conditions \(x_{1}(0) = x_{2}(0) = 1\), the response \(x(t)\) for the zero-input response is:
a. \(x_{1}(t) = (1 + t),x_{2}(t) = 1\) for \(t \geq 0\)
b. \(x_{1}(t) = (5 + t),x_{2}(t) = t\) for \(t \geq 0\)
c. \(x_{1}(t) = (5t + 1),x_{2}(t) = 1\) for \(t \geq 0\)
d. \(x_{1}(t) = x_{2}(t) = 1\) for \(t \geq 0\) -
A single-input, single-output system has the state variable representation
The transfer function of the system \(T(s) = Y(s)/U(s)\) is:
a. \(T(s) = \frac{- 50}{s^{3} + 5s^{2} + 50s}\)
b. \(T(s) = \frac{s + 10}{s^{2} + 10s - 5}\)
c. \(T(s) = \frac{- 5}{s + 5}\)
d. \(T(s) = \frac{- 50}{s^{2} + 5s + 5}\)
- The differential equation model for two first-order systems in series is
where \(u(t)\) is the input of the first system and \(x(t)\) is the output of the second system. The response \(x(t)\) of the system to a unit impulse \(u(t)\) is:
a. \(x(t) = e^{- t} - 2e^{- 2t}\)
b. \(x(t) = \frac{1}{2}e^{- 2t} - \frac{1}{3}e^{- 3t}\)
c. \(x(t) = \frac{1}{2}e^{- t} - \frac{1}{2}e^{- 3t}\)
d. \(x(t) = e^{- t} - e^{- 3t}\)
- A first-order dynamic system is represented by the differential equation
The corresponding transfer function and state-space representation are
a. $G(s) = \frac{1}{1 + 5s}\ $ and \(\ \begin{matrix}
\overset{˙}{x} = - 0.2x + 0.2u \\
y = x
\end{matrix}\)
b. $G(s) = \frac{10}{1 + 5s}\ $ and $\ \overset{˙}{x} = \begin{matrix}
- 0.2x + u \
y = x
\end{matrix}$
c. $G(s) = \frac{1}{s + 5}\ $ and \(\ \begin{matrix} \overset{˙}{x} = - 5x + u \\ y = x \end{matrix}\)
d. None of the above
Consider the block diagram in Figure 3.40 for Problems 12 through 14:
FIGURE 3.40 Block diagram for the Skills Check.
- The effect of the input \(R(s)\) and the disturbance \(T_{d}(s)\) on the output \(Y(s)\) can be considered independently of each other because:
a. This is a linear system, therefore we can apply the principle of superposition.
b. The input \(R(s)\) does not influence the disturbance \(T_{d}(s)\).
c. The disturbance \(T_{d}(s)\) occurs at high frequency, while the input \(R(s)\) occurs at low frequency.
d. The system is causal.
- The state-space representation of the closed-loop system from \(R(s)\) to \(Y(s)\) is:
a. \(\overset{˙}{x}(t) = - 10x(t) + 10Kr(t)\)
b. \(\overset{˙}{x}(t) = - (10 + 10K)x(t) + r(t)\)
c. \(\overset{˙}{x}(t) = - (10 + 10K)x(t) + 10Kr(t)\)
d. None of the above
-
The steady-state error \(E(s) = Y(s) - R(s)\) due to a unit step disturbance \(T_{d}(s) = 1/s\) is:
a. \(e_{ss} = \lim_{t \rightarrow \infty}\mspace{2mu} e(t) = \infty\)
b. \(e_{ss} = \lim_{t \rightarrow \infty}\mspace{2mu} e(t) = 1\)
c. \(e_{ss} = \lim_{t \rightarrow \infty}\mspace{2mu} e(t) = \frac{1}{K + 1}\)
d. \(e_{ss} = \lim_{t \rightarrow \infty}\mspace{2mu} e(t) = K + 1\) -
A system is represented by the transfer function
A state variable representation is:
a. $\overset{˙}{\mathbf{x}}(t) = \begin{bmatrix}
- 10 & - 20 & - 50 \
- 1 & 0 & 0 \
0 & - 1 & 0
\end{bmatrix}\mathbf{x}(t) + \begin{bmatrix}
1 \
1 \
0
\end{bmatrix}u(t)$
b. $\overset{˙}{\mathbf{x}}(t) = \begin{bmatrix}
- 10 & - 20 & - 50 \
- 1 & 0 & 0 \
0 & - 1 & 0
\end{bmatrix}\mathbf{x}(t) + \begin{bmatrix}
1 \
0 \
0
\end{bmatrix}u(t)$
c. $\overset{˙}{\mathbf{x}}(t) = \begin{bmatrix}
- 10 & - 20 & - 50 \
1 & 0 & 0 \
0 & 1 & 0
\end{bmatrix}\mathbf{x}(t) + \begin{bmatrix}
1 \
1 \
0
\end{bmatrix}u(t)$
d. $\overset{˙}{\mathbf{x}}(t) = \begin{bmatrix}
- 10 & - 20 \
0 & - 1
\end{bmatrix}\mathbf{x}(t) + \begin{bmatrix}
1 \
0
\end{bmatrix}u(t)$
In the following Word Match problems, match the term with the definition by writing the correct letter in the space provided.
a. State vector The differential equation for the state vector \(\overset{˙}{\mathbf{x}}(t) = \mathbf{Ax}(t) + \mathbf{B}u(t)\).
b. State of a The matrix exponential function that describes the unforced resystem sponse of the system.
c. Time-varying The mathematical domain that incorporates the time response system and the description of a system in terms of time, \(t\).
d. Transition Vector containing all \(n\) state variables, \(x_{1},x_{2},\ldots x_{n}\). matrix
e. State variables
A set of numbers such that the knowledge of these numbers and the input function will, with the equations describing the dynamics, provide the future state of the system.
f. State A system for which one or more parameters may vary with time.
differential
equation
g. Time domain The set of variables that describe the system.
130. EXERCISES
E3.1 For the spring-mass-damper model shown in Figure E3.1, identify a set of state variables.
FIGURE E3.1 Spring-mass-damper model.
E3.2 The voltage-current relationship for a series \(RLC\) circuit can be represented by the differential equation
where \(v(t)\) is the voltage source, \(i(t)\) is the current, and \(v_{c}(t)\) is the voltage across the capacitor. Put the equations in state variable form, and set up the matrix \(R = L = C = 1\). E3.3 The system in E3.3 can be represented by the state vector differential equation
where
Find the (a) characteristic equation and (b) characteristic roots of the system.
Answer: (a) \(\lambda^{2} + \lambda + 1 = 0;\) (b) \(- 0.5 + 0.866i\),
E3.4 Obtain a state variable matrix for a system with a differential equation
E3.5 A system is represented by a block diagram as shown in Figure E3.5. Write the state equations in the form
FIGURE E3.5 Block diagram.
E3.6 Consider the system
(a) Find the state transition matrix \(\mathbf{\Phi}(t)\). (b) For the initial conditions \(x_{1}(0) = x_{2}(0) = 1\), find \(\mathbf{x}(t)\).
Answer: (b) \(x_{1}(t) = 1 + t,x_{2}(t) = 1,t \geq 0\)
E3.7 Consider the spring and mass shown in Figure 3.3 where \(M = 1\text{ }kg,k = 100\text{ }N/m\), and \(b = 20Ns/m\).
(a) Find the state vector differential equation. (b) Find the roots of the characteristic equation for this system.
Find the characteristic equation, and the roots of the characteristic equation.
E3.9 A multi-loop block diagram is shown in Figure E3.9. The state variables are denoted by \(x_{1}(t)\) and \(x_{2}(t)\). (a) Determine a state variable representation of the closed-loop system where the output is denoted by \(y(t)\) and the input is \(r(t)\). (b) Determine the characteristic equation.
FIGURE E3.9 Multi-loop feedback control system.
E3.10 A hovering vehicle control system is represented by two state variables, and [13]
(a) Find the roots of the characteristic equation.
(b) Find the state transition matrix \(\mathbf{\Phi}(t)\).
Answer:
(a) \(s = - 3, - 2\)
(b) \(\mathbf{\Phi}(t) = \begin{bmatrix} 3e^{- 2t} - 2e^{- 3t} & - 6e^{- 3t} + 6e^{- 2t} \\ e^{- 3t} - e^{- 2t} & 3e^{- 3t} - 2e^{- 2t} \end{bmatrix}\)
E3.11 Determine a state variable representation for the system described by the transfer function
E3.12 Use a state variable model to describe the circuit of Figure E3.12. Obtain the response to an input unit step when the initial current is zero, and the initial capacitor voltage is zero.
FIGURE E3.12 RLC series circuit.
E3.13 A system is described by the two differential equations
and
where \(w(t)\) and \(y(t)\) are functions of time, and \(u(t)\) is an input. (a) Select a set of state variables. (b) Write the matrix differential equation and specify the elements of the matrices. (c) Find the characteristic roots of the system in terms of the parameters \(a\) and \(b\).
Answer: (c) \(s = - 1/2 \pm \sqrt{1 - 4ab}/2\)
E3.14 Develop the state-space representation of a radioactive material of mass \(M\) to which additional radioactive material is added at the rate \(r(t) = Ku(t)\), where \(K\) is a constant. Identify the state variables. Assume that the mass decay is proportional to the mass present.
E3.15 Consider the case of the two masses connected as shown in Figure E3.15. The sliding friction of each mass has the constant \(b\). Determine a state variable matrix differential equation.
FIGURE E3.15 Two-mass system.
E3.16 Two carts with negligible rolling friction are connected as shown in Figure E3.16. An input force is \(u(t)\). The output is the position of cart 2, that is, \(y(t) = q(t)\). Determine a state space representation of the system.
FIGURE E3.16 Two carts with negligible rolling friction.
E3.17 Determine a state variable differential matrix equation for the circuit shown in Figure E3.17:
FIGURE E3.17 \(RC\) circuit.
E3.18 Consider a system represented by the following differential equations:
where \(R,L_{1},L_{2}\), and \(C\) are given constants, and \(v_{a}(t)\) and \(v_{b}(t)\) are inputs. Let the state variables be defined as \(x_{1}(t) = i_{1}(t),x_{2}(t) = i_{2}(t)\), and \(x_{3}(t) = v(t)\). Obtain a state variable representation of the system where the output is \(x_{3}(t)\).
E3.19 A single-input, single-output system has the matrix equations
and
Determine the transfer function \(G(s) = Y(s)/U(s)\).
E3.20 For the simple pendulum shown in Figure E3.20, the nonlinear equations of motion are given by
where \(g\) is gravity, \(L\) is the length of the pendulum, \(m\) is the mass attached at the end of the pendulum (assume the rod is massless), and \(k\) is the coefficient of friction at the pivot point.
(a) Linearize the equations of motion about the equilibrium condition \(\theta_{0} = 0^{\circ}\).
(b) Obtain a state variable representation of the system. The system output is the angle \(\theta(t)\).
FIGURE E3.20 Simple pendulum. E3.21 A single-input, single-output system is described by
Obtain the transfer function \(G(s) = Y(s)/U(s)\) and determine the response of the system to a unit step input.
E3.22 Consider the system in state variable form
with
(a) Compute the transfer function \(G(s) = Y(s)/U(s)\). (b) Determine the poles and zeros of the system. (c) If possible, represent the system as a first-order system
where \(a,b,c\), and \(d\) are scalars such that the transfer function is the same as obtained in (a).
E3.23 Consider a system modeled via the third-order differential equation
Develop a state variable representation and obtain a block diagram of the system assuming the output is \(x(t)\) and the input is \(u(t)\).
131. PROBLEMS
P3.1 A spring-mass-damper system is shown in Figure P3.1. (a) Identify a suitable set of state variables. (b) Obtain the set of first-order differential equations in terms of the state variables. (c) Write the state differential equation.
FIGURE P3.1 Spring-mass-damper system.
P3.2 A balanced bridge network is shown in Figure P3.2.
(a) Show that the \(\mathbf{A}\) and \(\mathbf{B}\) matrices for this circuit are
(b) Sketch the block diagram. The state variables are \(\left( x_{1}(t),x_{2}(t) \right) = \left( v_{c}(t),i_{L}(t) \right)\).
P3.3 An \(RLC\) network is shown in Figure P3.3. Define the state variables as \(x_{1}(t) = i_{L}(t)\) and \(x_{2}(t) = v_{c}(t)\). Obtain the state differential equation.
Partial answer:
FIGURE P3.3 RLC circuit.
P3.4 The transfer function of a system is
Sketch the block diagram and obtain a state variable model.
P3.5 A closed-loop control system is shown in Figure P3.5. (a) Determine the closed-loop transfer function \(T(s) = Y(s)/R(s)\). (b) Determine a state variable model and sketch a block diagram model in phase variable form.
P3.6 Determine the state variable matrix equation for the circuit shown in Figure P3.6. Let \(x_{1}(t) = v_{1}(t),x_{2}(t) = v_{2}(t)\), and \(x_{3}(t) = i(t)\). FIGURE P3.5
Closed-loop
system.
132. \(RLC\) circuit.
FIGURE P3.7 Submarine depth control.
P3.7 An automatic depth-control system for a robot submarine is shown in Figure P3.7. The depth is measured by a pressure transducer. The gain of the stern plane actuator is \(K = 1\) when the vertical velocity is \(25\text{ }m/s\). The submarine has the transfer function
and the feedback transducer is \(H(s) = s + 3\). Determine a state variable representation for the system.
P3.8 The soft landing of a lunar module descending on the moon can be modeled as shown in Figure P3.8. Define the state variables as \(x_{1}(t) = y(t),x_{2}(t) =\) \(\overset{˙}{y}(t),x_{3}(t) = m(t)\), and the control as \(u = \overset{˙}{m}(t)\). Assume that \(g\) is the gravity constant on the moon. Find a state variable model for this system. Is this a linear model?
P3.9 A speed control system using fluid flow components is to be designed. The system is a pure fluid control system because it does not have any moving mechanical parts. The fluid may be a gas or a liquid. A system is desired that maintains the speed within \(0.5\%\) of the
FIGURE P3.8 Lunar module landing control.
desired speed by using a tuning fork reference and a valve actuator. Fluid control systems are insensitive and reliable over a wide range of temperature, electromagnetic and nuclear radiation, acceleration, and vibration. The amplification within the system is achieved by using a fluid jet deflection amplifier. The system can be designed for a \(500 - kW\) steam turbine with a speed of \(12,000rpm\). The block diagram of the system is shown in Figure P3.9. In dimensionless units, we have \(b = 0.1,J = 1\), and \(K_{1} = 0.5\). (a) Determine the closed-loop transfer function
(b) Determine a state variable representation. (c) Determine the characteristic equation obtained from the \(\mathbf{A}\) matrix. FIGURE P3.9
Steam turbine control.
P3.10 Many control systems must operate in two dimensions, for example, the \(x\) - and the \(y\)-axes. A two-axis control system is shown in Figure P3.10, where a set of state variables is identified. The gain of each axis is \(K_{1}\) and \(K_{2}\), respectively. (a) Obtain the state differential equation. (b) Find the characteristic equation from the A matrix. (c) Determine the state transition matrix for \(K_{1} = 1\) and \(K_{2} = 2\).
P3.11 A system is described by
where
and \(x_{1}(0) = 1\) and \(x_{2}(0) = 0\). Determine \(x_{1}(t)\) and \(x_{2}(t)\).
P3.12 A system is described by its transfer function
(a) Determine a state variable model.
(b) Determine the state transition matrix, \(\mathbf{\Phi}(t)\).
P3.13 Consider again the spring-mass-damper system of Problem P3.1 when \(M = 1,b = 6\), and \(k = 8\). (a) Determine whether the system is stable by finding the characteristic equation with the aid of the \(\mathbf{A}\) matrix. (b) Determine the state transition matrix of the system. (c) When the initial velocity is \(1\text{ }m/s\) and initial displacement is \(1\text{ }m/s\), determine the response of the system with \(F(t) = 0\).
FIGURE P3.10
Two-axis system. (a) Signal-flow graph. (b) Block diagram model.
(a)
(b) P3.14 Determine a state variable representation for a system with the transfer function
P3.15 Obtain a block diagram and a state variable representation of this system.
P3.16 The dynamics of a controlled submarine are significantly different from those of an aircraft, missile, or surface ship. This difference results primarily from the moment in the vertical plane due to the buoyancy effect. Therefore, it is interesting to consider the control of the depth of a submarine. The equations describing the dynamics of a submarine can be obtained by using Newton's laws and the angles defined in Figure P3.16. To simplify the equations, we will assume that \(\theta(t)\) is a small angle and the velocity \(v\) is constant and equal to \(25ft/s\). The state variables of the submarine, considering only vertical control, are \(x_{1}(t) = \theta(t),x_{2}(t) = \overset{˙}{\theta}(t)\), and \(x_{3}(t) = \alpha(t)\), where \(\alpha(t)\) is the angle of attack. Thus the state vector differential equation for this system, when the submarine has an Albacore type hull, is
where \(u(t) = \delta_{s}(t)\), the deflection of the stern plane. (a) Determine whether the system is stable. (b) Determine the response of the system to a stern plane step command of \({0.285}^{\circ}\) with the initial conditions equal to zero.
FIGURE P3.16 Submarine depth control.
P3.17 A system is described by the state variable equations
Determine \(G(s) = Y(s)/U(s)\).
P3.18 Consider the control of the robot shown in Figure P3.18. The motor turning at the elbow moves the wrist through the forearm, which has some flexibility as shown [16]. The spring has a spring constant \(k\) and friction-damping constant \(b\). Let the state variables be \(x_{1}(t) = \phi_{1}(t) - \phi_{2}(t)\) and \(x_{2}(t) = \omega_{1}(t)/\omega_{0}\), where
Write the state variable equation in matrix form when \(x_{3}(t) = \omega_{2}(t)/\omega_{0}\).
FIGURE P3.18 An industrial robot. (Courtesy of GCA Corporation.)
P3.19 Consider the system described by
where \(\mathbf{x}(t) = \left( x_{1}(t),x_{2}(t) \right)^{T}\). (a) Compute the state transition matrix \(\mathbf{\Phi}(t,0)\). (b) Using the state transition matrix from (a) and for the initial conditions \(x_{1}(0) = 1\) and \(x_{2}(0) = - 1\), find the solution \(\mathbf{x}(t)\) for \(t \geq 0\).
P3.20 A nuclear reactor that has been operating in equilibrium at a high thermal-neutron flux level is suddenly shut down. At shutdown, the density \(X\) of xenon 135 and the density \(I\) of iodine 135 are \(7 \times 10^{16}\) and \(3 \times 10^{15}\) atoms per unit volume, respectively. The half-lives of \(I_{135}\) and \({Xe}_{135}\) nucleides are 6.7 and 9.2 hours, respectively. The decay equations are \(\lbrack 15,19\rbrack\)
Determine the concentrations of \(I_{135}\) and \({Xe}_{135}\) as functions of time following shutdown by determining (a) the transition matrix and the system response.
(b) Verify that the response of the system is that shown in Figure P3.20.
P3.21 Consider the block diagram in Figure P3.21.
(a) Verify that the transfer function is
(b) Show that a state variable model is given by
FIGURE P3.20 Nuclear reactor response.
FIGURE P3.21 Model of second-order system.
P3.22 Determine a state variable model for the circuit shown in Figure P3.22. The state variables are \(x_{1}(t) =\) \(i(t),x_{2}(t) = v_{1}(t)\), and \(x_{3}(t) = v_{2}(t)\). The output variable is \(v_{0}(t)\).
FIGURE P3.22 RLC circuit.
P3.23 The two-tank system shown in Figure P3.23(a) is controlled by a motor adjusting the input valve and ultimately varying the output flow rate. The system has the transfer function
for the block diagram shown in Figure P3.23(b). Obtain a state variable model.
(a)
(b)
FIGURE P3.23 A two-tank system with the motor current controlling the output flow rate. (a) Physical diagram. (b) Block diagram. P3.24 It is desirable to use well-designed controllers to maintain building temperature with solar collector space-heating systems. One solar heating system can be described by [10]
and
where \(x_{1}(t) =\) temperature deviation from desired equilibrium, and \(x_{2}(t) =\) temperature of the storage material (such as a water tank). Also, \(u_{1}(t)\) and \(u_{2}(t)\) are the respective flow rates of conventional and solar heat, where the transport medium is forced air. A solar disturbance on the storage temperature (such as overcast skies) is represented by \(d(t)\). Write the matrix equations and solve for the system response when \(u_{1}(t) = 0,u_{2}(t) = 1\), and \(d(t) = 1\), with zero initial conditions.
P3.25 A system has the following differential equation:
Determine \(\mathbf{\Phi}(t)\) and its transform \(\mathbf{\Phi}(s)\) for the system.
P3.26 A system has a block diagram as shown in Figure P3.26. Determine a state variable model and the state transition matrix \(\mathbf{\Phi}(s)\).
FIGURE P3.26 Feedback system.
P3.27 A gyroscope with a single degree of freedom is shown in Figure P3.27. Gyroscopes sense the angular motion of a system and are used in automatic flight control systems. The gimbal moves about the output axis \(OB\). The input is measured around the input axis \(OA\). The equation of motion about the output axis is obtained by equating the rate of change of angular momentum to the sum of torques. Obtain a statespace representation of the gyro system.
P3.28 A two-mass system is shown in Figure P3.28. The rolling friction constant is \(b\). Determine a state
FIGURE P3.27 Gyroscope.
variable representation when the output variable is \(y_{2}(t)\).
FIGURE P3.28 Two-mass system.
P3.29 There has been considerable engineering effort directed at finding ways to perform manipulative operations in space-for example, assembling a space station and acquiring target satellites. To perform such tasks, space shuttles carry a remote manipulator system (RMS) in the cargo bay [4, 12,21]. The RMS has proven its effectiveness on recent shuttle missions, but now a new design approach can be considered-a manipulator with inflatable arm segments. Such a design might reduce manipulator weight by a factor of four while producing a manipulator that, prior to inflation, occupies only one-eighth as much space in the cargo bay as the present RMS.
The use of an RMS for constructing a space structure in the shuttle bay is shown in Figure P3.29(a), and a model of the flexible RMS arm is shown in Figure P3.29(b), where \(J\) is the inertia of the drive motor and \(L\) is the distance to the center of gravity of the load component. Derive the state equations for this system.
P3.30 Obtain the state equations for the two-input and one-output circuit shown in Figure P3.30, where the output is \(i_{2}(t)\).
P3.31 Extenders are robot manipulators that extend (that is, increase) the strength of the human arm in
(a)
(b)
FIGURE P3.29 Remote manipulator system.
FIGURE P3.30 Two-input RLC circuit.
load-maneuvering tasks (Figure P3.31) [19, 22]. The system is represented by the transfer function
where \(U(s)\) is the force of the human hand applied to the robot manipulator, and \(Y(s)\) is the force of the
FIGURE P3.31 Extender for increasing the strength of the human arm in load maneuvering tasks. robot manipulator applied to the load. Determine a state variable model and the state transition matrix for the system.
P3.32 A drug taken orally is ingested at a rate \(r(t)\). The mass of the drug in the gastrointestinal tract is denoted by \(m_{1}(t)\) and in the bloodstream by \(m_{2}(t)\). The rate of change of the mass of the drug in the gastrointestinal tract is equal to the rate at which the drug is ingested minus the rate at which the drug enters the bloodstream, a rate that is taken to be proportional to the mass present. The rate of change of the mass in the bloodstream is proportional to the amount coming from the gastrointestinal tract minus the rate at which mass is lost by metabolism, which is proportional to the mass present in the blood. Develop a state space representation of this system.
For the special case where the coefficients of \(\mathbf{A}\) are equal to 1 (with the appropriate sign), determine the response when \(m_{1}(0) = 1\) and \(m_{2}(0) = 0\). Plot the state variables versus time and on the \(x_{1} - x_{2}\) state plane.
P3.33 The attitude dynamics of a rocket are represented by
and state variable feedback is used where \(x_{1}(t) = y(t),x_{2}(t) = \overset{˙}{y}(t)\), and \(u(t) = - x_{2}(t) - 0.5x_{1}(t)\). Determine the roots of the characteristic equation of this system and the response of the system when the initial conditions are \(x_{1}(0) = 0\) and \(x_{2}(0) = 1\). The input \(U(s)\) is the applied torques, and \(Y(s)\) is the rocket attitude.
P3.34 A system has the transfer function
(a) Construct a state variable representation of the system.
(b) Determine the element \(\phi_{11}(t)\) of the state transition matrix for this system.
P3.35 Determine a state-space representation for the system shown in Figure P3.35. The motor inductance is negligible, the motor constant is \(K_{m} = 10\), the back electromagnetic force constant is \(K_{b} = 0.0706\), the motor friction is negligible. The motor and valve inertia is \(J = 0.006\), and the area of the tank is \(50{\text{ }m}^{2}\). Note that the motor is controlled by the armature current \(i_{a}(t)\). Let \(x_{1}(t) = h(t),x_{2}(t) = \theta(t)\), and \(x_{3}(t) = \overset{˙}{\theta}(t)\). Assume that \(q_{1}(t) = 80\theta(t)\), where \(\theta(t)\) is the shaft angle. The output flow is \(q_{0}(t) = 50h(t)\). FIGURE P3.35
One-tank system.
FIGURE P3.36 one damper.
FIGURE P3.37
A block diagram model of a thirdorder system.
P3.36 Consider the two-mass system in Figure P3.36. Find a state variable representation of the system. Assume the output is \(x(t)\).
P3.37 Consider the block diagram in Figure P3.37. Using the block diagram as a guide, obtain the state variable model of the system in the form
Using the state variable model as a guide, obtain a third-order differential equation model for the system.
Two-mass system with two springs and
133. ADVANCED PROBLEMS
AP3.1 Consider the electromagnetic suspension system shown in Figure AP3.1. An electromagnet is located at the upper part of the experimental system. Using the electromagnetic force \(f\), we want to suspend the iron ball. Note that this simple electromagnetic suspension system is essentially unworkable. Hence feedback control is indispensable. As a gap sensor, a standard induction probe of the type of eddy current is placed below the ball [20].
Assume that the state variables are \(x_{1}(t) = x(t)\), \(x_{2}(t) = \overset{˙}{x}(t)\), and \(x_{3}(t) = i(t)\). The electromagnet has an inductance \(L = 0.508H\) and a resistance \(R = 23.2\Omega\). Use a Taylor series approximation for the electromagnetic force. The current is \(i_{1}(t) = I_{0} + i(t)\), where \(I_{0} = 1.06\text{ }A\) is the operating point and \(i(t)\) is the variable. The mass \(m\) is equal to \(1.75\text{ }kg\). The gap is \(x_{g}(t) = X_{0} + x(t)\), where \(X_{0} = 4.36\text{ }mm\) is the operating point and \(x(t)\) is the variable. The electromagnetic force is \(f(t) = k\left( i_{1}(t)/x_{g}(t) \right)^{2}\), where \(k = 2.9 \times 10^{- 4}\text{ }N{\text{ }m}^{2}/A^{2}\). Determine the matrix differential equation and the equivalent transfer function \(X(s)/V(s)\).
FIGURE AP3.1 Electromagnetic suspension system.
AP3.2 A two-mass model of a robot arm is shown in Figure AP3.2. Determine the transfer function \(Y(s)/F(s)\), and use the transfer function to obtain a state-space representation of the system.
AP3.3 The control of an autonomous vehicle motion from one point to another point depends on accurate control of the position of the vehicle [16]. The control of the autonomous vehicle position \(Y(s)\) is obtained by the system shown in Figure AP3.3. Obtain a state variable representation of the system.
FIGURE AP3.2 The spring-mass-damper system of a robot arm.
FIGURE AP3.3 Position control.
AP3.4 Front suspensions have become standard equipment on mountain bikes. Replacing the rigid fork that attaches the bicycle's front tire to its frame, such suspensions absorb bump impact energy, shielding both frame and rider from jolts. Commonly used forks, however, use only one spring constant and treat bump impacts at high and low speeds - impacts that vary greatly in severity - essentially the same.
A suspension system with multiple settings that are adjustable while the bike is in motion would be attractive. One air and coil spring with an oil damper is available that permits an adjustment of the damping constant to the terrain as well as to the rider's weight [17]. The suspension system model is shown in Figure AP3.4, where \(b\) is adjustable. Select the appropriate value for \(b\) so that the bike accommodates (a) a large bump at high speeds and (b) a small bump at low speeds. Assume that \(k_{2} = 1\) and \(k_{1} = 2\).
FIGURE AP3.4 Shock absorber. AP3.5 Figure AP3.5 shows a mass \(M\) suspended from another mass \(m\) by means of a light rod of length \(L\). Obtain a state variable model using a linear model assuming a small angle for \(\theta(t)\). Assume the output is the angle, \(\theta(t)\).
FIGURE AP3.5 Mass suspended from cart.
AP3.6 Consider a crane moving in the \(x\) direction while the mass \(m\) moves in the \(z\) direction, as shown in Figure AP3.6. The trolley motor and the hoist motor are very powerful with respect to the mass of the trolley, the hoist wire, and the load \(m\). Consider the input control variables as the distances \(D(t)\) and \(R(t)\). Also assume that \(\theta(t) < 50^{\circ}\). Determine a linear model, and describe the state variable differential equation.
FIGURE AP3.6 A crane moving in the \(x\)-direction while the mass moves in the \(z\)-direction.
AP3.7 Consider the single-input, single-output system described by
where
Assume that the input is a linear combination of the states, that is,
where \(r(t)\) is the reference input. The matrix \(K = \begin{bmatrix} K_{1} & K_{2} \end{bmatrix}\) is known as the gain matrix. Substituting \(u(t)\) into the state variable equation gives the closed-loop system
The design process involves finding \(\mathbf{K}\) so that the eigenvalues of \(\mathbf{A - BK}\) are at desired locations in the left-half plane. Compute the characteristic polynomial associated with the closed-loop system and determine values of \(\mathbf{K}\) so that the closed-loop eigenvalues are in the left-half plane.
AP3.8 A system for dispensing radioactive fluid into capsules is shown in Figure AP3.8(a). The horizontal axis moving the tray of capsules is actuated by a linear motor. The \(x\)-axis control is shown in Figure AP3.8(b). (a) Obtain a state variable model of the closed-loop system with input \(r(t)\) and output \(y(t)\). (b) Determine the characteristic roots of the system and compute \(K\) such that the characteristic values are all co-located at \(s_{1} = - 3,s_{2} = - 3\), and \(s_{3} = - 3\). (c) Determine analytically the unit step-response of the closed-loop system.
(a)
(b)
FIGURE AP3.8 Automatic fluid dispenser.
134. DESIGN PROBLEMS
CDP3.1 The traction drive uses the capstan drive system shown in Figure CDP2.1. Neglect the effect of the motor inductance and determine a state variable model for the system. The parameters are given in Table CDP2.1. The friction of the slide is negligible.
DP3.1 A spring-mass-damper system, as shown in Figure 3.3, is used as a shock absorber for a large high-performance motorcycle. The original parameters selected are \(m = 1\text{ }kg,b = 9\text{ }N\text{ }s/m\), and \(k = 20\text{ }N/m\). (a) Determine the system matrix, the characteristic roots, and the transition matrix \(\Phi(t)\). The harsh initial conditions are assumed to be \(y(0) = 1\) and \(dy/\left. \ dt \right|_{t = 0} = 2\). (b) Plot the response of \(y(t)\) and \(\overset{˙}{y}(t)\) for the first two seconds. (c) Redesign the shock absorber by changing the spring constant and the damping constant in order to reduce the effect of a high rate of acceleration force \(\overset{˙}{y}(t)\) on the rider. The mass must remain constant at \(1\text{ }kg\).
DP3.2 A system has the state variable matrix equation in phase variable form
It is desired that the canonical diagonal form of the differential equation be
Determine the parameters \(a,b\), and \(d\) to yield the required diagonal matrix differential equation.
DP3.3 An aircraft arresting gear is used on an aircraft carrier as shown in Figure DP3.3. The linear model of each energy absorber has a drag force \(f_{D}(t) = K_{D}{\overset{˙}{x}}_{3}(t)\). It is desired to halt the airplane within \(30\text{ }m\) after engaging the arresting cable [13]. The speed of the aircraft on landing is \(60\text{ }m/s\). Select the required constant \(K_{D}\), and plot the response of the state variables.
DP3.4 The Mile-High Bungi Jumping Company wants you to design a bungi jumping system (that is a cord) so that the jumper cannot hit the ground when his or her mass is less than \(100\text{ }kg\), but greater than \(50\text{ }kg\). Also, the company wants a hang time (the time a jumper is moving up and down) greater than 25 seconds, but less than 40 seconds. Determine the characteristics of the cord. The jumper stands on a platform \(90\text{ }m\) above the ground, and the cord will be attached to a fixed beam secured \(10\text{ }m\) above the platform. Assume that the jumper is \(2\text{ }m\) tall and the cord is attached at the waist (1 m high).
FIGURE DP3.3
Aircraft arresting gear. DP3.5 Consider the single-input, single-output system described by
where
Assume that the input is a linear combination of the states, that is,
where \(r(t)\) is the reference input. Determine \(\mathbf{K} = \begin{bmatrix} K_{1} & K_{2} \end{bmatrix}\) so that the closed-loop system
possesses closed-loop eigenvalues at \(r_{1}\) and \(r_{2}\). Note that if \(r_{1} = \sigma + j\omega\) is a complex number, then \(r_{2} = \sigma - j\omega\) is its complex conjugate.
135. COMPUTER PROBLEMS
CP3.1 Determine a state variable representation for the following transfer functions (with unity feedback) using the ss function:
(a) \(G(s) = \frac{1}{s + 1}\)
(b) \(G(s) = \frac{s^{2} + s + 1}{2s^{2} + s + 1}\)
(c) \(G(s) = \frac{s + 1}{3s^{3} + 2s^{2} + s + 1}\)
CP3.2 By using the tf function, determine the transfer function representation for the state variable models based on your answers for CP3.1. Show that the transfer function is the same.
CP3.3 Consider the \(RLC\) circuit shown in Figure CP3.3. Determine the transfer function \(V_{2}(s)/V_{1}(s)\).
(a) Determine the state variable representation when \(R = 1\Omega,L = 0.5H\), and \(C = 1\text{ }F\).
(b) Using the state variable representation from part (a), plot the unit step response with the step function.
CP3.4 Consider the system
FIGURE CP3.3 RLC circuit.
(a) Using the tf function, determine the transfer function \(Y(s)/U(s)\).
(b) Plot the response of the system to the initial condition \(\mathbf{x}(0) = \begin{bmatrix} 0 & - 1 & 1 \end{bmatrix}^{T}\) for \(0 \leq t \leq 20\).
(c) Compute the state transition matrix using the expm function, and determine \(\mathbf{x}(t)\) at \(t = 20\) for the initial condition given in part (b). Compare the result with the system response obtained in part (b).
CP3.5 Consider the two systems
and
(a) Using the tf function, determine the transfer function \(Y(s)/U(s)\) for system (1).
(b) Repeat part (a) for system (2).
(c) Compare the results in parts (a) and (b) and comment.
CP3.6 Consider the closed-loop control system in Figure CP3.6.
(a) Determine a state variable representation of the controller.
(b) Repeat part (a) for the process.
(c) With the controller and process in state variable form, use the series and feedback functions to compute a closed-loop system representation in state variable form and plot the closed-loop system impulse response.
CP3.7 Consider the following system:
with
Using the Isim function obtain and plot the system response (for \(x_{1}(t)\) and \(x_{2}(t)\) ) when \(u(t) = 0\).
CP3.8 Consider the state variable model with parameter \(K\) given by
Plot the characteristic values of the system as a function of \(K\) in the range \(0 \leq K \leq 100\). Determine that range of \(K\) for which all the characteristic values lie in the left half-plane.
FIGURE CP3.6 A closed-loop feedback control system.
136. ANSWERS TO SKILLS CHECK
True or False: (1) True; (2) True; (3) False; (4) False; Word Match (in order, top to bottom): f, d, g, a, b, c, e (5) False
Multiple Choice: (6) a; (7) b; (8) c; (9) b; (10) c; (11) a; (12) a; (13) c; (14) c; (15) c
137. TERMS AND CONCEPTS
Canonical form A fundamental or basic form of the state variable model representation, including phase variable canonical form, input feedforward canonical form, diagonal canonical form, and Jordan canonical form.
Diagonal canonical form A decoupled canonical form displaying the \(n\) distinct system poles on the diagonal of the state variable representation \(\mathbf{A}\) matrix.
Fundamental matrix See Transition matrix.
Input feedforward canonical form A canonical form described by \(n\) feedback loops involving the \(a_{n}\) coefficients of the \(n\)th order denominator polynomial of the transfer function and feedforward loops obtained by feeding forward the input signal.
Jordan canonical form A block diagonal canonical form for systems that do not possess distinct system poles. Matrix exponential function An important matrix function, defined as \(e^{\mathbf{A}t} = I + \mathbf{A}t + (\mathbf{A}t)^{2}/2! + \cdots +\) \((\mathbf{A}t)^{k}/k! + \cdots\), that plays a role in the solution of linear constant coefficient differential equations.
Output equation The algebraic equation that relates the state vector \(\mathbf{x}\) and the inputs \(\mathbf{u}\) to the outputs \(\mathbf{y}\) through the relationship \(\mathbf{y} = \mathbf{Cx} + \mathbf{Du}\).
Phase variable canonical form A canonical form described by \(n\) feedback loops involving the \(a_{n}\) coefficients of the \(n\)th order denominator polynomial of the transfer function and \(m\) feedforward loops involving the \(b_{m}\) coefficients of the \(m\) th order numerator polynomial of the transfer function.
Phase variables The state variables associated with the phase variable canonical form.
Physical variables The state variables representing the physical variables of the system.
State differential equation The differential equation for the state vector: \(\overset{˙}{\mathbf{x}} = \mathbf{Ax} + \mathbf{Bu}\).
State of a system A set of numbers such that the knowledge of these numbers and the input function will, with the equations describing the dynamics, provide the future state of the system.
State-space representation A time-domain model comprising the state differential equation \(\overset{˙}{\mathbf{x}} = \mathbf{Ax} + \mathbf{Bu}\) and the output equation, \(\mathbf{y} = \mathbf{Cx} + \mathbf{Du}\).
State variables The set of variables that describe the system.
State vector The vector containing all \(n\) state variables, \(x_{1},x_{2},\ldots,x_{n}\).
Time domain The mathematical domain that incorporates the time response and the description of a system in terms of time \(t\).
Time-varying system A system for which one or more parameters may vary with time.
Transition matrix \(\mathbf{\Phi}(t)\) The matrix exponential function that describes the unforced response of the system.
138. CHAPTER
Feedback Control System Characteristics
4.1 Introduction 257
4.2 Error Signal Analysis 259
4.3 Sensitivity of Control Systems to Parameter Variations 261
4.4 Disturbance Signals in a Feedback Control System 264
4.5 Control of the Transient Response 269
4.6 Steady-State Error 272
4.7 The Cost of Feedback 274
4.8 Design Examples 275
4.9 Control System Characteristics Using Control Design Software 285
4.10 Sequential Design Example: Disk Drive Read System 291
4.11 Summary 295
139. PREVIEW
In this chapter, we explore the role of error signals to characterize feedback control system performance, including the reduction of sensitivity to model uncertainties, disturbance rejection, measurement noise attenuation, steady-state errors, and transient response characteristics. The error signal is employed in the feedback control system via negative feedback. We discuss the sensitivity of a system to parameter changes, since it is desirable to minimize the effects of parameter variations and uncertainties. We also wish to diminish the effect of unwanted disturbances and measurement noise on the ability of the system to track a desired input. We then describe the transient and steady-state performance of a feedback system and show how this performance can be readily improved with feedback. The chapter concludes with a system performance analysis of the Sequential Design Example: Disk Drive Read System.
140. DESIRED OUTCOMES
Upon completion of Chapter 4, students should be able to:
\(\square\) Explain the central role of error signals in analysis of control systems.
$\square\ $ Identify the improvements afforded by feedback control in reducing system sensitivity to parameter changes, disturbance rejection, and measurement noise attenuation.
\(\square\) Describe the differences between controlling the transient response and the steady-state response of a system.
$\square\ $ State the benefits and costs of feedback in the control design process.
140.1. INTRODUCTION
A control system is defined as an interconnection of components forming a system that will provide a desired system response. Because the desired system response is known, a signal proportional to the error between the desired and the actual response is generated. The use of this signal to control the process results in a closed-loop sequence of operations that is called a feedback system. This closed-loop sequence of operations is shown in Figure 4.1. The introduction of feedback to improve the control system is often necessary. It is interesting that this is also the case for systems in nature, such as biological and physiological systems; feedback is inherent in these systems. For example, the human heart rate control system is a feedback control system. To illustrate the characteristics and advantages of introducing feedback, we consider a single-loop feedback system. Although many control systems are multiloop, a thorough comprehension of the benefits of feedback can best be obtained from the single-loop system and then extended to multiloop systems.
An open-loop system, or a system without feedback, is shown in Figure 4.2. The disturbance, \(T_{d}(s)\), directly influences the output, \(Y(s)\). In the absence of feedback, the control system is highly sensitive to disturbances and to both knowledge of and variations in parameters of \(G(s)\).
If the open-loop system does not provide a satisfactory response, then a suitable cascade controller, \(G_{c}(s)\), can be inserted preceding the process, \(G(s)\), as shown in Figure 4.3. Then it is necessary to design the cascade transfer function, \(G_{c}(s)G(s)\), so that the resulting transfer function provides the desired transient response. This is known as open-loop control.
FIGURE 4.1
A closed-loop system.
FIGURE 4.2
An open-loop system with a disturbance input, \(T_{d}(s)\) (a) Signal-flow graph. (b) Block diagram.
(a)
(b)
(a)
(b)
FIGURE 4.3 Open-loop control system (without feedback). (a) Signal-flow graph. (b) Block diagram.
An open-loop system operates without feedback and directly generates the output in response to an input signal.
By contrast, a closed-loop negative feedback control system is shown in Figure 4.4.
A closed-loop system uses a measurement of the output signal and a comparison with the desired output to generate an error signal that is used by the controller to adjust the actuator.
FIGURE 4.4 A closed-loop control system. (a) Signal-flow graph. (b) Block diagram.
(a)
(b) Despite the cost and increased system complexity, closed-loop feedback control has the following advantages:
Decreased sensitivity of the system to variations in the parameters of the process.
-
Improved rejection of the disturbances.
-
Improved measurement noise attenuation.
-
Improved reduction of the steady-state error of the system.
-
Easy control and adjustment of the transient response of the system.
In this chapter, we examine how the application of feedback can result in the benefits listed above. Using the notion of a tracking error signal, it will be readily apparent that it is possible to utilize feedback with a controller in the loop to improve system performance.
140.2. ERROR SIGNAL ANALYSIS
The closed-loop feedback control system shown in Figure 4.4 has three inputs\(R(s),T_{d}(s)\), and \(N(s)\) - and one output, \(Y(s)\). The signals \(T_{d}(s)\) and \(N(s)\) are the disturbance and measurement noise signals, respectively. Define the tracking error as
For ease of discussion, we will consider a unity feedback system, that is, \(H(s) = 1\), in Figure 4.4. The influence of a nonunity feedback element in the loop will be considered later.
After some block diagram manipulation, we find that the output is given by
Therefore, with \(E(s) = R(s) - Y(s)\), we have
Define the function
The function, \(L(s)\), is known as the loop gain and plays a fundamental role in control system analysis [12]. In terms of \(L(s)\), the tracking error is given by
We can define the function
Then, in terms of \(F(s)\), we define the sensitivity function as
Similarly, in terms of the loop gain, we define the complementary sensitivity function as
In terms of the functions \(S(s)\) and \(C(s)\), we can write the tracking error as
Examining Equation (4.7), we see that (for a given \(G(s)\) ), if we want to minimize the tracking error, we want both \(S(s)\) and \(C(s)\) to be small. Remember that \(S(s)\) and \(C(s)\) are both functions of the controller, \(G_{c}(s)\), which the control engineer designs. However, the following special relationship between \(S(s)\) and \(C(s)\) holds
Clearly, we cannot simultaneously make \(S(s)\) and \(C(s)\) small, hence design compromises must be made.
To analyze the tracking error equation, we need to understand what it means for a transfer function to be "large" or to be "small." The discussion of magnitude of a transfer function is the subject of Chapters 8 and 9 on frequency response methods. However, for our purposes here, we describe the magnitude of the loop gain \(L(s)\) by considering the magnitude \(|L(j\omega)|\) over the range of frequencies, \(\omega\), of interest.
Considering the tracking error in Equation (4.4), it is evident that, for a given \(G(s)\), to reduce the influence of the disturbance, \(T_{d}(s)\), on the tracking error, \(E(s)\), we desire \(L(s)\) to be large over the range of frequencies that characterize the disturbances. That way, the transfer function \(G(s)/(1 + L(s))\) will be small, thereby reducing the influence of \(T_{d}(s)\). Since \(L(s) = G_{c}(s)G(s)\), this implies that we need to design the controller \(G_{c}(s)\) to have a large magnitude over the important range of frequencies. Conversely, to attenuate the measurement noise, \(N(s)\), and reduce the influence on the tracking error, we desire \(L(s)\) to be small over the range of frequencies that characterize the measurement noise. The transfer function \(L(s)/(1 + L(s)\) ) will be small, thereby reducing the influence of \(N(s)\). Again, since \(L(s) = G_{c}(s)G(s)\), that implies that we need to design the controller \(G_{c}(s)\) to have a small magnitude over the important range of frequencies. Fortunately, the apparent conflict between wanting to make \(G_{c}(s)\) large to reject disturbances and the wanting to make \(G_{c}(s)\) small to attenuate measurement noise can be addressed in the design phase by making the loop gain, \(L(s)\), large at low frequencies (generally associated with the frequency range of disturbances), and making \(L(s)\) small at high frequencies (generally associated with measurement noise).
More discussion on disturbance rejection and measurement noise attenuation follows in the subsequent sections. Next, we discuss how we can use feedback to reduce the sensitivity of the system to variations and uncertainty in parameters in the process, \(G(s)\). This is accomplished by analyzing the tracking error in Equation (4.2) when \(T_{d}(s) = N(s) = 0\).
140.3. SENSITIVITY OF CONTROL SYSTEMS TO PARAMETER VARIATIONS
A process, represented by the transfer function \(G(s)\) is subject to a changing environment, aging, uncertainty in the exact values of the process parameters, and other factors that affect a control process. In the open-loop system, all these errors and changes result in a changing and inaccurate output. However, a closed-loop system senses the change in the output due to the process changes and attempts to correct the output. The sensitivity of a control system to parameter variations is of prime importance. A primary advantage of a closed-loop feedback control system is its ability to reduce the system's sensitivity \(\lbrack 1 - 4,18\rbrack\).
For the closed-loop case, if \(G_{c}(s)G(s) \gg 1\) for all complex frequencies of interest, we can use Equation (4.2) to obtain (letting \(T_{d}(s) = 0\) and \(N(s) = 0\) )
The output is approximately equal to the input. However, the condition \(G_{c}(s)G(s) \gg 1\) may cause the system response to be highly oscillatory and even unstable. But the fact that increasing the magnitude of the loop gain reduces the effect of \(G(s)\) on the output is a useful result. Therefore, the first advantage of a feedback system is that the effect of the variation of the parameters of the process, \(G(s)\), is reduced.
Suppose the process (or plant) \(G(s)\) undergoes a change such that the true plant model is \(G(s) + \Delta G(s)\). The change in the plant may be due to a changing external environment or it may represent the uncertainty in certain plant parameters. We consider the effect on the tracking error \(E(s)\) due to \(\Delta G(s)\). Utilizing the principle of superposition, we let \(T_{d}(s) = N(s) = 0\) and consider only the reference input \(R(s)\). From Equation (4.3), it follows that
Then the change in the tracking error is
Since we usually find that \(G_{c}(s)G(s) \gg G_{c}(s)\Delta G(s)\), we have
We see that the change in the tracking error is reduced by the factor \(1 + L(s)\), which is generally greater than 1 over the range of frequencies of interest.
For large \(L(s)\), we have \(1 + L(s) \approx L(s)\), and we can approximate the change in the tracking error by
Larger magnitude \(L(s)\) translates into smaller changes in the tracking error (that is, reduced sensitivity to changes in \(\Delta G(s)\) in the process). Also, larger \(L(s)\) implies smaller sensitivity, \(S(s)\). The question arises, how do we define sensitivity? The system sensitivity is defined as the ratio of the percentage change in the system transfer function to the percentage change of the process transfer function. The system transfer function is
and therefore the sensitivity is defined as
In the limit, for small incremental changes, Equation (4.11) becomes
141. System sensitivity is the ratio of the change in the system transfer function to the change of a process transfer function (or parameter) for a small incremental change.
The sensitivity of the open-loop system to changes in the plant \(G(s)\) is equal to 1 . The sensitivity of the closed-loop is readily obtained by using Equation (4.12). The system transfer function of the closed-loop system is
Therefore, the sensitivity of the feedback system is
or
We find that the sensitivity of the system may be reduced below that of the openloop system by increasing \(L(s) = G_{c}(s)G(s)\) over the frequency range of interest. Note that \(S_{G}^{T}\) in Equation (4.12) is exactly the same as the sensitivity function \(S(s)\) given in Equation (4.5).
Often, we seek to determine \(S_{\alpha}^{T}\), where \(\alpha\) is a parameter within the transfer function, \(G(s)\). Using the chain rule yields
Very often, the transfer function of the system \(T(s)\) is a fraction of the form [1]
where \(\alpha\) is a parameter that may be subject to variation due to the environment. Then we may obtain the sensitivity to \(\alpha\) by rewriting Equation (4.11) as
where \(\alpha_{0}\) is the nominal value of the parameter.
An important advantage of feedback control systems is the ability to reduce the effect of the variation of parameters of a control system by adding a feedback loop. To obtain highly accurate open-loop systems, the components of the open-loop, \(G(s)\), must be selected carefully in order to meet the exact specifications. However, a closed-loop system allows \(G(s)\) to be less accurately specified because the sensitivity to changes or errors in \(G(s)\) is reduced by the loop gain \(L(s)\). This benefit of closed-loop systems is a profound advantage. A simple example will illustrate the value of feedback for reducing sensitivity.
142. EXAMPLE 4.1 Feedback amplifier
An amplifier used in many applications has a gain \(- K_{a}\), as shown in Figure 4.5(a). The output voltage is
We often add feedback using a potentiometer \(R_{p}\), as shown in Figure 4.5(b). The transfer function of the amplifier without feedback is
and the sensitivity to changes in the amplifier gain is
The block diagram model of the amplifier with feedback is shown in Figure 4.6, where
FIGURE 4.5
(a) Open-loop amplifier.
(b) Amplifier with feedback.
(a)
(b)
(a)
(b)
FIGURE 4.6 Block diagram model of feedback amplifier assuming \(R_{p} \gg R_{0}\) of the amplifier.
and
The closed-loop transfer function of the feedback amplifier is
The sensitivity of the closed-loop feedback amplifier is
If \(K_{a}\) is large, the sensitivity is low. For example, if
we have
or the magnitude is one-thousandth of the magnitude of the open-loop amplifier.
We shall return to the concept of sensitivity in subsequent chapters. These chapters will emphasize the importance of sensitivity in the design and analysis of control systems.
142.1. DISTURBANCE SIGNALS IN A FEEDBACK CONTROL SYSTEM
Many control systems are subject to extraneous disturbance signals that cause the system to provide an inaccurate output. Electronic amplifiers have inherent noise generated within the integrated circuits or transistors; radar antennas are subjected to wind gusts; and many systems generate unwanted distortion signals due to nonlinear elements. An important effect of feedback in a control system is the reduction of the effect of disturbance signals. A disturbance signal is an unwanted input signal that affects the output signal. The benefit of feedback systems is that the effect of distortion, noise, and unwanted disturbances can be effectively reduced.
143. Disturbance Rejection
When \(R(s) = N(s) = 0\), it follows from Equation (4.4) that
For a fixed \(G(s)\) and a given \(T_{d}(s)\), as the loop gain \(L(s)\) increases, the effect of \(T_{d}(s)\) on the tracking error decreases. In other words, the sensitivity function \(S(s)\) is small when the loop gain is large. We say that large loop gain leads to good disturbance rejection. More precisely, for good disturbance rejection, we require a large loop gain over the frequencies of interest associated with the expected disturbance signals.
In practice, the disturbance signals are often low frequency. When that is the case, we say that we want the loop gain to be large at low frequencies. This is equivalent to stating that we want to design the controller \(G_{c}(s)\) so that the sensitivity function \(S(s)\) is small at low frequencies.
As a specific example of a system with an unwanted disturbance, let us consider again the speed control system for a steel rolling mill [19]. The rolls, which process steel, are subjected to large load changes or disturbances. As a steel bar approaches the rolls (see Figure 4.7), the rolls are empty. However, when the bar engages in the rolls, the load on the rolls increases immediately to a large value. This loading effect can be approximated by a step change of disturbance torque.
The transfer function model of an armature-controlled DC motor with a load torque disturbance was determined in Example 2.5 and is shown in Figure 4.8, where it is assumed that \(L_{a}\) is negligible. Let \(R(s) = 0\) and examine \(E(s) = - \omega(s)\), for a disturbance \(T_{d}(s)\).
FIGURE 4.7
Steel rolling mill.
FIGURE 4.8 Open-loop speed control system (without tachometer feedback).
The change in speed due to the load disturbance is then
The steady-state error in speed due to the load torque, \(T_{d}(s) = D/s\), is found by using the final-value theorem. Therefore, for the open-loop system, we have
The closed-loop speed control system is shown in block diagram form in Figure 4.9. The closed-loop system is shown in signal-flow graph and block diagram form in Figure 4.10, where \(G_{1}(s) = K_{a}K_{m}/R_{a},G_{2}(s) = 1/\left( J_{s} + b \right)\), and \(H(s) = K_{t} + K_{b}/K_{a}\). The error, \(E(s) = - \omega(s)\), of the closed-loop system of Figure 4.10 is:
Then, if \(G_{1}G_{2}H(s)\) is much greater than 1 over the range of \(s\), we obtain the approximate result
Therefore, if \(G_{1}(s)H(s)\) is made sufficiently large, the effect of the disturbance can be decreased by closed-loop feedback. Note that
FIGURE 4.9
Closed-loop speed tachometer control system.
(a)
(b)
FIGURE 4.10 Closed-loop system. (a) Signal-flow graph model. (b) Block diagram model.
since \(K_{a} \gg K_{b}\). Thus, we seek a large amplifier gain, \(K_{a}\), while minimizing \(R_{a}\). The error for the system shown in Figure 4.10 is
and \(R(s) = \omega_{d}(s)\), the desired speed. Let \(R(s) = 0\) and examine \(\omega(s)\) yielding
The steady-state output is obtained by utilizing the final-value theorem, and we have
when the amplifier gain \(K_{a}\) is sufficiently high, we have
The ratio of closed-loop to open-loop steady-state speed output due to an undesired disturbance is
and is usually less than 0.02 .
144. Measurement Noise Attenuation
When \(R(s) = T_{d}(s) = 0\), it follows from Equation (4.4) that
As the loop gain \(L(s)\) decreases, the effect of \(N(s)\) on the tracking error decreases. In other words, the complementary sensitivity function \(C(s)\) is small when the loop gain \(L(s)\) is small. If we design \(G_{c}(s)\) such that \(L(s) \ll 1\), then the noise is attenuated because
We see that small loop gain leads to good noise attenuation. More precisely, for effective measurement noise attenuation, we need a small loop gain over the frequencies associated with the expected noise signals.
In practice, measurement noise signals are often high frequency. Thus we want the loop gain to be low at high frequencies. This is equivalent to a small complementary sensitivity function at high frequencies. The separation of disturbances (at low frequencies) and measurement noise (at high frequencies) is very fortunate because it gives the control system designer a way to approach the design process: the controller should be high gain at low frequencies and low gain at high frequencies. Remember that by low and high we mean that the loop gain magnitude is low/high at the various high/low frequencies. It is not always the case that the disturbances are low frequency or that the measurement noise is high frequency. If the frequency separation does not exist, the design process usually becomes more involved (for example, we may have to use notch filters to reject disturbances at known high frequencies). A noise signal that is prevalent in many systems is the noise generated by the measurement sensor. This noise, \(N(s)\), can be represented as shown in Figure 4.4. The effect of the noise on the output is
which is approximately
for large loop gain \(L(s) = G_{c}(s)G(s)\). This is consistent with the earlier discussion that smaller loop gain leads to measurement noise attentuation. Clearly, the designer must shape the loop gain appropriately.
The equivalency of sensitivity, \(S_{G}^{T}\), and the response of the closed-loop system tracking error to a reference input can be illustrated by considering Figure 4.4. The sensitivity of the system to \(G(s)\) is
The effect of the reference on the tracking error (with \(T_{d}(s) = 0\) and \(N(s) = 0\) ) is
In both cases, we find that the undesired effects can be alleviated by increasing the loop gain. Feedback in control systems primarily reduces the sensitivity of the system to parameter variations and the effect of disturbance inputs. Note that the measures taken to reduce the effects of parameter variations or disturbances are equivalent, and fortunately, they reduce simultaneously. As a final illustration, consider the effect of the noise on the tracking error,
We find that the undesired effects of measurement noise can be alleviated by decreasing the loop gain. Keeping in mind the relationship
the trade-off in the design process is evident.
144.1. CONTROL OF THE TRANSIENT RESPONSE
One of the most important characteristics of control systems is their transient response. The transient response is the response of a system as a function of time before steady-state. Because the purpose of control systems is to provide a desired response, the transient response often must be adjusted until it is satisfactory. If an open-loop control system does not provide a satisfactory response, then the loop transfer function, \(G_{c}(s)G(s)\), must be adjusted. To make this concept more comprehensible, consider a specific control system which may be operated in an open- or closed-loop manner. A speed control system, as shown in Figure 4.11, is often used in industrial processes to move materials and products. The transfer function of the open-loop system (without feedback) is given by
where
FIGURE 4.11
Open-loop speed control system (without feedback).
FIGURE 4.12
(a) Open-loop speed control system.
(b) Closed-loop speed control system.
(a)
(b)
In the case of a steel mill, the inertia of the rolls is quite large, and a large armature-controlled motor is required. If the steel rolls are subjected to a step command
the output response of the open-loop control system shown in Figure 4.12(a)
The transient speed change is then
If this transient response is too slow, we must choose another motor with a different time constant \(\tau_{1}\), if possible. However, because \(\tau_{1}\) is dominated by the load inertia, \(J\), it may not be possible to achieve much alteration of the transient response.
A closed-loop speed control system is easily obtained by using a tachometer to generate a voltage proportional to the speed, as shown in Figure 4.12(b). This voltage is subtracted from the potentiometer voltage and amplified as shown in Figure 4.12. The closed-loop transfer function is
The amplifier gain, \(K_{a}\), may be adjusted to meet the required transient response specifications. Also, the tachometer gain constant, \(K_{t}\), may be varied, if necessary.
The transient response to a step change in the input command is
FIGURE 4.13
The response of the open-loop and closed-loop speed control system when \(\tau = 10\) and \(K_{1}K_{a}K_{t} = 100\). The time to reach \(98\%\) of the final value for the open-loop and closed-loop system is 40 seconds and 0.4 seconds, respectively.
where \(p = \left( 1 + K_{a}K_{t}K_{1} \right)/\tau_{1}\). Because the load inertia is assumed to be very large, we alter the response by increasing \(K_{a}\). Thus, we have the approximate response
For a typical application, the open-loop pole might be \(1/\tau_{1} = 0.10\), whereas the closed-loop pole could be at least \(\left( K_{a}K_{t}K_{1} \right)/\tau_{1} = 10\), a factor of one hundred in the improvement of the speed of response. To attain the gain \(K_{a}K_{t}K_{1}\), the amplifier gain \(K_{a}\) must be reasonably large, and the armature voltage signal to the motor and its associated torque signal must be larger for the closed-loop than for the open-loop operation. Therefore, a higher-power motor will be required to avoid saturation of the motor. The responses of the closed-loop system and the open-loop system are shown in Figure 4.13. Note the rapid response of the closed-loop system relative to the open-loop system.
While we are considering this speed control system, it will be worthwhile to determine the sensitivity of the open- and closed-loop systems. As before, the sensitivity of the open-loop system to a variation in the motor constant or the potentiometer constant \(k_{2}\) is unity. The sensitivity of the closed-loop system to a variation in \(K_{m}\) is
Using the typical values given in the previous paragraph, we have
We find that the sensitivity is a function of \(s\) and must be evaluated for various values of frequency. This type of frequency analysis is straightforward but will be deferred until a later chapter. However, it is clearly seen that at a specific low frequency-for example, \(s = j\omega = j1\) - the magnitude of the sensitivity is approximately \(\left| S_{K_{m}}^{T} \right| \cong 0.1\).
144.2. STEADY-STATE ERROR
A feedback control system provides the engineer with the ability to adjust the transient response. In addition, as we have seen, the sensitivity of the system and the effect of disturbances can be reduced significantly. However, as a further requirement, we must examine and compare the final steady-state error for an open-loop and a closed-loop system. The steady-state error is the error after the transient response has decayed, leaving only the continuous response.
The error of the open-loop control system shown in Figure 4.3 is
when \(T_{d}(s) = 0\). Figure 4.4 shows the closed-loop system. When \(T_{d}(s) = 0\) and \(N(s) = 0\), and we let \(H(s) = 1\), the tracking error is given by
To calculate the steady-state error, we use the final-value theorem
Therefore, using a unit step input as a comparable input, we obtain for the openloop system
For the closed-loop system we have
The value of \(G_{c}(s)G(s)\) when \(s = 0\) is often called the DC gain and is normally greater than one. Therefore, the open-loop control system will usually have a steady-state error of significant magnitude. By contrast, the closed-loop system with a reasonably large DC loop gain \(L(0) = G_{c}(0)G(0)\) will have a small steady-state error.
Upon examination of Equation (4.49), we note that the open-loop control system can possess a zero steady-state error by adjusting and calibrating the systems DC gain, \(G_{c}(0)G(0)\), so that \(G_{c}(0)G(0) = 1\). Therefore, we may logically ask, What is the advantage of the closed-loop system in this case? To answer this question, we return to the concept of the sensitivity of the system to parameter uncertainty in \(G(s)\) and changes over time in those parameters. In the open-loop control system, we may calibrate the system so that \(G_{c}(0)G(0) = 1\), but during the operation of the system, it is inevitable that the parameters of \(G(s)\) will change due to environmental changes and that the DC gain of the system will no longer be equal to 1 . Because it is an open-loop control system, the steady-state error will not equal zero. By contrast, the closed-loop feedback system continually monitors the steady-state error and provides an actuating signal to reduce the steady-state error. Because systems are susceptible to parameter drift, environmental effects, and calibration errors, negative feedback provides benefits.
The advantage of the closed-loop system is that it reduces the steady-state error resulting from parameter changes and calibration errors. This may be illustrated by an example. Consider a unity feedback system with a process transfer function and controller
respectively. Which could represent a thermal control process, a voltage regulator, or a water-level control process. For a specific setting of the desired input variable, which may be represented by the normalized unit step input function, we have \(R(s) = 1/s\). Then the steady-state error of the open-loop system is, as in Equation (4.49),
when a consistent set of dimensional units is utilized for \(R(s)\) and \(KK_{a}\). The error for the closed-loop system is
where \(T(s) = G_{c}(s)G(s)/\left( 1 + G_{c}(s)G(s) \right)\). The steady-state error is
Then we have
For the open-loop control system, we would calibrate the system so that \(KK_{a} = 1\) and the steady-state error is zero. For the closed-loop system, we would set a large gain \(KK_{a}\). If \(KK_{a} = 100\), the closed-loop system steady-state error is \(e_{c}(\infty) = 1/101\).
If the process gain drifts or changes by \(\Delta K/K = 0.1\) (a \(10\%\) change), the magnitude of the open-loop steady-state error is \(\left| \Delta e_{o}(\infty) \right| = 0.1\). Then the percent change from the calibrated setting is
or \(10\%\). By contrast, the steady-state error of the closed-loop system, with \(\Delta K/K = 0.1\), is \(e_{c}(\infty) = 1/91\) if the gain decreases. Thus, the change is
and the relative change is
or \(0.11\%\). This is a significant improvement, since the closed-loop relative change is two orders of magnitude lower than that of the open-loop system.
144.3. THE COST OF FEEDBACK
The advantages of using feedback control have an attendant cost. The first cost of feedback is an increased number of components and complexity in the system. To add the feedback, it is necessary to consider several feedback components; the measurement component (sensor) is the key one. The sensor is often the most expensive component in a control system. Furthermore, the sensor introduces noise into the system.
The second cost of feedback is the loss of gain. For example, in a single-loop system, the loop gain is \(G_{c}(s)G(s)\) and is reduced to \(G_{c}(s)G(s)/\left( 1 + G_{c}(s)G(s) \right)\) in a unity negative feedback system. The closed-loop gain is smaller by a factor of \(1/\left( 1 + G_{c}(s)G(s) \right)\), which is exactly the factor that reduces the sensitivity of the system to parameter variations and disturbances. Usually, we have extra loop gain available, and we are more than willing to trade it for increased control of the system response.
The final cost of feedback is the introduction of the possibility of instability. Even when the open-loop system is stable, the closed-loop system may not be always stable.
The addition of feedback to dynamic systems causes more challenges for the designer. However, for most cases, the advantages far outweigh the disadvantages, and a feedback system is desirable. Therefore, it is necessary to consider the additional complexity and the problem of stability when designing a control system.
We want the output of the system, \(Y(s)\), to equal the input, \(R(s)\). We might ask, Why not set \(G_{c}(s)G(s) = 1\) ? (See Figure 4.3, assuming \(T_{d}(s) = 0\).) In other words, why not let \(G_{c}(s)\) be the inverse of the process \(G(s)\). The answer to this question becomes apparent once we recall that the process \(G(s)\) represents a real process and possesses dynamics that may not appear directly in the transfer function model. Additionally, the parameter in \(G(s)\) may be uncertain or vary with time. Hence, we cannot perfectly set \(G_{c}(s)G(s) = 1\) in practice. There are other issues that also arise, thus it is not advisable to design the open-loop control system in this fashion.
144.4. DESIGN EXAMPLES
In this section we present two illustrative examples: the English Channel boring machine and a blood pressure control problem during anesthesia. The English Channel boring machine example focuses on the closed-loop system response to disturbances. The example on blood pressure control is a more in-depth look at the control design problem. Since patient models in the form of transfer functions are difficult to obtain from basic biological and physical principles, a different approach using measured data is discussed. The positive impact of closed-loop feedback control is illustrated in the context of design.
145. EXAMPLE 4.2 English Channel boring machines
The tunnel under the English Channel from France to Great Britain is 23.5 miles long and is bored 250 feet below sea level at its lowest-point. The tunnel is a critical link between Europe and Great Britain, making it possible to travel from London to Paris in 2 hours and 15 minutes using the Channel Tunnel Rail Link (known as High Speed 1).
The boring machines, operating from both ends of the channel, bored toward the middle. To link up accurately in the middle of the channel, a laser guidance system kept the machines precisely aligned. A model of the boring machine control is shown in Figure 4.14, where \(Y(s)\) is the actual angle of direction of travel of the boring machine and \(R(s)\) is the desired angle. The effect of load on the machine is represented by the disturbance, \(T_{d}(s)\).
The design objective is to select the gain \(K\) so that the response to input angle changes is desirable while we maintain minimal error due to the disturbance. The output due to the two inputs is
Thus, to reduce the effect of the disturbance, we wish to set the gain greater than 10 . When we select \(K = 100\) and let the disturbance be zero, we have the step response for a unit step input \(r(t)\), as shown in Figure 4.15. When the input \(r(t) = 0\) and we determine the response to the unit step disturbance, we obtain \(y(t)\) as shown in Figure 4.15. The effect of the disturbance is quite small. If we set the gain \(K\) equal to 20 , we obtain the responses of \(y(t)\) due to a unit step input \(r(t)\) and disturbance
FIGURE 4.14 A block diagram model of a boring machine control system.
FIGURE 4.15
The response \(y(t)\) to a unit input step input (solid line) and a unit disturbance step (dashed line) with \(T_{d}(s) = 1/s\) for \(K = 100\).
FIGURE 4.16
The response \(y(t)\) for a unit step input (solid line) and for a unit step disturbance (dashed line) for \(K = 20\).
\(T_{d}(t)\) displayed together in Figure 4.16. When \(K = 100\), the percent overshoot is \(22\%\) and the settling time is \(0.7\text{ }s\). When \(K = 20\), the percent overshoot is \(3.9\%\) and the settling time is \(0.9\text{ }s\).
The steady-state error of the system to a unit step input \(R(s) = 1/s\) is
The steady-state value of \(y(t)\) when the disturbance is a unit step, \(T_{d}(s) = 1/s\), and the desired value is \(r(t) = 0\) is
Thus, the steady-state value is 0.01 and 0.05 for \(K = 100\) and 20, respectively. Finally, we examine the sensitivity of the system to a change in the process \(G(s)\) using Equation (4.12). Then
For low frequencies \((|s| < 1)\), the sensitivity can be approximated by
where \(K \geq 20\). Thus, the sensitivity of the system is reduced by increasing the gain, \(K\). In this case, we choose \(K = 20\) for a reasonable design compromise.
146. EXAMPLE 4.3 Blood pressure control during anesthesia
The objectives of anethesia are to eliminate pain, awareness, and natural reflexes so that surgery can be conducted safely. Before about 150 years ago, alcohol, opium, and cannabis were used to achieve these goals, but they proved inadequate [23]. Pain relief was insufficient both in magnitude and duration; too little pain medication and the patient felt great pain, too much medication and the patient died or became comatose. In the 1850 s ether was used successfully in the United States in tooth extractions, and shortly thereafter other means of achieving unconsciousness safely were developed, including the use of chloroform and nitrous oxide.
In a modern operating room, the depth of anesthesia is the responsibility of the anesthetist. Many vital parameters, such as blood pressure, heart rate, temperature, blood oxygenation, and exhaled carbon dioxide, are controlled within acceptable bounds by the anesthetist. Of course, to ensure patient safety, adequate anesthesia must be maintained during the entire surgical procedure. Any assistance that the anesthetist can obtain automatically will increase the safety margins by freeing the anesthetist to attend to other functions not easily automated. This is an example of human computer interaction for the overall control of a process. Clearly, patient safety is the ultimate objective. Our control goal then is to develop an automated system to regulate the depth of anesthesia. This function is amenable to automatic control and in fact is in routine use in clinical applications [24, 25].
We consider how to measure the depth of anesthesia. Many anesthetists regard mean arterial pressure (MAP) as the most reliable measure of the depth of anesthesia [26]. The level of the MAP serves as a guide for the delivery of inhaled anesthesia. Based on clinical experience and the procedures followed by the anesthetist, we determine that the variable to be controlled is the mean arterial pressure.
The elements of the control system design process emphasized in this example are illustrated in Figure 4.17. From the control system design perspective, the control goal can be stated in more concrete terms:
147. Control Goal
Regulate the mean arterial pressure to any desired set-point and maintain the prescribed set-point in the presence of unwanted disturbances. Topics emphasized in this example
FIGURE 4.17 Elements of the control system design process emphasized in the blood pressure control example.
Associated with the stated control goal, we identify the variable to be controlled:
148. Variable to Be Controlled
149. Mean arterial pressure (MAP).
Because it is our desire to develop a system that will be used in clinical applications, it is essential to establish realistic design specifications. In general terms the control system should have minimal complexity while satisfying the control specifications. Minimal complexity translates into increased system reliability and decreased cost.
The closed-loop system should respond rapidly and smoothly to changes in the MAP set-point (made by the anesthetist) without excessive overshoot. The closedloop system should minimize the effects of unwanted disturbances. There are two important categories of disturbances: surgical disturbances, such as skin incisions and measurement errors, such as calibration errors and random stochastic noise. For example, a skin incision can increase the MAP rapidly by \(10mmHg\) [26]. Finally, since we want to apply the same control system to many different patients and we cannot (for practical reasons) have a separate model for each patient, we must have a closed-loop system that is insensitive to changes in the process parameters (that is, it meets the specifications for many different people).
Based on clinical experience [24], we can explicitly state the control specifications as follows:
150. Control Design Specifications
DS1 Settling time less than 20 minutes for a 10% step change from the MAP set-point.
DS2 Percent overshoot less than \(15\%\) for a \(10\%\) step change from the MAP set-point.
DS3 Zero steady-state tracking error to a step change from the MAP set-point.
DS4 Zero steady-state error to a step surgical disturbance input (of magnitude \(|d(t)| \leq 50\) ) with a maximum response less than \(\pm 5\%\) of the MAP set-point.
DS5 Minimum sensitivity to process parameter changes.
We cover the notion of percent overshoot (DS1) and settling time (DS2) more thoroughly in Chapter 5. They fall more naturally in the category of system performance. The remaining three design specifications, DS3-DS5, covering steady-state tracking errors (DS3), disturbance rejection (DS4), and system sensitivity to parameter changes (DS5) are the main topics of this chapter. The last specification, DS5, is somewhat vague; however, this is a characteristic of many real-world specifications. In the system configuration, Figure 4.18, we identify the major system elements as the controller, anesthesia pump/vaporizer, sensor, and patient.
The system input \(R(s)\) is the desired mean arterial pressure change, and the output \(Y(s)\) is the actual pressure change. The difference between the desired and the measured blood pressure change forms a signal used by the controller to determine value settings to the pump/vaporizer that delivers anesthesia vapor to the patient.
FIGURE 4.18 Blood pressure control system configuration. The model of the pump/vaporizer depends directly on the mechanical design. We will assume a simple pump/vaporizer, where the rate of change of the output vapor is equal to the input valve setting, or
The transfer function of the pump is thus given by
This is equivalent to saying that, from an input-output perspective, the pump has the impulse response
Developing an accurate model of a patient is much more involved. Because the physiological systems in the patient (especially in a sick patient) are not easily modeled, a modeling procedure based on knowledge of the underlying physical processes is not practical. Even if such a model could be developed, it would, in general, be a nonlinear, time-varying, multi-input, multi-output model. This type of model is not directly applicable here in our linear, time-invariant, single-input, single-output system setting.
On the other hand, if we view the patient as a system and take an input-output perspective, we can use the familiar concept of an impulse response. Then if we restrict ourselves to small changes in blood pressure from a given set-point (such as \(100mmHg\) ), we might make the case that in a small region around the set-point the patient behaves in a linear time-invariant fashion. This approach fits well into our requirement to maintain the blood pressure around a given set-point (or baseline). The impulse response approach to modeling the patient response to anesthesia has been used successfully in the past [27].
Suppose that we take a black-box approach and obtain the impulse response in Figure 4.19 for a hypothetical patient. Notice that the impulse response initially has a time delay. This reflects the fact that it takes a finite amount of time for the patient MAP to respond to the infusion of anesthesia vapor. We ignore the timedelay in our design and analysis, but we do so with caution. In subsequent chapters we will learn to handle time delays. We keep in mind that the delay does exist and should be considered in the analysis at some point.
A reasonable fit of the data shown in Figure 4.19 is given by
where \(p = 2\) and time \((t)\) is measured in minutes. Different patients are associated with different values of the parameter \(p\). The corresponding transfer function is
For the sensor we assume a perfect noise-free measurement and
Therefore, we have a unity feedback system. FIGURE 4.19 Mean arterial pressure (MAP) impulse response for a hypothetical patient.
A good controller for this application is a proportional-integral-derivative (PID) controller:
where \(K_{P},K_{D}\), and \(K_{I}\) are the controller gains to be determined to satisfy all design specifications. The selected key parameters are as follows:
151. Select Key Tuning Parameters
Controller gains \(K_{P},K_{D}\), and \(K_{I}\).
We begin the analysis by considering the steady-state errors. The tracking error (shown in Figure 4.18 with \(T_{d}(s) = 0\) and \(N(s) = 0\) ) is
or
Using the final-value theorem, we determine that the steady-state tracking error is
where \(R(s) = R_{0}/s\) is a step input of magnitude \(R_{0}\). Therefore,
With a PID controller, we expect a zero steady-state tracking error (to a step input) for any nonzero values of \(K_{P},K_{D}\), and \(K_{I}\). The integral term, \(K_{I}/s\), in the PID controller is the reason that the steady-state error to a unit step is zero. Thus design specification DS3 is satisfied.
When considering the effect of a step disturbance input, we let \(R(s) = 0\) and \(N(s) = 0\). We want the steady-state output \(Y(s)\) to be zero for a step disturbance. The transfer function from the disturbance \(T_{d}(s)\) to the output \(Y(s)\) is
When
we find that
Therefore,
Thus a step disturbance of magnitude \(D_{0}\) will produce no output in the steady-state, as desired.
The sensitivity of the closed-loop transfer function to changes in \(p\) is given by
We compute \(S_{p}^{G}\) as follows:
and
Therefore,
Table 4.1 PID Controller Gains and System Performance Results
| PID | $$K_{\mathbf{P}}$$ | $$\mathbf{K}_{\mathbf{D}}$$ | $$\mathbf{K}_{\mathbf{I}}$$ | $$\begin |
\text{~}\text{Input response}\text{~} \\
\text{~}\text{overshoot (\textbackslash\%)}\text{~}
\end{matrix}$$ | $$\begin{matrix}
\text{~}\text{Settling}\text{~} \\
\text{~}\text{time (min)}\text{~}
\end{matrix}$$ | $$\begin{matrix}
\text{~}\text{Disturbance response}\text{~} \\
\text{~}\text{overshoot (\textbackslash\%)}\text{~}
\end{matrix}$$ |
| 1 | 6 | 4 | 1 | 14.0 | 10.9 | 5.25 |
| 2 | 5 | 7 | 2 | 14.2 | 8.7 | 4.39 |
| 3 | 6 | 4 | 4 | 39.7 | 11.1 | 5.16 |
We must evaluate the sensitivity function \(S_{p}^{T}\), at various values of frequency. For low frequencies we can approximate the system sensitivity \(S_{p}^{T}\) by
So at low frequencies and for a given \(p\) we can reduce the system sensitivity to variations in \(p\) by increasing the PID gain, \(K_{I}\). Suppose that three PID gain sets have been proposed, as shown in Table 4.1. With \(p = 2\) and the PID gains given as the cases 1-3 in Table 4.1, we can plot the magnitude of the sensitivity \(S_{p}^{T}\) as a function of frequency for each PID controller. The result is shown in Figure 4.20. We see that by using the PID 3 controller with the gains \(K_{P} = 6,K_{D} = 4\), and \(K_{I} = 4\), we have the smallest system sensitivity (at low frequencies) to changes in the process parameter, \(p\). PID 3 is the controller with the largest gain \(K_{I}\). As the frequency increases we see in Figure 4.20 that the sensitivity increases, and that PID 3 has the highest peak sensitivity.
FIGURE 4.20
System sensitivity to variations in the parameter \(p\).
FIGURE 4.21
Mean arterial pressure (MAP) step input response with \(R(s) = 10/s\).
Now we consider the transient response. Suppose we want to reduce the MAP by a \(10\%\) step change. The associated input is
The step response for each PID controller is shown in Figure 4.21. PID 1 and PID 2 meet the settling time and overshoot specifications; however PID 3 has excessive overshoot. The overshoot is the amount the system output exceeds the desired steadystate response. In this case the desired steady-state response is a \(10\%\) decrease in the baseline MAP. When a \(15\%\) overshoot is realized, the MAP is decreased by \(11.5\%\), as illustrated in Figure 4.21. The settling time is the time required for the system output to settle within a certain percentage (for example, \(2\%\) ) of the desired steady-state output amplitude. We cover the notions of overshoot and settling time more thoroughly in Chapter 5. The overshoot and settling times are summarized in Table 4.1.
We conclude the analysis by considering the disturbance response. From previous analysis we know that the transfer function from the disturbance input \(T_{d}(s)\) to the output \(Y(s)\) is
To investigate design specification DS4, we compute the disturbance step response with
FIGURE 4.22 Mean arterial pressure (MAP) disturbance step response.
This is the maximum magnitude disturbance \(\left( \left| T_{d}(t) \right| = D_{0} = 50 \right)\). Since any step disturbance of smaller magnitude (that is, \(\left| T_{d}(t) \right| = D_{0} < 50\) ) will result in a smaller maximum output response, we need only to consider the maximum magnitude step disturbance input when determining whether design specification DS4 is satisfied.
The unit step disturbance for each PID controller is shown in Figure 4.22. Controller PID 2 meets design specification DS4 with a maximum response less than \(\pm 5\%\) of the MAP set-point, while controllers PID 1 and 3 nearly meet the specification. The peak output values for each controller are summarized in Table 4.1.
In summary, given the three PID controllers, we would select PID 2 as the controller of choice. It meets all the design specifications while providing a reasonable insensitivity to changes in the plant parameter.
151.1. CONTROL SYSTEM CHARACTERISTICS USING CONTROL DESIGN SOFTWARE
In this section, the advantages of feedback will be illustrated with two examples. In the first example, we will introduce feedback control to a speed tachometer system in an effort to reject disturbances. The tachometer speed control system example can be found in Section 4.5 . The reduction in system sensitivity to process variations, adjustment of the transient response, and reduction in steady-state error will be demonstrated using the English Channel boring machine example of Section 4.8.
152. EXAMPLE 4.4 Speed control system
The open-loop block diagram description of the armature-controlled DC motor with a load torque disturbance \(T_{d}(s)\) is shown in Figure 4.8. The values for the various parameters are given in Table 4.2. We have two inputs to our system, \(V_{a}(s)\) and Table 4.2 Tachometer Control System Parameters
| $$\mathbf{R}_{\mathbf{a}}$$ | $$\mathbf{K}_{\mathbf{m}}$$ | $$J$$ | $$b$$ | $$\mathbf{K}_{b}$$ | $$\mathbf{K}_{\mathbf{a}}$$ | $$\mathbf{K}_{\mathbf{t}}$$ |
| $$1\Omega$$ | $$10Nm/A$$ | $$2\text{ }kg{\text{ }m}^{2}$$ | $$0.5Nms$$ | $$0.1Vs$$ | 54 | $$1Vs$$ |
\(T_{d}(s)\). Relying on the principle of superposition, which applies to our linear system, we consider each input separately. To investigate the effects of disturbances on the system, we let \(V_{a}(s) = 0\) and consider only the disturbance \(T_{d}(s)\). Conversely, to investigate the response of the system to a reference input, we let \(T_{d}(s) = 0\) and consider only the input \(V_{a}(s)\).
The closed-loop speed tachometer control system block diagram is shown in Figure 4.9. The values for \(K_{a}\) and \(K_{t}\) are given in Table 4.2.
If our system displays good disturbance rejection, then we expect the disturbance \(T_{d}(s)\) to have a small effect on the output \(\omega(s)\). Consider the open-loop system in Figure 4.8 first. We can compute the transfer function from \(T_{d}(s)\) to \(\omega(s)\) and evaluate the output response to a unit step disturbance (that is, \(T_{d}(s) = 1/s\) ). The time response to a unit step disturbance is shown in Figure 4.23(a). The script shown in Figure 4.23(b) is used to analyze the open-loop speed tachometer system.
(a)
FIGURE 4.23
Analysis of the open-loop speed control system.
(a) Response.
(b) m-file script.
The open-loop transfer function (from Equation (4.26)) is
where sys_o represents the open-loop transfer function in the script. Since our desired value of \(\omega(t)\) is zero (remember that \(V_{a}(s) = 0\) ), the steady-state error is just the final value of \(\omega(t)\), which we denote by \(\omega_{o}(t)\) to indicate open-loop. The steady-state error, shown on the plot in Figure 4.23(a), is approximately the value of the speed when \(t = 7\text{ }s\). We can obtain an approximate value of the steady-state error by looking at the last value in the output vector \(y_{o}\), which we computed in the process of generating the plot in Figure 4.23(a). The approximate steady-state value of \(\omega_{0}\) is
The plot verifies that we have reached steady state.
In a similar fashion, we begin the closed-loop system analysis by computing the closed-loop transfer function from \(T_{d}(s)\) to \(\omega(s)\) and then generating the timeresponse of \(\omega(t)\) to a unit step disturbance input. The output response and the script are shown in Figure 4.24. The closed-loop transfer function from the disturbance input (from Equation (4.30)) is
As before, the steady-state error is just the final value of \(\omega(t)\), which we denote by \(\omega_{c}(t)\) to indicate that it is a closed-loop. The steady-state error is shown on the plot in Figure 4.24(a). We can obtain an approximate value of the steady-state error by looking at the last value in the output vector \(y_{c}\), which we computed in the process of generating the plot in Figure 4.24(a). The approximate steady-state value of \(\omega\) is
We generally expect that \(\omega_{c}(\infty)/\omega_{o}(\infty) < 0.02\). In this example, the ratio of closedloop to open-loop steady-state speed output due to a unit step disturbance input is
We have achieved a remarkable improvement in disturbance rejection. It is clear that the addition of the negative feedback loop reduced the effect of the disturbance on the output. This demonstrates the disturbance rejection property of closed-loop feedback systems. FIGURE 4.24
Analysis of the closed-loop speed control system.
(a) Response.
(b) m-file script.
(a)
(b)
153. EXAMPLE 4.5 English Channel boring machines
The block diagram description of the English Channel boring machines is shown in Figure 4.14. The transfer function of the output due to the two inputs is (Equation (4.57))
The effect of the control gain, \(K\), on the transient response is shown in Figure 4.25 along with the script used to generate the plots. Comparing the two plots in parts (a) and (b), it is apparent that decreasing \(K\) decreases the overshoot. Although it is not as obvious from the plots in Figure 4.25, it is also true that decreasing \(K\) increases
(a)
(b)
FIGURE 4.25
The response to a step input when (a) \(K = 100\) and (b) \(K = 20\). (c) m-file script.
Closed-loop transfer functions.
Choose time interval.
Create subplots with \(x\) and \(y\) axis labels.
(c)
the settling time. This can be verified by taking a closer look at the data used to generate the plots. This example demonstrates how the transient response can be altered by feedback control gain, \(K\). Based on our analysis thus far, we would prefer to use \(K = 20\). Other considerations must be taken into account before we can establish the final design. Disturbance Response for \(K = 100\)
(a)
(b)
Before making the final choice of \(K\), it is important to consider the system response to a unit step disturbance, as shown in Figure 4.26. We see that increasing \(K\) reduces the steady-state response of \(y(t)\) to the step disturbance. The steady-state value of \(y(t)\) is 0.05 and 0.01 for \(K = 20\) and 100 , respectively. The steady-state Table 4.3 Response of the Boring Machine Control System
for \(K = 20\) and \(K = 100\)
| $$\mathbf{K} = \mathbf{20}$$ | $$\mathbf{K} = \mathbf{100}$$ | |
| Step Response | ||
| \(P.O\). | $$4%$$ | $$22%$$ |
| $$T_{S}$$ | $$1.0\text{ }s$$ | $$0.7\text{ }s$$ |
| Disturbance Response | ||
| $$\ e_{SS}$$ | $$5%$$ | $$1%$$ |
errors, percent overshoot, and settling times (2% criteria) are summarized in Table 4.3. The steady-state values are predicted from the final-value theorem for a unit disturbance input as follows:
If our only design consideration is disturbance rejection, we would prefer to use \(K = 100\).
We have just experienced a common trade-off situation in control system design. In this particular example, increasing \(K\) leads to better disturbance rejection, whereas decreasing \(K\) leads to better performance (that is, less overshoot). The final decision on how to choose \(K\) rests with the designer. Although control design software can certainly assist in the control system design, it cannot replace the engineer's decision-making capability and intuition.
The final step in the analysis is to look at the system sensitivity to changes in the process. The system sensitivity is given by (Equation 4.60),
We can compute the values of \(S_{G}^{T}(s)\) for different values of \(s\) and generate a plot of the system sensitivity. For low frequencies, we can approximate the system sensitivity by
Increasing the gain \(K\) reduces the system sensitivity. The system sensitivity plots when \(s = j\omega\) are shown in Figure 4.27 for \(K = 20\).
153.1. SEQUENTIAL DESIGN EXAMPLE: DISK DRIVE READ SYSTEM
The design of a disk drive system is an exercise in compromise and optimization. The disk drive must accurately position the head reader while being able to reduce the effects of parameter changes and external shocks and vibrations. FIGURE 4.27
(a) System sensitivity to plant variations \((s = j\omega)\). (b) m-file script.
(a)
(b)
The mechanical arm and flexure will resonate at frequencies that may be caused by excitations such as a shock to a notebook computer. Disturbances to the operation of the disk drive include physical shocks, wear or wobble in the spindle bearings, and parameter changes due to component changes. In this section, we will examine the performance of the disk drive system in response to disturbances and changes in system parameters. In addition, we examine the steady-state error of the system for a step command and the transient response as the amplifier gain \(K_{a}\) is adjusted.
Let us consider the system shown in Figure 4.28. This closed-loop system uses an amplifier with a variable gain as the controller, and the transfer functions are FIGURE 4.28
Control system for disk drive head reader.
FIGURE 4.29
Disk drive head control system with the typical parameters.
shown in Figure 4.29. First, we will determine the steady states for a unit step input, \(R(s) = 1/s\), when \(T_{d}(s) = 0\). When \(H(s) = 1\), we obtain
Therefore,
Then the steady-state error is \(e(\infty) = 0\) for a step input. This performance is obtained in spite of changes in the system parameters.
Now let us determine the transient performance of the system as \(K_{a}\) is adjusted. The closed-loop transfer function (with \(T_{d}(s) = 0\) ) is
Using the script shown in Figure 4.30(a), we obtain the response of the system for \(K_{a} = 10\) and \(K_{a} = 80\), shown in Figure 4.30(b). Clearly, the system is faster in responding to the command input when \(K_{a} = 80\), but the response is unacceptably oscillatory.
Now let us determine the effect of the disturbance \(T_{d}(s) = 1/s\) when \(R(s) = 0\). We wish to decrease the effect of the disturbance to an insignificant level. Using FIGURE 4.30
Closed-loop response. (a) m-file script. (b) Step response for \(K_{a} = 10\) and \(K_{a} = 80\).
(a)
(b)
the system of Figure 4.29, we obtain the response \(Y(s)\) for the input \(T_{d}(s)\) when \(K_{a} = 80\) as
Using the script shown in Figure 4.31(a), we obtain the response of the system when \(K_{a} = 80\) and \(T_{d}(s) = 1/s\), as shown in Figure 4.31(b). In order to further reduce the effect of the disturbance, we would need to raise \(K_{a}\) above 80 . However, the response to a step command \(r(t) = 1,t > 0\) is unacceptably oscillatory. In the next chapter, we attempt to determine the best value for \(K_{a}\), given our requirement for a quick, yet nonoscillatory response. FIGURE 4.31
Disturbance
step response.
(a) m-file script.
(b) Disturbance
response for
\(K_{a} = 80\).
\(Ka = 80\);
\(nf = \lbrack 5000\rbrack;df = \lbrack 111000\rbrack;\) sysf=tf(nf,df);
\(ng = \lbrack 1\rbrack\); dg=[1 20 0]; sysg=tf(ng,dg);
sys=feedback(sysg,Ka*sysf);
sys \(= -\) sys;
\(t = \lbrack 0:0.01:2\rbrack\);
\(y = step(\) sys, \(t)\);
plot(t,y), grid
ylabel('y(t)'), xlabel('Time (s)'), grid
Select \(K_{a}\).
Disturbance enters summer with a negative sign.
(a)
(b)
153.2. SUMMARY
The fundamental reasons for using feedback, despite its cost and additional complexity, are as follows:
-
Decrease in the sensitivity of the system to variations in the parameters of the process.
-
Improvement in the rejection of the disturbances.
-
Improvement in the attenuation of measurement noise.
-
Improvement in the reduction of the steady-state error of the system.
-
Ease of control and adjustment of the transient response of the system.
The loop gain \(L(s) = G_{c}(s)G(s)\) plays a fundamental role in control system analysis. Associated with the loop gain we can define the sensitivity and complementary sensitivity functions as
respectively. The tracking error is given by
In order to minimize the tracking error, \(E(s)\), we desire to make \(S(s)\) and \(C(s)\) small. Because the sensitivity and complementary sensitivity functions satisfy the constraint
we are faced with the fundamental trade-off in control system design between rejecting disturbances and reducing sensitivity to plant changes on the one hand, and attenuating measurement noise on the other hand.
Feedback control systems possess many beneficial characteristics. Thus, it is not surprising that there is a multitude of feedback control systems in industry, government, and nature.
154. 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 4.32 as specified in the various problem statements.
FIGURE 4.32 Block diagram for the Skills Check.
In the following True or False and Multiple Choice problems, circle the correct answer.
- One of the most important characteristics of control systems is their transient response.
True or False
- The system sensitivity is the ratio of the change in the system transfer function to the change of a process transfer function for a small incremental change.
True or False
- A primary advantage of an open-loop control system is the ability to reduce the system's sensitivity.
True or False
- A disturbance is a desired input signal that affects the system output signal.
True or False 5. An advantage of using feedback is a decreased sensitivity of the system to variations in the parameters of the process.
True or False
- The loop transfer function of the system in Figure 4.32 is
The sensitivity of the closed-loop system to small changes in \(\tau\) is:
a. \(S_{\tau}^{T}(s) = - \frac{\tau s}{\tau s + 60}\)
b. \(S_{\tau}^{T}(s) = \frac{\tau}{\tau s + 10}\)
c. \(S_{\tau}^{T}(s) = \frac{\tau}{\tau s + 60}\)
d. \(S_{\tau}^{T}(s) = - \frac{\tau s}{\tau s + 10}\)
- Consider the two systems in Figure 4.33 .
(i)
(ii)
FIGURE 4.33 Two feedback systems with gains \(K_{1}\) and \(K_{2}\).
These systems have the same transfer function when \(K_{1} = K_{2} = 100\). Which system is most sensitive to variations in the parameter \(K_{1}\) ? Compute the sensitivity using the nominal values \(K_{1} = K_{2} = 100\).
a. System (i) is more sensitive and \(S_{K_{1}}^{T} = 0.01\)
b. System (ii) is more sensitive and \(S_{K_{1}}^{T} = 0.1\)
c. System (ii) is more sensitive and \(S_{K_{1}}^{T} = 0.01\)
d. Both systems are equally sensitive to changes in \(K_{1}\).
- Consider the closed-loop transfer function
where \(A_{1},A_{2},A_{3}\), and \(A_{4}\) are constants. Compute the sensitivity of the system to variations in the parameter \(k\).
a. \(S_{k}^{T} = \frac{k\left( A_{2}A_{3} - A_{1}A_{4} \right)}{\left( A_{3} + kA_{4} \right)\left( A_{1} + kA_{2} \right)}\)
b. \(S_{k}^{T} = \frac{k\left( A_{2}A_{3} + A_{1}A_{4} \right)}{\left( A_{3} + kA_{4} \right)\left( A_{1} + kA_{2} \right)}\) c. \(S_{k}^{T} = \frac{k\left( A_{1} + kA_{2} \right)}{\left( A_{3} + kA_{4} \right)}\)
d. \(S_{k}^{T} = \frac{k\left( A_{3} + kA_{4} \right)}{\left( A_{1} + kA_{2} \right)}\)
Consider the block diagram in Figure 4.32 for Problems 9-12 where \(G_{c}(s) = K_{1}\) and \(G(s) = \frac{K}{s + K_{1}K_{2}}\). \(T_{d}(s) = 0\).
-
The closed-loop transfer function is:
a. \(T(s) = \frac{KK_{1}^{2}}{s + K_{1}\left( K + K_{2} \right)}\)
b. \(T(s) = \frac{KK_{1}}{s + K_{1}\left( K + K_{2} \right)}\)
c. \(T(s) = \frac{KK_{1}}{s - K_{1}\left( K + K_{2} \right)}\)
d. \(T(s) = \frac{KK_{1}}{s^{2} + K_{1}Ks + K_{1}K_{2}}\) -
The sensitivity \(S_{K_{1}}^{T}\) of the closed-loop system to variations in \(K_{1}\) is:
a. \(S_{K_{1}}^{T}(s) = \frac{Ks}{\left( s + K_{1}\left( K + K_{2} \right) \right)^{2}}\)
b. \(S_{K_{1}}^{T}(s) = \frac{2s}{s + K_{1}\left( K + K_{2} \right)}\)
c. \(S_{K_{1}}^{T}(s) = \frac{s}{s + K_{1}\left( K + K_{2} \right)}\)
d. \(S_{K_{1}}^{T}(s) = \frac{K_{1}\left( s + K_{1}K_{2} \right)}{\left( s + K_{1}\left( K + K_{2} \right) \right)^{2}}\) -
The sensitivity \(S_{K}^{T}\) of the closed-loop system to variations in \(K\) is:
a. \(S_{K}^{T}(s) = \frac{s + K_{1}K_{2}}{s + K_{1}\left( K + K_{2} \right)}\)
b. \(S_{K}^{T}(s) = \frac{Ks}{\left( s + K_{1}\left( K + K_{2} \right) \right)^{2}}\)
c. \(S_{K}^{T}(s) = \frac{s + KK_{1}}{s + K_{1}K_{2}}\)
d. \(S_{K}^{T}(s) = \frac{K_{1}\left( s + K_{1}K_{2} \right)}{\left( s + K_{1}\left( K + K_{2} \right) \right)^{2}}\) -
The steady-state tracking error to a unit step input \(R(s) = 1/s\) with \(T_{d}(s) = 0\) is:
a. \(e_{ss} = \frac{K}{K + K_{2}}\)
b. \(e_{ss} = \frac{K_{2}}{K + K_{2}}\)
c. \(e_{ss} = \frac{K_{2}}{K_{1}\left( K + K_{2} \right)}\)
d. \(e_{ss} = \frac{K_{1}}{K + K_{2}}\) Consider the block diagram in Figure 4.32 for Problems 13-14 with \(G_{c}(s) = K\) and \(G(s) = \frac{b}{s + 1}\). -
The sensitivity \(S_{b}^{T}\) is:
a. \(S_{b}^{T} = \frac{1}{s + Kb + 1}\)
b. \(S_{b}^{T} = \frac{s + 1}{s + Kb + 1}\)
c. \(S_{b}^{T} = \frac{s + 1}{s + Kb + 2}\)
d. \(S_{b}^{T} = \frac{s}{s + Kb + 2}\) -
Compute the minimal value of \(K\) so that the steady-state error due to a unit step disturbance is less than \(10\%\).
a. \(K = 1 - \frac{1}{b}\)
b. \(K = b\)
c. \(K = 10 - \frac{1}{b}\)
d. The steady-state error is \(\infty\) for any \(K\). -
A process is designed to follow a desired path described by
where \(r(t)\) is the desired response and \(u(t)\) is a unit step function. Consider the unity feedback system in Figure 4.32. Compute the steady-state error \((E(s) = R(s) - Y(s)\) with \(T_{d}(s) = 0\) ) when the loop transfer function is
a. \(e_{ss} = \lim_{t \rightarrow \infty}\mspace{2mu} e(t) \rightarrow \infty\)
b. \(e_{ss} = \lim_{t \rightarrow \infty}\mspace{2mu} e(t) = 1\)
c. \(e_{ss} = \lim_{t \rightarrow \infty}\mspace{2mu} e(t) = 0.5\)
d. \(e_{ss} = \lim_{t \rightarrow \infty}\mspace{2mu} e(t) = 0\)
In the following Word Match problems, match the term with the definition by writing the correct letter in the space provided.
a. Instability
b. Steady-state error
c. System sensitivity
d. Components
An unwanted input signal that affects the system output signal.
The difference between the desired output, \(R(s)\), and the actual output, \(Y(s)\).
A system without feedback that directly generates the output in response to an input signal.
The error when the time period is large and the transient response has decayed leaving the continuous response. e. Disturbance signal
f. Transient response
g. Complexity
h. Error signal
i. Closed-loop system
j. Loss of gain
k. Open-loop system
The ratio of the change in the system transfer function to the change of a process transfer function (or parameter) for a small incremental change.
The response of a system as a function of time.
A system with a measurement of the output signal and a comparison with the desired output to generate an error signal that is applied to the actuator.
A measure of the structure, intricateness, or behavior of a system that characterizes the relationships and interactions between various components.
The parts, subsystems, or subassemblies that comprise a total system.
An attribute of a system that describes a tendency of the system to depart from the equilibrium condition when initially displaced.
A reduction in the amplitude of the ratio of the output signal to the input signal through a system, usually measured in decibels.
155. EXERCISES
E4.1 A digital audio system is designed to minimize the effect of disturbances as shown in Figure E4.1. As an approximation, we may represent \(G(s) = K_{2}\).
(a) Calculate the sensitivity of the system due to \(K_{2}\). (b) Calculate the effect of the disturbance noise \(T_{d}(s)\) on \(V_{o}(s)\). (c) What value would you select for \(K_{1}\) to minimize the effect of the disturbance?
E4.2 A closed-loop system is used to track the sun to obtain maximum power from a photovoltaic array. The tracking system may be represented by a unity feedback control system and
where $\tau = 3\ $ s nominally. (a) Calculate the sensitivity of this system for a small change in \(\tau\). (b) Calculate the time constant of the closed-loop system response.
Answers: \(S = - 3s/(3s + 101);\tau_{c} = 3/101\text{ }s\)
E4.3 A robotic arm and camera could be used to pick fruit, as shown in Figure E4.3(a). The camera is used to close the feedback loop to a microcomputer, which controls the arm \(\lbrack 8,9\rbrack\). The transfer function for the process is
(a) Calculate the expected steady-state error of the gripper for a step command \(A\) as a function of \(K\). (b) Name a possible disturbance signal for this system.
Answers: (a) \(e_{ss} = \frac{A}{1 + K/25}\)
E4.4 A magnetic disk drive requires a motor to position a read/write head over tracks of data on a spinning disk, as shown in Figure E4.4. The motor and head may be represented by the transfer function
where \(\tau = 0.001\text{ }s\). The controller takes the difference of the actual and desired positions and generates an error. This error is multiplied by an amplifier \(K\). (a) What is the steady-state position error for a step change in the desired input? (b) Calculate
FIGURE E4.1
Digital audio system.
(a)
(b)
FIGURE E4.3 Robot fruit picker.
FIGURE E4.4 Disk drive control.
the required \(K\) in order to yield a steady-state error of \(0.1\text{ }mm\) for a ramp input of \(10\text{ }cm/s\).
Answers: \(e_{ss} = 0;K = 100\)
E4.5 A unity feedback system has the loop transfer function
Determine the relationship between the steady-state error to a ramp input and the gain \(K\) and system parameter \(b\). For what values of \(K\) and \(b\) can we guarantee that the magnitude of the steady-state error to a ramp input is less than 0.1 ?
E4.6 A feedback system has the closed-loop transfer function given by
Compute the sensitivity of the closed-loop transfer function to changes in the parameter \(p\), where \(p > 0\). Compute the steady-state error to a unit step input as a function of the parameter \(p\).
E4.7 In laser cutting, it is important for the focusing lens to be placed at an angle perpendicular to the laser, so that the laser beam can be focused. An unfocused laser beam causes an elliptical beam shape, a visual example of a steady-state error. Draw the block diagram of an autofocus system, and describe how the system works.
E4.8 Four-wheel drive automobiles are popular in regions where winter road conditions are often slippery due to snow and ice. A four-wheel drive vehicle with antilock brakes uses a sensor to keep each wheel rotating to maintain traction. One such system is shown in Figure E4.8. Find the closed-loop response of this system as it attempts to maintain a constant speed of the wheel. Determine the response when \(R(s) = A/s\).
FIGURE E4.8 Four-wheel drive auto.
E4.9 Submersibles with clear plastic hulls have the potential to revolutionize underwater leisure. One small submersible vehicle has a depth-control system as illustrated in Figure E4.9.
(a) Determine the closed-loop transfer function \(T(s) = Y(s)/R(s)\)
(b) Determine the sensitivity \(S_{K_{1}}^{T}\) and \(S_{K}^{T}\).
(c) Determine the steady-state error due to a disturbance \(T_{d}(s) = 1/s\).
(d) Calculate the response \(y(t)\) for a step input \(R(s) = 1/s\) when \(K = 2\) and \(K_{2} = 3\) and \(1 < K_{1} < 10\). Select \(K_{1}\) for the fastest response.
E4.10 Consider the feedback control system shown in Figure E4.10. (a) Determine the steady-state error for a step input in terms of the gain, \(K\). (b) Determine the overshoot for the step response for \(40 \leq K \leq 400\). (c) Plot the overshoot and the steady-state error versus \(K\).
E4.11 Consider the closed-loop system in Figure E4.11, where
FIGURE E4.9
Depth control system.
FIGURE E4.10
Feedback control system.
FIGURE E4.11 Closed-loop system with nonunity feedback.
(a) Compute the transfer function \(T(s) = Y(s)/\) \(R(s)\).
(b) Define the tracking error to be \(E(s) = R(s)\) \(Y(s)\). Compute \(E(s)\) and determine the steadystate tracking error due to a unit step input, that is, let \(R(s) = 1/s\).
(c) Compute the transfer function \(Y(s)/T_{d}(s)\) and determine the steady-state error of the output due to a unit step disturbance input, that is, let \(T_{d}(s) = 1/s\).
(d) Compute the sensitivity \(S_{K}^{T}\).
E4.12 In Figure E4.12, consider the closed-loop system with measurement noise \(N(s)\), where
$G(s) = \frac{100}{s + 100},\ G_{c}(s) = K_{1},\ $ and \(\ H(s) = \frac{K_{2}}{s + 5}\).
In the following analysis, the tracking error is defined to be \(E(s) = R(s) - Y(s)\) :
(a) Compute the transfer function \(T(s) = Y(s)/R(s)\) and determine the steady-state tracking error due
FIGURE E4.12 Closed-loop system with nonunity feedback and measurement noise.
to a unit step response, that is, let \(R(s) = 1/s\) and assume that \(N(s) = 0\).
(b) Compute the transfer function \(Y(s)/N(s)\) and determine the steady-state tracking error due to a unit step disturbance response, that is, let \(N(s) = 1/s\) and assume that \(R(s) = 0\). Remember, in this case, the desired output is zero.
(c) If the goal is to track the input while rejecting the measurement noise (in other words, while minimizing the effect of \(N(s)\) on the output), how would you select the parameters \(K_{1}\) and \(K_{2}\) ?
E4.13 A closed-loop system is used in a high-speed steel rolling mill to control the accuracy of the steel strip thickness. The transfer function for the process shown in Figure E4.13 can be represented as
Calculate the sensitivity of the closed-loop transfer function to changes in the controller gain \(K\). FIGURE E4.13
Control system for a steel rolling mill. (a) Signal flow graph. (b) Block diagram.
(a)
(b)
E4.14 Consider the unity feedback system shown in Figure E4.14. The system has two parameters, the controller gain \(K\) and the constant \(K_{1}\) in the process.
(a) Calculate the sensitivity of the closed-loop transfer function to changes in \(K_{1}\).
(b) How would you select a value for \(K\) to minimize the effects of external disturbances, \(T_{d}(s)\) ?
E4.15 Reconsider the unity feedback system discussed in E4.14. This time select \(K = 120\) and \(K_{1} = 10\). The closed-loop system is depicted in Figure E4.15.
(a) Calculate the steady-state error of the closedloop system due to a unit step input, \(R(s) = 1/s\), with \(T_{d}(s) = 0\). Recall that the tracking error is defined as \(E(s) = R(s) - Y(s)\).
(b) Calculate the steady-state response, \(y_{ss} =\) \(\lim_{t \rightarrow \infty}\mspace{2mu} y(t)\), when \(T_{d}(s) = 1/s\) and \(R(s) = 0\).
FIGURE E4.14
Closed-loop feedback system with two parameters, \(K\) and \(K_{1}\).
FIGURE E4.15 Closed-loop feedback system with \(K = 120\) and \(K_{1} = 10\).
156. PROBLEMS
P4.1 The open-loop transfer function of a fluid-flow system can be written as
where \(\tau = RC,R\) is a constant equivalent to the resistance offered by the orifice so that \(1/R = 1/2kH_{0}\ ^{- 1/2}\), and \(C =\) the cross-sectional area of the tank. Since \(\Delta H = R\Delta Q_{2}\), we have the following for the transfer function relating the head to the input change:
For a closed-loop feedback system, a float-level sensor and valve may be used as shown in Figure P4.1. Assuming the float is a negligible mass, the valve is controlled so that a reduction in the flow rate, \(\Delta Q_{1}\), is proportional to an increase in head, \(\Delta H\), or \(\Delta Q_{1} = - K\Delta H\). Draw a signal flow graph or block diagram. Determine and compare the open-loop and closed-loop systems for (a) sensitivity to changes in the equivalent coefficient \(R\) and the feedback coefficient \(K\), (b) the ability to reduce the effects of a disturbance in the level \(\Delta H(s)\), and (c) the steady-state error of the level (head) for a step change of the input \(\Delta Q_{1}(s)\).
P4.2 It is important to ensure passenger comfort on ships by stabilizing the ship's oscillations due to
waves [13]. Most ship stabilization systems use fins or hydrofoils projecting into the water to generate a stabilization torque on the ship. A simple diagram of a ship stabilization system is shown in Figure P4.2. The rolling motion of a ship can be regarded as an oscillating pendulum with a deviation from the vertical of \(\theta(t)\) degrees and a typical period of \(3\text{ }s\). The transfer function of a typical ship is
where \(\omega_{n} = 3.5rad/s\) and \(\zeta = 0.3\). With this low damping factor \(\zeta\), the oscillations continue for several cycles, and the rolling amplitude can reach \(18^{\circ}\) for the expected amplitude of waves in a normal sea. Determine and compare the open-loop and closedloop system for (a) sensitivity to changes in the actuator constant \(K_{a}\) and the roll sensor \(K_{1}\), and (b) the ability to reduce the effects of step disturbances of the waves. Note that the desired roll \(\theta_{d}(s)\) is zero degrees. (c) Find a range of \(K_{1}\) and \(K_{a}\) such that the steady-state tracking error reduced by \(90\%\) or more than a step disturbance magnitude where \(T_{d}(s) = A/s\).
P4.3 One of the most important variables that must be controlled in industrial and chemical systems is temperature. A simple representation of a thermal control system is shown in Figure P4.3 [14]. The temperature \(\mathcal{T}\) of the process is controlled by the heater with a resistance \(R\). An approximate representation of the dynamic linearly relates the heat loss from the process to the temperature difference \(\mathcal{T} - \mathcal{T}_{e}\). This relation holds if the temperature difference is relatively small and the energy storage of the heater and the vessel walls is negligible. It is assumed that \(E_{h}(s) = k_{a}E_{b}E(s)\), where \(k_{a}\) is the constant of the
FIGURE P4.1 Tank level control.
FIGURE P4.2
Ship stabilization system. The effect of the waves is a torque \(T_{d}(s)\) on the ship.
(a)
(b) FIGURE P4.3
Temperature control system.
actuator. The linearized open-loop response of the system is
where
Determine and compare the open-loop and closed-loop systems for (a) sensitivity to changes in the constant \(K = k_{1}k_{a}E_{b}\); (b) the ability to reduce the effects of a step disturbance in the environmental temperature \(\Delta\mathcal{T}_{e}(s)\); and (c) the steady-state error of the temperature controller for a step change in the input, \(E_{des}(s)\).
P4.4 A control system has two forward paths, as shown in Figure P4.4. (a) Determine the overall transfer function \(T(s) = Y(s)/R(s)\). (b) Calculate the sensitivity, \(S_{G}^{T}\), using Equation (4.16). (c) Does the sensitivity depend on \(U(s)\) or \(M(s)\) ?
P4.5 Large microwave antennas have become increasingly important for radio astronomy and satellite tracking. A large antenna with a diameter of \(60ft\), for example, is susceptible to large wind gust torques. A proposed antenna is required to have an error of less than \(20^{\circ}\) in a \(35mph\) wind. Experiments show that this wind force exerts a maximum disturbance at the antenna of \(200,000ft\) \(lb\) at \(35mph\), or the equivalent to 10 volts at the input \(T_{d}(s)\) to the amplidyne. One problem of driving large antennas is the form of the system transfer function that possesses a structural resonance. The antenna servosystem is shown in Figure P4.5. The transfer function of the antenna, drive motor, and amplidyne is approximated by
FIGURE P4.4
Two-path system.
FIGURE P4.5
Antenna control system.
where \(\zeta = 0.707\) and \(\omega_{n} = 10\). The transfer function of the power amplifier is approximately
where \(\tau = 0.2\text{ }s\). (a) Determine the sensitivity of the system to a change of the parameter \(k_{a}\). (b) The system is subjected to a disturbance \(T_{d}(s) = 1/s\). Determine the required magnitude of \(k_{a}\) in order to maintain the steady-state error of the system less than \(20^{\circ}\) when the input \(R(s)\) is zero. (c) Determine the error of the system when subjected to a disturbance \(T_{d}(s) = 10/s\) when it is operating as an open-loop system \(\left( k_{s} = 0 \right)\) with \(R(s) = 0\).
P4.6 An automatic speed control system will be necessary for passenger cars traveling on the automatic highways of the future. A model of a feedback speed control system for a standard vehicle is shown in Figure P4.6. The load disturbance due to a percent grade \(\Delta T_{d}(s)\) is also shown. The engine gain \(K_{e}\) varies within the range of 10 to 1000 for various models of automobiles. The engine time constant \(\tau_{e}\) is 20 seconds. (a) Determine the sensitivity of the system to changes in the engine gain \(K_{e}\). (b) Determine the effect of the load torque on the speed. (c) Determine the constant percent grade \(\Delta T_{d}(s) = \Delta d/s\) for which the vehicle stalls (velocity \(V(s) = 0\) ) in terms of the gain factors. Note that since the grade is constant, the steady-state solution is sufficient. Assume that \(R(s) = 30/skm/hr\) and that \(K_{e}K_{1} \gg 1\). When \(K_{g}/K_{1} = 2\), what percent grade \(\Delta d\) would cause the automobile to stall?
P4.7 A robot uses feedback to control the orientation of each joint axis. The load effect varies due to varying load objects and the extended position of the arm. The system will be deflected by the load carried in the gripper. Thus, the system may be represented by Figure P4.7, where the load torque is \(T_{d}(s) = D/s\). Assume \(R(s) = 0\) at the index position. (a) What is the effect of \(T_{d}(s)\) on \(Y(s)\) ? (b) Determine the sensitivity of the closed loop to \(k_{2}\). (c) What is the steady-state error when \(R(s) = 1/s\) and \(T_{d}(s) = 0\) ?
P4.8 Extreme temperature changes result in many failures of electronic circuits [1]. Temperature control feedback systems reduce the change of temperature by using a heater to overcome outdoor low temperatures. A block diagram of one system is shown in Figure P4.8. The effect of a drop in environmental temperature is a step decrease in \(T_{d}(s)\). The actual temperature of the electronic circuit is \(Y(s)\). The dynamics of the electronic circuit temperature change are represented by the transfer function.
(a) Determine the sensitivity of the system to \(K\). (b) Obtain the effect of the disturbance \(T_{d}(s)\) on the output \(Y(s)\).
(c) Find the range of \(K\) such that the output \(Y(s)\) is less than \(10\%\) of the step disturbance input with magnitude \(A\) (that is, \(\left. \ T_{d}(s) = A/s \right)\).
FIGURE P4.6
Automobile speed control.
Load disturbance
angle
FIGURE P4.7
Robot control system. FIGURE P4.8
Temperature control system.
P4.9 A useful unidirectional sensing device is the photoemitter sensor [15]. A light source is sensitive to the emitter current flowing and alters the resistance of the photosensor. Both the light source and the photoconductor are packaged in a single four-terminal device. This device provides a large gain and total isolation. A feedback circuit utilizing this device is shown in Figure P4.9(a), and a typical nonlinear resistancecurrent characteristic is shown in Figure P4.9(b). The resistance curve can be represented by the equation
where \(i\) is the lamp current. The normal operating point is obtained when \(v_{\text{in}\text{~}} = 2.0\text{ }V\), and \(v_{o} = 35\text{ }V\). (a) Determine the closed-loop transfer function of the system. (b) Determine the sensitivity of the system to changes in the gain, \(K\).
P4.10 For a paper processing plant, it is important to maintain a constant tension on the continuous sheet of paper between the wind-off and wind-up rolls. The tension varies as the widths of the rolls change, and an adjustment in the take-up motor speed is necessary, as shown in Figure P4.10. If the wind-up motor speed is uncontrolled, as the paper transfers from the wind-off roll to the wind-up roll, the velocity \(v_{0}(t)\) decreases and the tension of the paper drops \(\lbrack 10,14\rbrack\). The three-roller and spring combination provides a measure of the tension of the paper. The spring force is equal to \(k_{1}Y(s)\), and the linear differential transformer, rectifier, and amplifier may be represented by \(E_{0}(s) = - k_{2}Y(s)\). Therefore, the
(b)
157. Photosensor
system.
(a)
FIGURE P4.10
Paper tension control.
(a)
FIGURE P4.11
Paper-making control.
(b)
measure of the tension is described by the relation \(2T(s) = k_{1}Y(s)\), where \(Y(s)\) is the deviation from the equilibrium condition, and \(T(s)\) is the vertical component of the deviation in tension from the equilibrium condition. The time constant of the motor is \(\tau = L_{a}/R_{a}\), and the linear velocity of the wind-up roll is twice the angular velocity of the motor, that is, \(V_{0}(s) = 2\omega_{0}(s)\). The equation of the motor is then
where \(\Delta T(s) =\) a tension disturbance. (a) Draw the closed-loop block diagram for the system, including the disturbance \(\Delta T(s)\). (b) Add the effect of a disturbance in the wind-off roll velocity \(\Delta V_{1}(s)\) to the block diagram. (c) Determine the sensitivity of the system to the motor constant \(K_{m}\). (d) Determine the steady-state error in the tension when a step disturbance in the input velocity, \(\Delta V_{1}(s) = A/s\), occurs.
P4.11 One important objective of the paper-making process is to maintain uniform consistency of the stock output as it progresses to drying and rolling. A diagram of the thick stock consistency dilution control system is shown in Figure P4.11(a). The amount of water added determines the consistency. The block diagram of the system is shown in Figure P4.11(b). Let \(H(s) = 1\) and
Determine (a) the closed-loop transfer function \(T(s) = Y(s)/R(s)\), (b) the sensitivity \(S_{K}^{T}\), and (c) the steady-state error for a step change in the desired consistency \(R(s) = A/s\). (d) Calculate the value of \(K\) required for an allowable steady-state error of \(2\%\).
P4.12 Two feedback systems are shown in Figures P4.12(a) and (b). (a) Evaluate the closed-loop transfer functions \(T_{1}\) and \(T_{2}\) for each system. (b) Compare the sensitivities of the two systems with respect to the parameter \(K_{1}\) for the nominal values of \(K_{1} = K_{2} = 1\).
(a)
(b)
FIGURE P4.12 Two feedback systems. FIGURE P4.13
Closed-loop
system.
FIGURE P4.14
Hypersonic airplane speed control.
P4.13 One form of a closed-loop transfer function is
(a) Show that
(b) Determine the sensitivity of the system shown in Figure P4.13, using the equation verified in part (a).
P4.14 A proposed hypersonic plane would climb to 80,000 feet, fly 3800 miles per hour, and cross the Pacific in 2 hours. Control of the aircraft speed could be represented by the model in Figure P4.14. (a) Find the sensitivity of the closed-loop transfer function \(T(s)\) to a small change in the parameter \(a\). (b) What is the range of the parameter \(a\) for a stable closed-loop system?
P4.15 Figure P4.15 shows the model of a two-tank system containing a heated liquid, where \(T_{0}(s)\) is the temperature of the fluid flowing into the first tank and \(T_{2}(s)\) is the temperature of the liquid flowing out of the second tank. The system of two tanks has a heater in the first tank with a controllable heat input \(Q\). The time constants are \(\tau_{1} = 10\text{ }s\) and \(\tau_{2} = 50\text{ }s\). (a) Determine \(T_{2}(s)\) in terms of \(T_{0}(s)\) and \(T_{2d}(s)\). (b) If \(T_{2d}(s)\), the desired output temperature, is changed instantaneously from \(T_{2d}(s) = A/s\) to \(T_{2d}(s) = 2A/s\), where \(T_{0}(s) = A/s\), determine the transient response of \(T_{2}(s)\) when \(G_{c}(s) = K = 500\).
(c) Find the steady-state error \(e_{ss}\) for the system of part (b), where \(E(s) = T_{2d}(s) - T_{2}(s)\).
P4.16 The steering control of a modern ship may be represented by the system shown in Figure P4.16 \(\lbrack 16,20\rbrack\). (a) Find the steady-state effect of a constant wind force represented by \(T_{d}(s) = 1/s\) for \(K = 10\) and \(K = 25\). Assume that the rudder input \(R(s)\)
FIGURE P4.15
Two-tank temperature control.
FIGURE P4.16
Ship steering control.
is zero, without any disturbance, and has not been adjusted. (b) Show that the rudder can then be used to bring the ship deviation back to zero.
P4.17 A robot gripper, shown in part (a) of Figure \(P4.17\), is to be controlled so that it closes to an angle \(\theta\) by using a DC motor control system, as shown in part (b). The model of the control system is shown in part (c), where \(K_{m} = 30\), \(R_{f} = 1\Omega,K_{f} = K_{i} = 1,J = 0.1\), and \(b = 1\). (a) Determine the response \(\theta(t)\) of the system to a step change in \(\theta_{d}(t)\) when \(K = 20\). (b) Assuming \(\theta_{d}(t) = 0\), find the effect of a load disturbance \(T_{d}(s) = A/s\). (c) Determine the steady-state error \(e_{ss}\) when the input is \(r(t) = t,t > 0\). (Assume that \(T_{d}(s) = 0\).)
(a)
(b)
(c)
FIGURE P4.17 Robot gripper control.
158. ADVANCED PROBLEMS
AP4.1 A tank level regulator control is shown in Figure AP4.1(a). It is desired to regulate the level \(H(s)\) in response to a disturbance change \(Q_{3}(s)\). The block diagram shows small variable changes about the equilibrium conditions so that the desired \(H_{d}(s) = 0\). Determine the equation for the error \(E(s)\), and determine the steady-state error for a unit step disturbance when (a) \(G(s) = K\) and (b) \(G(s) = K/s\).
AP4.2 The shoulder joint of a robotic arm uses a DC motor with armature control and a set of gears on the output shaft. The model of the system is shown in Figure AP4.2 with a disturbance torque \(T_{d}(s)\) which represents the effect of the load. Determine the steady-state error when the desired angle input is a step so that \(\theta_{d}(s) = A/s,G_{c}(s) = K\), and the disturbance input is zero. When \(\theta_{d}(s) = 0\) and the load FIGURE AP4.1
A tank level
regulator.
FIGURE AP4.2
Robot joint control.
(a)
(b)
Load disturbance
FIGURE AP4.3
Machine tool feedback.
effect is \(T_{d}(s) = M/s\), determine the steady-state error when (a) \(G_{c}(s) = K\) and (b) \(G_{c}(s) = K/s\).
AP4.3 A machine tool is designed to follow a desired path so that
where \(u(t)\) is the unit step function. The machine tool control system is shown in Figure AP4.3. (a) Determine the steady-state error when \(R(s)\) is the desired path as given and \(T_{d}(s) = 0\).
(b) Plot the error \(e(t)\) for the desired path for part (a) for \(0 < t \leq 10\text{ }s\).
(c) If \(R(s) = 0\), find the steady-state error when \(T_{d}(s) = 1/s\).
(d) Plot the error \(e(t)\) for part (c) for \(0 < t \leq 10\text{ }s\). FIGURE AP4.4
DC motor with feedback.
Surgical
disturbance
FIGURE AP4.5
Blood pressure control.
FIGURE AP4.6 A proportional integral controller using an operational amplifier.
(c) Determine and plot the transient response \(V_{2}(s)\) for a step input \(V_{1}(s) = 1/s\).
AP4.7 A feedback control system with sensor noise and a disturbance input is shown in Figure AP4.7. The goal is to reduce the effects of the noise and the disturbance. Let \(R(s) = 0\).
(a) Determine the effect of the disturbance on \(Y(s)\).
(b) Determine the effect of the noise on \(Y(s)\).
(c) Choose gains \(K\) and \(K_{1}\) so that the effect of steady-state error due to the disturbance and the noise is minimized. Assume \(T_{d}(s) = A/s\), and \(N(s) = B/s\).
AP4.8 The block diagram of a machine-tool control system is shown in Figure AP4.8.
(a) Determine the transfer function \(T(s) = Y(s)/\) \(R(s)\).
(b) Determine the sensitivity \(S_{b}^{T}\).
(c) Select \(K\) and \(K_{1}\) so that the effects of a unit step disturbance are minimized. FIGURE AP4.7
Feedback system with noise.
FIGURE AP4.8
Machine-tool
control.
Sensor noise
159. DESIGN PROBLEMS
CDP4.1 A capstan drive for a table slide is described in CDP2.1. The position of the slide \(x\) is measured with a capacitance gauge, as shown in Figure CDP4.1, which is very linear and accurate. Sketch the model of the feedback system and determine the response of the system when the controller is an amplifier and \(H(s) = 1\). Determine the step response for several selected values of the amplifier gain \(G_{c}(s) = K_{a}\).
DP4.1 A closed-loop speed control system is subjected to a disturbance due to a load, as shown in Figure DP4.1. The desired speed is \(\omega_{d}(t) = 100rad/s\), and the load disturbance is a unit step input \(T_{d}(s) = 1/s\). Assume that the speed has attained the no-load speed of \(100rad/s\) and is in steady state. (a) Determine the steady-state effect of the load disturbance, and (b) plot \((t)\) for the step disturbance for various values of gain \(K\). Comment on the effect
FIGURE CDP4.1
The model of the feedback system with a capacitance measurement sensor. The tachometer may be mounted on the motor (optional), and the switch will normally be open.
FIGURE DP4.1
Speed control system.
of the gain \(K\) on the steady-state error due to the load disturbance.
DP4.2 The control of the roll angle of an airplane is achieved by using the torque developed by the ailerons. A linear model of the roll control system for a small experimental aircraft is shown in Figure DP4.2, where
The goal is to maintain a small roll angle due to disturbances. Select an appropriate gain \(KK_{1}\) that will reduce the effect of the disturbance while attaining a desirable transient response to a step disturbance. To obtain a desirable transient response, let \(KK_{1} < 50\).
DP4.3 Consider the system shown in Figure DP4.3.
(a) Determine the range of \(K_{1}\) allowable so that the steady state tracking error is \(e_{ss} \leq 1\%\).
(b) Determine a suitable value for \(K_{1}\) and \(K\) so that the magnitude of the steady-state error to a wind disturbance \(T_{d}(t) = 2tmrad/s,0 \leq t < 5\text{ }s\), is less than \(0.1mrad\).
DP4.4 Lasers have been used in eye surgery for many years. They can cut tissue or aid in coagulation [17]. The laser allows the ophthalmologist to apply heat to a location in the eye in a controlled manner.
Many procedures use the retina as a laser target. The retina is the thin sensory tissue that rests on the Inner surface of the back of the eye and is the actual transducer of the eye, converting light energy into electrical pulses. On occasion, this layer will detach from the wall, resulting in death of the detached area from lack of blood and leading to partial or total blindness in that eye. A laser can be used to "weld" the retina into its proper place on the inner wall.
Automated control of position enables the ophthalmologist to indicate to the controller where lesions should be inserted. The controller then monitors the retina and controls the laser's position so that each lesion is placed at the proper location. A wide-angle video-camera system is required to monitor the movement of the retina, as shown in Figure DP4.4(a). If the eye moves during the irradiation, the laser must be either redirected or turned off. The position-control system is shown in Figure DP4.4(b). Select an appropriate gain for the controller so that the transient response to a step change in \(R(s)\) is satisfactory and the effect of the disturbance due to noise in the system is minimized. Also, ensure that the steady-state error for a step input command is zero. Determine the largest value of \(K > 0\) to ensure closed-loop stability.
FIGURE DP4.2
Control of the roll angle of an airplane.
FIGURE DP4.3
Speed control system. FIGURE DP4.4
Laser eye surgery system.
(a)
(b)
DP4.5 An op-amp circuit can be used to generate a rapidly exponentially decaying signal. The circuit shown in Figure DP4.5 can generate the signal \(v_{2}(t) = - e^{- 1000t},t > 0\), when the input \(v_{1}(t)\) is a unit step input [6]. Select appropriate values for the resistors and capacitor. Assume an ideal op-amp.
FIGURE DP4.5 Op-amp circuit.
DP4.6 A hydrobot is under consideration for remote exploration under the ice of Europa, a moon of the giant planet Jupiter. Figure DP4.6(a) shows one artistic version of the mission. The hydrobot is a selfpropelled underwater vehicle that would analyze the chemical composition of the water in a search for signs of life. An important aspect of the vehicle is a controlled vertical descent to depth in the presence of underwater currents. A simplified control feedback system is shown in Figure DP4.6(b). The parameter \(J > 0\) is the pitching moment of inertia. (a) Suppose that \(G_{c}(s) = K\). For what range of \(K\) is the system stable? (b) What is the steady-state error to a unit step disturbance when \(G_{c}(s) = K\) ? (c) Suppose that \(G_{c}(s) = K_{p} + K_{D}s\). For what range of \(K_{p}\) and \(K_{D}\) is the system stable? (d) What is the steady-state error to a unit step disturbance when \(G_{c}(s) = K_{p} + K_{D}s\) ?
DP4.7 Interest in unmanned underwater vehicles (UUVs) has been increasing recently, with a large number of possible applications being considered. These include intelligence-gathering, mine detection, and surveillance applications. Regardless of the intended mission, a strong need exists for reliable and robust control of the vehicle. The proposed vehicle is shown in Figure DP4.7 (a) [28].
We want to control the vehicle through a range of operating conditions. The vehicle is 30 feet long with a vertical sail near the front. The control inputs are stern plane, rudder, and shaft speed commands. In this case, we wish to control the vehicle roll by using the stern planes. The control system is shown in Figure DP4.7(b), where \(R(s) = 0\), the desired roll angle, and \(T_{d}(s) = 1/s\). Suppose that the controller is
(a)
FIGURE DP4.6
(a) Europa exploration under the ice. (Used with permission. Courtesy of NASA.)
(b) Feedback system.
(b)
(a)
FIGURE DP4.7
Control of an underwater vehicle.
(b) (a) Design the controller gain \(K\) such that the maximum roll angle error due the unit step disturbance input is less than 0.05. (b) Compute the steady-state roll angle error to the disturbance input and explain the result.
DP4.8 A new suspended, mobile, remote-controlled video camera system to bring three-dimensional mobility to professional football is shown in Figure DP4.8(a) [29]. The camera can be moved over the field, as well as up and down. The motor control on
160. FIGURE DP4.8
161. Remote-controlled
TV camera.
(a)
(b)
each pulley is represented by the system in Figure DP4.8(b), where the nominal values are \(\tau_{1} = 20\) \(ms\) and \(\tau_{2} = 2\text{ }ms\). (a) Compute the sensitivity
\(S_{\tau_{1}}^{T}\) and the sensitivity \(S_{\tau_{2}}^{T}\). (b) Design the controller gain \(K\) such that the steady-state tracking error to a unit step disturbance is less than 0.05 .
162. COMPUTER PROBLEMS
CP4.1 Consider a system with the following closed-loop transfer function
Obtain the step response, and determine the percent overshoot. What is the steady-state error?
CP4.2 Consider the closed-loop transfer function
When the input is a unit ramp input, the desired steady-state error of the output is zero. Using the Isim function, show that the steady-state error to a unit ramp input is one.
CP4.3 Consider a unity feedback system with
Show that the system will have a finite steady-state error to a parabolic input. Why does the system have a finite steady-state error?
CP4.4 Consider the feedback system in Figure CP4.4. Obtain the step responses for controller gain \(K = 1,5\), and 10. (a) Develop an m-file to compute the closed-loop transfer function \(T(s) = Y(s)/R(s)\), and plot the unit step response. (b) In the same \(m\)-file, compute the transfer function from the disturbance
FIGURE CP4.4
Unity feedback system with controller gain \(K\).
\(T_{d}(s)\) to the output \(Y(s)\) and plot the unit step disturbance response. (c) From the plots in (a) and (b) above, plot the combined steady-state tracking error to the unit step input and the steady-state tracking error to the unit step disturbance input for controller gain \(K = 1,5\), and 10. (d) From the plots in (c) above, estimate the maximum tracking error to the combined unit step input and unit step disturbance input. At what gain does the maximum tracking error occur?
CP4.5 Consider the closed-loop control system shown in Figure CP4.5. Develop an \(m\)-file script to assist in the search for a value of \(k\) so that the percent overshoot to a unit step input is approximately P.O. \(\approx 10\%\). The script should compute the closed-loop transfer function \(T(s) = Y(s)/R(s)\) and generate the step response. Verify graphically that the steady-state error to a unit step input is zero.
CP4.6 Consider the closed-loop control system shown in Figure CP4.6. The controller gain is \(K = 2\). The nominal value of the plant parameter is \(a = 1\).
The nominal value is used for design purposes only, since in reality the value is not precisely known. The objective of our analysis is to investigate the sensitivity of the closed-loop system to the parameter \(a\).
(a) When \(a = 1\), show analytically that the steady-state value of \(Y(s)\) is equal to 2 when \(R(s)\) is a unit step. Verify that the unit step response is within \(2\%\) of the final value after 4 seconds.
(b) The sensitivity of the system to changes in the parameter \(a\) can be investigated by studying the effects of parameter changes on the transient response. Plot the unit step response for \(a = 0.5,2\), and 5. Discuss the results.
CP4.7 Consider the torsional mechanical system in Figure CP4.7(a). The torque due to the twisting of the shaft is \(- k\theta(s)\); the damping torque due to the braking device is \(- b\theta(s)\); the disturbance torque is \(T_{d}(s)\); the input torque is \(R(s)\); and the moment of inertia of the mechanical system is \(J\). The transfer function of the torsional mechanical system is
FIGURE CP4.5
A closed-loop negative feedback control system.
FIGURE CP4.6 A closed-loop control system with uncertain parameter \(a\).
FIGURE CP4.7
(a) A torsional mechanical system. (b) The torsional mechanical system feedback control system.
(a)
(b) A closed-loop control system for the system is shown in Figure CP4.7(b). Suppose the desired angle \(\theta_{d} = 0^{\circ},k = 5,b = 0.9\), and \(J = 1\).
(a) Determine the open-loop response \(\theta(s)\) of the system for a unit step disturbance.
(b) With the controller gain \(K_{0} = 50\), determine the closed-loop response, \(\theta(s)\), to a unit step disturbance.
(c) Plot the open-loop versus the closed-loop response to the disturbance input. Discuss your results and make an argument for using closed-loop feedback control to improve the disturbance rejection properties of the system.
CP4.8 A negative feedback control system is depicted in Figure CP4.8. Suppose that our design objective is to find a controller \(G_{c}(s)\) of minimal complexity such that our closed-loop system can track a unit step input with a steady-state error of zero.
(a) As a first try, consider a simple proportional controller
where \(K\) is a fixed gain. Let \(K = 2\). Plot the unit step response and determine the steadystate error from the plot.
(b) Now consider a more complex controller
where \(K_{0} = 2\) and \(K_{1} = 20\). This controller is known as a proportional, integral (PI) controller. Plot the unit step response, and determine the steady-state error from the plot.
(c) Compare the results from parts (a) and (b), and discuss the trade-off between controller complexity and steady-state tracking error performance.
CP4.9 Consider the closed-loop system in Figure \(CP4.9\), whose transfer function is
(a) Obtain the closed-loop transfer function \(T(s) = Y(s)/R(s)\) and the unit step response; that is, let \(R(s) = 1/s\) and assume that \(N(s) = 0\).
FIGURE CP4.9 Closed-loop system with nonunity feedback and measurement noise.
(b) Obtain the disturbance response when
is a sinusoidal input of frequency \(\omega = 10rad/s\). Assume that \(R(s) = 0\).
(c) In the steady-state, what is the frequency and peak magnitude of the disturbance response from part (b)?
CP4.10 Consider the closed-loop system is depicted in Figure CP4.10. The controller gain \(K\) can be modified to meet the design specifications.
(a) Determine the closed-loop transfer function \(T(s) = Y(s)/R(s)\).
(b) Plot the response of the closed-loop system for \(K = 5,10\), and 50 .
(c) When the controller gain is \(K = 10\), determine the steady-state value of \(y(t)\) when the disturbance is a unit step, that is, when \(T_{d}(s) = 1/s\) and \(R(s) = 0\).
CP4.11 Consider the nonunity feedback system is depicted in Figure CP4.11.
(a) Determine the closed-loop transfer function \(T(s) = Y(s)/R(s)\).
(b) For \(K = 10,12\), and 15, plot the unit step responses. Determine the steady-state errors and the settling times from the plots.
For parts (a) and (b), develop a m-file that computes the closed-loop transfer function and generates the plots for varying \(K\).
FIGURE CP4.8
A simple single-loop feedback control system.
FIGURE CP4.10
Closed-loop
feedback system
with external
disturbances.
FIGURE CP4.11
Closed-loop system with a sensor in the feedback loop.
163. ANSWERS TO SKILLS CHECK
True or False: (1) True; (2) True; (3) False; (4) False; (5) True
Multiple Choice: (6) a; (7) b; (8) a; (9) b; (10) c; (11) a; (12) b; (13) b; (14) c; (15) c
Word Match (in order, top to bottom): e, h, k, b, c, f, i, \(g,d,a,j\)
164. TERMS AND CONCEPTS
Closed-loop system A system with a measurement of the output signal and a comparison with the desired output to generate an error signal that is applied to the actuator.
Complexity A measure of the structure, intricateness, or behavior of a system that characterizes the relationships and interactions between various components.
Components The parts, subsystems, or subassemblies that comprise a total system.
Disturbance signal An unwanted input signal that affects the system's output signal.
Error signal The difference between the desired output \(R(s)\) and the actual output \(Y(s)\). Therefore, \(E(s) = R(s) - Y(s)\).
Instability An attribute of a system that describes a tendency of the system to depart from the equilibrium condition when initially displaced.
Loop gain The ratio of the feedback signal to the controller actuating signal. For a unity feedback system we have \(L(s) = G_{c}(s)G(s)\).
Loss of gain A reduction in the amplitude of the ratio of the output signal to the input signal through a system, usually measured in decibels.
Open-loop system A system without feedback that directly generates the output in response to an input signal.
Steady-state error The error when the time period is large and the transient response has decayed, leaving the continuous response.
System sensitivity The ratio of the change in the system transfer function to the change of a process transfer function (or parameter) for a small incremental change.
165. Tracking error See error signal.
Transient response The response of a system as a function of time before steady-state.
$5.1\ $ Introduction 322
5.2 Test Input Signals 322
5.3 Performance of Second-Order Systems 325
5.4 Effects of a Third Pole and a Zero on the Second-Order System
Response 330
5.5 The s-Plane Root Location and the Transient Response 335
5.6 The Steady-State Error of Feedback Control Systems 337
5.7 Performance Indices 344
5.8 The Simplification of Linear Systems 349
5.9 Design Examples 352
5.10 System Performance Using Control Design Software 364
5.11 Sequential Design Example: Disk Drive Read System 370
5.12 Summary 372
PREVIEW
The ability to adjust the transient and steady-state response of a control system is a beneficial outcome of the design of control systems. In this chapter, we introduce the time-domain performance specifications and use key input signals to test the response of the control system. The correlation between the system performance and the location of the transfer function poles and zeros is discussed. We develop relationships between the performance specifications and the natural frequency and damping ratio for second-order systems. Relying on the notion of dominant poles, we can extrapolate the ideas associated with second-order systems to those of higher order. The concept of a performance index is also considered. We present a set of quantitative performance indices that adequately represent the performance of the control system. The chapter concludes with a performance analysis of the Sequential Design Example: Disk Drive Read System.
166. DESIRED OUTCOMES
Upon completion of Chapter 5, students should be able to:
$\square\ $ Identify key test signals used in controls and describe the resulting transient response characteristics of second-order systems to test signal inputs.
$\square\ $ Recognize the direct relationship between the pole locations of second-order systems and the transient response.
$\square\ $ Identify the design formulas that relate the second-order pole locations to percent overshoot, settling time, rise time, and time to peak.
\(\square\) Explain the impact of a zero and a third pole on the second-order system response.
$\square\ $ Describe optimal control as measured with performance indices.
166.1. INTRODUCTION
The ability to adjust the transient and steady-state performance is a distinct advantage of feedback control systems. To analyze and design a control system, we must define and measure its performance. Based on the desired performance of the control system, the controller parameters are adjusted to provide the desired response. Because control systems are inherently dynamic, their performance is usually specified in terms of both the transient response and the steady-state response. The transient response is the response that disappears with time. The steady-state response is the response that exists for a long time following an input signal initiation.
The design specifications for control systems normally include several time-response indices for a specified input command, as well as a desired steadystate accuracy. In the course of the design, the specifications are often revised to effect a compromise. Therefore, specifications are seldom a rigid set of requirements, but rather an attempt to quantify the desired performance. The effective compromise and adjustment of specifications are graphically illustrated in Figure 5.1. The parameter \(p\) may minimize the performance measure \(M_{2}\) if we select \(p\) as a very small value. However, this results in large measure \(M_{1}\), an undesirable situation. If the performance measures are equally important, the crossover point at \(p_{\min}\) provides the best compromise. This type of compromise is often encountered in feedback control system design. It is clear that if the original specifications called for both \(M_{1}\) and \(M_{2}\) to be minimized, the specifications could not be simultaneously met; they would then have to be altered to allow for the compromise resulting with \(p_{\min}\lbrack 1,10,15,20\rbrack\).
The specifications, which are stated in terms of the measures of performance, indicate the quality of the system to the designer. In other words, the performance measures help to answer the question, How well does the system perform the task for which it was designed?
166.2. TEST INPUT SIGNALS
Time-domain performance specifications are important indices because control systems are inherently time-domain systems. The transient response is of prime interest for control system designers. It is necessary to determine initially whether
FIGURE 5.1
Two performance measures versus parameter \(p\).
167. Table 5.1 Test Signal Inputs
| Test Signal | $$r(t)$$ | $$R(s)$$ |
| Step | $$\begin |
r(t) & \ = A,t > 0 \\
& \ = 0,t < 0
\end{matrix}$$ | $$R(s) = A/s$$ |
| Ramp | $$\begin{matrix}
r(t) & \ = At,t > 0 \
& \ = 0,t < 0
\end{matrix}$$ | $$R(s) = A/s^{2}$$ |
| Parabolic | $$\begin{matrix}
r(t) & \ = At^{2},t > 0 \
& \ = 0,t < 0
\end{matrix}$$ | $$R(s) = 2A/s$$ |
the system is stable; we can achieve this goal by using the techniques of ensuing chapters. If the system is stable, the response to a specific input signal will provide several measures of the performance. However, because the actual input signal of the system is usually unknown, a standard test input signal is normally chosen. This approach is quite useful because there is a reasonable correlation between the response of a system to a standard test input and the system's ability to perform under normal operating conditions. Furthermore, using a standard input allows the designer to compare several competing designs. Many control systems experience input signals that are very similar to the standard test signals.
The standard test input signals commonly used are the step input, the ramp input, and the parabolic input. These inputs are shown in Figure 5.2. The equations representing these test signals are given in Table 5.1, where the Laplace transform can be obtained by using Table 2.3 and a more complete list of Laplace transform pairs can be found at the MCS website. The ramp signal is the integral of the step input, and the parabola is the integral of the ramp input. A unit impulse function is also useful for test signal purposes. The unit impulse is based on a rectangular function
where \(\epsilon > 0\). As \(\in\) approaches zero, the function \(f_{\epsilon}(t)\) approaches the unit impulse function \(\delta(t)\), which has the following properties
FIGURE 5.2
Test input signals: (a) step, (b) ramp, and (c) parabolic.
(a)
(b)
(c) The impulse input is useful when we consider the convolution integral for the output \(y(t)\) in terms of an input \(r(t)\), which is written as
The relationship in Equation (5.2) represents the open-loop input-output relationship of a system \(G(s)\). If the input is a unit impulse function, we have
The integral has a value only at \(\tau = 0\); therefore,
the impulse response of the system \(G(s)\). The impulse response test signal can often be used for a dynamic system by subjecting the system to a large-amplitude, narrow-width pulse of area \(A\).
The standard test signals are of the general form
and the Laplace transform is
Hence, the response to one test signal may be related to the response of another test signal of the form of Equation (5.4). The step input signal is the easiest to generate and evaluate and is usually chosen for performance tests.
Consider the response of a system \(G(s)\) for a unit step input, \(R(s) = 1/s\), when
Then the output is
the response during the transient period is
and the steady-state response is
If the error is \(E(s) = R(s) - Y(s)\), then the steady-state error is
167.1. PERFORMANCE OF SECOND-ORDER SYSTEMS
Let us consider a single-loop second-order system and determine its response to a unit step input. A closed-loop feedback control system is shown in Figure 5.3. The closed-loop transfer function is
We may rewrite Equation (5.6) as
With a unit step input, we obtain
from which it follows that
where \(\beta = \sqrt{1 - \zeta^{2}},\theta = \cos^{- 1}\zeta\), and \(0 < \zeta < 1\). The response of this secondorder system for various values of the damping ratio \(\zeta\) is shown in Figure 5.4. As \(\zeta\) decreases, the closed-loop poles approach the imaginary axis, and the response becomes increasingly oscillatory.
The Laplace transform of the unit impulse is \(R(s) = 1\), and therefore the output for an impulse is
The response for an impulse function input is then
FIGURE 5.3
Second-order closed-loop control system.
(b) FIGURE 5.4
Transient response of a second-order system for a step input.
FIGURE 5.5
Response of a second-order system for an impulse input.
which is the derivative of the response to a step input. The impulse response of the second-order system is shown in Figure 5.5 for several values of the damping ratio \(\zeta\).
Standard performance measures are often defined in terms of the step response of the closed-loop system as shown in Figure 5.6. The swiftness of the response is measured by the rise time \(T_{r}\) and the peak time \(T_{p}\). For underdamped systems with an overshoot, the \(0 - 100\%\) rise time is a useful index. If the system is overdamped, then the peak time is not defined, and the \(10 - 90\%\) rise time \(T_{r1}\) is normally used. FIGURE 5.6
Step response of a second-order system.
The similarity with which the actual response matches the step input is measured by the percent overshoot and settling time \(T_{s}\). The percent overshoot is defined as
for a unit step input, where \(M_{pt}\) is the peak value of the time response, and \(fv\) is the final value of the response. Normally, \(fv\) is the magnitude of the input, but many systems have a final value significantly different from the desired input magnitude. For the system with a unit step represented by Equation (5.8), we have \(fv = 1\).
The settling time, \(T_{s}\), is defined as the time required for the system to settle within a certain percentage \(\delta\) of the input amplitude. This band of \(\pm \delta\) is shown in Figure 5.6. For the second-order system with closed-loop damping constant \(\zeta\omega_{n}\) and a response described by Equation (5.9), we seek to determine the time \(T_{s}\) for which the response remains within \(2\%\) of the final value. This occurs approximately when
or
Therefore, we have
Hence, we define the settling time as four time constants (that is, \(\tau = 1/\zeta\omega_{n}\) ) of the dominant roots of the characteristic equation. The steady-state error of the system may be measured on the step response of the system as shown in Figure 5.6. The transient response of the system may be described in terms of two factors:
-
The swiftness of response, as represented by the rise time and the peak time.
-
The closeness of the response to the desired response, as represented by the overshoot and settling time.
As it turns out, these are often contradictory requirements; thus, a compromise must be obtained. To obtain an explicit relation for \(M_{pt}\) and \(T_{p}\) as a function of \(\zeta\), one can differentiate Equation (5.9) and set it equal to zero yielding
which is equal to zero when \(\omega_{n}\beta t = n\pi\), where \(n = 0,1,2,\ldots\). The first nonzero time this is equal to zero is when \(n = 1\). Thus, we find that the peak time relationship for this second-order system is
and the peak response is
Therefore, the percent overshoot is
The percent overshoot versus the damping ratio, \(\zeta\), is shown in Figure 5.7. Also, the normalized peak time, \(\omega_{n}T_{p}\), is shown versus the damping ratio, \(\zeta\), in Figure 5.7. Upon inspection of Figure 5.7, we see that we are confronted with a necessary compromise between the swiftness of response and the allowable percent overshoot.
FIGURE 5.7
Percent overshoot and normalized peak time versus damping ratio \(\zeta\) for a secondorder system (Equation 5.8).
FIGURE 5.8
Normalized rise time, \(T_{r11}\), versus \(\zeta\) for a second-order system.
The swiftness of step response can be measured as the time it takes to rise from \(10\%\) to \(90\%\) of the magnitude of the step input. This is the definition of the rise time, \(T_{r1}\), shown in Figure 5.6. The normalized rise time, \(\omega_{n}T_{r1}\), versus \(\zeta(0.05 \leq \zeta \leq 0.95)\) is shown in Figure 5.8. Although it is difficult to obtain exact analytic expressions for \(T_{r1}\), we can utilize the linear approximation
which is accurate for \(0.3 \leq \zeta \leq 0.8\). This linear approximation is shown in Figure 5.8.
The swiftness of a response to a step input as described by Equation (5.17) is dependent on \(\zeta\) and \(\omega_{n}\). For a given \(\zeta\), the response is faster for larger \(\omega_{n}\), as shown in Figure 5.9. Note that the percent overshoot is independent of \(\omega_{n}\).
168. FIGURE 5.9
The step response for \(\zeta = 0.2\) for \(\omega_{n} = 1\) and \(\omega_{n} = 10\). FIGURE 5.10 The step response for \(\omega_{n} = 5\) with \(\zeta = 0.7\) and \(\zeta = 1\).
For a given \(\omega_{n}\), the response is faster for lower \(\zeta\), as shown in Figure 5.10. The swiftness of the response, however, will be limited by the overshoot that can be accepted.
168.1. EFFECTS OF A THIRD POLE AND A ZERO ON THE SECOND-ORDER SYSTEM RESPONSE
The curves presented in Figure 5.7 are exact only for the second-order system of Equation (5.8). However, they provide important information because many systems possess a dominant pair of roots and the step response can be estimated by utilizing Figure 5.7. This approach, although an approximation, avoids the evaluation of the inverse Laplace transformation in order to determine the percent overshoot and other performance measures. For example, for a third-order system with a closed-loop transfer function
the \(s\)-plane diagram is shown in Figure 5.11. This third-order system is normalized with \(\omega_{n} = 1\). The performance (as indicated by the percent overshoot, P.O., and the settling time, \(T_{s}\) ), is adequately represented by the second-order system approximation when [4]
In other words, the response of a third-order system can be approximated by the dominant roots of the second-order system as long as the real part of the dominant roots is less than one tenth of the real part of the third root \(\lbrack 15,20\rbrack\).
Consider the third-order system
FIGURE 5.11
An s-plane diagram of a third-order system.
FIGURE 5.12
Comparison of two third-order systems with a second-order system (dashed line) illustrating the concept of dominant poles when \(|1/\gamma| \geq 10\zeta\omega_{n}\)
where \(\omega_{n} = 1.0,\zeta = 0.45\), and \(\gamma = 1.0\). In this case, \(|1/\gamma|\). The system poles are at \(s_{1,2} = - 0.45 \pm 0.89i\) and \(s_{3} = - 1.0\). As illustrated in Figure 5.12, the percent overshoot is P.O. \(= 10.9\%\), the settling time (to within \(2\%\) of the final value) is \(T_{s} = 8.84\text{ }s\), and the rise time \(T_{r1} = 2.16\text{ }s\). Suppose that we have another third-order system with \(\omega_{n} = 1.0,\zeta = 0.45\), and \(\gamma = 0.22\). Then the system poles are at \(s_{1,2} = - 0.45 \pm 0.89i\) (the same as the first system) and \(s_{3} = - 4.5\). In this case, \(|1/\gamma| \geq 10\zeta\omega_{n}\) and the complex poles pair are the dominant poles. As illustrated in Figure 5.12, the percent overshoot is P.O. \(= 20.0\%\), the settling time is \(T_{s} = 8.56\text{ }s\), and the rise time \(T_{r1} = 1.6\text{ }s\). When the complex pair of poles are the dominant poles, we can create the second-order system approximation
and we would expect the percent overshoot, settling time, and rise time to be P.O. = \(100e^{- \zeta\pi/\sqrt{1 - \zeta^{2}}} = 20.5\%,T_{s} = 4/\zeta\omega_{n} = 8.89\text{ }s\), and \(T_{r1} = (2.16\zeta + 0.6)/\omega_{n} = 1.57\text{ }s\), respectively. In Figure 5.12, it is evident that for the third-order system satisfying the condition \(|1/\gamma| \geq 10\zeta\omega_{n}\), the step response more closely matches the response of the second-order system, as expected.
The performance measures associated with the second-order system in Equation (5.10) are precise only for transfer functions without finite zeros. If the transfer function of a system possesses a finite zero and it is located relatively near the dominant complex poles, then the zero will materially affect the transient response of the system. In other words, the transient response of a system with one zero and two poles may be affected by the location of the zero [5]. Consider a system with the system transfer function
We can investigate the response of the system compared to a second-order system without the finite zero. Suppose that \(\zeta = 0.45\) and let \(a/\zeta\omega_{n} = 0.5,1,2\), and 10.0. The resulting unit step responses are shown in Figure 5.13. As the ratio \(a/\zeta\omega_{n}\) increases, the finite zero moves farther into the left half-plane and away from the poles, and the step response approaches the second-order system response, as expected.
The correlation of the time-domain response of a system with the \(s\)-plane location of the poles of the closed-loop transfer function is a key concept in understanding system performance in the closed-loop.
169. EXAMPLE 5.1 Parameter selection
A single-loop feedback control system is shown in Figure 5.14. We need to select the controller gain \(K\) and the parameter \(p\) so that the time-domain specifications are satisfied. The transient response to a unit step is specified to have a percent overshoot of P.O. \(\leq 5\%\) and a settling time to within \(2\%\) of the final value of \(T_{s} \leq 4\text{ }s\). For second-order systems, we know the relationship in Equation (5.16) between percent overshoot and \(\zeta\), and the relationship in Equation (5.13) between settling time and \(\zeta\omega_{n}\). Solving for P.O. \(\leq 5\%\) yields \(\zeta \geq 0.69\) and solving for \(T_{s} \leq 4\text{ }s\) yields \(\zeta\omega_{n} \geq 1\).
The region that will satisfy both time-domain requirements is shown on the \(s\)-plane of Figure 5.15. To meet the performance specifications, we can choose \(\zeta = 0.707(P.O. = 4.3\%)\) and \(\zeta\omega_{n} = 1\left( T_{s} = 4\text{ }s \right)\). Hence, the desired closed-loop FIGURE 5.13
The response for the second-order transfer function with a zero for four values of the ratio \(a/\zeta\omega_{n} = 0.5,1,2\), and 10.0 when \(\zeta = 0.45\).
FIGURE 5.14
Single-loop feedback control system.
poles are \(r_{1} = - 1 + j1\) and \({\widehat{r}}_{1} = - 1 - j1\). Therefore, \(\zeta = 1/\sqrt{2}\) and \(\omega_{n} = 1/\zeta = \sqrt{2}\). The closed-loop transfer function is
Solving for \(K\) and \(p\) yields \(K = \omega_{n}^{2} = 2\) and \(p = 2\zeta\omega_{n} = 2\). Since this is exactly a second-order system of the form in Equation (5.7), the time-domain performance specifications will be precisely satisfied. FIGURE 5.15
Specifications and root locations on the s-plane.
170. EXAMPLE 5.2 Impact of a zero and an additional pole
Consider a system with a closed-loop transfer function
Both the zero and the real pole may affect the transient response. If \(a \gg \zeta\omega_{n}\) and \(\tau \ll 1/\zeta\omega_{n}\), then the pole and zero will have minimal effect on the step response.
Suppose that we have
Note that the DC gain is \(T(0) = 1\), and we expect a zero steady-state error for a step input. Comparing the two transfer functions, we determine that \(\zeta\omega_{n} = 3,\tau = 0.16\), and \(a = 2.5\). The poles and the zero are shown on the \(s\)-plane in Figure 5.16. As an approximation, we neglect the real pole and zero to obtain
FIGURE 5.16
The poles and zeros on the s-plane for a third-order system. We now have \(\zeta = 0.6\) and \(\omega_{n} = 5\) for dominant poles. For this second-order system we expect
For the actual third-order system, we find that the P.O. \(= 38\%\) and the \(T_{s} = 1.6\text{ }s\). Thus, the effect of the third pole and zero of \(T(s)\) cannot be neglected. This is expected since \(a \gg \zeta\omega_{n}\) and \(\tau \ll 1/\zeta\omega_{n}\).
The damping ratio plays a fundamental role in closed-loop system performance. As seen in the design formulas for settling time, percent overshoot, peak time, and rise time, the damping ratio is a key factor in determining the overall performance. In fact, for second-order systems, the damping ratio is the only factor determining the value of the percent overshoot to a step input. As it turns out, the damping ratio can be estimated from the response of a system to a step input [12]. The step response of a second-order system for a unit step input is given in Equation (5.9) yields the frequency of the damped sinusoidal term (for \(\zeta < 1\) ) of
and the number of cycles in 1 second is \(\omega/(2\pi)\).
The time constant for the exponential decay is \(\tau = 1/\left( \zeta\omega_{n} \right)\) in seconds. The number of cycles of the damped sinusoid during one time constant is
Assuming that the response decays in \(n\) visible time constants, we have
For the second-order system, the response remains within \(2\%\) of the steady-state value after four time constants \((4\tau)\). Hence, \(n = 4\), and
for \(0.2 \leq \zeta \leq 0.6\). From the step response, you count the number of cycles visible up to the settling time, and use Equation (5.20) to estimate \(\zeta\),
An alternative method of estimating \(\zeta\) is to determine the percent overshoot for the step response and use Equation (5.16) to estimate \(\zeta\).
170.1. THE s-PLANE ROOT LOCATION AND THE TRANSIENT RESPONSE
The transient response of a closed-loop feedback control system can be described in terms of the location of the poles of the transfer function. The closed-loop transfer function is written in general as
where \(\Delta(s) = 0\) is the characteristic equation of the system. For a unity negative feedback control system the characteristic equation reduces to \(1 + G_{c}(s)G(s) = 0\). It is the poles and zeros of \(T(s)\) that determine the transient response. However, for a closed-loop system, the poles of \(T(s)\) are the roots of the characteristic equation \(\Delta(s) = 0\). The output of a system (with DC gain \(= 1\) ) without repeated roots and a unit step input can be formulated as a partial fraction expansion as
where the \(A_{i},B_{k}\), and \(C_{k}\) are constants. The roots of the system must be either \(s = - \sigma_{i}\) or complex conjugate pairs such as \(s = - \alpha_{k} \pm j\omega_{k}\). Then the inverse transform results in the transient response as the sum of terms
where \(D_{k}\) is a constant and depends on \(B_{k},C_{k},\alpha_{k}\), and \(\omega_{k}\). The transient response is composed of the steady-state output, exponential terms, and damped sinusoidal terms. For the response to be stable-that is, bounded for a step input-the real part of the poles must be in the left-hand portion of the \(s\)-plane. The impulse response for various root locations is shown in Figure 5.17. The information imparted by the location of the roots is very descriptive.
It is important for the control system designer to understand the complete relationship of the frequency domain representation of a linear system, the poles and zeros of its transfer function, and its time-domain response to step and other inputs. In such areas as signal processing and control, many of the analysis and design calculations are done in the \(s\)-plane, where a system model is represented in terms of the poles and zeros of its transfer function \(T(s)\). On the other hand, system performance is often analyzed by examining time-domain responses, particularly when dealing with control systems.
FIGURE 5.17 Impulse response for various root locations in the s-plane. (The conjugate root is not shown.)
The control system designer will envision the effects on the step and impulse responses of adding, deleting, or moving poles and zeros of \(T(s)\) in the \(s\)-plane. Likewise, the designer should visualize the necessary changes for the poles and zeros of \(T(s)\), in order to effect desired changes in the step and impulse responses.
An experienced designer is aware of the effects of zero locations on system response, as well. The poles of \(T(s)\) determine the particular response modes that will be present, and the zeros of \(T(s)\) establish the relative weightings of the individual mode functions. For example, moving a zero closer to a specific pole will reduce the relative contribution to the output response. In other words, the zeros have a direct impact on the values of \(A_{i}\) and \(D_{k}\) in Equation (5.22). For example, if there is a zero near the pole at \(s = - \sigma_{i}\), then \(A_{i}\) will be much smaller in magnitude.
170.2. THE STEADY-STATE ERROR OF FEEDBACK CONTROL SYSTEMS
One of the fundamental reasons for using feedback, despite its cost and increased complexity, is the attendant improvement in the reduction of the steady-state error of the system. The steady-state error of a stable closed-loop system is usually several orders of magnitude smaller than the error of an open-loop system. The system actuating signal, which is a measure of the system error, is denoted as \(E_{a}(s)\). Consider a unity negative feedback system. In the absence of external disturbances, \(T_{d}(s) = 0\), and measurement noise, \(N(s) = 0\), the tracking error is
Using the final value theorem and computing the steady-state tracking error yields
It is useful to determine the steady-state error of the system for the three standard test inputs for the unity feedback system. Later in this section we will consider steady-state tracking errors for nonunity feedback systems.
Step Input. The steady-state error for a step input of magnitude \(A\) is therefore
It is the form of the loop transfer function \(G_{c}(s)G(s)\) that determines the steadystate error. The loop transfer function is written in general form as
where \(\prod\) denotes the product of the factors and \(z_{i} \neq 0,p_{k} \neq 0\) for any \(1 \leq i \leq M\) and \(i \leq k \leq Q\). Therefore, the loop transfer function as \(s\) approaches zero depends on the number of integrations, \(N\). If \(N\) is greater than zero, then \(\lim_{s \rightarrow 0}\mspace{2mu} G_{c}(s)G(s)\) approaches infinity, and the steady-state error approaches zero. The number of integrations is often indicated by labeling a system with a type number that is equal to \(N\).
Consequently, for a type-zero system, \(N = 0\), the steady-state error is
The constant \(G_{c}(0)G(0)\) is denoted by \(K_{p}\), the position error constant, and is given by
The steady-state tracking error for a step input of magnitude \(A\) is thus given by
Hence, the steady-state error for a unit step input with one integration or more, \(N \geq 1\), is zero because
Ramp Input. The steady-state error for a ramp (velocity) input with a slope \(A\) is
Again, the steady-state error depends upon the number of integrations, \(N\). For a type-zero system, \(N = 0\), the steady-state error is infinite. For a type-one system, \(N = 1\), the error is
or
where \(K_{v}\) is designated the velocity error constant. The velocity error constant is computed as
When the transfer function possesses two or more integrations, \(N \geq 2\), we obtain a steady-state error of zero. When \(N = 1\), a steady-state error exists. However,
171. Table 5.2 Summary of Steady-State Errors
| $$\begin | |||
| \text{~}\text{Number of}\text{~} \ | |||
| \text{}\text{Integrations}\text{} \ | |||
| \text{}\text{in}\text{}G_{C}(s)G(s), | |||
| \end{matrix}$$ | Step, \(r(t) = A\), | Ramp, \(r(t) = At\), | Parabola, \(r(t) = At^{2}/2\), |
| Type Number | $$R(s) = A/s$$ | $$R(s) = A/s^{2}$$ | $$R(s) = A/s^{3}$$ |
| 0 | $$e_{ss} = \frac{A}{1 + K_{p}}$$ | $$\infty$$ | $$\infty$$ |
| 1 | $$e_{ss} = 0$$ | $$\frac{A}{K_{v}}$$ | $$\infty$$ |
| 2 | $$e_{Ss} = 0$$ | 0 | $$\frac{A}{K_{a}}$$ |
the steady-state velocity of the output is equal to the velocity of input, as we shall see shortly.
Acceleration Input. When the system input is \(r(t) = At^{2}/2\), the steady-state error is
The steady-state error is infinite for one integration. For two integrations, \(N = 2\), and we obtain
where \(K_{a}\) is designated the acceleration error constant. The acceleration error constant is
When the number of integrations equals or exceeds three, then the steady-state error of the system is zero.
Control systems are often described in terms of their type number and the error constants, \(K_{p},K_{v}\), and \(K_{a}\). Definitions for the error constants and the steady-state error for the three inputs are summarized in Table 5.2.
172. EXAMPLE 5.3 Mobile robot steering control
A mobile robot may be designed as an assisting device or servant for a severely disabled person [7]. The steering control system for such a robot can be represented by the block diagram shown in Figure 5.18. The steering controller is
FIGURE 5.18 Block diagram of steering control system for a mobile robot.
FIGURE 5.19 Triangular wave response.
Therefore, the steady-state error of the system for a step input when \(K_{2} = 0\) and \(G_{c}(s) = K_{1}\) is
where \(K_{p} = KK_{1}\). When \(K_{2}\) is greater than zero, we have a type- 1 system,
and the steady-state error is zero for a step input.
If the steering command is a ramp input, the steady-state error is
where
The transient response of the vehicle to a triangular wave input when \(G_{c}(s) = \left( K_{1}s + K_{2} \right)/s\) is shown in Figure 5.19. The transient response clearly shows the effect of the steady-state error, which may not be objectionable if \(K_{v}\) is sufficiently large. Note that the output attains the desired velocity as required by the input, but it exhibits a steady-state error.
The control system error constants, \(K_{p},K_{v}\), and \(K_{a}\), describe the ability of a system to reduce or eliminate the steady-state error. Therefore, they are utilized as numerical measures of the steady-state performance. The designer determines the error constants for a given system and attempts to determine methods of increasing the error constants while maintaining an acceptable transient response. In the case of the steering control system, we want to increase the gain factor \(KK_{2}\) in order to increase \(K_{v}\) and reduce the steady-state error. However, an increase in \(KK_{2}\) results in an attendant decrease in the system damping ratio \(\zeta\) and therefore a more oscillatory response to a step input. Thus, we seek a compromise that provides the largest \(K_{v}\) based on the smallest \(\zeta\) allowable.
In the preceding discussions, we assumed that we had a unity feedback system. Now we consider nonunity feedback systems. For a system in which the feedback is not unity, the units of the output \(Y(s)\) are usually different from the output of the sensor. For example, a speed control system is shown in Figure 5.20. The constants \(K_{1}\) and \(K_{2}\) account for the conversion of one set of units to another set of units (here we convert rad/s to volts). We can select \(K_{1}\), and thus we set \(K_{1} = K_{2}\) and move the block for \(K_{1}\) and \(K_{2}\) past the summing node. Then we obtain the equivalent block diagram shown in Figure 5.21. Thus, we obtain a unity feedback system as desired.
Consider a nonunity negative feedback system with the system \(H(s)\) in the feedback loop given by
which has a DC gain of
If we set \(K_{2} = K_{1}\), then the system is transformed to that of Figure 5.21 for the steady-state calculation. To see this, consider error of the system \(E(s)\), where
since \(Y(s) = T(s)R(s)\). Note that
FIGURE 5.20
A speed control system.
FIGURE 5.21
The speed control system of Figure 5.20 with \(K_{1} = K_{2}\).
and therefore,
Then the steady-state error for a unit step input is
We assume here that
173. EXAMPLE 5.4 Steady-state error
Let us determine the appropriate value of \(K_{1}\) and calculate the steady-state error for a unit step input for the system shown in Figure 4.4 when
Selecting \(K_{1} = K_{2} = 2\), we can use Equation (5.36) to determine
or \(5.9\%\) of the magnitude of the step input.
174. EXAMPLE 5.5 Nonunity feedback control system
Let us consider the system of Figure 5.22, where we assume we cannot insert a gain \(K_{1}\) following \(R(s)\) as we did for the system of Figure 5.20. Then the actual error is given by Equation (5.35), which is
Let us determine an appropriate gain \(K\) so that the steady-state error to a step input is minimized. The steady-state error is
where
Then we have
FIGURE 5.22
A system with a feedback \(H(s)\).
The steady-state error for a unit step input is
Thus, to achieve a zero steady-state error, we require that
or \(8 + 2K = 4K\). Thus, \(K = 4\) will yield a zero steady-state error. It is unlikely that meeting a steady-state error specification is the only requirement of the feedback control system, so choosing the control as a gain with only one parameter to adjust is probably not practical.
The determination of the steady-state error is simpler for unity feedback systems. However, it is possible to extend the notion of error constants to nonunity feedback systems by first appropriately rearranging the block diagram to obtain an equivalent unity feedback system. Remember that the underlying system must be stable, otherwise our use of the final value theorem will be compromised. Consider the nonunity feedback system in Figure 5.21 and assume that \(K_{1} = 1\). The closedloop transfer function is
By manipulating the block diagram appropriately we can obtain the equivalent unity feedback system with
The loop transfer function of the equivalent unity feedback system is \(Z(s)\). It follows that the error constants for nonunity feedback systems are given as:
Note that when \(H(s) = 1\), then \(Z(s) = G_{c}(s)G(s)\) and we maintain the unity feedback error constants. For example, when \(H(s) = 1\), then \(K_{p} = \lim_{s \rightarrow 0}\mspace{2mu} Z(s) =\) \(\lim_{s \rightarrow 0}\mspace{2mu} G_{c}(s)G(s)\), as expected.
174.1. PERFORMANCE INDICES
Modern control theory assumes that we can specify quantitatively the required system performance. Then a performance index can be calculated or measured and used to evaluate the system performance. Quantitative measures of the performance of a system are very valuable in the design and operation of control systems.
A system is considered an optimum control system when the system parameters are adjusted so that the index reaches an extremum, commonly a minimum value. To be useful, a performance index must be a number that is always positive or zero. Then the best system is defined as the system that minimizes this index.
175. A performance index is a quantitative measure of the performance of a system and is chosen so that emphasis is given to the important system specifications.
A common performance index is the integral of the square of the error, ISE, which is defined as
The upper limit \(T\) is a finite time selected by the control system designer. It is convenient to choose \(T\) as the settling time \(T_{s}\). The step response for a specific feedback control system is shown in Figure 5.23(b), and the error in Figure 5.23(c). The error squared is shown in Figure 5.23(d), and the integral of the error squared in Figure 5.23(e). This criterion will discriminate between excessively overdamped and excessively underdamped systems. The minimum value of the integral occurs for a compromise value of the damping. The performance index of Equation (5.37) is mathematically convenient for analytical and computational purposes.
Three other performance indices we might consider include
and
The ITAE is able to reduce the contribution of any large initial errors, as well as to emphasize errors occurring later in the response [6]. The performance index ITAE provides the best selectivity of the performance indices; that is, the minimum value of the integral is readily discernible as the system parameters are varied. FIGURE 5.23
The calculation of the Integral squared error. (a)
The general form of the performance integral is
where \(f\) is a function of the error, input, output, and time. We can obtain numerous indices based on various combinations of the system variables and time.
176. EXAMPLE 5.6 Space telescope control system
Consider a space telescope pointing control system shown in Figure 5.24 [9]. We desire to select the magnitude of the gain, \(K_{3}\), to minimize the effect of the disturbance, \(T_{d}(s)\). The closed-loop transfer function from the disturbance to the output is
FIGURE 5.24
A space telescope pointing control system. (a) Block diagram. (b) Signalflow graph.
(a)
(b)
Typical values for the constants are \(K_{1} = 0.5\) and \(K_{1}K_{2}K_{p} = 2.5\). In this case, the goal is to minimize \(y(t)\), where, for a unit step disturbance, the minimum ISE can be analytically calculated. The attitude is
where \(\beta = \sqrt{10 - K_{3}^{2}/4}\). Squaring \(y(t)\) and integrating the result yields
Differentiating \(I\) and equating the result to zero, and solving for \(K_{3}\), we obtain
FIGURE 5.25
The performance indices of the telescope control system versus \(K_{3}\).
Therefore, the minimum ISE is obtained when \(K_{3} = \sqrt{10} = 3.2\). This value of \(K_{3}\) corresponds to a damping ratio \(\zeta = 0.50\). The values of ISE and IAE for this system are plotted in Figure 5.25. The minimum for the IAE performance index is obtained when \(K_{3} = 4.2\) and \(\zeta = 0.665\). While the ISE criterion is not as selective as the IAE criterion, it is clear that it is possible to solve analytically for the minimum value of ISE. The minimum of IAE is obtained by computing the actual value of IAE for several values of the parameter of interest.
A control system is optimum when the selected performance index is minimized. However, the optimum value of the parameters depends directly on the definition of optimum, that is, the performance index. Therefore, in Example 5.6, we found that the optimum setting varied for different performance indices.
The coefficients that will minimize the ITAE performance criterion for a step input have been determined for the general closed-loop transfer function [6]
This transfer function has a steady-state error equal to zero for a step input. Note that the transfer function has \(n\) poles and no zeros. The optimum coefficients for the ITAE criterion are given in Table 5.3. The responses using optimum coefficients for a step input are given in Figure 5.26 for ISE, IAE, and ITAE. The responses are provided for normalized time \(\omega_{n}t\). Other standard forms based on different performance indices are available and can be useful in aiding the designer to determine the range of coefficients for a specific problem.
Table 5.3 The Optimum Coefficients of \(\mathbf{T}(\mathbf{s})\) Based on the ITAE Criterion for a Step Input
(a)
Step responses of a normalized transfer function using optimum coefficients for (a) ISE, (b) IAE, and (c) ITAE. The response is for normalized time, \(w_{n}^{t}\).
(b)
For a ramp input, the coefficients have been determined that minimize the ITAE criterion for the general closed-loop transfer function [6]
FIGURE 5.26 (Continued)
(c)
177. Table 5.4 The Optimum Coefficients of \(\mathbf{T}(s)\) Based
on the ITAE Criterion for a Ramp Input
This transfer function has a steady-state error equal to zero for a ramp input. The optimum coefficients for this transfer function are given in Table 5.4. The transfer function, Equation (5.50), implies that the process \(G(s)\) has two or more pure integrations, as required to provide zero steady-state error.
177.1. THE SIMPLIFICATION OF LINEAR SYSTEMS
It is quite useful to study complex systems with high-order transfer functions by using lower-order approximate models. Several methods are available for reducing the order of a systems transfer function. One relatively simple way to delete a certain insignificant pole of a transfer function is to note a pole that has a negative real part that is much more negative than the other poles. Thus, that pole is expected to affect the transient response insignificantly.
For example, if we have a system with transfer function
we can safely neglect the impact of the pole at \(s = - 30\). However, we must retain the steady-state response of the system, so we reduce the system to
A more sophisticated approach attempts to match the frequency response of the reduced-order transfer function with the original transfer function frequency response as closely as possible. Although frequency response methods are covered in Chapter 8, the associated approximation method strictly relies on algebraic manipulation and is presented here. Consider the high-order system be described by the transfer function
in which the poles are in the left-hand \(s\)-plane and \(m \leq n\). The lower-order approximate transfer function is
where \(p \leq g < n\). Notice that the gain constant, \(K\), is the same for the original and approximate system; this ensures the same steady-state response. The method outlined in Example 5.7 is based on selecting \(c_{i}\) and \(d_{i}\) in such a way that \(G_{L}(s)\) has a frequency response very close to that of \(G_{H}(s)\). This is equivalent to stating that \(G_{H}(j\omega)/G_{L}(j\omega)\) is required to deviate the least amount from unity for various frequencies. The \(c\) and \(d\) coefficients are obtained via
and
where \(M(s)\) and \(\Delta(s)\) are the numerator and denominator polynomials of \(G_{H}(s)/G_{L}(s)\), respectively. We also define
and an analogous equation for \(\Delta_{2q}\). The solutions for the \(c\) and \(d\) coefficients are obtained by equating
for \(q = 1,2,\ldots\) up to the number required to solve for the unknown coefficients.
178. EXAMPLE 5.7 A simplified model
Consider the third-order system
Using the second-order model
we determine that
Then we know that
and \(M^{(0)}(0) = 1\). Similarly, we have
Therefore, \(M^{(1)}(0) = d_{1}\). Continuing this process, we find that
We now equate \(M_{2q} = \Delta_{2q}\) for \(q = 1\) and 2. We find that, for \(q = 1\),
Since the equation for \(\Delta_{2}\) is similar, we have
Equation (5.53) with \(q = 1\) requires that \(M_{2} = \Delta_{2}\); therefore,
Completing the process for \(M_{4} = \Delta_{4}\), we obtain
Solving Equations (5.61) and (5.62) yields \(d_{1} = 1.615\) and \(d_{2} = 0.624\). (The other sets of solutions are rejected because they lead to unstable poles.) The lower-order system transfer function is
It is interesting to see that the poles of \(G_{H}(s)\) are \(s = - 1, - 2, - 3\), whereas the poles of \(G_{L}(s)\) are \(s = - 1.024\) and -1.565 . Because the lower-order model has two poles, we estimate that we would obtain a slightly overdamped step response with a settling time to within \(2\%\) of the final value in approximately 3 seconds.
It is sometimes desirable to retain the dominant poles of the original system, \(G_{H}(s)\), in the low-order model. This can be accomplished by specifying the denominator of \(G_{L}(s)\) to be the dominant poles of \(G_{H}(s)\) and allowing the numerator of \(G_{L}(s)\) to be subject to approximation.
Another novel and useful method for reducing the order is the Routh approximation method based on the idea of truncating the Routh table used to determine stability. The Routh approximants can be computed by a finite recursive algorithm [19].
178.1. DESIGN EXAMPLES
In this section we present two illustrative examples. The first example is a simplified view of the Hubble space telescope pointing control problem. The Hubble space telescope problem highlights the process of computing controller gains to achieve desired percent overshoot specifications, as well as meeting steady-state error specifications. The second example considers the control of the bank angle of an airplane. The airplane attitude motion control example represents a more in-depth look at the control design problem. Here we consider a complex fourthorder model of the lateral dynamics of the aircraft motion that is approximated by a second-order model using the approximation methods of Section 5.8. The simplified model can be used to gain insight into the controller design and the impact of key controller parameters on the transient performance.
179. EXAMPLE 5.8 Hubble space telescope control
The orbiting Hubble space telescope is the most complex and expensive scientific instrument that has ever been built. The telescope's 2.4 meter mirror has the
(a)
(b)
FIGURE 5.27
(a) The Hubble telescope pointing system, (b) reduced block diagram, and (c) system response to a unit step input command and a unit step disturbance input.
(c)
smoothest surface of any mirror made, and its pointing system can center it on a dime 400 miles away [18, 21]. Consider the model of the telescope-pointing system shown in Figure 5.27.
The goal of the design is to choose \(K_{1}\) and \(K\) so that (1) the percent overshoot of the output to a step command, \(r(t)\), is \(P.O. \leq 10\%,(2)\) the steady-state error to a ramp command is minimized, and (3) the effect of a step disturbance is reduced. Since the system has an inner loop, block diagram reduction can be used to obtain the simplified system of Figure 5.27(b). The output due to the two inputs of the system of Figure 5.27(b) is given by
where
The tracking error is
First, let us select \(K\) and \(K_{1}\) to meet the percent overshoot requirement for a step input, \(R(s) = A/s\). Setting \(T_{d}(s) = 0\), we have
To set the percent overshoot to P.O. \(\leq 10\%\), we select \(\zeta = 0.6\). We can use Equation (5.16) to determine that P.O. \(= 9.5\%\) for \(\zeta = 0.6\). We next examine the steady-state error for a ramp, \(r(t) = Bt,t \geq 0\). Using Equation 5.28 we find
The steady-state error due to the ramp disturbance is reduced by increasing \(KK_{1}\). The steady-state error due to a unit step disturbance is equal to \(- 1/K\). The steadystate error due to the step disturbance input can thus be reduced by increasing \(K\). In summary, we seek a large \(K\) and a large value of \(K/K_{1}\) to obtain low steady-state errors due to a step and ramp disturbance, respectively. We also require \(\zeta = 0.6\) to limit the percent overshoot.
With \(\zeta = 0.6\), the characteristic equation of the system is
Therefore, \(\omega_{n} = \sqrt{K}\), and the second term of the denominator of Equation (5.69) requires \(K_{1} = 2(0.6)\omega_{n}\). Then \(K_{1} = 1.2\sqrt{K}\), so the ratio \(K/K_{1}\) is
If we select \(K = 100\), we have \(K_{1} = 12\) and \(K/K_{1} = 8.33\). The responses of the system to a unit step input command and a unit step disturbance input are shown in Figure 5.27(c). Note how the effect of the disturbance is relatively insignificant. Finally, we note that the steady-state error for a ramp input is
This design, using \(K = 100\), provides acceptable results.
180. EXAMPLE 5.9 Attitude control of an airplane
Each time we fly on a commercial airliner, we experience first-hand the benefits of automatic control systems. These systems assist pilots by improving the handling qualities of the aircraft over a wide range of flight conditions and by providing pilot relief (for such emergencies as going to the restroom) during extended flights. The special relationship between flight and controls began in the early work of the Wright brothers. Using wind tunnels, the Wright brothers applied systematic design techniques to make their dream of powered flight a reality. This systematic approach to design contributed to their success.
Another significant aspect of their approach was their emphasis on flight controls; the brothers insisted that their aircraft be pilot-controlled. Observing birds control their rolling motion by twisting their wings, the Wright brothers built aircraft with mechanical mechanisms that twisted their airplane wings. Today we no longer use wing warping as a mechanism for performing a roll maneuver; instead we control rolling motion by using ailerons, as shown in Figure 5.28. The Wright brothers also used elevators (located forward) for longitudinal control (pitch motion) and rudders for lateral control (yaw motion). Today's aircraft still use both elevators and rudders, although the elevators are generally located on the tail (rearward).
The first controlled, powered, unassisted take-off flight occurred in 1903 with the Wright Flyer I (a.k.a. Kitty Hawk). The first practical airplane, the Flyer III, could fly figure eights and stay aloft for half an hour. Three-axis flight control was a major (and often overlooked) contribution of the Wright brothers. A concise historical perspective is presented in Stevens and Lewis [24]. The continuing desire to fly faster, lighter, and longer fostered further developments in automatic flight control.
The main topic of this chapter is control of the automatic rolling motion of an airplane. The elements of the design process emphasized in this chapter are illustrated in Figure 5.29.
We begin by considering the model of the lateral dynamics of an airplane moving along a steady, wings-level flight path. By lateral dynamics, we mean the attitude motion of the aircraft about the forward velocity. An accurate mathematical model describing the motion (translational and rotational) of an aircraft is a complicated set of highly nonlinear, time-varying, coupled differential equations. A good description of the process of developing such a mathematical model appears in Etkin and Reid [25].
For our purposes a simplified dynamic model is required for the autopilot design process. A simplified model might consist of a transfer function describing
FIGURE 5.28
Control of the bank angle of an airplane using differential deflections of the ailerons.
Topics emphasized in this example
FIGURE 5.29 Elements of the control system design process emphasized in the airplane attitude control example.
the input-output relationship between the aileron deflection and the aircraft bank angle. Obtaining such a transfer function would require many prudent simplifications to the original high-fidelity, nonlinear mathematical model.
Suppose we have a rigid aircraft with a plane of symmetry. The airplane is assumed to be cruising at subsonic or low supersonic \((\) Mach \(< 3)\) speeds. This allows us to make a flat-Earth approximation. We ignore any rotor gyroscopic effects due to spinning masses on the aircraft (such as propellors or turbines). These assumptions allow us to decouple the longitudinal rotational (pitching) motion from the lateral rotational (rolling and yawing) motion.
Of course, we also need to consider a linearization of the nonlinear equations of motion. To accomplish this, we consider only steady-state flight conditions such as
$\square\ $ Steady, wings-level flight
\(\square\) Steady, level turning flight
$\square\ $ Steady, symmetric pull-up
$\square\ $ Steady roll. For this example we assume that the airplane is flying at low speed in a steady, wings-level attitude, and we want to design an autopilot to control the rolling motion. We can state the control goal as follows:
181. Control Goal
Regulate the airplane bank angle to zero degrees (steady, wings level) and maintain the wings-level orientation in the presence of unpredictable external disturbances.
We identify the variable to be controlled as
182. Variable to Be Controlled
Airplane bank angle (denoted by \(\phi)\).
Defining system specifications for aircraft control is complicated, so we do not attempt it here. It is a subject in and of itself, and many engineers have spent significant efforts developing good, practical design specifications. The goal is to design a control system such that the dominant closed-loop system poles have satisfactory natural frequency and damping [24]. We must define satisfactory and choose test input signals on which to base our analysis.
The Cooper-Harper pilot opinion ratings provide a way to correlate the feel of the airplane with control design specifications [26]. These ratings address the handling qualities issues. Many flying qualities requirements are specified by government agencies, such as the United States Air Force [27]. The USAF MIL-F\(8785C\) is a source of time-domain control system design specifications.
For example we might design an autopilot control system for an aircraft in steady, wings-level flight to achieve a P.O. \(\leq 20\%\) to a step input with minimal oscillatory motion and rapid response time (that is, a short time-to-peak). Subsequently we implement the controller in the aircraft control system and conduct flight tests or high-fidelity computer simulations, after which the pilots tell us whether they liked the performance of the aircraft. If the overall performance was not satisfactory, we change the time-domain specification (in this case a percent overshoot specification) and redesign until we achieve a feel and performance that pilots (and ultimately passengers) will accept. Despite the simplicity of this approach and many years of research, precise-control system design specifications that provide acceptable airplane flying characteristics in all cases are still not available [24].
The control design specifications given in this example may seem somewhat contrived. In reality the specifications would be much more involved and, in many ways, less precisely known. But we must begin the design process somewhere. With that approach in mind, we select simple design specifications and begin the iterative design process. The design specifications are
183. Control Design Specifications
DS1 Percent overshoot is P.O. \(\leq 20\%\) for a unit step input.
DS2 Fast response time as measured by time-to-peak.
By making the simplifying assumptions discussed above and linearizing about the steady, wings-level flight condition, we can obtain a transfer function model describing the bank angle output, \(\phi(s)\), to the aileron deflection input, \(\delta_{a}(s)\). The transfer function has the form
The lateral (roll/yaw) motion has three main modes: Dutch roll mode, spiral mode, and roll subsidence mode. The Dutch roll mode, which gets its name from its similarities to the motion of an ice speed skater, is characterized by a rolling and yawing motion. The airplane center of mass follows nearly a straightline path, and a rudder impulse can excite this mode. The spiral mode is characterized by a mainly yawing motion with some roll motion. This is a weak mode, but it can cause an airplane to enter a steep spiral dive. The roll subsidence motion is almost a pure roll motion. This is the motion we are concerned with for our autopilot design. The denominator of the transfer function in Equation (5.69) shows two first-order modes (spiral and roll subsidence modes) and a second-order mode (Dutch roll mode).
In general the coefficients \(c_{0},b_{0},b_{1},d_{0},e_{0},f_{0},f_{1}\) and the gain \(k\) are complicated functions of stability derivatives. The stability derivatives are functions of the flight conditions and the aircraft configuration; they differ for different aircraft types. The coupling between the roll and yaw is included in Equation (5.69).
In the transfer function in Equation (5.69), the pole at \(s = - d_{0}\) is associated with the spiral mode. The pole at \(s = - e_{0}\) is associated with the roll subsidence mode. Generally, \(e_{0} \gg d_{0}\). For an F-16 flying at \(500ft/s\) in steady, wings-level flight, we have \(e_{0} = 3.57\) and \(d_{0} = 0.0128\) [24]. The complex conjugate poles given by the term \(s^{2} + f_{1}s + f_{0}\) represent the Dutch roll motion.
For low angles of attack (such as with steady, wings-level flight), the Dutch roll mode generally cancels out of the transfer function with the \(s^{2} + b_{1}s + b_{0}\) term. This is an approximation, but it is consistent with our other simplifying assumptions. Also, we can ignore the spiral mode since it is essentially a yaw motion only weakly coupled to the roll motion. The zero at \(s = c_{0}\) represents a gravity effect that causes the aircraft to sideslip as it rolls. We assume that this effect is negligible, since it is most pronounced in a slow roll maneuver in which the sideslip is allowed to build up, and we assume that the aircraft sideslip is small or zero. Therefore we can simplify the transfer function in Equation (5.69) to obtain a single-degree-of-freedom approximation:
For our aircraft we select \(e_{0} = 1.4\) and \(k = 11.4\). The associated time-constant of the roll subsidence is \(\tau = 1/e_{0} = 0.7\text{ }s\). These values represent a fairly fast rolling motion response.
For the aileron actuator model, we typically use a simple first-order system model,
where \(e(s) = \phi_{d}(s) - \phi(s)\). In this case we select \(p = 10\). This corresponds to a time constant of \(\tau = 1/p = 0.1\text{ }s\). This is a typical value consistent with a fast response. We need to have an actuator with a fast response so that the dynamics of the actively controlled airplane will be the dominant component of the system response. A slow actuator is akin to a time delay that can cause performance and stability problems.
For a high-fidelity simulation, we would need to develop an accurate model of the gyro dynamics. The gyro, typically an integrating gyro, is usually characterized by a very fast response. To remain consistent with our other simplifying assumptions, we ignore the gyro dynamics in the design process. This means we assume that the sensor measures the bank angle precisely. The gyro model is given by a unity transfer function,
Thus our physical system model is given by Equations (5.70), (5.71), and (5.72).
The controller we select for this design is a proportional controller,
The system configuration is shown in Figure 5.30. The select key parameter is as follows:
184. Select Key Tuning Parameter
Controller gain \(K\).
The closed-loop transfer function is
We want to determine analytically the values of \(K\) that will give us the desired response, namely, a percent overshoot less than \(20\%\) and a fast time-to-peak. The analytic analysis would be simpler if our closed-loop system were a second-order system (since we have valuable relationships between settling time, percent overshoot, natural frequency and damping ratio); however we have a third-order system, given by \(T(s)\) in Equation (5.73). We could consider approximating the third-order transfer function by a second-order transfer function - this is sometimes a very good engineering approach to analysis. There are many methods available to
FIGURE 5.30
Bank angle control autopilot.
obtain approximate transfer functions. Here we use the algebraic method described in Section 5.8 that attempts to match the frequency response of the approximate system as closely as possible to the actual system.
Our transfer function can be rewritten as
by factoring the constant term out of the numerator and denominator. Suppose our approximate transfer function is given by the second-order system
The objective is to find appropriate values of \(d_{1}\) and \(d_{2}\). As in Section 5.8, we define \(M(s)\) and \(\Delta(s)\) as the numerator and denominator of \(T(s)/G_{L}(s)\). We also define
and
Then, forming the set of algebraic equations
we can solve for the unknown parameters of the approximate function. The index \(q\) is incremented until sufficient equations are obtained to solve for the unknown coefficients of the approximate function. In this case, \(q = 1,2\) since we have two parameters \(d_{1}\) and \(d_{2}\) to compute.
We have
Thus evaluating at \(s = 0\) yields
Similarly,
Evaluating at \(s = 0\), it follows that
Using Equation (5.77) for \(q = 1\) and \(q = 2\) yields
and
Similarly using Equation (5.78), we find that
Thus forming the set of algebraic equations in Equation (5.79),
we obtain
Solving for \(d_{1}\) and \(d_{2}\) yields
where we always choose the positive values of \(d_{1}\) and \(d_{2}\) so that \(G_{L}(s)\) has poles in the left half-plane. Thus (after some manipulation) the approximate transfer function is
We require that \(K < 0.65\) so that the coefficient of the \(s\) term remains a real number.
Our desired second-order transfer function can be written as
Comparing coefficients in Equations (5.79) and (5.80) yields
The design specification that the percent overshoot P.O. is to be less than \(20\%\) implies that we want \(\zeta \geq 0.45\). Setting \(\zeta = 0.45\) in Equation (5.81) and solving for \(K\) yields
With \(K = 0.16\) we compute
Then we can estimate the time-to-peak \(T_{p}\) from Equation (5.14) to be
We might be tempted at this point to select \(\zeta > 0.45\) so that we reduce the percent overshoot even further than \(20\%\). What happens if we decide to try this approach? From Equation (5.81) we see that \(K\) decreases as \(\zeta\) increases. Then, since
FIGURE 5.31 Step response comparison of third-order aircraft model versus second-order approximation.
as \(K\) decreases, then \(\omega_{n}\) also decreases. But the time-to-peak increases as \(\omega_{n}\) decreases. Since our goal is to meet the specification of percent overshoot less than \(20\%\) while minimizing the time-to-peak, we use the initial selection of \(\zeta = 0.45\) so that we do not increase \(T_{p}\) unnecessarily.
The second-order system approximation has allowed us to gain insight into the relationship between the parameter \(K\) and the system response, as measured by percent overshoot and time-to-peak. Of course, the gain \(K = 0.16\) is only a starting point in the design because we in fact have a third-order system and must consider the effect of the third pole (which we have ignored so far).
A comparison of the third-order aircraft model in Equation (5.73) with the second-order approximation in Equation (5.79) for a unit step input is shown in Figure 5.31. The step response of the second-order system is a good approximation of the original system step response, so we would expect that the analytic analysis using the simpler second-order system to provide accurate indications of the relationship between \(K\) and the percent overshoot and time-to-peak.
With the second-order approximation, we estimate that with \(K = 0.16\) the percent overshoot is P.O. \(= 20\%\) and the time-to-peak is \(T_{p} = 2.62\text{ }s\). As shown in Figure 5.32 the percent overshoot of the original third-order system is P.O. \(= 20.5\%\) and the time-to-peak is \(T_{p} = 2.73\text{ }s\). Thus, we see that that analytic analysis using the approximate system is an excellent predictor of the actual response. For comparison purposes, we select two variations in the gain and observe the response. For \(K = 0.1\), the percent overshoot is \(P.O. = 9.5\%\) and the time-to-peak is \(T_{p} = 3.74\text{ }s\). FIGURE 5.32
Step response of the third-order aircraft model with \(K = 0.10,0.16\), and 0.20 showing that, as predicted, as \(K\) decreases percent overshoot decreases while the time-to-peak increases.
For \(K = 0.2\), the percent overshoot is P.O. \(= 26.5\%\) and the time-to-peak is \(T_{p} = 2.38\text{ }s\). So as predicted, as \(K\) decreases the damping ratio increases, leading to a reduction in the percent overshoot. Also as predicted, as the percent overshoot decreases the time-to-peak increases.
184.1. SYSTEM PERFORMANCE USING CONTROL DESIGN SOFTWARE
In this section, we investigate time-domain performance specifications given in terms of transient response to a given input signal and the resulting steady-state tracking errors. We conclude with a discussion of the simplification of linear systems. The function introduced in this section is impulse. We discuss the Isim function and see how these functions are used to simulate a linear system.
Time-Domain Specifications. Time-domain performance specifications are generally given in terms of the transient response of a system to a given input signal. Because the actual input signals are generally unknown, a standard test input signal is used. Consider the second-order system shown in Figure 5.3. The closed-loop output is
We have already discussed the use of the step function to compute the step response of a system. Now we address another important test signal: the impulse.
FIGURE 5.33 The impulse function.
The impulse response is the time derivative of the step response. We compute the impulse response with the impulse function shown in Figure 5.33.
We can obtain a plot similar to that of Figure 5.4 with the step function, as shown in Figure 5.34. Using the impulse function, we can obtain a plot similar to that of Figure 5.5. The response of a second-order system for an impulse function input is shown in Figure 5.35. In the script, we set \(\omega_{n} = 1\), which is equivalent to computing the step response versus \(\omega_{n}t\). This gives us a more general plot valid for any \(\omega_{n} > 0\).
In many cases, it may be necessary to simulate the system response to an arbitrary but known input. In these cases, we use the Isim function. The Isim function is shown in Figure 5.36.
185. EXAMPLE 5.10 Mobile robot steering control
The block diagram for a steering control system for a mobile robot is shown in Figure 5.18. Suppose the transfer function of the steering controller is
When the input is a ramp, the steady-state error is
where
FIGURE 5.34
(a) Response of a second-order system to a step input. (b) m-file script.
(a)
(b)
The effect of the controller constant, \(K_{2}\), on the steady-state error is evident from Equation (5.83). Whenever \(K_{2}\) is large, the steady-state error is small.
We can simulate the closed-loop system response to a ramp input using the Isim function. The controller gains, \(K_{1}\) and \(K_{2}\), and the system gain \(K\) can be represented symbolically in the script so that various values can be selected and simulated. The results are shown in Figure 5.37 for \(K_{1} = K = 1,K_{2} = 2\), and \(\tau = 1/10\). FIGURE 5.35
(a) Response of a second-order system to an impulse function input. (b) m-file script.
(a)
(b)
Simplification of Linear Systems. It may be possible to develop a lower-order approximate model that closely matches the input-output response of a high-order model. A procedure for approximating transfer functions is given in Section 5.8. We can use computer simulation to compare the approximate model to the actual model, as illustrated in the following example. FIGURE 5.36
The Isim function.
FIGURE 5.37
(a) Transient response of the mobile robot steering control system to a ramp input. (b) \(m\)-file script.
(a)
%Compute the response of the Mobile Robot Control \(\%\) System to a triangular wave input \(\%\) numg=[10 20]; deng=[1 10 0]; sysg=tf(numg,deng); [sys]=feedback(sysg, [1]); \(t = \lbrack 0:0.1:8.2\rbrack\); \(v1 = \lbrack 0:0.1:2\rbrack^{'};v2 = \lbrack 2: - 0.1: - 2\rbrack^{'};v3 = \lbrack - 2:0.1:0\rbrack^{'};\) \(u = \lbrack v1;v2;v3\rbrack\);
\(\lbrack y,T\rbrack = Isim(sys,u,t)\); \(plot\left( T,y,t,u,'^{- -} - \right)\), xlabel('Time (s)'), ylabel('theta (rad)'), grid
Compute triangular wave input.
Linear simulation.
186. EXAMPLE 5.11 A simplified model
Consider the third-order system
A second-order approximation is
A comparison of their respective step responses is given in Figure 5.38.
(a)
(b)
FIGURE 5.38 (a) Step response comparison for an approximate transfer function versus the actual transfer function. (b) m-file script.
186.1. SEQUENTIAL DESIGN EXAMPLE: DISK DRIVE READ SYSTEM
In this section, we further consider the design process of the disk drive read system. We will specify the desired performance for the system. Then we will attempt to adjust the amplifier gain \(K_{a}\) in order to obtain the best performance possible.
Our goal is to achieve the fastest response to a step input \(r(t)\) while (1) limiting the percent overshoot and oscillatory nature of the response and (2) reducing the effect of a disturbance on the output position of the read head. The specifications are summarized in Table 5.5.
Let us consider the second-order model of the motor and arm, which neglects the effect of the coil inductance. We then have the closed-loop system shown in Figure 5.39. Then the output when \(T_{d}(s) = 0\) is
Therefore, \(\omega_{n}^{2} = 5K_{a}\), and \(2\zeta\omega_{n} = 20\). We then determine the response of the system as shown in Figure 5.40. Table 5.6 shows the performance measures for selected values of \(K_{a}\).
When \(K_{a}\) is increased to 60 , the effect of a disturbance is reduced by a factor of 2 . We can show this by plotting the output, \(y(t)\), as a result of a unit step disturbance
| Table 5.5 | Specifications for the Transient Response |
| Performance Measure | Desired Value |
| Percent overshoot | Less than \(5\%\) |
| Settling time | Less than \(250\text{ }ms\) |
| $$\begin | |
| \text{~}\text{Maximum value of response}\text{~} \ | |
| \text{~}\text{to a unit step disturbance}\text | |
| \end{matrix}$$ | Less than \(5 \times 10^{- 3}\) |
FIGURE 5.39
Control system model with a second-order model of the motor and load.
FIGURE 5.40 Response of the system to a unit step input, \(r(t) = 1,t > 0\). (a) \(m\)-file script. (b) Response for \(K_{a} = 30\) and 60 .
(a)
(b)
187. Table 5.6 Response for the Second-Order Model for a Step Input
| $$K_{a}$$ | 20 | 30 | 40 | 60 | 80 |
| Percent overshoot | 0 | $$1.2%$$ | $$4.3%$$ | $$10.8%$$ | $$16.3%$$ |
| Settling time \((s)\) | 0.55 | 0.40 | 0.40 | 0.40 | 0.40 |
| Damping ratio | 1 | 0.82 | 0.707 | 0.58 | 0.50 |
| $$\begin | |||||
| \text{~}\text{Maximum value of the}\text{~} \ | |||||
| \text{}\text{response}\text{}y(t)\text{~}\text{to a unit}\text | |||||
| \end{matrix}$$ | $$- 10 \times 10^{- 3}$$ | $$- 6.6 \times 10^{- 3}$$ | $$- 5.2 \times 10^{- 3}$$ | $$- 3.7 \times 10^{- 3}$$ | $$- 2.9 \times 10^{- 3}$$ |
| disturbance |
input, as shown in Figure 5.41. Clearly, if we wish to meet our goals with this system, we need to select a compromise gain. In this case, we select \(K_{a} = 40\) as the best compromise. However, this compromise does not meet all the specifications. In the next chapter, we consider again the design process and change the configuration of the control system. FIGURE 5.41
Response of the system to a unit step disturbance, \(T_{d}(s) = 1/s\).
(a) m-file script.
(b) Response for \(K_{a} = 30\) and 60 .
(a)
(b)
187.1. SUMMARY
In this chapter, we have considered the definition and measurement of the performance of a feedback control system. The concept of a performance measure or index was discussed, and the usefulness of standard test signals was outlined. Then, several performance measures for a standard step input test signal were delineated. For example, the overshoot, peak time, and settling time of the response of the system under test for a step input signal were considered. The fact that the specifications on the desired response are often contradictory was noted, and the concept of a design compromise was proposed. The relationship between the location of the \(s\)-plane root of the system transfer function and the system response was discussed. A most important measure of system performance is the steady-state error for specific test input signals. Thus, the relationship of the steady-state error of a system in terms of the system parameters was developed by utilizing the final-value theorem. Finally, the utility of an integral performance index was outlined, and several design examples that minimized a system's performance index were completed. Thus, we have been concerned with the definition and usefulness of quantitative measures of the performance of feedback control systems.
188. 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 5.42 as specified in the various problem statements.
FIGURE 5.42 Block diagram for the Skills Check.
In the following True or False and Multiple Choice problems, circle the correct answer.
- In general, a third-order system can be approximated by a second-order system's dominant roots if the real part of the dominant roots is less than \(1/10\) of the real part of the third root.
True or False
- The number of zeros of the forward path transfer function at the origin is called the type number.
True or False
- The rise time is defined as the time required for the system to settle within a certain percentage of the input amplitude.
True or False
- For a second-order system with no zeros, the percent overshoot to a unit step is a function of the damping ratio only.
True or False
- A type-1 system has a zero steady-state tracking error to a ramp input.
True or False Consider the closed-loop control system in Figure 5.42 for Problems 6 and 7 with
-
The steady-state error to a unit step input \(R(s) = 1/s\) is:
a. \(e_{ss} = \lim_{t \rightarrow \infty}\mspace{2mu} e(t) = 0\)
b. \(e_{ss} = \lim_{t \rightarrow \infty}\mspace{2mu} e(t) = 1/2\)
c. \(e_{ss} = \lim_{t \rightarrow \infty}\mspace{2mu} e(t) = 1/6\)
d. \(e_{ss} = \lim_{t \rightarrow \infty}\mspace{2mu} e(t) = \infty\) -
The percent overshoot of the output to a unit step input is:
a. \(P.O. = 9\%\)
b. \(P.O. = 1\%\)
c. \(P.O. = 20\%\)
d. No overshoot
Consider the block diagram of the control system shown in Figure 5.42 in Problems 8 and 9 with the loop transfer function
-
Find the value of \(K\) so that the system provides an optimum ITAE response for a step input.
a. \(K = 1.10\)
b. \(K = 12.56\)
c. \(K = 51.02\)
d. \(K = 104.7\) -
Compute the expected percent overshoot to a unit step input for the value of \(K\) in Problem 8.
a. \(P.O. = 1.4\%\)
b. \(P.O. = 4.6\%\)
c. \(P.O. = 10.8\%\)
d. No overshoot expected -
A system has the closed-loop transfer function \(T(s)\) given by
Using the notion of dominant poles, estimate the expected percent overshoot.
a. \(P.O. \approx 5\%\)
b. \(P.O. \approx 20\%\)
c. \(P.O. \approx 50\%\)
d. No overshoot expected
- Consider the unity feedback control system in Figure 5.42 where
The design specifications are:
i. Peak time \(T_{p} \leq 1.0\)
ii. Percent overshoot P.O. \(\leq 10\%\).
With \(K\) as the design parameter, it follows that
a. Both specifications can be satisfied.
b. Only the first specification \(T_{p} \leq 1.0\) can be satisfied.
c. Only the second specification \(P.O. \leq 10\%\) can be satisfied.
d. Neither specification can be satisfied.
- Consider the feedback control system in Figure 5.43 where \(G(s) = \frac{K}{s + 10}\).
FIGURE 5.43 Feedback system with integral controller and derivative measurement. The nominal value of \(K = 10\). Using a \(2\%\) criterion, compute the settling time, \(T_{s}\), for a unit step disturbance, \(T_{d}(s) = 1/s\), when \(R(s) = 0\).
a. \(T_{s} = 0.02\text{ }s\)
b. \(T_{s} = 0.2\text{ }s\)
c. \(T_{s} = 1.03\text{ }s\)
d. \(T_{s} = 4.83\text{ }s\)
- A plant has the transfer function given by
and is controlled by a proportional controller \(G_{c}(s) = K\), as shown in the block diagram in Figure 5.42. The value of \(K\) that yields a steady-state error \(E(s) = Y(s) - R(s)\) with a magnitude equal to 0.01 for a unit step input is:
a. \(K = 49\)
b. \(K = 99\)
c. \(K = 169\)
d. None of the above
In Problems 14 and 15, consider the control system in Figure 5.42, where
-
A second-order approximate model of the loop transfer function is:
a. \({\widehat{G}}_{c}(s)\widehat{G}(s) = \frac{(3/25)K}{s^{2} + 7s + 10}\)
b. \({\widehat{G}}_{c}(s)\widehat{G}(s) = \frac{(1/25)K}{s^{2} + 7s + 10}\)
c. \({\widehat{G}}_{c}(s)\widehat{G}(s) = \frac{(3/25)K}{s^{2} + 7s + 500}\)
d. \({\widehat{G}}_{c}(s)\widehat{G}(s) = \frac{6K}{s^{2} + 7s + 10}\) -
Using the second-order system approximation (see Problem 14), estimate the gain \(K\) so that the percent overshoot is approximately \(P.O. \approx 15\%\).
a. \(K = 10\)
b. \(K = 300\)
c. \(K = 1000\)
d. None of the above
In the following Word Match problems, match the term with the definition by writing the correct letter in the space provided.
a. Unit impulse
b. Rise time
c. Settling time
The time for a system to respond to a step input and rise to a peak response.
The roots of the characteristic equation that cause the dominant transient response of the system.
The number \(N\) of poles of the transfer function, \(G(s)\), at the origin. d. Type number
e. Percent overshoot
f. Position error constant, \(K_{P}\)
g. Velocity error constant, \(K_{v}\)
h. Steady-state response
i. Peak time
j. Dominant roots
k. Test input signal
- Acceleration error constant, \(K_{a}\)
m. Transient response
n. Design specifications
o. Performance index
p. Optimum control system
The constant evaluated as \(\lim_{s \rightarrow 0}\mspace{2mu} sG(s)\).
An input signal used as a standard test of a system's ability to respond adequately.
