Modern Control Systems_P3
The time required for the system output to settle within a certain percentage of the input amplitude.
A set of prescribed performance criteria.
A system whose parameters are adjusted so that the performance index reaches an extremum value.
A quantitative measure of the performance of a system.
The time for a system to respond to a step input and attain a response equal to a percentage of the magnitude of the input.
The amount by which the system output response proceeds beyond the desired response.
The constant evaluated as \(\lim_{s \rightarrow 0}\mspace{2mu} s^{2}G(s)\).
The constant evaluated as \(\lim_{s \rightarrow 0}\mspace{2mu} G(s)\).
The constituent of the system response that exists a long time following any signal initiation.
The constituent of the system response that disappears with time.
A test input consisting of an impulse of infinite amplitude and zero width, and having an area of unity.
189. EXERCISES
E5.1 A laser cutter is used to cut a parabolic path on sheet metal. We want to have a finite steady-state error for the laser beam positioning control system. (a) Which type of number system is required? (How many integrations?) (b) If we want to achieve a zero steady-state error, which type of number system is required?
E5.2 The engine, body, and tires of a racing vehicle affect the acceleration and speed attainable [9]. The speed control of the car is represented by the model shown in Figure E5.2. (a) Calculate the steady-state error of the car to a step command in speed. (b) Calculate overshoot of the speed to a step command.
Answer: (a) \(e_{ss} = A/11\); (b) P.O. \(= 36\%\)
E5.3 New passenger rail systems that could profitably compete with air travel are under development. Two of these systems, the French TGV and the Japanese Shinkansen, reach speeds of \(160mph\) [17]. The Trans-rapid, a magnetic levitation train, is shown in Figure E5.3(a).
The use of magnetic levitation and electromagnetic propulsion to provide contactless vehicle
FIGURE E5.2 Racing car speed control.
movement makes the Transrapid technology radically different. The underside of the carriage (where the wheel trucks would be on a conventional car) wraps around a guideway. Magnets are attached to the wraparound and pull the train to the reaction rail at the bottom of the guideway.
The levitation control is represented by Figure E5.3(b). (a) Select \(K\) so that the system provides an optimum ITAE response. (b) Determine the expected percent overshoot to a step input of \(I(s)\).
Answer: \(K = 100;4.6\%\)
(a)
(b)
FIGURE E5.3 Levitated train control. (Bernd Mellmann/ Alamy Stock Photo.)
E5.4 A feedback system with negative unity feedback has a loop transfer function
(a) Determine the closed-loop transfer function \(T(s) = Y(s)/R(s)\). (b) Find the time response, \(y(t)\), for a step input \(r(t) = A\) for \(t > 0\). (c) Determine the percent overshoot of the response. (d) Using the final-value theorem, determine the steady-state value of \(y(t)\).
Answer: (b) \(y(t) = 1 - 1.07e^{- 3t}sin(\sqrt{7}t + 1.2)\)
E5.5 Consider the feedback system in Figure E5.5. Find \(K\) such that the closed-loop system minimizes the ITAE performance criterion for a step input.
E5.6 Consider the block diagram shown in Figure E5.6 [16]. (a) Calculate the steady-state error for a ramp input. (b) Select a value of \(K\) that will result in zero percent overshoot to a step input. Provide rapid response.
Plot the poles and zeros of this system and discuss the dominance of the complex poles. What overshoot for a step input do you expect?
Position feedback
FIGURE E5.6 Block diagram with position and velocity feedback.
E5.7 Effective control of insulin injections can result in better lives for diabetic persons. Automatically controlled insulin injection by means of a pump and a sensor that measures blood sugar can be very effective. A pump and injection system has a feedback control as shown in Figure E5.7. Calculate the suitable gain \(K\) so that the percent overshoot of the step response due to the drug injection is P.O. \(= 7\%.R(s)\) is the desired blood-sugar level and \(Y(s)\) is the actual blood-sugar level.
Answer: \(K = 1.67\)
E5.8 A control system for positioning the head of a floppy disk drive has the closed-loop transfer function
Plot the poles and zeros of this system, and discuss the dominance of the third pole. What percent overshoot for a step input do you expect?
E5.9 A unity negative feedback control system has the loop transfer function
(a) Determine the percent overshoot and settling time (using a \(2\%\) settling criterion) due to a unit step input.
(b) For what range of \(K\) is the settling time is approximately \(T_{s} \leq 1\text{ }s\) ?
FIGURE E5.5
Feedback system with proportional controller \(G_{c}(s) = K\).
FIGURE E5.7
Blood-sugar level control.
E5.10 A second-order control system has the closedloop transfer function \(T(s) = Y(s)/R(s)\). The system specifications for a step input follow:
-
Percent overshoot P.O. \(\leq 5\%\).
-
Settling time \(T_{s} < 4\text{ }s\).
-
Peak time \(T_{p} < 1\text{ }s\).
Show the desired region for the poles of \(T(s)\) in order to achieve the desired response. Use a \(2\%\) settling criterion to determine settling time.
E5.11 A system with unity feedback is shown in Figure E5.11. Determine the steady-state error for a step and a ramp input when
FIGURE E5.11 Unity feedback system.
E5.12 The Ferris wheel is often featured at state fairs and carnivals. George Ferris was born in Galesburg, Illinois, in 1859; he later moved to Nevada and then graduated from Rensselaer Polytechnic Institute in 1881. By 1891, Ferris had considerable experience with iron, steel, and bridge construction. He conceived and constructed his famous wheel for the 1893 Columbian Exposition in Chicago [8]. Consider the requirement that the steady-state speed must be controlled to within \(5\%\) of the desired speed for the Ferris wheel speed control system shown in Figure E5.12.
(a) Determine the required gain \(K\) to achieve the steady-state requirement. (b) For the gain of part (a), determine and plot the tracking error for a unit step disturbance. Does the speed change more than \(5\%\) ? (Set \(R(s) = 0\) and recall that the tracking error \(E(s) = R(s) - T(s)\).)
E5.13 For the system with unity feedback shown in Figure E5.11, determine the steady-state error for a step and a ramp input when
Answer: \(e_{ss} = 0.71\) for a step and \(e_{ss} = \infty\) for a ramp.
E5.14 A feedback system is shown in Figure E5.14.
(a) Determine the steady-state error for a unit step when \(K = 0.6\) and \(G_{p}(s) = 1\).
(b) Select an appropriate value for \(G_{p}(s)\) so that the steady-state error is equal to zero for the unit step input.
FIGURE E5.14 Feedback system.
E5.15 A closed-loop control system has a transfer function \(T(s)\) as follows:
190. Disturbance
FIGURE E5.12
Speed control of a Ferris wheel. Plot \(y(t)\) for a unit step input when (a) the actual \(T(s)\) is used, and (b) using the dominant complex poles. Compare the results.
E5.16 A second-order system is
Consider the case where \(1 < z < 8\). Obtain the partial fraction expansion, and plot the output for a unit step input for \(z = 2,4\), and 6 .
E5.17 A closed-loop control system transfer function \(T(s)\) has two dominant complex conjugate poles. Sketch the region in the left-hand \(s\)-plane where the complex poles should be located to meet the given specifications.
(a) \(0.6 \leq \zeta \leq 0.8,\ \omega_{n} \leq 10\)
(b) \(0.5 \leq \zeta \leq 0.707,\ \omega_{n} \geq 10\)
(c) \(\zeta \geq 0.5,\ 5 \leq \omega_{n} \leq 10\)
(d) \(\zeta \leq 0.707,\ 5 \leq \omega_{n} \leq 10\)
(c) \(\zeta \geq 0.6,\ \omega_{n} \leq 6\)
E5.18 A system is shown in Figure E5.18(a). The response to a unit step, when \(K = 1\), is shown in Figure E5.18(b). Determine the value of \(K\) so that the steady-state error is equal to zero.
Answer: \(K = 1.25\).
(a)
(b)
191. FIGURE E5.18 Feedback system with prefilter.
E5.19 A second-order system has the closed-loop transfer function
(a) Estimate the percent overshoot P.O., the time to peak \(T_{p}\), and the settling time \(T_{s}\) of the unit step response.
(b) Obtain the system response to a unit step, and verify the results in part (a).
E5.20 Consider the closed-loop system in Figure E5.20, where
(a) Determine the closed-loop transfer function \(T(s)\) \(= Y(s)/R(s)\).
(b) Determine the steady-state error of the closedloop system response to a unit ramp input.
(c) Select a value for \(K_{a}\) so that the steady-state error of the system response to a unit step input is zero.
FIGURE E5.20 Nonunity closed-loop feedback control system with parameter \(K_{a}\).
192. PROBLEMS
P5.1 An important problem for television systems is the jumping or wobbling of the picture due to the movement of the camera. This effect occurs when the camera is mounted in a moving truck or airplane. The Dynalens system has been designed to reduce the effect of rapid scanning motion; see Figure P5.1. A maximum scanning motion of \(25\%\) is expected.
Let \(K_{g} = K_{t} = 1\) and assume that \(\tau_{g}\) is negligible. (a) Determine the error of the system \(E(s)\). (b) Determine the necessary loop gain \(K_{a}K_{m}K_{t}\) when a \(1\%\) steady-state error is allowable. (c) The motor time constant is \(\tau_{m} = 0.40\text{ }s\). Determine the necessary loop gain so that the settling time (to within \(2\%\) of the final value of \(v_{b}\) ) is \(T_{s} \leq 0.03\text{ }s\).
(a)
(b)
FIGURE P5.1 Camera wobble control.
P5.2 A specific closed-loop control system is to be designed for an underdamped response to a step input. The specifications for the system are as follows:
(a) Identify the desired area for the dominant roots of the system. (b) Determine the smallest value of a third root \(r_{3}\) if the complex conjugate roots are to represent the dominant response. (c) The closed-loop system transfer function \(T(s)\) is third-order, and the feedback has a unity gain. Determine the loop transfer function \(G(s) = Y(s)/E(s)\) when the settling time to within \(2\%\) of the final value is \(T_{s} = 0.6\text{ }s\) and the percent overshoot is P.O. \(= 20\%\).
P5.3 A laser beam can be used to weld, drill, etch, cut, and mark metals, as shown in Figure P5.3(a) [14]. Assume we have a work requirement for an accurate laser to mark a linear path with a closed-loop control system, as shown in Figure P5.3(b). Calculate the necessary gains \(K\) and \(K_{1}\) to result in a steady-state error of \(5\text{ }mm\) for \(r(t) = t\text{ }mm\).
P5.4 The loop transfer function of a unity negative feedback system
(a)
(b)
FIGURE P5.3 Laser beam control.
A system response to a step input is specified as follows:
(a) Determine whether both specifications can be met simultaneously. (b) If the specifications cannot be met simultaneously, determine a compromise value for \(K\) so that the peak time and percent overshoot specifications are relaxed by the same percentage.
P5.5 A space telescope is to be launched to carry out astronomical experiments [8]. The pointing control system is desired to achieve 0.01 minute of arc and track solar objects with apparent motion up to 0.21 arc minute per second. The system is illustrated in Figure P5.5(a). The control system is shown in Figure P5.5(b). Assume that \(\tau_{1} = 1\text{ }s\) and \(\tau_{2} = 0\). (a) Determine the gain \(K = K_{1}K_{2}\) required so that the response to a unit step command is as rapid as reasonable with a percent overshoot of P.O. \(\leq 5\%\). (b) Determine the steadystate error of the system for a step and a ramp input.
P5.6 A robot is programmed to have a tool or welding torch follow a prescribed path [7,11]. Consider a robot tool that is to follow a sawtooth path, as shown in Figure P5.6(a). The loop transfer function of the plant is
FIGURE P5.5
(a) The space telescope. (b) The space telescope pointing control system.
(a)
(1)
(a)
(b)
FIGURE P5.6
Robot path control. in Figure P5.7(b). The moment of inertia of the equipment and astronaut is \(I = 25\text{ }kg{\text{ }m}^{2}\). (a) Determine the necessary gain \(K_{3}\) to maintain a steady-state error equal to \(1\text{ }cm\) when the input is a unit ramp. (b) With this gain \(K_{3}\), determine the necessary gain \(K_{1}K_{2}\) in order to restrict the percent overshoot to P.O. \(\leq 10\%\).
(a)
FIGURE P5.7
(a) Astronaut Bruce McCandless II is shown a few meters away from the Earth-orbiting space shuttle. He used a nitrogenpropelled hand-controlled device called the manned maneuvering unit. (Courtesy of NASA.) (b) Block diagram.
(b)
P5.8 Photovoltaic arrays generate a DC voltage that can be used to drive DC motors or that can be converted to \(AC\) power and added to the distribution network. It is desirable to maintain the power out of the array at its maximum available as the solar incidence changes during the day. One such closed-loop system is shown in Figure P5.8. The transfer function for the process is
where \(K = 40\). (a) Compute the closed-loop transfer function, and (b) determine the settling time to within \(2\%\) of the final value of the system to a unit step disturbance.
P5.9 Antennas that receive and transmit signals to communication satellites generally include an extremely large horn antenna. The microwave antenna can be \(175ft\) long and weigh 340 tons. A photo of an antenna is shown in Figure P5.9. Suppose that the communication satellite is \(3ft\) in diameter and moves at about
\(16,000mph\) at an altitude of 2500 miles. The antenna must be positioned accurately to \({0.1}^{\circ}\) because the microwave beam is \({0.2}^{\circ}\) wide and highly attenuated by the large distance. If the antenna is following the moving satellite, determine the Kv necessary for the system.
FIGURE P5.9 A large antenna receives and transmits signals to a communication satellite. (Gary Woods/Alamy Stock Photo.)
P5.10 A speed control system of an armature-controlled DC motor uses the back emf voltage of the motor as a feedback signal. (a) Draw the block diagram of this system (see Example 2.5). (b) Calculate the steady-state error of this system to a step input command setting the speed to a new level. Assume that \(R_{a} = L_{a} = J = b = 1\), the motor constant is \(K_{m} = 1\), and \(K_{b} = 1\). (c) Select a feedback gain for the back emf signal to yield a step response with a percent overshoot of P.O. \(= 15\%\).
P5.11 A unity feedback control system has a process transfer function
The system input is a step function with an amplitude \(A\). The initial condition of the system at time \(t_{0}\) is \(y\left( t_{0} \right) = Q\), where \(y(t)\) is the output of the system. The performance index is defined as
(a) Show that \(I = (A - Q)^{2}/(2K)\). (b) Determine the gain \(K\) that will minimize the performance index \(I\). Is this gain a practical value? (c) Select a practical value of gain and determine the resulting value of the performance index.
P5.12 Train travel between cities will increase as trains are developed that travel at high speeds, making the travel time from city center to city center equivalent to airline travel time. The Japanese National Railway has a train called the Shinkansen train that travels an average speed of \(320\text{ }km/hr\) [17]. To maintain a desired speed, a speed control system is proposed that yields a zero steady-state error to a ramp input. A third-order system is sufficient. Determine the optimum system transfer function \(T(s)\) for an ITAE performance criterion. Estimate the settling time (with a \(2\%\) criterion) and percent overshoot for a step input when \(\omega_{n} = 10\).
P5.13 We want to approximate a fourth-order system by a lower-order model. The transfer function of the original system is
Show that if we obtain a second-order model by the method of Section 5.8, and we do not specify the poles and the zero of \(G_{L}(s)\), we have
P5.14 For the original system of Problem P5.13, we want to find the lower-order model when the poles of the second-order model are specified as -1 and -2 and the model has one unspecified zero. Show that this low-order model is
P5.15 Consider a unity feedback system with loop transfer function
Determine the value of the gain \(K\) such that the percent overshoot to a unit step is minimized.
P5.16 A magnetic amplifier with a low-output impedance is shown in Figure P5.16 in cascade with a low-pass filter and a preamplifier. The amplifier has a high-input impedance and a gain of 1 and is used for adding the signals as shown. Select a value for the capacitance \(C\) so that the transfer function \(V_{0}(s)/V_{\text{in}\text{~}}(s)\) has a damping ratio of \(1/\sqrt{2}\). The time constant of the magnetic amplifier is equal to 1 second, and the gain is \(K = 10\). Calculate the settling time (with a \(2\%\) criterion) of the resulting system. FIGURE P5.16
Feedback amplifier.
FIGURE P5.17
Heart pacemaker.
P5.17 Electronic pacemakers for human hearts regulate the speed of the heart pump. A proposed closed-loop system that includes a pacemaker and the measurement of the heart rate is shown in Figure P5.17 \(\lbrack 2,3\rbrack\). The transfer function of the heart pump and the pacemaker is found to be
Design the amplifier gain to yield a system with a settling time to a step disturbance of less than 1 second. The percent overshoot to a step in desired heart rate should be P.O. \(\leq 10\%\). (a) Find a suitable range of \(K\). (b) If the nominal value of \(K\) is \(K = 10\), find the sensitivity of the system to small changes in \(K\). (c) Evaluate the sensitivity of part (b) at \(DC\) (set \(s = 0\) ). (d) Evaluate the magnitude of the sensitivity at the normal heart rate of 60 beats/minute.
P5.18 Consider the third-order system
Determine a first-order model with one pole unspecified and no zeros that will represent the third-order system.
P5.19 A closed-loop control system with negative unity feedback has a loop transfer function
(a) Determine the closed-loop transfer function \(T(s)\).
(b) Determine a second-order approximation for \(T(s)\).
(c) Plot the response of \(T(s)\) and the second-order approximation to a unit step input and compare the results.
P5.20 A system is shown in Figure P5.20.
(a) Determine the steady-state error for a unit step input in terms of \(K\) and \(K_{1}\), where \(E(s) = R(s) - Y(s)\).
(b) Select \(K_{1}\), so that the steady-state error is zero.
FIGURE P5.20 System with pregain, \(K_{1}\).
P5.21 Consider the closed-loop system in Figure P5.21. Determine values of the parameters \(k\) and \(a\) so that the following specifications are satisfied:
(a) The steady-state error to a unit step input is zero.
(b) The closed-loop system has a percent overshoot of P.O. \(\leq 5\%\).
FIGURE P5.21 Closed-loop system with parameters \(k\) and \(a\). P5.22 Consider the closed-loop system in Figure P5.22, where
(a) If \(\tau = 2.43\), determine the value of \(K\) such that the steady-state error of the closed-loop system response to a unit step input, is zero.
(b) Determine the percent overshoot and the time to peak of the unit step response when \(K\) is as in part (a).
FIGURE P5.22 Nonunity closed-loop feedback control system.
193. ADVANCED PROBLEMS
AP5.1 Consider the following closed-loop transfer functions
and
(a) Determine the steady-state error for a unit step input.
(b) Assume that the complex poles dominate, and determine the percent overshoot and settling time to within \(2\%\) of the final value.
(c) Plot the actual system response, and compare it with the estimates of part (b).
AP5.2 A closed-loop system is shown in Figure AP5.2. process.
Plot the response to a unit step input for the system for \(\tau_{z} = 0,0.05,0.1\), and 0.5 . Record the percent overshoot, rise time, and settling time (with a \(2\%\) criterion) as \(\tau_{z}\) varies. Describe the effect of varying \(\tau_{z}\). Compare the location of the zero \(- 1/\tau_{z}\) with the location of the closed-loop poles.
FIGURE AP5.2 System with a variable zero.
AP5.3 A closed-loop system is shown in Figure AP5.3. Plot the response to a unit step input for the system with \(\tau_{p} = 0,0.2,1\), and 4 . Record the percent overshoot, rise time, and settling time (with a \(2\%\) criterion) as \(\tau_{p}\) varies. Describe the effect of varying \(\tau_{p}\). Compare the location of the open-loop pole \(- 1/\tau_{p}\) with the location of the closed-loop poles.
FIGURE AP5.3 System with a variable pole in the
AP5.4 The speed control of a high-speed train is represented by the system shown in Figure AP5.4 [17]. Determine the equation for steady-state error for \(K\) for a unit step input. Consider the three values for \(K\) equal to 1,10 , and 100 .
(a) Determine the steady-state error.
(b) Determine and plot the response \(y(t)\) for (i) a unit step input \(R(s) = 1/s\) and (ii) a unit step disturbance input \(T_{d}(s) = 1/s\).
(c) Create a table showing percent overshoot, settling time (with a \(2\%\) criterion), \(e_{ss}\) for \(r(t)\), and \(\left| y/t_{d} \right|_{\max}\) for the three values of \(K\). Select the best compromise value.
FIGURE AP5.4
Speed control.
FIGURE AP5.5
System with control parameter \(\alpha\).
FIGURE AP5.6
DC motor control.
(a) Assume that the complex poles dominate, and estimate the settling time and percent overshoot to a unit step input for \(K = 1,10,25\), and 50 .
AP5.5 A system with a controller is shown in Figure AP5.5. The zero of the controller may be varied.
(a) Determine the steady-state error for a unit step input for \(\alpha = 0\) and \(\alpha \neq 0\).
(b) Let \(\alpha = 1,15\), and 75. Plot the response of the system to a unit step input disturbance for the three values of \(\alpha\). Compare the results, and select the best value of the three values of \(\alpha\).
AP5.6 The block diagram model of an armature-current controlled DC motor is shown in Figure AP5.6.
(a) Determine the steady-state tracking error to a ramp input in terms of \(K,K_{b}\), and \(K_{m}\).
(b) Let \(K_{m} = 12\) and \(K_{b} = 0.01\), and select \(K\) so that steady-state tracking error is equal to 1 .
(c) Plot the response to a unit step input and a unit ramp input for 30 seconds. Are the responses acceptable?
AP5.7 Consider the closed-loop system in Figure AP5.7 with transfer functions
(b) Determine the actual settling time and percent overshoot to a unit step for the values of \(K\) in part (a).
(c) Co-plot the results of (a) and (b) and comment.
AP5.8 A unity negative feedback system has an open loop transfer function
Determine the gain \(K\) that results in the fastest response without overshoot. What are the corresponding poles?
FIGURE AP5.7
Closed-loop system with unity feedback.
FIGURE AP5.9
Feedback control system with a proportional plus integral controller.
The controller is a proportional plus integral controller with gains \(K_{p}\) and \(K_{I}\). The objective is to design the controller gains such that the dominant roots have a damping ratio \(\zeta\) equal to 0.707 . Determine the resulting peak time and settling time (with a \(2\%\) criterion) of the system to a unit step input.
194. DESIGN PROBLEMS
CDP5.1 The capstan drive system of the previous problems (see CDP1.1-CDP4.1) has a disturbance due to changes in the part that is being machined as material is removed. The controller is an amplifier \(G_{c}(s) = K_{a}\). Evaluate the effect of a unit step disturbance, and determine the best value of the amplifier gain so that the percent overshoot to a step command \(r(t) = A,t > 0\) is \(P.O. \leq 5\%\), while reducing the effect of the disturbance as much as possible.
DP5.1 The roll control autopilot of an aircraft is shown in Figure DP5.1. The goal is to select a suitable \(K\) so that the response to a step command \(\phi_{d}(t) = A,t \geq 0\), will provide a response \(\phi(t)\) that is a fast response and has an percent overshoot of P.O. \(\leq 20\%\). (a) Determine the closed-loop transfer function \(\phi(s)/\phi_{d}(s)\). (b) Determine the roots of the characteristic equation for \(K = 0.7,3\), and 6. (c) Using the concept of dominant roots, find the expected percent overshoot and peak time for the approximate second-order system. (d) Plot the actual response and compare with the approximate results of part (c). (e) Select the gain \(K\) so that the percent overshoot is \(P.O. = 16\%\). What is the resulting peak time?
DP5.2 The design of the control for a welding arm with a long reach requires the careful selection of the parameters [13]. The system is shown in Figure DP5.2. The damping ratio \(\zeta\), the gain \(K\), and the natural frequency \(\omega_{n}\) can be selected. (a) Determine \(K\), and \(\omega_{n}\) so that the response to a unit step input achieves \(T_{p} \leq 1\text{ }s\) and P.O. \(\leq 10\%\). (b) Plot the response of the system designed in part (a) to a step input.
FIGURE DP5.1
Roll angle control.
FIGURE DP5.2 Welding tip position control. FIGURE DP5.3
Active suspension system.
DP5.3 Active suspension systems for modern automobiles provide a comfortable firm ride. The design of an active suspension system adjusts the valves of the shock absorber so that the ride fits the conditions. A small electric motor, as shown in Figure DP5.3, changes the valve settings [13]. The controller is a proportional plus integral controller with gains \(K_{P}\) and \(K_{I}\). Select a design value for \(K_{P},K_{I}\), and the parameter \(q\) in order to satisfy the ITAE performance for a step command \(R(s)\) with a natural frequency, \(\omega_{n} = 1rad/s\). Upon completion of your design, assuming that the complex poles dominate, determine the natural frequency of the system for a step input.
DP5.4 The space satellite, as shown in Figure DP5.4 (a), uses a control system to readjust its orientation, as shown in Figure DP5.4 (b).
(a) Determine a second-order model for the closedloop system.
(b) Using the second-order model, select a gain \(K\) so that the percent overshoot is \(\leq 10\%\), and the steady-state error to a step is less than \(8\%\). (c) Verify your design by determining the actual performance of the third-order system.
DP5.5 A deburring robot can be used to smooth off machined parts by following a preplanned path (input command signal). In practice, errors occur due to robot inaccuracy, machining errors, large tolerances, and tool wear. These errors can be eliminated using force feedback to modify the path online \(\lbrack 8,11\rbrack\).
While force control has been able to address the problem of accuracy, it has been more difficult to solve the contact stability problem. In fact, by closing the force loop and introducing a compliant wrist force sensor (the most common type of force control), one can add to the stability problem.
A model of a robot deburring system is shown in Figure DP5.5. Determine the region of stability for the system for \(K_{1}\) and \(K_{2}\). Assume both adjustable gains are greater than zero.
DP5.6 The model for a position control system using a DC motor is shown in Figure DP5.6. The goal is to select \(K_{1}\) and \(K_{2}\) so that the peak time is \(T_{p} \leq 0.7\text{ }s\),
FIGURE DP5.4
Control of a space satellite.
(a)
(b) FIGURE DP5.5 Deburring robot.
FIGURE DP5.6
Position control robot.
and the percent overshoot for a step input is \(\leq 5\%\).
DP5.7 A three-dimensional cam for generating a function of two variables is shown in Figure DP5.7(a). Both \(x\) and \(y\) may be controlled using a position control system [31]. The control of \(x\) may be achieved with a DC motor and position feedback of the form shown in Figure DP5.7(b), with the DC motor and load represented by
where \(K = 2\) and \(p = 2\). Design a proportional plus derivative controller
to achieve a percent overshoot \(P.O. \leq 5\%\) to a unit step input and a settling time \(T_{s} \leq 2\text{ }s\).
DP5.8 Computer control of a robot to spray-paint an automobile is accomplished by the system shown in Figure DP5.8(a) [7]. We wish to investigate the system when \(K = 1,10\), and 20 . The feedback control
FIGURE DP5.7
(a) Threedimensional cam and (b) \(x\)-axis control system.
(a)
(b) FIGURE DP5.8
Spray-paint robot.
(a)
(b) block diagram is shown in Figure DP5.8(b). (a) For the three values of \(K\), determine the percent overshoot, the settling time (with a \(2\%\) criterion), and the steadystate error for a unit step input. Record your results in a table. (b) Choose one of the three values of \(K\) that provides acceptable performance. (c) For the value selected in part (b), determine the output for a disturbance \(T_{d}(s) = 1/s\) when \(R(s) = 0\).
195. COMPUTER PROBLEMS
CP5.1 Consider the closed-loop transfer function
Obtain the impulse response analytically, and compare the result to one obtained using the impulse function.
CP5.2 A unity negative feedback system has the loop transfer function
Using Isim, obtain the response of the closed-loop system to a unit ramp input,
Consider the time interval \(0 \leq t \leq 50\). What is the steady-state error?
CP5.3 A working knowledge of the relationship between the pole locations of a second-order system and the transient response is important in control design. With that in mind, consider the following five pole location cases:
(a) \(s_{1,2} = \pm j\)
(b) \(s_{1,2} = - 1, - 1\)
(c) \(s_{1,2} = - 1, - 2\),
(d) \(s_{1,2} = - 1 \pm j\)
(e) \(s_{1,2} = 1 \pm j\).
Using the impulse and subplot functions, create a plot containing two subplots, with each subplot depicting the pole location and impulse response of each of the five cases listed. Discuss the results. FIGURE CP5.4
Step response of a simple secondorder system.
FIGURE CP5.5
Feedback control system with controller and prefilter.
CP5.4 Consider the unit step response of the simple second-order closed-loop control system shown in Figure CP5.4.
(a) Determine analytically the damping ratio \(\xi\) and natural frequency \(\omega_{n}\) of the closed-loop system response to a unit step input, and the corresponding closed-loop system transfer function.
(b) Develop an m-file to plot the unit step response of the closed-loop system in (a), and estimate the percent overshoot from the plot. Compare the plot with the step response in Figure CP5.4, and discuss the results.
CP5.5 Consider the feedback system in Figure CP5.5. Develop an m-file to design a controller and prefilter
such that the ITAE performance criterion is minimized. For \(\omega_{n} = 0.45\) and \(\zeta = 0.59\),plot the unit step response and determine the percent overshoot and settling time.
CP5.6 The closed-loop transfer function of a simple second-order system is
Consider the following cases:
-
\(\omega_{n} = 1,\xi = 0.5\),
-
\(\omega_{n} = 2,\xi = 0.5\),
-
\(\omega_{n} = 3,\xi = 0.5\),
-
\(\omega_{n} = 4,\xi = 0.5\),
Develop an m-file to plot the unit step response, and determine the values of peak overshoot \(M_{p}\), time to peak \(T_{p}\), and settling time \(T_{s}\) (with a \(2\%\) criterion) for each of the four cases listed. Discuss the results.
CP5.7 An autopilot designed to hold an aircraft in straight and level flight is shown in Figure CP5.7.
(a) Suppose the controller is a constant gain controller given by \(G_{c}(s) = 5\). Using the Isim function, compute and plot the ramp response for \(\theta_{d}(t) = at\), where \(a = {1.5}^{\circ}/s\). Determine the attitude error after 10 seconds.
(b) If we increase the complexity of the controller, we can reduce the steady-state tracking error. With this objective in mind, suppose we replace the constant gain controller with the more sophisticated controller
This type of controller is known as a proportional plus integral (PI) controller. Repeat the simulation of part (a) with the PI controller, and compare the steady-state tracking errors of the constant gain controller versus the PI controller. FIGURE CP5.7
An aircraft autopilot block diagram.
FIGURE CP5.8
A missile rate loop autopilot.
CP5.8 The block diagram of a rate loop for a missile autopilot is shown in Figure CP5.8. Using the analytic formulas for second-order systems, predict \(M_{pt},T_{p}\), and \(T_{s}\) for the closed-loop system due to a unit step input. Compare the predicted results with the actual unit step response obtained with the step function. Explain any differences.
CP5.9 Develop an \(m\)-file that can be used to analyze the closed-loop system in Figure CP5.9. Drive the system with a step input, and display the output on a graph. What is the settling time and the percent overshoot?
CP5.10 Develop an \(m\)-file to simulate the response of the system in Figure CP5.10 to a parabolic input \(R(s) = 1/s^{3}\). What is the steady-state error? Display the output on an \(x - y\) graph.
FIGURE CP5.9 Nonunity feedback system.
CP5.11 A closed-loop transfer function is given by
(a) Obtain the response of the closed-loop transfer function \(T(s) = Y(s)/R(s)\) to a unit step input.
(b) By consecutively adding zeros at \(0, - 0.5, - 1.5\), and -2.5 , determine the step response. Compare the results with the step response in part (a). What conclusions can be drawn regarding the effect of adding a zero to a second-order system?
CP5.12 A closed-loop transfer function is given by
(a) Obtain the response of the closed-loop transfer function \(T(s) = Y(s)/R(s)\) to a unit step input. What is the settling time \(T_{s}\) (use a \(2\%\) criterion) and percent overshoot P.O.?
(b) Neglecting the real pole at \(s = - 10\), determine the settling time \(T_{s}\) and percent overshoot P.O. Compare the results with the actual system response in part (a). What conclusions can be drawn regarding neglecting the pole?
FIGURE CP5.10 Closed-loop system for m-file.
196. ANSWERS TO SKILLS CHECK
True or False: (1) True; (2) False; (3) False; (4) True; (5) False
Multiple Choice: (6) a; (7) a; (8) c; (9) b; (10) b; (11) a; (12) b; (13) b; (14) a; (15) b
Word Match (in order, top to bottom): i, j, d, g, k, c, n, \(p,o,b,e,l,f,h,m,a\)
197. TERMS AND CONCEPTS
Acceleration error constant, \(\mathbf{K}_{\mathbf{a}}\) The constant evaluated as \(\lim\left\lbrack s^{2}G_{c}(s)G(s) \right\rbrack\). The steady-state error for a parabolic input, \(r(t) = At^{2}/2\), is equal to \(A/K_{a}\).
Design specifications A set of prescribed performance criteria.
Dominant roots The roots of the characteristic equation that cause the dominant transient response of the system.
Optimum control system A system whose parameters are adjusted so that the performance index reaches an extremum value.
Peak time The time for a system to respond to a step input and rise to a peak response.
Percent overshoot The amount by which the system output response proceeds beyond the desired response.
Performance index A quantitative measure of the performance of a system.
Position error constant, \(\mathbf{K}_{\mathbf{p}}\) The constant evaluated as \(\lim_{s \rightarrow 0}\mspace{2mu} G_{c}(s)G(s)\). The steady-state error for a step input (of magnitude \(A\) ) is equal to \(A/\left( 1 + K_{p} \right)\).
Rise time The time for a system to respond to a step input and attain a response equal to a percentage of the magnitude of the input. The \(0 - 100\%\) rise time,
\(T_{r}\), measures the time to \(100\%\) of the magnitude of the input. Alternatively, \(T_{r_{1}}\) measures the time from \(10\%\) to \(90\%\) of the response to the step input.
Settling time The time required for the system output to settle within a certain percentage of the input amplitude.
Steady-state response The constituent of the system response that exists a long time following any signal initiation.
Test input signal An input signal used as a standard test of a system's ability to respond adequately.
Transient response The constituent of the system response that disappears with time.
Type number The number \(N\) of poles of the transfer function, \(G_{c}(s)G(s)\), at the origin. \(G_{c}(s)G(s)\) is the loop transfer function.
Unit impulse A test input consisting of an impulse of infinite amplitude and zero width, and having an area of unity. The unit impulse is used to determine the impulse response.
Velocity error constant, \(\mathbf{K}_{v}\) The constant evaluated as \(\lim_{s \rightarrow 0}\mspace{2mu}\left\lbrack G_{c}(s)G(s) \right\rbrack\). The steady-state error for a ramp input (of slope \(A\) ) for a system is equal to \(A/K_{v}\).
198. CHAPTER
The Stability of Linear Feedback Systems
6.1 The Concept of Stability 395
6.2 The Routh-Hurwitz Stability Criterion 399
6.3 The Relative Stability of Feedback Control Systems 407
6.4 The Stability of State Variable Systems 408
6.5 Design Examples 411
6.6 System Stability Using Control Design Software 419
6.7 Sequential Design Example: Disk Drive Read System 425
6.8 Summary \(\ 427\)
199. PREVIEW
Stability of closed-loop feedback systems is central to control system design. A stable system should exhibit a bounded output if the input is bounded. This is known as bounded-input, bounded-output stability. The stability of a feedback system is directly related to the location of the roots of the characteristic equation of the system transfer function and to the location of the eigenvalues of the system matrix for a system in state variable format. The Routh-Hurwitz method is introduced as a useful tool for assessing system stability. The technique allows us to compute the number of roots of the characteristic equation in the right half plane without actually computing the values of the roots. This gives us a design method for determining values of certain system parameters that will lead to closed-loop stability. For stable systems, we will introduce the notion of relative stability which allows us to characterize the degree of stability. The chapter concludes with a stabilizing controller design based on the Routh-Hurwitz method for the Sequential Design Example: Disk Drive Read System.
200. DESIRED OUTCOMES
Upon completion of Chapter 6, students should be able to:
\(\square\) Explain the concept of stability of dynamic systems.
- Describe the key concepts of absolute and relative stability.
\(\square\) Explain bounded-input, bounded-output stability.
$\square\ $ Describe the relationship of the \(s\)-plane pole locations (for transfer function models) and of the eigenvalue locations (for state variable models) to system stability.
\(\square\) Construct a Routh array and employ the Routh-Hurwitz stability criterion to determine stability.
200.1. THE CONCEPT OF STABILITY
When considering the design and analysis of feedback control systems, stability is of the utmost importance. From a practical point of view, a closed-loop feedback system that is unstable is of minimal value. As with all such general statements, there are exceptions; but for our purposes, we will declare that all our control designs must result in a closed-loop stable system. Many physical systems are inherently open-loop unstable, and some systems are even designed to be open-loop unstable. Most modern fighter aircraft are open-loop unstable by design, and without active feedback control assisting the pilot, they cannot fly. Active control is introduced by engineers to stabilize the unstable system - that is, the aircraft - so that other considerations, such as transient performance, can be addressed. Using feedback, we can stabilize unstable systems and then with a judicious selection of controller parameters, we can adjust the transient performance. For open-loop stable systems, we still use feedback to adjust the closed-loop performance to meet the design specifications. These specifications take the form of steady-state tracking errors, percent overshoot, settling time, time to peak, and the other indices.
We can say that a closed-loop feedback system is either stable or it is not stable. This type of stable/not stable characterization is referred to as absolute stability. A system possessing absolute stability is called a stable system-the label of absolute is dropped. Given that a closed-loop system is stable, we can further characterize the degree of stability. This is referred to as relative stability. The pioneers of aircraft design were familiar with the notion of relative stability - the more stable an aircraft was, the more difficult it was to maneuver (that is, to turn). One outcome of the relative instability of modern acrobatic aircraft is high maneuverability. A acrobatic aircraft is less stable than a commercial transport; hence it can maneuver more quickly. As we will discuss later in this section, we can determine that a system is stable (in the absolute sense) by determining that all transfer function poles lie in the left-half \(s\)-plane, or equivalently, that all the eigenvalues of the system matrix A lie in the left-half \(s\)-plane. Given that all the poles (or eigenvalues) are in the left-half \(s\)-plane, we investigate relative-stability by examining the relative locations of the poles (or eigenvalues).
A stable system is defined as a system with a bounded (limited) system response. That is, if the system is subjected to a bounded input or disturbance and the response is bounded in magnitude, the system is said to be stable.
201. A stable system is a dynamic system with a bounded response to a bounded input.
The concept of stability can be illustrated by considering a right circular cone placed on a plane horizontal surface. If the cone is resting on its base and is tipped slightly, it returns to its original equilibrium position. This position and response are said to be stable. If the cone rests on its side and is displaced slightly, it rolls with no tendency to leave the position on its side. This position is designated as the neutral stability. FIGURE 6.1
Illustration of stability.
FIGURE 6.2
Stability in the s-plane.
(a) Stable
(b) Neutral
(c) Unstable
On the other hand, if the cone is placed on its tip and released, it falls onto its side. This position is said to be unstable. These three positions are illustrated in Figure 6.1.
The stability of a dynamic system is defined in a similar manner. The response to a displacement, or initial condition, will result in either a decreasing, neutral, or increasing response. Specifically, it follows from the definition of stability that a linear system is stable if and only if the absolute value of its impulse response \(g(t)\), integrated over an infinite range, is finite. That is, in terms of the convolution integral Equation (5.2) for a bounded input, \(\int_{0}^{\infty}\mspace{2mu}|g(t)|dt\) must be finite.
The location in the \(s\)-plane of the poles of a system indicates the resulting transient response. The poles in the left-hand portion of the s-plane result in a decreasing response for disturbance inputs. Similarly, poles on the \(j\omega\)-axis and in the right-hand plane result in a neutral and an increasing response, respectively, for a disturbance input. This division of the \(s\)-plane is shown in Figure 6.2. Clearly, the poles of desirable dynamic systems must lie in the left-hand portion of the s-plane [1-3].
A common example of the potential destabilizing effect of feedback is that of feedback in audio amplifier and speaker systems used for public address in auditoriums. In this case, a loudspeaker produces an audio signal that is an amplified version of the sounds picked up by a microphone. In addition to other audio inputs, the sound coming from the speaker itself may be sensed by the microphone. The strength of this particular signal depends upon the distance between the loudspeaker and the microphone. Because of the attenuating properties of air, a larger distance will cause a weaker signal to reach the microphone. Due to the finite propagation speed of sound waves, there will also be a time delay between the signal produced by the loudspeaker and the signal sensed by the microphone. In this case, the output from the feedback path is added to the external input. This is an example of positive feedback.
As the distance between the loudspeaker and the microphone decreases, we find that if the microphone is placed too close to the speaker, then the system will be unstable. The result of this instability is an excessive amplification and distortion of audio signals and an oscillatory squeal. In terms of linear systems, we recognize that the stability requirement may be defined in terms of the location of the poles of the closed-loop transfer function. A closed-loop system transfer function can be written as
where \(q(s) = \Delta(s) = 0\) is the characteristic equation whose roots are the poles of the closed-loop system. The output response for an impulse function input (when \(N = 0)\) is then
where \(A_{k}\) and \(B_{m}\) are constants that depend on \(\sigma_{k},z_{i},\alpha_{m},K\), and \(\omega_{m}\). To obtain a bounded response, the poles of the closed-loop system must be in the left-hand portion of the \(s\)-plane. Thus, a necessary and sufficient condition for a feedback system to be stable is that all the poles of the system transfer function have negative real parts. A system is stable if all the poles of the transfer function are in the lefthand \(s\)-plane. A system is not stable if not all the roots are in the left-hand plane. If the characteristic equation has simple roots on the imaginary axis ( \(j\omega\)-axis) with all other roots in the left half-plane, the steady-state output will be sustained oscillations for a bounded input, unless the input is a sinusoid (which is bounded) whose frequency is equal to the magnitude of the \(j\omega\)-axis roots. For this case, the output becomes unbounded. Such a system is called marginally stable, since only certain bounded inputs (sinusoids of the frequency of the poles) will cause the output to become unbounded. For an unstable system, the characteristic equation has at least one root in the right half of the \(s\)-plane or repeated \(j\omega\) roots; for this case, the output will become unbounded for any input.
For example, if the characteristic equation of a closed-loop system is
then the system is said to be marginally stable. If this system is excited by a sinusoid of frequency \(\omega = 4\), the output becomes unbounded.
An example of how mechanical resonance can cause large displacements occurred in a 39-story shopping mall in Seoul, Korea. The Techno-Mart building, shown in Figure 6.3, hosts activities such as physical aerobics, in addition to shopping. After a Tae Bo workout session on the \(12^{\text{th}\text{~}}\) floor with about twenty participants, the building shook for 10 minutes triggering an evacuation for two days [5]. A team of experts concluded that the building was likely excited to mechanical resonance by the vigorous exercise.
To ascertain the stability of a feedback control system, we could determine the roots of the characteristic polynomial \(q(s)\). However, we are first interested in determining the answer to the question, Is the system stable? If we calculate the roots of the characteristic equation in order to answer this question, we have determined much more information than is necessary. Therefore, several FIGURE 6.3
Vigorous exercising on the 12th floor likely led to mechanical resonance of the building triggering a two-day evacuation. (Photo courtesy of Truth Leem/ Reuters.)
methods have been developed that provide the required yes or no answer to the stability question. The three approaches to the question of stability are (1) the \(s\)-plane approach, (2) the frequency \((j\omega)\) approach, and (3) the time-domain approach.
Industrial robot sales were the highest level ever recorded for a single year in 2013. In fact, since the introduction of industrial robots at the end of the 1960s until 2013, there have been over 2.5 million operational industrial robots sold. The worldwide stock of operational industrial robots at the end of 2013 was in the range of 1.3-1.6 million units. The projections are that from 2015-2017, industrial robot installations will increase by \(12\%\) on average per year [10]. Clearly, the market for industrial robots is dynamic. The worldwide market for service robots is similarly active. The projections for the period 2014-2017 are that approximately 31 million new service robots for personal use (such as vacuum cleaners and lawn mowers) and approximately 134,500 new service robots for professional use will be put into service [10]. As the capability of robots increases, it is reasonable to assume that the numbers in service will continue to rise. Especially interesting are robots with human characteristics, particularly those that can walk upright [21]. The IHMC robot depicted in Figure 6.4 competed in the recent DARPA Robotics Challenge [24]. Examining the IHMC robot in Figure 6.4, one can imagine that it is not inherently stable and that active control is required to keep it upright during the walking motion. In the next sections we present the Routh-Hurwitz stability criterion to investigate system stability by analyzing the characteristic equation without direct computation of the roots. FIGURE 6.4
Team IHMC on the rubble on the first day of the DARPA Robotics Challenge 2015.
(DOD Photo/Alamy Stock Photo.)
201.1. THE ROUTH-HURWITZ STABILITY CRITERION
The discussion and determination of stability has occupied the interest of many engineers. Maxwell and Vyshnegradskii first considered the question of stability of dynamic systems. In the late 1800s, A. Hurwitz and E. J. Routh independently published a method of investigating the stability of a linear system [6, 7]. The Routh-Hurwitz stability method provides an answer to the question of stability by considering the characteristic equation of the system. The characteristic equation is written as
To ascertain the stability of the system, it is necessary to determine whether any one of the roots of \(q(s)\) lies in the right half of the \(s\)-plane. If Equation (6.3) is written in factored form, we have
where \(r_{i} = i\) th root of the characteristic equation. Multiplying the factors together, we find that
In other words, for an \(n\) th-degree equation, we obtain
Examining Equation (6.5), we note that all the coefficients of the polynomial will have the same sign if all the roots are in the left-hand plane. Also, it is necessary that all the coefficients for a stable system be nonzero. These requirements are necessary but not sufficient. That is, we immediately know the system is unstable if they are not satisfied; yet if they are satisfied, we must proceed further to ascertain the stability of the system. For example, when the characteristic equation is
the system is unstable, and yet the polynomial possesses all positive coefficients.
The Routh-Hurwitz criterion is a necessary and sufficient criterion for the stability of linear systems. The method was originally developed in terms of determinants, but we shall use the more convenient array formulation. The RouthHurwitz criterion is based on ordering the coefficients of the characteristic equation
into an array as follows [4]:
Further rows of the array, known as the Routh array, are then completed as
where
and so on. The algorithm for calculating the entries in the array can be followed on a determinant basis or by using the form of the equation for \(b_{n - 1}\).
The Routh-Hurwitz criterion states that the number of roots of \(q(s)\) with positive real parts is equal to the number of changes in sign of the first column of the Routh array. This criterion requires that there be no changes in sign in the first column for a stable system. This requirement is both necessary and sufficient.
Four distinct cases or configurations of the first column array must be considered, and each must be treated separately and requires suitable modifications of the array calculation procedure: (1) No element in the first column is zero; (2) there is a zero in the first column, but some other elements of the row containing the zero in the first column are nonzero; (3) there is a zero in the first column, and the other elements of the row containing the zero are also zero; and (4) as in the third case, but with repeated roots on the \(j\omega\)-axis.
To illustrate this method clearly, several examples will be presented for each case.
202. Case 1. No element in the first column is zero.
203. EXAMPLE 6.1 Second-order system
The characteristic polynomial of a second-order system is
The Routh array is written as
where
Therefore, the requirement for a stable second-order system is that all the coefficients be positive or all the coefficients be negative.
204. EXAMPLE 6.2 Third-order system
The characteristic polynomial of a third-order system is
The Routh array is
| $$s^{3}$$ | $$a_{3}$$ | $$a_{1}$$ |
| $$s^{2}$$ | $$a_{2}$$ | $$a_{0}$$ |
| $$s^{1}$$ | $$b_{1}$$ | 0 |
| $$s^{0}$$ | $$c_{1}$$ | 0 |
where
For the third-order system to be stable, it is necessary and sufficient that the coefficients be positive and \(a_{2}a_{1} > a_{0}a_{3}\). The condition when \(a_{2}a_{1} = a_{0}a_{3}\) results in a marginal stability case, and one pair of roots lies on the imaginary axis in the \(s\)-plane. This marginal case is recognized as Case 3 because there is a zero in the first column when \(a_{2}a_{1} = a_{0}a_{3}\). It will be discussed under Case 3 .
As a final example of characteristic equations that result in no zero elements in the first row, let us consider the polynomial
The polynomial satisfies all the necessary conditions because all the coefficients exist and are positive. Therefore, utilizing the Routh array, we have
Because two changes in sign appear in the first column, we find that two roots of \(q(s)\) lie in the right-hand plane, and our prior knowledge is confirmed.
Case 2. There is a zero in the first column, but some other elements of the row containing the zero in the first column are nonzero. If only one element in the array is zero, it may be replaced with a small positive number, \(\epsilon\), that is allowed to approach zero after completing the array. For example, consider the following characteristic polynomial:
The Routh array is then
where
When \(0 < \epsilon \ll 1\), we find that \(c_{1} < 0\) and \(d_{1} > 0\). Therefore, there are two sign changes in the first column; hence the system is unstable with two roots in the right half-plane.
205. EXAMPLE 6.3 Unstable system
As a final example of the type of Case 2, consider the characteristic polynomial
where we desire to determine the gain \(K\) that results in marginal stability. The Routh array is then
| $$s^{4}$$ | 1 | 1 | $$K$$ |
| $$s^{3}$$ | 1 | 1 | 0 |
| $$s^{2}$$ | $$\in$$ | $$K$$ | 0 |
| $$s^{1}$$ | $$c_{1}$$ | 0 | 0 |
| $$s^{0}$$ | $$K$$ | 0 | 0 |
where
When \(0 < \epsilon \ll 1\) and \(K > 0\), we find that \(c_{1} < 0\). Therefore, there are two sign changes in the first column; hence, the system is unstable with two roots in the right half-plane. When \(0 < \epsilon \ll 1\) and \(K < 0\), we find that \(c_{1} > 0\), but because the last term in the first column is equal to \(K\), we have a sign change in the first column; hence, the system is unstable with one root in the right half-plane. Consequently, the system is unstable for all values of gain \(K\).
Case 3. There is a zero in the first column, and the other elements of the row containing the zero are also zero. Case 3 occurs when all the elements in one row are zero or when the row consists of a single element that is zero. This condition occurs when the polynomial contains singularities that are symmetrically located about the origin of the \(s\)-plane. Therefore, Case 3 occurs when factors such as \((s + \sigma)(s - \sigma)\) or \((s + j\omega)(s - j\omega)\) occur. This problem is circumvented by utilizing the auxiliary polynomial, \(U(s)\), which immediately precedes the zero entry in the Routh array. The order of the auxiliary polynomial is always even and indicates the number of symmetrical root pairs.
To illustrate this approach, let us consider a third-order system with the characteristic polynomial
where \(K\) is an adjustable loop gain. The Routh array is then
For a stable system, we require that
When \(K = 8\), we have two roots on the \(j\omega\)-axis and a marginal stability case. Note that we obtain a row of zeros (Case 3 ) when \(K = 8\). The auxiliary polynomial, \(U(s)\), is the equation of the row preceding the row of zeros. The equation of the row preceding the row of zeros is, in this case, obtained from the \(s^{2}\)-row. We recall that this row contains the coefficients of the even powers of \(s\), and therefore we have
When \(K = 8\), the factors of the characteristic polynomial are
Case 4. Repeated roots of the characteristic equation on the \(j\omega\)-axis. If the \(j\omega\)-axis roots of the characteristic equation are simple, the system is neither stable nor unstable; it is instead called marginally stable, since it has an undamped sinusoidal mode. If the \(j\omega\)-axis roots are repeated, the system response will be unstable with a form \(tsin(\omega t + \phi)\). The Routh-Hurwitz criteria will not reveal this form of instability [20].
Consider the system with a characteristic polynomial
The Routh array is
| $$s^{5}$$ | 1 | 2 | 1 |
| $$s^{4}$$ | 1 | 2 | 1 |
| $$s^{3}$$ | $$\in$$ | $$\in$$ | 0 |
| $$s^{2}$$ | 1 | 1 | |
| $$s^{1}$$ | $$\in$$ | 0 | |
| $$s^{0}$$ | 1 |
When \(0 < \epsilon \ll 1\), we note the absence of sign changes in the first column. However, as \(\in \rightarrow 0\), we obtain a row of zero at the \(s^{3}\) line and a now of zero at the \(s^{1}\) line. The auxiliary polynomial at the \(s^{2}\) line is \(s^{2} + 1\), and the auxiliary polynomial at the \(s^{4}\) line is \(s^{4} + 2s^{2} + 1 = \left( s^{2} + 1 \right)^{2}\), indicating the repeated roots on the \(j\omega\)-axis. Hence, the system is unstable.
206. EXAMPLE 6.4 Fifth-order system with roots on the \(\mathbf{j\omega}\)-axis
Consider the characteristic polynomial
The Routh array is
| $$s^{5}$$ | 1 | 4 | 3 |
| $$s^{4}$$ | 1 | 24 | 63 |
| $$s^{3}$$ | -20 | -60 | 0 |
| $$s^{2}$$ | 21 | 63 | 0 |
| $$s^{1}$$ | 0 | 0 | 0 |
Therefore, the auxiliary polynomial is
which indicates that two roots are on the imaginary axis. To examine the remaining roots, we divide by the auxiliary polynomial to obtain
Establishing a Routh array for this equation, we have
The two changes in sign in the first column indicate the presence of two roots in the right-hand plane, and the system is unstable. The roots in the right-hand plane are \(s = + 1 \pm j\sqrt{6}\).
207. EXAMPLE 6.5 Welding control
Large welding robots are used in today's auto plants. The welding head is moved to different positions on the auto body, and a rapid, accurate response is required. A block diagram of a welding head positioning system is shown in Figure 6.5.
FIGURE 6.5
Welding head position control.
We desire to determine the range of \(K\) and \(a\) for which the system is stable. The characteristic equation is
Therefore, \(q(s) = s^{4} + 6s^{3} + 11s^{2} + (K + 6)s + Ka = 0\). Establishing the Routh array, we have
| $$s^{4}$$ | 1 | 11 | $$Ka$$ |
| $$s^{3}$$ | 6 | $$K + 6$$ | |
| $$s^{2}$$ | $$b_{3}$$ | $$Ka$$ | |
| $$s^{1}$$ | $$c_{3}$$ | ||
| $$s^{0}$$ | $$Ka$$ |
where
The coefficient \(c_{3}\) sets the acceptable range of \(K\) and \(a\), while \(b_{3}\) requires that \(K\) be less than 60. Requiring \(c_{3} \geq 0\), we obtain
The required relationship between \(K\) and \(a\) is then
when \(a\) is positive. Therefore, if \(K = 40\), we require \(a \leq 0.639\).
Suppose we write the characteristic equation of an \(n\) th-order system as
We divide through by \(\omega_{n}^{n}\) and use \(\overset{*}{s} = s/\omega_{n}\) to obtain the normalized form of the characteristic equation:
For example, we normalize
by dividing through by \(8 = \omega_{n}^{3}\), obtaining
or
208. Table 6.1 The Routh-Hurwitz Stability Criterion
209. \(n\) Characteristic Equation
Note: The equations are normalized by \(\left( \omega_{n} \right)^{n}\).
where \(\overset{*}{s} = s/\omega_{n}\). In this case, \(b = 2.5\) and \(c = 0.5\). Using this normalized form of the characteristic equation, we summarize the stability criterion for up to a sixth-order characteristic equation, as provided in Table 6.1. Note that \(bc = 1.25\) and the system is stable.
209.1. THE RELATIVE STABILITY OF FEEDBACK CONTROL SYSTEMS
The verification of stability using the Routh-Hurwitz criterion provides only a partial answer to the question of stability. The Routh-Hurwitz criterion ascertains the absolute stability of a system by determining whether any of the roots of the characteristic equation lie in the right half of the \(s\)-plane. However, if the system satisfies the Routh-Hurwitz criterion and is stable, it is desirable to determine the relative stability; that is, it is interesting to investigate the relative damping of each root of the characteristic equation. The relative stability of a system can be defined as the property that is measured by the relative real part of each root or pair of roots. Thus, root \(r_{2}\) is relatively more stable than the roots \(r_{1},{\widehat{r}}_{1}\), as shown in Figure 6.6. The relative stability of a system can also be defined in terms of the relative damping coefficients \(\zeta\) of each complex root pair and, therefore, in terms of the speed of response and overshoot instead of settling time.
The investigation of the relative stability of each root is important because the location of the closed-loop poles in the \(s\)-plane determines the performance of the system. Thus, we reexamine the characteristic polynomial \(q(s)\) and consider several methods for the determination of relative stability.
FIGURE 6.6
Root locations in the s-plane.
Because the relative stability of a system is determined by the location of the roots of the characteristic equation, a first approach using an \(s\)-plane formulation is to extend the Routh-Hurwitz criterion to ascertain relative stability. This can be accomplished by utilizing a change of variable, which shifts the \(s\)-plane axis in order to utilize the Routh-Hurwitz criterion. Examining Figure 6.6, we notice that a shift of the vertical axis in the \(s\)-plane to \(- \sigma_{1}\) will result in the roots \(r_{1},{\widehat{r}}_{1}\) appearing on the shifted axis. The correct magnitude to shift the vertical axis must be obtained on a trial-and-error basis. Then, without solving the fifth-order polynomial \(q(s)\), we may determine the real part of the dominant roots \(r_{1},{\widehat{r}}_{1}\).
210. EXAMPLE 6.6 Axis shift
Consider the third-order characteristic equation
Setting the shifted variable \(s_{n}\) equal to \(s + 1\), we obtain
Then the Routh array is established as
| $$s_{n}^{3}$$ | 1 | 1 |
| $$s_{n}^{2}$$ | 1 | 1 |
| $$s_{n}^{1}$$ | 0 | 0 |
| $$s_{n}^{0}$$ | 1 | 0 |
.
There are roots on the shifted imaginary axis that can be obtained from the auxiliary polynomial
The shifting of the \(s\)-plane axis to ascertain the relative stability of a system is a very useful approach, particularly for higher-order systems with several pairs of closed-loop complex conjugate roots.
210.1. THE STABILITY OF STATE VARIABLE SYSTEMS
The stability of a system modeled by a state variable flow graph model can be readily ascertained. If the system we are investigating is represented by a signal-flow graph state model, we obtain the characteristic equation by evaluating the flow graph determinant. If the system is represented by a block diagram model we obtain the characteristic equation using the block diagram reduction methods.
211. EXAMPLE 6.7 Stability of a second-order system
A second-order system is described by the two first-order differential equations
(a)
212. FIGURE 6.7
(a) Flow graph model for state variable equations of Example 6.7.
(b) Block diagram model.
(b)
where \(u(t)\) is the input. The flow graph model of this set of differential equations is shown in Figure 6.7(a) and the block diagram model is shown in Figure 6.7(b).
Using Mason's signal-flow gain formula, we note three loops:
where \(L_{1}\) and \(L_{2}\) do not share a common node. Therefore, the determinant is
We multiply by \(s^{2}\) to obtain the characteristic equation
Since all coefficients must be positive, we require \(K > 3\) for stability. A similar analysis can be undertaken using the block diagram. Closing the two feedback loops yields the two transfer functions
as illustrated in Figure 6.7(b). The closed loop transfer function is thus
Therefore, the characteristic equation is
or
This confirms the results obtained using signal-flow graph techniques.
A method of obtaining the characteristic equation directly from the vector differential equation is based on the fact that the solution to the unforced system is an exponential function. The vector differential equation without input signals is
where \(\mathbf{x}\) is the state vector. The solution is of exponential form, and we can define a constant \(\lambda\) such that the solution of the system for one state can be of the form \(x_{i}(t) = k_{i}e^{\lambda_{i}t}\). The \(\lambda_{i}\) are called the characteristic roots or eigenvalues of the system, which are simply the roots of the characteristic equation. If we let \(\mathbf{x} = \mathbf{k}e^{\lambda t}\) and substitute into Equation (6.22), we have
or
Equation (6.24) can be rewritten as
where \(\mathbf{I}\) equals the identity matrix and \(\mathbf{0}\) equals the null matrix. This set of simultaneous equations has a nontrivial solution if and only if the determinant vanishesthat is, only if
The \(n\) th-order equation in \(\lambda\) resulting from the evaluation of this determinant is the characteristic equation, and the stability of the system can be readily ascertained.
213. EXAMPLE 6.8 Closed epidemic system
The vector differential equation of the epidemic system is given in Equation (3.63) and repeated here as
The characteristic equation is then
Thus, we obtain the characteristic equation of the system. The additional root \(\lambda = 0\) results from the definition of \(x_{3}\) as the integral of \(\alpha x_{1} + \gamma x_{2}\), and \(x_{3}\) does not affect the other state variables. Thus, the root \(\lambda = 0\) indicates the integration connected with \(x_{3}\). The characteristic equation indicates that the system is marginally stable when \(\alpha + \gamma > 0\) and \(\alpha\gamma + \beta^{2} > 0\).
213.1. DESIGN EXAMPLES
In this section we present two illustrative examples. The first example is a tracked vehicle control problem. In this first example, stability issues are addressed employing the Routh-Hurwitz stability criterion and the outcome is the selection of two key system parameters. The second example illustrates the stability problem robot-controlled motorcycle and how Routh-Hurwitz can be used in the selection of controller gains during the design process. The robot-controlled motorcycle example highlights the design process with special attention to the impact of key controller parameters on stability.
214. EXAMPLE 6.9 Tracked vehicle turning control
The design of a turning control for a tracked vehicle involves the selection of two parameters [8]. In Figure 6.8, the system shown in part (a) has the model shown in part (b). The two tracks are operated at different speeds in order to turn the vehicle. We must select \(K\) and \(a\) so that the system is stable and the steady-state error for a ramp command is less than or equal to \(24\%\) of the magnitude of the command.
The characteristic equation of the feedback system is
(a)
FIGURE 6.8
(a) Turning control system for a two-track vehicle. (b) Block diagram.
(b) or
Therefore, we have
or
To determine the stable region for \(K\) and \(a\), we establish the Routh array as
| $$s^{4}$$ | 1 | 17 | $$Ka$$ |
| $$s^{3}$$ | 8 | $$K + 10$$ | 0 |
| $$s^{2}$$ | $$b_{3}$$ | $$Ka$$ | |
| $$s^{1}$$ | $$c_{3}$$ | ||
| $$s^{0}$$ | $$Ka$$ |
where
For the elements of the first column to be positive, we require that \(Ka,b_{3}\), and \(c_{3}\) be positive. Therefore, we require that
The region of stability for \(K > 0\) is shown in Figure 6.9. The steady-state error to a ramp input \(r(t) = At,t > 0\) is
The stable region. where
Therefore, we have
When \(e_{ss}\) is equal to \(23.8\%\) of \(A\), we require that \(Ka = 42\). This can be satisfied by the selected point in the stable region when \(K = 70\) and \(a = 0.6\), as shown in Figure 6.9. Another acceptable design would be attained when \(K = 50\) and \(a = 0.84\). We can calculate a series of possible combinations of \(K\) and \(a\) that can satisfy \(Ka = 42\) and that lie within the stable region, and all will be acceptable design solutions. However, not all selected values of \(K\) and \(a\) will lie within the stable region. Note that \(K\) cannot exceed 126.
215. EXAMPLE 6.10 Robot-controlled motorcycle
Consider the robot-controlled motorcycle shown in Figure 6.10. The motorcycle will move in a straight line at constant forward speed \(v\). Let \(\phi(t)\) denote the angle between the plane of symmetry of the motorcycle and the vertical. The desired angle \(\phi_{d}(t)\) is equal to zero, thus
The design elements highlighted in this example are illustrated in Figure 6.11. Using the Routh-Hurwitz stability criterion will allow us to get to the heart of the matter, that is, to develop a strategy for computing the controller gains while ensuring closed-loop stability. The control goal is
216. Control Goal
Control the motorcycle in the vertical position, and maintain the prescribed position in the presence of disturbances.
The variable to be controlled is
217. Variable to Be Controlled
The motorcycle position from vertical, \(\phi(t)\).
FIGURE 6.10
The robotcontrolled motorcycle.
Topics emphasized in this example
FIGURE 6.11 Elements of the control system design process emphasized in this robot-controlled motorcycle example.
Since our focus here is on stability rather than transient response characteristics, the control specifications will be related to stability only; transient performance is an issue that we need to address once we have investigated all the stability issues. The control design specification is
218. Design Specification
DS1 The closed-loop system must be stable.
The main components of the robot-controlled motorcycle are the motorcycle and robot, the controller, and the feedback measurements. The main subject of the chapter is not modeling, so we do not concentrate on developing the motorcycle dynamics model. We rely instead on the work of others (see [22]). The motorcycle model is given by
where \(\alpha_{1} = g/h,g = 9.806\text{ }m/s^{2}\), and \(h\) is the height of the motorcycle center of gravity above the ground (see Figure 6.10). The motorcycle is unstable with poles at \(s = \pm \sqrt{\alpha_{1}}\). The controller is given by where
and
The forward speed of the motorcycle is denoted by \(v\), and \(c\) denotes the wheelbase (the distance between the wheel centers). The length, \(L\), is the horizontal distance between the front wheel axle and the motorcycle center of gravity. The time-constant of the controller is denoted by \(\tau\). This term represents the speed of response of the controller; smaller values of \(\tau\) indicate an increased speed of response. Many simplifying assumptions are necessary to obtain the simple transfer function models in Equations (6.31) and (6.32).
Control is accomplished by turning the handlebar. The front wheel rotation about the vertical is not evident in the transfer functions. Also, the transfer functions assume a constant forward speed \(v\) which means that we must have another control system at work regulating the forward speed. Nominal motorcycle and robot controller parameters are given in Table 6.2.
Assembling the components of the feedback system gives us the system configuration shown in Figure 6.12. Examination of the configuration reveals that the robot controller block is a function of the physical system \((h,c\), and \(L)\), the operating conditions \((v)\), and the robot time-constant \((\tau)\). No parameters need adjustment unless we physically change the motorcycle parameters and/or speed. In fact, in this example the parameters we want to adjust are in the feedback loop:
219. Select Key Tuning Parameters
Feedback gains \(K_{P}\) and \(K_{D}\).
The key tuning parameters are not always in the forward path; in fact they may exist in any subsystem in the block diagram.
We want to use the Routh-Hurwitz technique to analyze the closed-loop system stability. What values of \(K_{P}\) and \(K_{D}\) lead to closed-loop stability? A related question that we can pose is, given specific values of \(K_{P}\) and \(K_{D}\) for the nominal system (that is, nominal values of \(\alpha_{1},\alpha_{2},\alpha_{3}\), and \(\tau\) ), how can the parameters themselves vary while still retaining closed-loop stability?
| Table 6.2 | Physical Parameters |
| $$\tau$$ | $$0.2\text{ }s$$ |
| $$\alpha_{1}$$ | $$91/s^{2}$$ |
| $$\alpha_{2}$$ | $$2.71/s^{2}$$ |
| $$\alpha_{3}$$ | $$1.351/s$$ |
| $$h$$ | $$1.09\text{ }m$$ |
| $$V$$ | $$2.0\text{ }m/s$$ |
| $$L$$ | $$1.0\text{ }m$$ |
| $$c$$ | $$1.36\text{ }m$$ |
FIGURE 6.12
The robot-controlled motorcyle feedback system block diagram.
The closed-loop transfer function from \(\phi_{d}(s)\) to \(\phi(s)\) is
where
The question that we need to answer is for what values of \(K_{P}\) and \(K_{D}\) does the characteristic equation \(\Delta(s) = 0\) have all roots in the left half-plane?
We can set up the following Routh array:
where
By inspecting column 1, we determine that for stability we require
Choosing \(K_{D} > 0\) satisfies the second inequality (note that \(\alpha_{3} > 0\) ). In the event \(\tau = 0\), we would reformulate the characteristic equation and rework the Routh array. We need to determine the conditions on \(K_{P}\) and \(K_{D}\) such that \(a > 0\). We find that \(a > 0\) implies that the following relationship must be satisfied:
Using the nominal values of the parameters \(\alpha_{1},\alpha_{2},\alpha_{3}\), and \(\tau\) (see Table 6.2), for all \(K_{D} > 0\) and \(K_{P} > 3.33\), the left hand-side of Equation (6.33) is positive, hence \(a > 0\). Taking into account all the inequalities, a valid region for selecting the gains is \(K_{D} > 0\) and \(K_{P} > \alpha_{1}/\alpha_{2} = 3.33\). Selecting any point \(\left( K_{P},K_{D} \right)\) in the stability region yields a valid (that is, stable) set of gains for the feedback loop. For example, selecting
yields a stable closed-loop system. The closed-loop poles are
Since all the poles have negative real parts, we know the system response to any bounded input will be bounded.
For this robot-controlled motorcycle, we do not expect to have to respond to nonzero command inputs (that is, \(\phi_{d}(t) \neq 0\) ) since we want the motorcyle to remain upright, and we certainly want to remain upright in the presence of external disturbances. The transfer function for the disturbance \(T_{d}(s)\) to the output \(\phi(s)\) without feedback is
The characteristic equation is
The system poles are
Thus we see that the motorcycle is unstable; it possesses a pole in the right half-plane. Without feedback control, any external disturbance will result in the motorcycle falling over. Clearly the need for a control system (usually provided by the human rider) is necessary. With the feedback and robot controller in the loop, the closed-loop transfer function from the disturbance to the output is
The response to a step disturbance is shown in Figure 6.13; the response is stable. The control system manages to keep the motorcycle upright, although it is tilted at about \(\phi = 0.055rad = 3.18deg\).
It is important to give the robot the ability to control the motorcycle over a wide range of forward speeds. Is it possible for the robot, with the feedback gains as selected $\left( K_{P} = 10 \right.\ $ and \(\left. \ K_{D} = 5 \right)\), to control the motorcycle as the velocity varies? From experience we know that at slower speeds a bicycle becomes more difficult to control. We expect to see the same characteristics in the stability analysis of our system. Whenever possible, we try to relate the engineering problem at hand to real-life experiences. This helps to develop intuition that can be used as a reasonableness check on our solution.
A plot of the roots of the characteristic equation as the forward speed \(v\) varies is shown in Figure 6.14. The data in the plot were generated using the nominal values of the feedback gains, \(K_{P} = 10\) and \(K_{D} = 5\). We selected these gains for the case where \(v = 2\text{ }m/s\). Figure 6.14 shows that as \(v\) increases, the roots of the FIGURE 6.13
Disturbance response with \(K_{P} = 10\) and \(K_{D} = 5\).
FIGURE 6.14
Roots of the characteristic equation as the motorcycle velocity varies.
characteristic equation remain stable (that is, in the left half-plane) with all points negative. But as the motorcycle forward speed decreases, the roots move toward zero, with one root becoming positive at \(v = 1.15\text{ }m/s\). At the point where one root is positive, the motorcycle is unstable.
219.1. SYSTEM STABILITY USING CONTROL DESIGN SOFTWARE
In this section we will see how the computer can assist us in the stability analysis by providing an easy and accurate method for computing the poles of the characteristic equation. For the case of the characteristic equation as a function of a single parameter, it will be possible to generate a plot displaying the movement of the poles as the parameter varies. The section concludes with an example.
The function introduced in this section is the function for, which is used to repeat a number of statements a specific number of times.
Routh-Hurwitz Stability. As stated earlier, the Routh-Hurwitz criterion is a necessary and sufficient criterion for stability. Given a characteristic equation with fixed coefficients, we can use Routh-Hurwitz to determine the number of roots in the right half-plane. For example, consider the characteristic equation
associated with the closed-loop control system shown in Figure 6.15. The corresponding Routh-Hurwitz array is shown in Figure 6.16. The two sign changes in the first column indicate that there are two roots of the characteristic polynomial in the right half-plane; hence, the closed-loop system is unstable. We can verify the Routh-Hurwitz result by directly computing the roots of the characteristic equation, as shown in Figure 6.17, using the pole function. Recall that the pole function computes the system poles.
Whenever the characteristic equation is a function of a single parameter, the Routh-Hurwitz method can be utilized to determine the range of values that the parameter may take while maintaining stability. Consider the closed-loop feedback system in Figure 6.18. The characteristic equation is
Using a Routh-Hurwitz approach, we find that we require \(0 < K < 8\) for stability (see Equation 6.12). We can verify this result graphically. As shown in Figure 6.19(b), we
FIGURE 6.15
Closed-loop control system with
$Y(s)/R(s) = 1/\left( s^{3} + \right.\ $ \(\left. \ s^{2} + 2s + 24 \right)\).
FIGURE 6.16
Routh array for the closed-loop control system with \(T(s) =\) \(Y(s)/R(s) =\) $1/\left( s^{3} + s^{2} + \right.\ $ \(2s + 24)\).
FIGURE 6.17
Using the pole function to compute the closed-loop control system poles of the system shown in Figure 6.16.
FIGURE 6.18
Closed-loop control system with
$Y(s)/R(s) = K/\left( s^{3} + \right.\ $ \(\left. \ 2s^{2} + 4s + 4 \right)\).
establish a vector of values for \(K\) at which we wish to compute the roots of the characteristic equation. Then using the roots function, we calculate and plot the roots of the characteristic equation, as shown in Figure 6.19(a). It can be seen that as \(K\) increases, the roots of the characteristic equation move toward the right half-plane as the gain tends toward \(K = 8\), and eventually into the right half-plane when \(K > 8\).
The script in Figure 6.19 contains the for function. This function provides a mechanism for repeatedly executing a series of statements a given number of times. The for function connected to an end statement sets up a repeating calculation loop. Figure 6.20 describes the for function format and provides an illustrative example of its usefulness. The example sets up a loop that repeats ten times. During the \(i\) th iteration, where \(1 \leq i \leq 10\), the \(i\) th element of the vector \(\mathbf{a}\) is set equal to 20 , and the scalar \(b\) is recomputed.
The Routh-Hurwitz method allows us to make definitive statements regarding absolute stability of a linear system. The method does not address the issue of relative stability, which is directly related to the location of the roots of the characteristic equation. Routh-Hurwitz tells us how many poles lie in the right half-plane, but not the specific location of the poles. With control design software, we can easily calculate the poles explicitly, thus allowing us to comment on the relative stability.
220. EXAMPLE 6.11 Tracked vehicle control
The block diagram of the control system for the two-track vehicle is shown in Figure 6.8. The design objective is to find \(a\) and \(K\) such that the system is stable and the steady-state error for a ramp input is less than or equal to \(24\%\) of the command.
We can use the Routh-Hurwitz method to aid in the search for appropriate values of \(a\) and \(K\). The closed-loop characteristic equation is
Using the Routh array, we find that, for stability, we require that
FIGURE 6.19
(a) Plot of root locations of \(q(s) = s^{3} + 2s^{2} +\) \(4\text{ }s + K\) for \(0 \leq K \leq 20\). (b) \(m\)-file script.
FIGURE 6.20 The for function and an illustrative example.
(a)
(b)
FIGURE 6.21
(a) Stability region for \(a\) and \(K\) for two-track vehicle turning control.
(b) m-file script.
(a)
(b)
For positive \(K\), it follows that we can restrict our search to \(0 < K < 126\) and \(a > 0\). Our approach will be to use the computer to help find a parameterized \(a\) versus \(K\) region in which stability is assured. Then we can find a set of \((a,K)\) belonging to the stable region such that the steady-state error specification is met. This procedure, shown in Figure 6.21, involves selecting a range of values for \(a\) and \(K\) and computing the roots of the characteristic polynomial for specific values of \(a\) and \(K\). For each value of \(K\), we find the first value of \(a\) that results in at least one root of the characteristic equation in the right half-plane. The process is repeated until the entire selected range of \(a\) and \(K\) is exhausted. The plot of the \((a,K)\) pairs defines the separation between the stable and unstable regions. The region to the left of the plot of \(a\) versus \(K\) in Figure 6.21 is the stable region.
If we assume that \(r(t) = At,t > 0\), then the steady-state error is
where we have used the fact that
Given the steady-state specification, \(e_{ss} < 0.24A\), we find that the specification is satisfied when
FIGURE 6.22
(a) Ramp response for \(a = 0.6\) and \(K = 70\) for twotrack vehicle turning control. (b) m-file script.
(a)
\(\%\) Two-track vehicle turning control ramp response
\(\%\) with \(a = 0.6\) and \(K = 70\).
\(t = \lbrack 0:0.01:16\rbrack;u = t\);
\(u =\) unit ramp input
numgc=[1 0.6\(\rbrack;\) dengc=[1 1 \(\rbrack\); sysgc=tf(numgc, dengc);
numg=[70]; deng=[1 \(\left. \ 7\begin{matrix} 1 & 10 & 0 \end{matrix} \right\rbrack\); sysg=tf(numg,deng);
sysa=series(sysgc,sysg);
sys=feedback(sysa,[1]);
\(y = Isim(sys,u,t)\);
Linear simulation
plot(t,y,t,u,'--'), grid
\(a = 0.6\) and \(K = 70\)
(b)
or
Any values of \(a\) and \(K\) that lie in the stable region in Figure 6.21 and satisfy Equation (6.34) will lead to an acceptable design. For example, \(K = 70\) and \(a = 0.6\) will satisfy all the design requirements. The closed-loop transfer function (with \(a = 0.6\) and \(K = 70)\) is
The associated closed-loop poles are
The corresponding unit ramp input response is shown in Figure 6.22. The steadystate error is less than 0.24 , as desired. The Stability of State Variable Systems. Now let us turn to determining the stability of systems described in state variable form. Suppose we have a system in state-space form as in Equation (6.22). The stability of the system can be evaluated with the characteristic equation associated with the system matrix \(\mathbf{A}\). The characteristic equation is
The left-hand side of the characteristic equation is a polynomial in \(s\). If all of the roots of the characteristic equation have negative real parts (i.e., \(Re\left( s_{i} \right) < 0\) ), then the system is stable.
When the system model is given in state variable form, we must calculate the characteristic polynomial associated with the A matrix. In this regard, we have several options. We can calculate the characteristic equation directly from Equation (6.35) by manually computing the determinant of \(s\mathbf{I} - \mathbf{A}\). Then, we can compute the roots using the roots function to check for stability, or alternatively, we can use the Routh-Hurwitz method to detect any unstable roots. Unfortunately, the manual computations can become lengthy, especially if the dimension of \(\mathbf{A}\) is large. We would like to avoid this manual computation if possible. As it turns out, the computer can assist in this endeavor.
The poly function can be used to compute the characteristic equation associated with \(\mathbf{A}\). The poly is used to form a polynomial from a vector of roots. It can also be used to compute the characteristic equation of \(\mathbf{A}\), as illustrated in Figure 6.23. The input matrix \(\mathbf{A}\) is
and the associated characteristic polynomial is
FIGURE 6.23 Computing the characteristic polynomial of \(\mathbf{A}\) with the poly function.
If \(\mathbf{A}\) is an \(n \times n\) matrix, \(poly(\mathbf{A})\) is an \(n + 1\) element row vector whose elements are the coefficients of the characteristic equation \(det(s\mathbf{I} - \mathbf{A}) = 0\).
220.1. SEQUENTIAL DESIGN EXAMPLE: DISK DRIVE READ SYSTEM
In this section, we will examine the stability of the disk drive read system as \(K_{a}\) is adjusted and then reconfigure the system.
Let us consider the system as shown in Figure 6.24. Initially, we consider the case where the switch is open. Then the closed-loop transfer function is
where
The characteristic equation is
The Routh array is
where
The case \(b_{1} = 0\) results in marginal stability when \(K_{a} = 4080\). Using the auxiliary equation, we have
FIGURE 6.24
The closed-loop disk drive head system with an optional velocity feedback.
FIGURE 6.25
Equivalent system with the velocity feedback switch closed.
or the roots of the \(j\omega\)-axis are \(s = \pm j141.4\). In order for the system to be stable, \(K_{a} < 4080\).
Now let us add the velocity feedback by closing the switch in the system of Figure 6.24. The closed-loop transfer function for the system is then
since the feedback factor is equal to \(1 + K_{1}s\), as shown in Figure 6.25.
The characteristic equation is
or
Therefore, we have
The Routh array is
where
To guarantee stability, it is necessary to select the pair \(\left( K_{a},K_{1} \right)\) such that \(b_{1} > 0\), where \(K_{a} > 0\). When \(K_{1} = 0.05\) and \(K_{a} = 100\), we can determine the system response using the script shown in Figure 6.26. The settling time (with a \(2\%\) criterion) is approximately \(T_{s} = 260\text{ }ms\), and the percent overshoot is P.O. \(= 0\%\). The system performance is summarized in Table 6.3. The performance specifications are nearly satisfied, and some iteration of \(K_{1}\) is necessary to obtain the desired \(T_{s} = 250\text{ }ms\). FIGURE 6.26
Response of the system with velocity feedback. (a) m-file script. (b) Response with \(K_{a} = 100\) and \(K_{1} = 0.05\).
\(Ka = 100;K1 = 0.05\);
\(ng1 = \lbrack 5000\rbrack;dg1 = \lbrack 1\) 1000]; sys \(1 = tf(ng1,dg1)\); ng2=[1]; dg2=[1 20 0]; sys2=tf(ng2,dg2);
\(nc = \begin{bmatrix} K1 & 1 \end{bmatrix};dc = \begin{bmatrix} 0 & 1 \end{bmatrix};sysc = tf(nc,dc)\);
Select the velocity feedback gain \(K_{1}\) and amplifier gain \(K_{a}\).
syso=series $\left( {Ka}^{*} \right.\ $ sys 1 , sys 2\()\);
sys=feedback(syso,sysc); sys=minreal(sys);
\(t = \lbrack 0:0.001:0.5\rbrack\);
\(y =\) step(sys,t); plot(t,y)
ylabel('y(t)'), xlabel('Time (s)'), grid
(a)
(b)
Table 6.3 Performance of the Disk Drive System Compared to the Specifications
| Performance Measure | Desired Value | Actual Response |
| $$\begin | ||
| \text{~}\text{Percent overshoot}\text{~} \ | ||
| \text{~}\text{Settling time}\text | ||
| \end{matrix}$$ | Less than \(5\%\) | $$0%$$ |
| $$\begin | ||
| \text{~}\text{Maximum response}\text{~} \ | ||
| \text{~}\text{to a unit disturbance}\text | ||
| \end{matrix}$$ | Less than \(250\text{ }ms\) | $$260\text{ }ms$$ |
220.2. SUMMARY
In this chapter, we have considered the concept of the stability of a feedback control system. A definition of a stable system in terms of a bounded system response was outlined and related to the location of the poles of the system transfer function in the \(s\)-plane. The Routh-Hurwitz stability criterion was introduced, and several examples were considered. The relative stability of a feedback control system was also considered in terms of the location of the poles and zeros of the system transfer function in the \(s\)-plane. The stability of state variable systems was considered.
221. 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 6.27 as specified in the various problem statements.
FIGURE 6.27 Block diagram for the Skills Check.
In the following True or False and Multiple Choice problems, circle the correct answer.
- A stable system is a dynamic system with a bounded output response for any input.
True or False
- A marginally stable system has poles on the \(j\omega\)-axis.
True or False
- A system is stable if all poles lie in the right half-plane.
True or False
- The Routh-Hurwitz criterion is a necessary and sufficient criterion for determining the stability of linear systems.
True or False
- Relative stability characterizes the degree of stability.
True or False
- A system has the characteristic equation
The range of \(K\) for a stable system is:
a. \(K \geq 0.46\)
b. \(K < 0.46\)
c. \(0 < K < 0.46\)
d. Unstable for all \(K\)
- Utilizing the Routh-Hurwitz criterion, determine whether Systems 1 and 2 with the following polynomials are stable or unstable:
a. System 1 is stable, System 2 is stable
b. System 1 is unstable, System 2 is stable
c. System 1 is stable, System 2 is unstable
d. System 1 is unstable, System 2 is unstable 8. Consider the feedback control system block diagram in Figure 6.27. Investigate closedloop stability for \(G_{c}(s) = K(s + 1)\) and \(G(s) = \frac{1}{(s + 2)(s - 1)}\), for the two cases where \(K = 1\) and \(K = 3\).
a. Unstable for \(K = 1\) and stable for \(K = 3\)
b. Unstable for \(K = 1\) and unstable for \(K = 3\)
c. Stable for \(K = 1\) and unstable for \(K = 3\)
d. Stable for \(K = 1\) and stable for \(K = 3\)
- Consider a unity negative feedback system in Figure 6.27 with loop transfer function where
Determine the value of \(K\) for which the closed-loop system is marginally stable.
a. \(K = 10\)
b. \(K = 3\)
c. The system is unstable for all \(K\)
d. The system is stable for all \(K\)
- A system is represented by \(\overset{˙}{\mathbf{x}} = \mathbf{Ax}\), where
The values of \(K\) for a stable system are
a. \(K < 1/2\)
b. \(K > 1/2\)
c. \(K = 1/2\)
d. The system is stable for all \(K\)
- Use the Routh array to assist in computing the roots of the polynomial
a. \(s_{1} = - 1;s_{2,3} = \pm j\frac{\sqrt{2}}{2}\)
b. \(s_{1} = 1;s_{2,3} = \pm j\frac{\sqrt{2}}{2}\)
c. \(s_{1} = - 1;s_{2,3} = 1 \pm j\frac{\sqrt{2}}{2}\)
d. \(s_{1} = - 1;s_{2,3} = 1\)
- Consider the following unity feedback control system in Figure 6.27 where
The range of \(K\) for stability is
a. \(K < 260.68\)
b. \(50.06 < K < 123.98\)
c. \(100.12 < K < 260.68\)
d. The system is unstable for all \(K > 0\) In Problems 13 and 14, consider the system represented in a state-space form
-
The characteristic equation is:
a. \(q(s) = s^{3} + 5s^{2} - 10s - 6\)
b. \(q(s) = s^{3} + 5s^{2} + 10s + 5\)
c. \(q(s) = s^{3} - 5s^{2} + 10s - 5\)
d. \(q(s) = s^{2} - 5s + 10\) -
Using the Routh-Hurwitz criterion, determine whether the system is stable, unstable, or marginally stable.
a. Stable
b. Unstable
c. Marginally stable
d. None of the above -
A system has the block diagram representation as shown in Figure 6.27, where \(G(s) = \frac{10}{(s + 15)^{2}}\) and \(G_{c}(s) = \frac{K}{s + 80}\), where \(K\) is always positive. The limiting gain for a stable system is:
a. \(0 < K < 28875\)
b. \(0 < K < 27075\)
c. \(0 < K < 25050\)
d. Stable for all \(K > 0\)
In the following Word Match problems, match the term with the definition by writing the correct letter in the space provided.
a. Routh-Hurwitz criterion
b. Auxiliary polynomial
c. Marginally stable
d. Stable system
e. Stability
f. Relative stability
g. Absolute stability
A performance measure of a system.
A dynamic system with a bounded system response to a bounded input.
The property that is measured by the relative real part of each root or pair of roots of the characteristic equation.
A criterion for determining the stability of a system by examining the characteristic equation of the transfer function.
The equation that immediately precedes the zero entry in the Routh array.
A system description that reveals whether a system is stable or not stable without consideration of other system attributes such as degree of stability.
A system possesses this type of stability if the zero input response remains bounded as \(t \rightarrow \infty\).
222. EXERCISES
E6.1 A system has a characteristic equation \(s^{3} + 5Ks^{2} +\) \((2 + K)s + 15 = 0\). Determine the range of \(K\) for a stable system.
Answer: \(K > 0\)
E6.2 A system has a characteristic equation \(s^{3} + 10s^{2} +\) \(2s + 30 = 0\). Using the Routh-Hurwitz criterion, show that the system is unstable.
E6.3 A system has the characteristic equation \(s^{4} + 10s^{3} +\) \(32s^{2} + 37s + 20 = 0\). Using the Routh-Hurwitz criterion, determine if the system is stable.
E6.4 A control system has the structure shown in Figure E6.4. Determine the gain at which the system will become unstable.
Answer: \(K = 20/7\)
E6.5 A unity feedback system has a loop transfer function
where \(K = 15\). Find the roots of the closed-loop system's characteristic equation.
E6.6 A negative feedback system has a loop transfer function
(a) Find the value of the gain when the \(\zeta\) of the closed-loop roots is equal to 0.5. (b) Find the value of the gain when the closed-loop system has two roots on the imaginary axis.
E6.7 For the feedback system of Exercise E6.5, find the value of \(K\) when two roots lie on the imaginary axis. Determine the value of the roots.
Answer: \(s = \pm j2.5131\)
E6.8 Designers have developed small, fast, vertical-takeoff fighter aircraft that are invisible to radar (stealth aircraft). This aircraft concept uses quickly turning jet nozzles to steer the airplane [16]. The control system for the heading or direction control is shown in Figure E6.8. Determine the maximum gain of the system for stable operation.
E6.9 A system has a characteristic equation
Find the range of \(K\) for a stable system.
Answer: \(K > 3\)
E6.10 Consider a feedback system with closed-loop transfer function
Is the system stable?
E6.11 A system with a transfer function \(Y(s)/R(s)\) is
Determine the steady-state error to a unit step input. Is the system stable?
E6.12 A system has the second-order characteristic equation
where \(a\) and \(b\) are constant parameters. Determine the necessary and sufficient conditions for the system to be stable. Is it possible to determine stability of a second-order system just by inspecting the coefficients of the characteristic equation?
FIGURE E6.4
Feedforward system.
FIGURE E6.8
Aircraft heading control.
FIGURE E6.13 Closed-loop system with a proportional plus derivative controller \(G_{C}(s) = K_{P} + K_{D}s\).
E6.13 Consider the feedback system in Figure E6.13. Determine the range of \(K_{P}\) and \(K_{D}\) for stability of the closed-loop system.
E6.14 By using magnetic bearings, a rotor is supported contactless. The technique of contactless support for rotors becomes more important in light and heavy industrial applications [14]. The matrix differential equation for a magnetic bearing system is
where \(\mathbf{x}^{T}(t) = (y(t),\overset{˙}{y}(t),i(t)),y(t) =\) bearing gap, and \(i(t)\) is the electromagnetic current. Determine whether the system is stable.
Answer: The system is stable.
E6.15 A system has a characteristic equation
(a) Determine whether the system is stable, using the Routh-Hurwitz criterion. (b) Determine the roots of the characteristic equation.
Answer: (a) The system is marginally stable.
(b) \(s = - 3, - 4, - 1 \pm 2j, \pm 0.5j\)
E6.16 A system has a characteristic equation
(a) Determine whether the system is stable, using the Routh-Hurwitz criterion. (b) Determine the roots of the characteristic equation.
E6.17 The matrix differential equation of a state variable model of a system is
(a) Determine the characteristic equation. (b) Determine whether the system is stable. (c) Determine the roots of the characteristic equation.
Answer: (a) \(q(s) = s^{3} + 16s^{2} + 68s + 80 = 0\)
E6.18 A system has a characteristic equation
(a) Determine whether the system is stable, using the Routh-Hurwitz criterion. (b) Determine the roots of the characteristic equation.
E6.19 Determine whether the systems with the following characteristic equations are stable or unstable:
(a) \(s^{3} + 4s^{2} + 6s + 100 = 0\),
(b) \(s^{4} + 6s^{3} + 10s^{2} + 17s + 6 = 0\), and
(c) \(s^{2} + 6s + 3 = 0\).
E6.20 Find the roots of the following polynomials:
(a) \(s^{3} + 5s^{2} + 8s + 4 = 0\) and
(b) \(s^{3} + 9s^{2} + 27s + 27 = 0\).
E6.21 A system has a transfer function \(Y(s)/R(s) =\) \(T(s) = 1/s(s + 1)\). (a) Is this system stable? (b) If \(r(t)\) is a unit step input, determine the response \(y(t)\).
E6.22 A system has the characteristic equation
Shift the vertical axis to the right by 1 by using \(s = s_{n} - 1\), and determine the value of gain \(K\) so that the complex roots are \(s = - 1 \pm \sqrt{3}j\).
E6.23 The matrix differential equation of a state variable model of a system is
Find the range of \(k\) where the system is stable.
E6.24 Consider the system represented in state variable form
where
(a) What is the system transfer function? (b) For what E6.26 Consider the closed-loop system in Figure E6.26, values of \(k\) is the system stable?
where
E6.25 A closed-loop feedback system is shown in Figure E6.25. For what range of values of the parameters \(K\) and \(p\) is the system stable?
FIGURE E6.25 Closed-loop system with parameters \(K\) and \(p\).
(a) Determine the characteristic equation associated with the closed-loop system.
(b) Determine the values of \(K\) for which the closedloop system is stable.
(a)
FIGURE E6.26
Closed-loop feedback control system with parameter \(K\).
(b)
223. PROBLEMS
P6.1 Utilizing the Routh-Hurwitz criterion, determine the stability of the following polynomials:
(a) \(s^{2} + 5s + 2 = 0\)
(b) \(s^{3} + 4s^{2} + 8s + 4 = 0\)
(c) \(s^{3} + 2s^{2} - 6s + 20 = 0\)
(d) \(s^{4} + s^{3} + 2s^{2} + 12s + 10 = 0\)
(e) \(s^{4} + s^{3} + 3s^{2} + 2s + K = 0\)
(f) \(s^{5} + s^{4} + 2s^{3} + s + 6 = 0\)
(g) \(s^{5} + s^{4} + 2s^{3} + s^{2} + s + K = 0\)
Determine the number of roots, if any, in the righthand plane. If it is adjustable, determine the range of \(K\) that results in a stable system. P6.2 An antenna control system was analyzed in Problem P4.5, and it was determined that, to reduce the effect of wind disturbances, the gain of the magnetic amplifier, \(k_{a}\), should be as large as possible. (a) Determine the limiting value of gain for maintaining a stable system. (b) We want to have a system settling time equal to 1.5 seconds. Using a shifted axis and the Routh-Hurwitz criterion, determine the value of the gain that satisfies this requirement. Assume that the complex roots of the closed-loop system dominate the transient response. (Is this a valid approximation in this case?)
P6.3 Arc welding is one of the most important areas of application for industrial robots [11]. In most manufacturing welding situations, uncertainties in dimensions of the part, geometry of the joint, and the welding process itself require the use of sensors for maintaining weld quality. Several systems use a vision system to measure the geometry of the puddle of melted metal, as shown in Figure P6.3. This system uses a constant rate of feeding the wire to be melted. (a) Calculate the maximum value for \(K\) for the system that will result in a stable system. (b) For half of the maximum value of \(K\) found in part (a), determine the roots of the characteristic equation. (c) Estimate the overshoot of the system of part (b) when it is subjected to a step input.
P6.4 A feedback control system is shown in Figure P6.4. The controller and process transfer functions are given by
and the feedback transfer function is \(H(s) = 1\) / \((s + 50)\). (a) Determine the limiting value of gain \(K\) for a stable system. (b) For the gain that results in marginal stability, determine the magnitude of the imaginary roots. (c) Reduce the gain to one-third of the magnitude of the marginal value, and determine the relative stability of the system (1) by shifting the axis and using the Routh-Hurwitz criterion and (2) by determining the root locations. Show the roots are between -4 and -5 .
P6.5 Determine the relative stability of the systems with the following characteristic equations (1) by shifting the axis in the \(s\)-plane and using the Routh-Hurwitz criterion, and (2) by determining the location of the complex roots in the \(s\)-plane:
(a) \(s^{3} + 3s^{2} + 4s + 2 = 0\).
(b) \(s^{4} + 9s^{3} + 30s^{2} + 42s + 20 = 0\).
(c) \(s^{3} + 19s^{2} + 110s + 200 = 0\).
P6.6 A unity-feedback control system is shown in Figure P6.6. Determine the stability of the system
FIGURE P6.3
Welder control.
FIGURE P6.4 Nonunity feedback system.
FIGURE P6.6 Unity feedback system.
with the following loop transfer functions using the Routh-Hurwitz criterion:
(a) \(G_{c}(s)G(s) = \frac{(10s + 30)(s + 1)}{s^{2}(s - 2)}\)
(b) \(G_{c}(s)G(s) = \frac{10}{s\left( s^{3} + 2s^{2} + 5s + 2 \right)}\)
(c) \(G_{c}(s)G(s) = \frac{s^{2} + s + 3}{s(s + 1)(s + 2)}\)
P6.7 The linear model of a phase detector (phase-lock loop) can be represented by Figure P6.7 [9]. The phase-lock systems are designed to maintain zero difference in phase between the input carrier signal and a local voltage-controlled oscillator. The filter for a particular application is chosen as
We want to minimize the steady-state error of the system for a ramp change in the phase information signal. (a) Determine the limiting value of the gain \(K_{a}K = K_{v}\) in order to maintain a stable system. (b) A steady-state error equal to \(3^{\circ}\) is acceptable for a ramp signal of \(120rad/s\). For that value of gain \(K_{v}\), determine the location of the roots of the system.
P6.8 A very interesting and useful velocity control system has been designed for a wheelchair control system. A proposed system utilizing velocity sensors mounted in a headgear is shown in Figure P6.8. The headgear sensor provides an output proportional to the magnitude of the head movement. There is a sensor mounted at \(90^{\circ}\) intervals so that forward, left, right, or reverse can be commanded. Typical values for the time constants are \(\tau_{1} = 0.5\text{ }s,\tau_{2} = 1\text{ }s\), and \(\tau_{3} = 1/4\text{ }s\).
(a) Determine the limiting gain \(K = K_{1}K_{2}K_{3}\) for a stable system.
(b) When the gain \(K\) is set equal to one-third of the limiting value, determine whether the settling time (to within \(2\%\) of the final value of the system) is \(T_{s} \leq 4\text{ }s\).
(c) Determine the value of gain that results in a system with a settling time of \(T_{s} \leq 4\text{ }s\). Also, obtain the value of the roots of the characteristic equation when the settling time is \(T_{s} \leq 4\text{ }s\).
P6.9 A cassette tape storage device has been designed for mass-storage [1]. It is necessary to control the velocity of the tape accurately. The speed control of the tape drive is represented by the system shown in Figure P6.9.
FIGURE P6.7
Phase-lock loop system.
FIGURE P6.8 Wheelchair control system.
FIGURE P6.9
Tape drive control. (a) Determine the limiting gain for a stable system.
(b) Determine a suitable gain so that the percent overshoot to a step command is P.O. \(= 5\%\).
P6.10 Robots can be used in manufacturing and assembly operations that require accurate, fast, and versatile manipulation \(\lbrack 10,11\rbrack\). The loop transfer function of a direct-drive arm is
(a) Determine the value of gain \(K\) when the system oscillates. (b) Calculate the roots of the closed-loop system for the \(K\) determined in part (a).
P6.11 A feedback control system has a characteristic equation
The parameter \(K\) must be positive. What is the maximum value \(K\) can assume before the system becomes unstable? When \(K\) is equal to the maximum value, the system oscillates. Determine the frequency of oscillation.
P6.12 A system has the third-order characteristic equation
where \(a,b\), and c are constant parameters. Determine the necessary and sufficient conditions for the system to be stable. Is it possible to determine stability of the system by just inspecting the coefficients of the characteristic equation?
P6.13 Consider the system in Figure P6.13. Determine the conditions on \(K,p\), and \(z\) that must be satisfied for closed-loop stability. Assume that \(K > 0,\zeta > 0\), and \(\omega_{n} > 0\).
P6.14 A feedback control system has a characteristic equation
Determine whether the system is stable, and determine the values of the roots.
P6.15 The stability of a motorcycle and rider is an important area for study \(\lbrack 12,13\rbrack\). The handling characteristics of a motorcycle must include a model of the rider as well as one of the vehicle. The dynamics of one motorcycle and rider can be represented by the loop transfer function
(a) As an approximation, calculate the acceptable range of \(K\) for a stable unity negative feedback system when the numerator polynomial (zeros) and the denominator polynomial \(\left( s^{2} + 80s + 2000 \right)\) are neglected. (b) Calculate the actual range of acceptable \(K\), account for all zeros and poles.
P6.16 A system has a closed-loop transfer function
(a) Determine whether the system is stable. (b) Determine the roots of the characteristic equation. (c) Plot the response of the system to a unit step input.
P6.17 The elevator in Yokohama's 70-story Landmark Tower operates at a peak speed of \(45\text{ }km/hr\). To reach such a speed without inducing discomfort in passengers, the elevator accelerates for longer periods, rather than more precipitously. Going up, it reaches full speed only at the 27 th floor; it begins decelerating 15 floors later. The result is a peak acceleration similar to that of other skyscraper elevators - a bit less than a tenth of the force of gravity. Admirable ingenuity has gone into making this safe and comfortable. Special ceramic brakes had to be developed; iron ones would melt. Computer-controlled systems damp out vibrations. The lift has been streamlined to reduce the wind noise as it speeds up and down [19]. One proposed control system for the elevator vertical position is shown in Figure P6.17. Determine the range of \(K\) for a stable system, where \(K > 0\).
FIGURE P6.13
Control system with controller with three parameters \(K,p\), and \(z\).
FIGURE P6.17 Elevator control system.
P6.18 Consider the case of rabbits and foxes. The number of rabbits is \(x_{1}\) and, if left alone, it would grow indefinitely (until the food supply was exhausted) so that
However, with foxes present, we have
where \(x_{2}\) is the number of foxes. Now, if the foxes must have rabbits to exist, we have
Determine whether this system is stable and thus decays to the condition \(x_{1}(t) = x_{2}(t) = 0\) at \(t = \infty\). What are the requirements on \(a,b,h\), and \(k\) for a stable system? What is the result when \(k\) is greater than \(h\) ?
P6.19 The goal of vertical takeoff and landing (VTOL) aircraft is to achieve operation from relatively small airports and yet operate as a normal aircraft in level flight [16]. An aircraft taking off in a form similar to a rocket is inherently unstable. A control system using adjustable jets can control the vehicle, as shown in Figure P6.19. (a) Determine the range of gain for which the system is stable. (b) Determine the gain \(K\) for which the system is marginally stable and the roots of the characteristic equation for this value of \(K\).
P6.20 A personal vertical take-off and landing (VTOL) aircraft is shown in Figure P6.20(a). A possible control system for aircraft altitude is shown in Figure P6.20(b). (a) For \(K = 17\), determine whether the system is stable.
(b) Determine a range of stability, if any, for \(K > 0\).
P6.21 Consider the system described in state variable form by
where
and where \(k_{1} \neq k_{2}\) and both \(k_{1}\) and \(k_{2}\) are real numbers.
(a) Compute the state transition matrix \(\Phi(t,0)\).
(b) Compute the eigenvalues of the system matrix \(\mathbf{A}\).
(c) Compute the roots of the characteristic polynomial. (d) Discuss the results of parts (a)-(c) in terms of stability of the system.
FIGURE P6.19 Control of a jump-jet aircraft.
FIGURE P6.20
(a) Personal VTOL aircraft. (Cheskyw/123RF.)
(b) Control system.
(a)
(b)
224. ADVANCED PROBLEMS
AP6.1 A teleoperated control system incorporates both a person (operator) and a remote machine. The normal teleoperation system is based on a one-way link to the machine and limited feedback to the operator. However, two-way coupling using bilateral information exchange enables better operation [18]. In the case of remote control of a robot, force feedback plus position feedback is useful. The characteristic equation for a teleoperated system, as shown in Figure AP6.1, is
where \(K_{1}\) and \(K_{2}\) are feedback gain factors. Determine and plot the region of stability for this system for \(K_{1}\) and \(K_{2}\).
AP6.2 Consider the case of a navy pilot landing an aircraft on an aircraft carrier. The pilot has three basic tasks. The first task is guiding the aircraft's approach to the ship along the extended centerline of the runway. The second task is maintaining the aircraft on the correct glideslope. The third task is maintaining the correct speed. A model of a lateral position control system is shown in Figure AP6.2. Determine the range of stability for \(K \geq 0\).
FIGURE AP6.1 Model of a teleoperated machine.
AP6.3 A control system is shown in Figure AP6.3. We want the system to be stable and the steady-state error for a unit step input to be less than or equal to 0.1 . (a) Determine the range of \(\alpha\) that satisfies the error requirement. (b) Determine the range of \(\alpha\) that satisfies the stability requirement. (c) Select an \(\alpha\) that meets both requirements.
AP6.4 A bottle-filling line uses a feeder screw mechanism, as shown in Figure AP6.4. The tachometer feedback is used to maintain accurate speed control. Determine and plot the range of \(K\) and \(p\) that permits stable operation.
AP6.5 Consider the closed-loop system in Figure AP6.5. Suppose that all gains are positive, that is, \(K_{1} > 0,K_{2} > 0,K_{3} > 0,K_{4} > 0\), and \(K_{5} > 0\).
(a) Determine the closed-loop transfer function \(T(s) = Y(s)/R(s)\).
(b) Obtain the conditions on selecting the gains \(K_{1},K_{2},K_{3},K_{4}\), and \(K_{5}\), so that the closedloop system is guaranteed to be stable.
(c) Using the results of part (b), select values of the five gains so that the closed-loop system is stable, and plot the unit step response.
AP6.6 A spacecraft with a camera is shown in Figure AP6.6(a). The camera slews about \(16^{\circ}\) in a canted plane relative to the base. Reaction jets stabilize the base against the reaction torques from the slewing motors. Suppose that the rotational speed control for the camera slewing has a plant transfer function
FIGURE AP6.2
Lateral position control for landing on an aircraft carrier.
FIGURE AP6.3 Third-order unity feedback system.
FIGURE AP6.4 Speed control of a bottle-filling line.
(a) System layout.
(b) Block diagram.
(a)
(b)
(a)
(b)
FIGURE AP6.5 Multiloop feedback control system. (a) Signal flow graph. (b) Block diagram. FIGURE AP6.6
(a) Spacecraft with a camera.
(b) Feedback control system.
(a)
(b)
A proportional plus derivative controller is used in a system as shown in Figure AP6.6(b), where
and where \(K_{p} > 0\) and \(K_{D} > 0\). Obtain and plot the relationship between \(K_{p}\) and \(K_{D}\) that results in a stable closed-loop system.
AP6.7 A human's ability to perform physical tasks is limited not by intellect but by physical strength. If, in an appropriate environment, a machine's mechanical power is closely integrated with a human arm's mechanical strength under the control of the human intellect, the resulting system will be superior to a loosely integrated combination of a human and a fully automated robot.
Extenders are defined as a class of robot manipulators that extend the strength of the human arm while maintaining human control of the task [23]. The defining characteristic of an extender is the transmission of both power and information signals.
The extender is worn by the human; the physical contact between the extender and the human allows the direct transfer of mechanical power and information signals. Because of this unique interface, control of the extender trajectory can be accomplished without any type of joystick or keyboard. The human provides a control system for the extender, while the extender actuators provide most of the strength necessary for the task. The human becomes a part of the extender and "feels" a scaled-down version of the load that the extender is carrying. An extender is shown in Figure AP6.7(a) [23]. The block diagram of the system is shown in Figure AP6.7(b). Consider the proportional plus derivative controller
Determine the range of values of the controller gains \(K_{P}\) and \(K_{D}\) such that the closed-loop system is stable. FIGURE AP6.7
Extender robot control.
(a)
(b)
225. DESIGN PROBLEMS
CDP6.1 The capstan drive system of problem CDP5.1 uses the amplifier as the controller. Determine the maximum value of the gain \(K_{a}\) before the system becomes unstable.
DP6.1 The control of the spark ignition of an automotive engine requires constant performance over a wide range of parameters [15]. The control system is shown in Figure DP6.1, with a controller gain \(K\) to be selected. The parameter \(p\) is equal to 2 for many autos but can equal zero for those with high performance.
Select a gain \(K\) that will result in a stable system for both values of \(p\).
DP6.2 An automatically guided vehicle on Mars is represented by the system in Figure DP6.2. The system has a steerable wheel in both the front and back of the vehicle, and the design requires that \(H(s) = Ks + 1\). Determine (a) the value of \(K\) required for stability, (b) the value of \(K\) when one root of the characteristic equation is equal to \(s = - 1\), and (c) the value of the two remaining roots for the gain selected in
FIGURE DP6.1
Automobile engine control.
FIGURE DP6.2
Mars guided vehicle control.
part (b). (d) Find the response of the system to a step command for the gain selected in part (b).
DP6.3 A unity negative feedback system with
has two parameters to be selected. (a) Determine and plot the regions of stability for this system. (b) Select \(\tau\) and \(K\) so that the steady-state error to a unit ramp input is less than or equal to 0.1. (c) Determine the percent overshoot for a step input for the design selected in part (b).
DP6.4 The attitude control system of a rocket is shown in Figure DP6.4 [17]. (a) Determine the range of gain \(K\) and parameter \(m\) so that the system is stable, and plot the region of stability. (b) Select the gain and parameter values so that the steady-state error to a ramp input is less than or equal to \(10\%\) of the input magnitude. (c) Determine the percent overshoot for a step input for the design selected in part (b).
DP6.5 A traffic control system is designed to control the distance between vehicles, as shown in Figure DP6.5 [15]. (a) Determine the range of gain \(K\) for which the system is stable. (b) If \(K_{m}\) is the maximum value of \(K\) so that the characteristic roots are on the \(j\omega\)-axis, then let \(K = K_{m}/N\), where \(N\) is to be selected. We want the peak time to be \(T_{p} \leq 2\text{ }s\) and the percent overshoot to be P.O. \(\leq 20\%\). Determine an appropriate value for \(N\).
DP6.6 Consider the single-input, single-output system as 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 \(\mathbf{K} =\) \(\begin{bmatrix} K_{1} & K_{2} \end{bmatrix}\) is known as the gain matrix. If you substitute \(u(t)\) into the state variable equation you obtain the closed-loop system
For what values of \(\mathbf{K}\) is the closed-loop system stable? Determine the region of the left half-plane where the desired closed-loop eigenvalues should be placed
FIGURE DP6.4
Rocket attitude
control.
FIGURE DP6.5
Traffic distance control. FIGURE DP6.7
Feedback system with inner and outer loop.
226. FIGURE DP6.8
A marginally stable plant with a PD controller in the loop.
so that the percent overshoot to a unit step input, \(R(s) = 1/s\), is \(P.O. < 5\%\) and the settling time is \(T_{s} < 4\text{ }s\). Select a gain matrix, \(\mathbf{K}\), so that the system step response meets the specifications P.O. \(< 5\%\) and \(T_{s} < 4s\).
DP6.7 Consider the feedback control system in Figure DP6.7. The system has an inner loop and an outer loop. The inner loop must be stable and have a quick speed of response. (a) Consider the inner loop first. Determine the range of \(K_{1}\) resulting in a stable inner loop. That is, the transfer function \(Y(s)/U(s)\) must be stable. (b) Select the value of \(K_{1}\) in the stable range leading to the fastest step response. (c) For the value of \(K_{1}\) selected in (b), determine the range of \(K_{2}\) such that the closed-loop system \(T(s) = Y(s)/R(s)\) is stable.
DP6.8 Consider the feedback system shown in Figure DP6.8. The process transfer function is marginally stable. The controller is the proportional-derivative (PD) controller
Determine if it is possible to find values of \(K_{P}\) and \(K_{D}\) such that the closed-loop system is stable. If so, obtain values of the controller parameters such that the steadystate tracking error \(E(s) = R(s) - Y(s)\) to a unit step input \(R(s) = 1/s\) is \(e_{ss} = \lim_{t \rightarrow \infty}\mspace{2mu} e(t) \leq 0.01\) and the damping of the closed-loop system is \(\zeta = \sqrt{2}/2\).
227. COMPUTER PROBLEMS
CP6.1 Determine the roots of the following characteristic equations:
(a) \(q(s) = s^{3} + 3s^{2} + 10s + 14 = 0\).
(b) \(q(s) = s^{4} + 8s^{3} + 24s^{2} + 32s + 16 = 0\).
(c) \(q(s) = s^{4} + 2s^{2} + 1 = 0\).
CP6.2 Consider a unity negative feedback system with
Develop an m-file to compute the roots of the closedloop transfer function characteristic polynomial for \(K = 1,2\), and 5 . For which values of \(K\) is the closedloop system stable?
CP6.3 A unity negative feedback system has the loop transfer function
Develop an m-file to determine the closed-loop transfer function, and show that the roots of the characteristic equation are \(s_{1} = - 9.79\) and \(s_{2,3} = - 0.104\) \(\pm j1.7178\).
CP6.4 Consider the closed-loop transfer function
(a) Using the Routh-Hurwitz method, determine whether the system is stable. If it is not stable, how many poles are in the right half-plane? (b) Compute the poles of \(T(s)\), and verify the result in part (a).
(c) Plot the unit step response, and discuss the results.
CP6.5 A "paper-pilot" model is sometimes utilized in aircraft control design and analysis to represent the pilot in the loop. A block diagram of an aircraft with a pilot "in the loop" is shown in Figure CP6.5. The variable \(\tau\) represents the pilot's time delay. Assume that we have a fast pilot with \(\tau = 0.1\) and \(K = 1\). Develop an m-file to obtain the region of stability for \(\tau_{1}\) and \(\tau_{2}\), changing in the range of 0 to 5 . Show step responses for two points: one inside and one outside of this region.
CP6.6 Consider the feedback control system in Figure CP6.6. Using the for function, develop an m-file script to compute the closed-loop transfer function poles for \(0 \leq K \leq 5\) and plot the results denoting the poles with the " \(x\) " symbol. Determine the maximum range of \(K\) for stability with the Routh-Hurwitz method. Compute the roots of the characteristic equation when \(K\) is the minimum value allowed for stability.
CP6.7 Consider a system in state variable form:
(a) Compute the characteristic equation using the poly function. (b) Compute the roots of the characteristic equation, and determine whether the system is stable. (c) Obtain the response plot of \(y(t)\) when \(u(t)\) is a unit step and when the system has zero initial conditions.
CP6.8 Consider the feedback control system in Figure CP6.8. (a) Using the Routh-Hurwitz method, determine the range of \(K_{1}\) resulting in closed-loop stability. (b) Develop an m-file to plot the pole locations as a function of \(0 < K_{1} < 30\) and comment on the results.
CP6.9 Consider a system represented in state variable form
where
(a) For what values of \(k\) is the system stable?
(b) Develop an m-file to plot the pole locations as a function of \(0 < k < 50\), and comment on the results.
FIGURE CP 6.5
An aircraft with a pilot in the loop.
FIGURE CP6.6 A single-loop feedback control system with parameter \(K\).
FIGURE CP6.8 Nonunity feedback system with parameter \(K_{1}\).
228. ANSWERS TO SKILLS CHECK
True or False: (1) False; (2) True; (3) False; (4) True; (5) True
Multiple Choice: (6) a; (7) c; (8) a; (9) b; (10) b; (11) a; (12) c; (13) b; (14) a; (15) b
Word Match (in order, top to bottom): e, d, f, a, b, \(g,c\)
229. TERMS AND CONCEPTS
Absolute stability A system description that reveals whether a system is stable or not stable without consideration of other system attributes such as degree of stability.
Auxiliary polynomial The equation that immediately precedes the zero entry in the Routh array.
Marginally stable A system is marginally stable if and only if the zero input response remains bounded as \(t \rightarrow \infty\).
Relative stability The property that is measured by the relative real part of each root or pair of roots of the characteristic equation.
Routh-Hurwitz criterion A criterion for determining the stability of a system by examining the characteristic equation of the transfer function. The criterion states that the number of roots of the characteristic equation with positive real parts is equal to the number of changes of sign of the coefficients in the first column of the Routh array.
Stability A performance measure of a system. A system is stable if all the poles of the transfer function have negative real parts.
Stable system A dynamic system with a bounded system response to a bounded input.
CHAPTER
230. The Root Locus Method
7.1 Introduction 447
7.2 The Root Locus Concept 447
7.3 The Root Locus Procedure 452
7.4 Parameter Design by the Root Locus Method 466
7.5 Sensitivity and the Root Locus 472
7.6 PID Controllers 477
7.7 Negative Gain Root Locus 488
7.8 Design Examples 493
7.9 The Root Locus Using Control Design Software 502
7.10 Sequential Design Example: Disk Drive Read System 508
\(\mathbf{7.11}\) Summary \(\mathbf{510}\)
231. PREVIEW
The performance of a feedback system can be described in terms of the location of the roots of the characteristic equation in the \(s\)-plane. A graph showing how the roots of the characteristic equation move around the \(s\)-plane as a single parameter varies is known as a root locus plot. The root locus is a powerful tool for designing and analyzing feedback control systems. We will discuss practical techniques for obtaining a sketch of a root locus plot. We also consider computer-generated root locus plots and illustrate their effectiveness in the design process. We show that it is possible to use root locus methods for controller design when more than one parameter varies. This is important because we know that the response of a closed-loop feedback system can be adjusted to achieve the desired performance by judicious selection of one or more controller parameters. The popular PID controller is introduced as a practical controller structure. We also define a measure of sensitivity of a specified root to a small incremental change in a system parameter. The chapter concludes with a controller design based on root locus methods for the Sequential Design Example: Disk Drive Read System.
232. DESIRED OUTCOMES
Upon completion of Chapter 7, students should be able to:
$\square\ $ Describe the powerful concept of the root locus and its role in control system design.
$\square\ $ Create a root locus plot by sketching or using computers.
$\square\ $ Identify the PID controller as a key element of many feedback systems.
\(\square\) Explain the role of root locus plots in parameter design and system sensitivity analysis.
\(\square\) Design controllers to meet desired specifications using root locus methods.
232.1. INTRODUCTION
The relative stability and the transient performance of a closed-loop control system are directly related to the location of the closed-loop roots of the characteristic equation in the \(s\)-plane. It is frequently necessary to adjust one or more system parameters in order to obtain suitable root locations. Therefore, it is worthwhile to determine how the roots of the characteristic equation of a given system migrate about the \(s\)-plane as the parameters are varied; that is, it is useful to determine the locus of roots in the \(s\)-plane as a parameter is varied. The root locus method was introduced by Evans in 1948 and has been developed and utilized extensively in control engineering practice [1-3]. The root locus technique is a graphical method for sketching the locus of roots in the \(s\)-plane as a parameter is varied. The root locus method provides the engineer with a measure of the sensitivity of the roots of the system to a variation in the parameter being considered. The root locus technique may be used to great advantage in conjunction with the Routh-Hurwitz criterion.
The root locus method provides graphical information, and therefore an approximate sketch can be used to obtain qualitative information concerning the stability and performance of the system. Furthermore, the locus of roots of the characteristic equation of a multiloop system may be investigated as readily as for a single-loop system. If the root locations are not satisfactory, the necessary parameter adjustments often can be readily ascertained from the root locus [4].
232.2. THE ROOT LOCUS CONCEPT
The dynamic performance of a closed-loop control system is described by the closed-loop transfer function
where \(p(s)\) and \(q(s)\) are polynomials in \(s\). The roots of the characteristic equation \(q(s)\) determine the modes of response of the system. In the case of the simple single-loop system shown in Figure 7.1, we have the characteristic equation
where \(K\) is a variable parameter and \(0 \leq K < \infty\). The characteristic roots of the system must satisfy Equation (7.2), where the roots lie in the \(s\)-plane. Because \(s\) is a complex variable, Equation (7.2) may be rewritten in polar form as
FIGURE 7.1
Closed-loop control system with a variable parameter \(K\).
and therefore it is necessary that
and
where \(k = 0, \pm 1, \pm 2, \pm 3,\ldots\)
233. The root locus is the path of the roots of the characteristic equation traced
out in the
-plane as a system parameter varies from zero to infinity.
Consider the second-order system shown in Figure 7.2. The characteristic equation is
or, alternatively,
The locus of the roots as the gain \(K\) is varied is found by requiring that
and
The gain \(K\) is varied from zero to infinity. For a second-order system, the roots are
and for \(\zeta < 1\), we know that \(\theta = \cos^{- 1}\zeta\). Graphically, for two open-loop poles as shown in Figure 7.3, the locus of roots is a vertical line for \(\zeta \leq 1\) in order to satisfy the angle requirement, Equation (7.7). For example, as shown in Figure 7.4, at a root \(s_{1}\), the angles are
FIGURE 7.2
Unity feedback control system. The gain \(K\) is a variable parameter.
FIGURE 7.3
Root locus for a second-order system when \(K_{e} < K_{1} < K_{2}\). The locus is shown as heavy lines, with arrows indicating the direction of increasing \(K\). Note that roots of the characteristic equation are denoted by " \(\square\) " on the root locus.
This angle requirement is satisfied at any point on the vertical line that is a perpendicular bisector of the line 0 to -2 . Furthermore, the gain \(K\) at the particular points is found by using Equation (7.6) as
and thus
where \(\left| s_{1} \right|\) is the magnitude of the vector from the origin to \(s_{1}\), and \(\left| s_{1} + 2 \right|\) is the magnitude of the vector from -2 to \(s_{1}\).
For a multiloop closed-loop system, using the Mason's signal-flow gain formula yields
FIGURE 7.4
Evaluation of the angle and gain at \(s_{1}\) for gain \(K = K_{1}\).
where \(L_{n}\) equals the value of the \(n\)th self-loop transmittance. Hence, we have a characteristic equation, which may be written as
To find the roots of the characteristic equation, we set Equation (7.13) equal to zero and obtain
Equation (7.14) may be rewritten as
and the roots of the characteristic equation must also satisfy this relation.
In general, the function \(F(s)\) may be written as
Then the magnitude and angle requirement for the root locus are
and
where \(k\) is an integer. The magnitude requirement in Equation (7.16) enables us to determine the value of \(K\) for a given root location \(s_{1}\). A test point in the \(s\)-plane, \(s_{1}\), is verified as a root location when Equation (7.17) is satisfied. All angles are measured in a counterclockwise direction from a horizontal line.
To further illustrate the root locus procedure, let us consider the second-order system of Figure 7.5(a) where \(a > 0\). The effect of varying the parameter \(a\) can
234. FIGURE 7.5
(a) Single-loop system. (b) Root locus as a function of the parameter a, where \(a > 0\).
(b) be effectively portrayed by rewriting the characteristic equation for the root locus form with \(a\) as the multiplying factor in the numerator. Then the characteristic equation is
or, alternatively,
Dividing by the factor \(s^{2} + K\), we obtain
Then the magnitude criterion is satisfied when
at the root \(s_{1}\). The angle criterion is
In principle, we could construct the root locus by determining the points in the \(s\)-plane that satisfy the angle criterion. In the next section, we develop a multistep procedure to sketch the root locus. The root locus for the characteristic equation in Equation (7.18) is shown in Figure 7.5(b). Specifically at the root \(s_{1}\), the magnitude of the parameter \(a\) is found from Equation (7.19) as
The roots of the system merge on the real axis at the point \(s_{2}\) and provide a critically damped response to a step input. The parameter \(a\) has a magnitude at the critically damped roots, \(s_{2} = \sigma_{2}\), equal to
where \(\sigma_{2}\) is evaluated from the \(s\)-plane vector lengths as \(\sigma_{2} = \sqrt{K}\). As \(a\) increases beyond the critical value, the roots are both real and distinct; one root is larger than \(\sigma_{2}\), and one is smaller.
In general, we desire an orderly process for locating the locus of roots as a parameter varies. In the next section, we will develop such an orderly approach to sketching a root locus diagram.
234.1. THE ROOT LOCUS PROCEDURE
The roots of the characteristic equation of a system provide valuable insight concerning the response of the system. To locate the roots of the characteristic equation in a graphical manner on the \(s\)-plane, we develop an orderly procedure of seven steps that facilitates the rapid sketching of the locus. tion as
Step 1: Prepare the root locus sketch. Begin by writing the characteristic equa-
Rearrange the equation, if necessary, so that the parameter of interest, \(K\), appears as the multiplying factor in the form,
We are interested in determining the locus of roots when \(K\) varies as \(0 \leq K < \infty\). In Section 7.7, we consider the case when \(K\) varies as \(- \infty < K \leq 0\).
Factor \(P(s)\), and write the polynomial in the form of poles and zeros as follows:
Locate the poles \(- p_{i}\) and zeros \(- z_{i}\) on the \(s\)-plane with selected symbols. By convention, we use " \(x\) " to denote poles and "o" to denote zeros.
Rewriting Equation (7.24), we have
Note that Equation (7.25) is another way to write the characteristic equation. When \(K = 0\), the roots of the characteristic equation are the poles of \(P(s)\). To see this, consider Equation (7.25) with \(K = 0\). Then, we have
When solved, this yields the values of \(s\) that coincide with the poles of \(P(s)\). Conversely, as \(K \rightarrow \infty\), the roots of the characteristic equation are the zeros of \(P(s)\). To see this, first divide Equation (7.25) by \(K\). Then, we have
which, as \(K \rightarrow \infty\), reduces to
When solved, this yields the values of \(s\) that coincide with the zeros of \(P(s)\). Therefore, we note that the locus of the roots of the characteristic equation \(1 + KP(s) = 0\) begins at the poles of \(P(s)\) and ends at the zeros of \(P(s)\) as \(K\) increases from zero to infinity. For most functions \(P(s)\) that we will encounter, several of the zeros of \(P(s)\) lie at infinity in the \(s\)-plane. This is because most of our functions have more poles than zeros. With \(n\) poles and \(M\) zeros and \(n > M\), we have \(n - M\) branches of the root locus approaching the \(n - M\) zeros at infinity.
Step 2: Locate the segments of the real axis that are root loci. The root locus on the real axis always lies in a section of the real axis to the left of an odd number of poles and zeros. This fact is ascertained by examining the angle criterion of Equation (7.17). These two useful steps in plotting a root locus will be illustrated by a suitable example.
235. EXAMPLE 7.1 Second-order system
A feedback control system possesses the characteristic equation
Step 1: The characteristic equation can be written as
where
The transfer function, \(P(s)\), is rewritten in terms of poles and zeros as
To determine the locus of roots for the gain \(0 \leq K < \infty\), we locate the poles and zeros on the real axis as shown in Figure 7.6(a).
FIGURE 7.6
(a) The zero and poles of a secondorder system, (b) the root locus segments, and (c) the magnitude of each vector at \(s_{1}\).
(a)
(b)
(c) Step 2: The angle criterion is satisfied on the real axis between the points 0 and -2 , because the angle from pole \(p_{1}\) at the origin is \(180^{\circ}\), and the angle from the zero and pole \(p_{2}\) at \(s = - 4\) is zero degrees. The locus begins at the pole and ends at the zeros, and therefore the locus of roots appears as shown in Figure 7.6(b), where the direction of the locus as \(K\) is increasing \((K \uparrow )\) is shown by an arrow. We note that because the system has two real poles and one real zero, the second locus segment ends at a zero at negative infinity. To evaluate the gain \(K\) at a specific root location on the locus, we use the magnitude criterion, Equation (7.16). For example, the gain \(K\) at the root \(s = s_{1} = - 1\) is found from (7.16) as
or
This magnitude can also be evaluated graphically, as shown in Figure 7.6(c). For the gain of \(K = 3/2\), one other root exists, located on the locus to the left of the pole at -4 . The location of the second root is found graphically to be located at \(s = - 6\), as shown in Figure 7.6(c).
Now, we determine the number of separate loci, \(SL\). Because the loci begin at the poles and end at the zeros, the number of separate loci is equal to the number of poles since the number of poles is greater than or equal to the number of zeros. Therefore, as we found in Figure 7.6, the number of separate loci is equal to two because there are two poles and one zero.
Note that the root loci must be symmetrical with respect to the horizontal real axis because the complex roots must appear as pairs of complex conjugate roots.
We now return to developing a general list of root locus steps.
Step 3: The loci proceed to the zeros at infinity along asymptotes centered at \(\sigma_{A}\) and with angles \(\phi_{A}\). When the number of finite zeros of \(P(s),M\), is less than the number of poles \(n\) by the number \(N = n - M\), then \(N\) sections of loci must end at zeros at infinity. These sections of loci proceed to the zeros at infinity along asymptotes as \(K\) approaches infinity. These linear asymptotes are centered at a point on the real axis given by
The angle of the asymptotes with respect to the real axis is
where \(k\) is an integer index [3]. The usefulness of this rule is obvious for sketching the approximate form of a root locus. Equation (7.30) can be readily derived by considering a point on a root locus segment at a remote distance from the finite poles and zeros in the \(s\)-plane. The net phase angle at this remote point is \(180^{\circ}\), because it is a point on a root locus segment. The finite poles and zeros of \(P(s)\) are a great distance from the remote point, and so the angles from each pole and zero, \(\phi\), are essentially equal, and therefore the net angle is simply \((n - M)\phi\), where \(n\) and \(M\) are the number of finite poles and zeros, respectively. Thus, we have
or, alternatively,
Accounting for all possible root locus segments at remote locations in the s-plane, we obtain Equation (7.30).
The center of the linear asymptotes, often called the asymptote centroid, is determined by considering the characteristic equation in Equation (7.24). For large values of \(s\), only the higher-order terms need be considered, so that the characteristic equation reduces to
However, this relation, which is an approximation, indicates that the centroid of \(n - M\) asymptotes is at the origin, \(s = 0\). A better approximation is obtained if we consider a characteristic equation of the form
with a centroid at \(\sigma_{A}\).
The centroid is determined by considering the first two terms of Equation (7.24), which may be found from the relation
We note that
Considering only the first two terms of this expansion, we have
The first two terms of
are
Equating the term for \(s^{n - M - 1}\), we obtain
or
which is Equation (7.29).
For example, reexamine the system shown in Figure 7.2 and discussed in Section 7.2. The characteristic equation is written as
Because \(n - M = 2\), we expect two loci to end at zeros at infinity. The asymptotes of the loci are located at a center
and at angles of
The root locus is readily sketched, and the locus shown in Figure 7.3 is obtained. An example will further illustrate the process of using the asymptotes.
236. EXAMPLE 7.2 Fourth-order system
A unity negative feedback control system has a characteristic equation as follows:
We wish to sketch the root locus in order to determine the effect of the gain \(K\). The poles and zeros are located in the \(s\)-plane, as shown in Figure 7.7(a). The root loci on the real axis must be located to the left of an odd number of poles and zeros; they are shown as heavy lines in Figure 7.7(a). The intersection of the asymptotes is
FIGURE 7.7 A fourth-order system with (a) a zero and (b) root locus.
(a)
(b)
The angles of the asymptotes are
where there are three asymptotes, since \(n - M = 3\). Also, we note that the root loci must begin at the poles; therefore, two loci must leave the double pole at \(s = - 4\). Then with the asymptotes sketched in Figure 7.7(b), we may sketch the form of the root locus as shown in Figure 7.7(b). The actual shape of the locus in the area near \(\sigma_{A}\) would be graphically evaluated, if necessary.
We now proceed to develop more steps for the process of determining the root loci.
Step 4: Determine where the locus crosses the imaginary axis (if it does so), using the Routh-Hurwitz criterion. The actual point at which the root locus crosses the imaginary axis is readily evaluated by using the criterion.
Step 5: Determine the breakaway point on the real axis (if any). The root locus in Example 7.2 left the real axis at a breakaway point. The locus breakaway from the real axis occurs where the net change in angle caused by a small displacement is zero. The locus leaves the real axis where there is a multiplicity of roots (typically, two). The breakaway point for a simple second-order system is shown in Figure 7.8(a) and, for a special case of a fourth-order system, is shown in Figure 7.8(b). In general, due to the phase criterion, the tangents to the loci at the breakaway point are equally spaced over \(360^{\circ}\). Therefore, in Figure 7.8(a), we find that the two loci at the breakaway point are spaced \(180^{\circ}\) apart, whereas in Figure 7.8(b), the four loci are spaced \(90^{\circ}\) apart.
The breakaway point on the real axis can be evaluated graphically or analytically. The most straightforward method of evaluating the breakaway point FIGURE 7.8
Illustration of the breakaway point (a) for a simple second-order system and (b) for a fourth-order system.
(a)
(b)
involves the rearranging of the characteristic equation to isolate the multiplying factor \(K\). Then the characteristic equation is written as
For example, consider a unity feedback closed-loop system with a loop transfer function
which has the characteristic equation
Alternatively, the equation may be written as
The root loci for this system are shown in Figure 7.8(a). We expect the breakaway point to be near \(s = \sigma = - 3\) and plot \(\left. \ p(s) \right|_{s = \sigma}\) near that point, as shown in Figure 7.9. In this case, \(p(s)\) equals zero at the poles \(s = - 2\) and \(s = - 4\). The plot of \(p(s)\) versus \(s - \sigma\) is symmetrical, and the maximum point occurs at \(s = \sigma = - 3\), the breakaway point.
FIGURE 7.9
A graphical evaluation of the breakaway point.
Analytically, the very same result may be obtained by determining the maximum of \(K = p(s)\). To find the maximum analytically, we differentiate, set the differentiated polynomial equal to zero, and determine the roots of the polynomial. Therefore, we may evaluate
in order to find the breakaway point. Equation (7.36) is an analytical expression of the graphical procedure outlined in Figure 7.9 and will result in an equation of only one degree less than the total number of poles and zeros \(n + M - 1\).
The proof of Equation (7.36) is obtained from a consideration of the characteristic equation
which may be written as
For a small increment in \(K\), we have
Dividing by \(X(s) + KY(s)\) yields
Because the denominator is the original characteristic equation, a multiplicity \(m\) of roots exists at a breakaway point, and
Then we may write Equation (7.38) as
or, alternatively,
Therefore, as we let \(\Delta s\) approach zero, we obtain
at the breakaway points. Now, considering again the specific case where
we obtain
Then, when we differentiate, we have
or the breakaway point occurs at \(s = - 3\). A more complicated example will illustrate the approach and demonstrate the use of the graphical technique to determine the breakaway point.
237. EXAMPLE 7.3 Third-order system
A feedback control system is shown in Figure 7.10. The characteristic equation is
The number of poles \(n\) minus the number of zeros \(M\) is equal to 2, and so we have two asymptotes at \(\pm 90^{\circ}\) with a center at \(\sigma_{A} = - 2\). The asymptotes and the sections of loci on the real axis are shown in Figure 7.11(a). A breakaway point occurs between \(s = - 2\) and \(s = - 3\). To evaluate the breakaway point, we rewrite the characteristic equation so that \(K\) is separated; thus,
or
Then, evaluating \(p(s)\) at various values of \(s\) between \(s = - 2\) and \(s = - 3\), we obtain the results of Table 7.1, as shown in Figure 7.11(b). Alternatively, we differentiate
238. Table 7.1
| $$p(s)$$ | 0 | 0.411 | 0.419 | 0.417 | +0.390 | 0 |
| $$s$$ | -2.00 | -2.40 | -2.46 | -2.50 | -2.60 | -3.0 |
Equation (7.46) and set it equal to zero to obtain
Now to locate the maximum of \(p(s)\), we locate the roots of Equation (7.47) to obtain \(s = - 2.46, - 0.77 \pm 0.79j\). The only value of \(s\) on the real axis in the interval \(s = - 2\) to \(s = - 3\) is \(s = - 2.46\); hence this must be the breakaway point. It is evident from this one example that the numerical evaluation of \(p(s)\) near the expected breakaway point provides an effective method of evaluating the breakaway point.
Step 6: Determine the angle of departure of the locus from a pole and the angle of arrival of the locus at a zero, using the phase angle criterion. The angle of locus departure from a pole is the difference between the net angle due to all other poles and zeros and the criterion angle of \(\pm 180^{\circ}(2k + 1)\), and similarly for the locus angle of arrival at a zero. The angle of departure (or arrival) is particularly of interest for complex poles (and zeros) because the information is helpful in completing the root locus. For example, consider the third-order loop transfer function
The pole locations and the vector angles at one complex pole \(- p_{1}\) are shown in Figure 7.12(a). The angles at a test point \(s_{1}\), an infinitesimal distance from \(- p_{1}\), must meet the angle criterion. Therefore, since \(\theta_{2} = 90^{\circ}\), we have
FIGURE 7.11
Evaluation of the (a) asymptotes and (b) breakaway point.
(a)
(b) FIGURE 7.12
Illustration of the angle of departure. (a) Test point infinitesimal distance from \(- p_{1}\). (b) Actual departure vector at \(- p_{1}\).
(a)
(b)
or the angle of departure at pole \(p_{1}\) is
as shown in Figure 7.12(b). The departure at pole \(- p_{2}\) is the negative of that at \(- p_{1}\), because \(- p_{1}\) and \(- p_{2}\) are complex conjugates. Another example of a departure angle is shown in Figure 7.13. In this case, the departure angle is found from
Since \(\theta_{2} - \theta_{3} = \gamma\) in the diagram, we find that the departure angle is \(\theta_{1} = 90^{\circ} + \gamma\).
Step 7: The final step in the root locus sketching procedure is to complete the sketch. This entails sketching in all sections of the locus not covered in the previous six steps.
FIGURE 7.13
Evaluation of the angle of departure. In some situation, we may want to determine a root location \(s_{x}\) and the value of the parameter \(K_{x}\) at that root location. Determine the root locations that satisfy the phase criterion at the root \(s_{x},x = 1,2,\ldots,n\), using the phase criterion. The phase criterion, given in Equation (7.17), is
To determine the parameter value \(K_{x}\) at a specific root \(s_{x}\), we use the magnitude requirement (Equation 7.16). The magnitude requirement at \(s_{x}\) is
The seven steps utilized in the root locus method are summarized in Table 7.2.
239. Table 7.2 Seven Steps for Sketching a Root Locus
240. Step
- Prepare the root locus sketch.
(a) Write the characteristic equation so that the parameter of interest, \(K\), appears as a multiplier.
(b) Factor \(P(s)\) in terms of \(n\) poles and \(M\) zeros.
(c) Locate the open-loop poles and zeros of \(P(s)\) in the \(s\)-plane with selected symbols.
(d) Determine the number of separate loci, \(SL\).
(e) The root loci are symmetrical with respect to the horizontal real axis.
-
Locate the segments of the real axis that are root loci.
-
The loci proceed to the zeros at infinity along asymptotes centered at \(\sigma_{A}\) and with angles \(\phi_{A}\).
-
Determine the points at which the locus crosses the imaginary axis (if it does so).
-
Determine the breakaway point on the real axis (if any).
-
Determine the angle of locus departure from complex poles and the angle of locus arrival at complex zeros, using the phase criterion.
-
Complete the root locus sketch.
241. Related Equation or Rule
\(1 + KP(s) = 0\).
\(1 + K\frac{}{} = 0\).
\(\times =\) poles, \(O =\) zeros
Locus begins at a pole and ends at a zero.
\(SL = n\) when \(n \geq M;n =\) number of finite poles, \(M =\) of finite zeros.
Locus lies to the left of an odd number of poles and zeros.
\(\sigma_{A} = \frac{\sum\left( - p_{j} \right) - \sum\left( - z_{i} \right)}{n - M}\).
\(\phi_{A} = \frac{2k + 1}{n - M}180^{\circ},k = 0,1,2,\ldots,(n - M - 1)\).
Use Routh-Hurwitz criterion.
a) Set \(K = p(s)\).
b) Determine roots of \(dp(s)/ds = 0\) or use graphical method to find maximum of \(p(s)\). \(\angle P(s) = 180^{\circ} + k360^{\circ}\) at \(s = - p_{j}\) or \(- z_{i}\).
242. EXAMPLE 7.4 Fourth-order system
- (a) Consider the root locus for the characteristic equation of a system as \(K\) varies for \(0 \leq K < \infty\) when
(b) Determining the poles, we have
This system has no finite zeros.
(c) The poles are located on the \(s\)-plane as shown in Figure 7.14(a).
(d) Because the number of poles \(n\) is equal to 4, we have four separate loci.
(e) The root loci are symmetrical with respect to the real axis.
-
A segment of the root locus exists on the real axis between \(s = 0\) and \(s = - 4\).
-
The angles of the asymptotes are
The center of the asymptotes is
Then the asymptotes are drawn as shown in Figure 7.14(a).
FIGURE 7.14 The root locus for Example 7.4. Locating (a) the poles and (b) the asymptotes.
(a)
(b) 4. The characteristic equation is rewritten as
Therefore, the Routh array is
| $$s^{4}$$ | 1 | 64 | $$K$$ |
| $$s^{3}$$ | 12 | 128 | |
| $$s^{2}$$ | $$b_{1}$$ | $$K$$ | |
| $$s^{1}$$ | $$c_{1}$$ | ||
| $$s^{0}$$ | $$K$$ |
where
Hence, the limiting value of gain for stability is \(K = 568.89\), and the roots of the auxiliary equation are
The points where the locus crosses the imaginary axis are shown in Figure 7.14(a). Therefore, when \(K = 568.89\), the root locus crosses the \(j\omega\)-axis at \(s = \pm j3.266\).
- The breakaway point is estimated by evaluating
between \(s = - 4\) and \(s = 0\). We expect the breakaway point to lie between \(s = - 3\) and \(s = - 1\), so we search for a maximum value of \(p(s)\) in that region. The resulting values of \(p(s)\) for several values of \(s\) are given in Table 7.3. The maximum of \(p(s)\) is found to lie at approximately \(s = - 1.577\), as indicated in the table. A more accurate estimate of the breakaway point is normally not necessary. The breakaway point is then indicated on Figure 7.14(a).
- The angle of departure at the complex pole \(p_{1}\) can be estimated by utilizing the angle criterion as follows:
Here, \(\theta_{3}\) is the angle subtended by the vector from pole \(p_{3}\). The angles from the pole at \(s = - 4\) and \(s = - 4 - j4\) are each equal to \(90^{\circ}\). Since \(\theta_{3} = 135^{\circ}\), we find that
as shown in Figure 7.14(a).
- Complete the sketch as shown in Figure 7.14(b).
Table 7.3
| $$p(s)$$ | 0 | 51.0 | 68.44 | 80.0 | 83.57 | 75.0 | 0 |
| $$s$$ | -4.0 | -3.0 | -2.5 | -2.0 | -1.577 | -1.0 | 0 |
Using the information derived from the seven steps of the root locus method, the complete root locus sketch is obtained by filling in the sketch as well as possible by visual inspection. The root locus for this system is shown in Figure 7.14(b). When the complex roots near the origin have a damping ratio of \(\zeta = 0.707\), the gain \(K\) can be determined graphically as shown in Figure 7.14(b). The vector lengths to the root location \(s_{1}\) from the open-loop poles are evaluated and result in a gain at \(s_{1}\) of
The remaining pair of complex roots occurs at \(s_{2}\) and \({\widehat{s}}_{2}\), when \(K = 126\). The effect of the complex roots at \(s_{2}\) and \({\widehat{s}}_{2}\) on the transient response will be negligible compared to the roots \(s_{1}\) and \({\widehat{s}}_{1}\). This fact can be ascertained by considering the damping of the response due to each pair of roots. The damping due to \(s_{1}\) and \({\widehat{s}}_{1}\) is
and the damping factor due to \(s_{2}\) and \({\widehat{s}}_{2}\) is
where \(\sigma_{2}\) is approximately five times as large as \(\sigma_{1}\). Therefore, the transient response term due to \(s_{2}\) will decay much more rapidly than the transient response term due to \(s_{1}\). Thus, the response to a unit step input may be written as
The complex conjugate roots near the origin of the \(s\)-plane relative to the other roots of the closed-loop system are labeled the dominant roots of the system because they represent or dominate the transient response. The relative dominance of the complex roots, in a third-order system with a pair of complex conjugate roots, is determined by the ratio of the real root to the real part of the complex roots and will result in approximate dominance for ratios exceeding 5.
The dominance of the second term of Equation (7.53) also depends upon the relative magnitudes of the coefficients \(c_{1}\) and \(c_{2}\). These coefficients, which are the residues evaluated at the complex roots, in turn depend upon the location of the zeros in the \(s\)-plane. Therefore, the concept of dominant roots is useful for estimating the response of a system, but must be used with caution and with a comprehension of the underlying assumptions.
242.1. PARAMETER DESIGN BY THE ROOT LOCUS METHOD
Originally, the root locus method was developed to determine the locus of roots of the characteristic equation as the system gain, \(K\), is varied from zero to infinity. However, as we have seen, the effect of other system parameters may be readily investigated by using the root locus method. Fundamentally, the root locus method is concerned with a characteristic equation (Equation 7.22), which may be written as
Then the standard root locus method we have studied may be applied. The question arises: How do we investigate the effect of two parameters, \(\alpha\) and \(\beta\) ? It appears that the root locus method is a single-parameter method; fortunately, it can be readily extended to the investigation of two or more parameters. This method of parameter design uses the root locus approach to select the values of the parameters.
The characteristic equation of a dynamic system may be written as
Hence, the effect of varying \(0 \leq a_{1} < \infty\) may be ascertained from the root locus equation
If the parameter of interest, \(\alpha\), does not appear solely as a coefficient, the parameter may be isolated as
For example, a third-order equation of interest might be
To ascertain the effect of the parameter \(\alpha\), we isolate the parameter and rewrite the equation in root locus form, as shown in the following steps:
Then, to determine the effect of two parameters, we must repeat the root locus approach twice. Thus, for a characteristic equation with two variable parameters, \(\alpha\) and \(\beta\), we have
The two variable parameters have been isolated, and the effect of \(\alpha\) will be determined. Then, the effect of \(\beta\) will be determined. For example, for a certain third-order characteristic equation with \(\alpha\) and \(\beta\) as parameters, we obtain
In this particular case, the parameters appear as the coefficients of the characteristic equation. The effect of varying \(\beta\) from zero to infinity is determined from the root locus equation
We note that the denominator of Equation (7.63) is the characteristic equation of the system with \(\beta = 0\). Therefore, we must first evaluate the effect of varying \(\alpha\) from zero to infinity by using the equation
rewritten as
where \(\beta\) has been set equal to zero in Equation (7.62). Then, upon evaluating the effect of \(\alpha\), a value of \(\alpha\) is selected and used with Equation (7.63) to evaluate the effect of \(\beta\). This two-step method of evaluating the effect of \(\alpha\) and then \(\beta\) may be carried out as two root locus procedures. First, we obtain a locus of roots as \(\alpha\) varies, and we select a suitable value of \(\alpha\); the results are satisfactory root locations. Then, we obtain the root locus for \(\beta\) by noting that the poles of Equation (7.63) are the roots evaluated by the root locus of Equation (7.64). A limitation of this approach is that we will not always be able to obtain a characteristic equation that is linear in the parameter under consideration.
To illustrate this approach, let us obtain the root locus for \(\alpha\) and then \(\beta\) for Equation (7.62). A sketch of the root locus as \(\alpha\) varies for Equation (7.64) is shown in Figure 7.15(a), where the roots for two values of gain \(\alpha\) are shown. If the gain \(\alpha\) is selected as \(\alpha_{1}\), then the resultant roots of Equation (7.64) become the poles of Equation (7.63). The root locus of Equation (7.63) as \(\beta\) varies is shown in Figure 7.15(b), and a suitable \(\beta\) can be selected on the basis of the desired root locations.
Using the root locus method, we will further illustrate this parameter design approach by a specific design example.
FIGURE 7.15
Root loci as a function of \(\alpha\) and \(\beta\). (a) Loci as \(\alpha\) varies. (b) Loci as \(\beta\) varies for one value of \(\alpha = \alpha_{1}\).
(a)
(b)
243. EXAMPLE 7.5 Welding head control
A welding head for an auto body requires an accurate control system for positioning the welding head [4]. The feedback control system is to be designed to satisfy the following specifications:
-
Steady-state error for a ramp input is \(e_{ss} \leq 35\%\) of input slope
-
Damping ratio of dominant roots is \(\zeta \geq 0.707\)
-
Settling time to within \(2\%\) of the final value is \(T_{s} \leq 3\text{ }s\).
The structure of the feedback control system is shown in Figure 7.16, where the amplifier gain \(K_{1}\) and the derivative feedback gain \(K_{2}\) are to be selected. The steadystate error specification can be written as
where \(G_{2}(s) = G(s)/(1 + G(s)H(s))\). Therefore, the steady-state error requirement is
Thus, we will select a small value of \(K_{2}\) to achieve a low value of steady-state error. The damping ratio specification requires that the roots of the closed-loop system be below the line at \(45^{\circ}\) in the left-hand \(s\)-plane, as illustrated in Figure 7.17. The settling time specification can be rewritten in terms of the real part of the dominant roots as
Therefore, it is necessary that \(\sigma \geq 4/3\); this area in the left-hand \(s\)-plane is indicated along with the \(\zeta\) - requirement in Figure 7.17. Note that \(\sigma \geq 4/3\) implies that we want the dominant roots to lie to the left of the line defined by \(\sigma = - 4/3\). To satisfy the specifications, all the roots must lie within the shaded area of the left-hand plane.
The parameters to be selected are \(\alpha = K_{1}\) and \(\beta = K_{2}K_{1}\). The characteristic equation is
FIGURE 7.16
Block diagram of welding head control system.
FIGURE 7.17
A region in the \(s\)-plane for desired root location.
The locus of roots as \(\alpha = K_{1}\) varies (set \(\beta = 0\) ) is determined from the equation
as shown in Figure 7.18(a). For a gain of \(K_{1} = \alpha = 20\), the roots are \(s = - 1 \pm j4.36\) as indicated on the locus. Then the effect of varying \(\beta = 20K_{2}\) is determined from the locus equation
The root locus for Equation (7.70) is shown in Figure 7.18(b), and roots with \(\zeta = 0.707\) are obtained when \(\beta = 4.3 = 20K_{2}\) or when \(K_{2} = 0.215\). The real part
(a)
(b) tion of (a) \(\alpha\) and (b) \(\beta\).
FIGURE 7.18
Root loci as a func- of these roots is \(\sigma = - 3.15\); therefore, the time to settle (to within \(2\%\) of the final value) is \(T_{s} = 1.27\text{ }s\), which is considerably less than the specification of \(T_{s} \leq 3\text{ }s\).
We can extend the root locus method to more than two parameters by extending the number of steps in the method outlined in this section. Furthermore, a family of root loci can be generated for two parameters in order to determine the total effect of varying two parameters. For example, let us determine the effect of varying \(\alpha\) and \(\beta\) of the following characteristic equation:
The root locus equation as a function of \(\alpha\) is \((set\beta = 0)\)
The root locus as a function of \(\beta\) is
The root locus for Equation (7.72) as a function of \(\alpha\) is shown in Figure 7.19 (unbroken lines). The roots of this locus, indicated by slashes, become the poles for the locus of Equation (7.73). Then the locus of Equation (7.73) is continued on Figure 7.19 (dotted lines), where the locus for \(\beta\) is shown for several selected values of \(\alpha\). This family of loci, often called root contours, illustrates the effect of \(\alpha\) and \(\beta\) on the roots of the characteristic equation of a system [3].
FIGURE 7.19
Two-parameter root locus. The loci for \(\alpha\) varying are solid; the loci for \(\beta\) varying are dashed.
243.1. SENSITIVITY AND THE ROOT LOCUS
One of the prime reasons for the use of negative feedback in control systems is the reduction of the effect of parameter variations. The effect of parameter variations can be described by a measure of the sensitivity of the system performance to specific parameter changes. We define the logarithmic sensitivity originally suggested by Bode as
where the system transfer function is \(T(s)\) and the parameter of interest is \(K\).
It is useful to define a sensitivity measure in terms of the positions of the roots of the characteristic equation [7-9]. Because these roots represent the dominant modes of transient response, the effect of parameter variations on the position of the roots is an important and useful measure of the sensitivity. The root sensitivity of a system \(T(s)\) can be defined as
where \(r_{i}\) equals the \(i\) th root of the system, so that
and \(K\) is a parameter affecting the roots. The root sensitivity relates the changes in the location of the root in the \(s\)-plane to the change in the parameter. The root sensitivity is related to the logarithmic sensitivity by the relation
when the zeros of \(T(s)\) are independent of the parameter \(K\), so that
This logarithmic sensitivity can be readily obtained by determining the derivative of \(T(s)\) in Equation (7.76) with respect to \(K\). For this particular case, when the gain of the system is independent of the parameter \(K\), we have
and the two sensitivity measures are directly related. The evaluation of the root sensitivity for a control system can be readily accomplished by utilizing the root locus methods of the preceding section. The root sensitivity \(S_{K}^{r_{i}}\) may be evaluated at root \(- r_{i}\) by examining the root contours for the parameter \(K\). We can change \(K\) by a small finite amount \(\Delta K\) and determine the modified root \(- \left( r_{i} + \Delta r_{i} \right)\) at \(K + \Delta K\). Then, using Equation (7.75), we have
Equation (7.79) is an approximation that approaches the actual value of the sensitivity as \(\Delta K \rightarrow 0\). An example will illustrate the process of evaluating the root sensitivity.
244. EXAMPLE 7.6 Root sensitivity of a control system
The characteristic equation of the feedback control system shown in Figure 7.20 is
or, alternatively,
The gain \(K\) will be considered to be the parameter \(\alpha\). Then the effect of a change in each parameter can be determined by utilizing the relations
where \(\alpha_{0}\) and \(\beta_{0}\) are the nominal or desired values for the parameters \(\alpha\) and \(\beta\), respectively. We shall consider the case when the nominal value is \(\beta_{0} = 1\) and the desired gain is \(\alpha_{0} = K = 0.5\). Then the root locus can be obtained as a function of \(\alpha = K\) by utilizing the root locus equation
as shown in Figure 7.21. The nominal value of gain \(K = \alpha_{0} = 0.5\) results in two complex roots, \(- r_{1} = - 0.5 + j0.5\) and \(- r_{2} = - {\widehat{r}}_{1}\), as shown in Figure 7.21. To evaluate the effect of changes in the gain, the characteristic equation with \(\alpha = \alpha_{0} \pm \Delta\alpha\) becomes
Therefore, the effect of changes in the gain can be evaluated from the root locus of Figure 7.21. For a \(20\%\) change in \(\alpha\), we have \(\Delta\alpha = \pm 0.1\). The root locations for
FIGURE 7.20
A feedback control system.
FIGURE 7.21
The root locus for \(K\).
a gain \(\alpha = 0.4\) and \(\alpha = 0.6\) are readily determined by root locus methods, and the root locations for \(\Delta\alpha = \pm 0.1\) are shown in Figure 7.21. When \(\alpha = K = 0.6\), the root in the second quadrant of the \(s\)-plane is
and the change in the root is \(\Delta r_{1} = + j0.09\). When \(\alpha = K = 0.4\), the root in the second quadrant is
and the change in the root is \(- \Delta r_{1} = - j0.11\). Thus, the root sensitivity for \(r_{1}\) is
for positive changes of gain. For negative increments of gain, the sensitivity is
For infinitesimally small changes in the parameter \(K\), the sensitivity will be equal for negative or positive increments in \(K\). The angle of the root sensitivity indicates the direction the root moves as the parameter varies. The angle of movement for \(+ \Delta\alpha\) is always \(180^{\circ}\) from the angle of movement for \(- \Delta\alpha\) at the point \(\alpha = \alpha_{0}\).
The pole \(\beta\) variation is represented by \(\beta = \beta_{0} + \Delta\beta\), where \(\beta_{0} = 1\). Then the effect of variation of the poles is represented by the characteristic equation
or, in root locus form,
The denominator of the second term is the unchanged characteristic equation when \(\Delta\beta = 0\). The root locus for the unchanged system \((\Delta\beta = 0)\) is shown in Figure 7.21 as a function of \(K\). For a design specification requiring \(\zeta = 0.707\), the complex roots lie at
Then, because the roots are complex conjugates, the root sensitivity for \(r_{1}\) is the conjugate of the root sensitivity for \({\widehat{r}}_{1} = r_{2}\). Using the parameter root locus techniques discussed in the preceding section, we obtain the root locus for \(\Delta\beta\) as shown in Figure 7.22. We are normally interested in the effect of a variation for the parameter so that \(\beta = \beta_{0} \pm \Delta\beta\), for which the locus as \(\beta\) decreases is obtained from the root locus equation
We note that the equation is of the form
Comparing this equation with Equation (7.23) in Section 7.3, we find that the sign preceding the gain \(\Delta\beta\) is negative in this case. In a manner similar to the development of the root locus method in Section 7.3, we require that the root locus satisfy the equations
FIGURE 7.22
The root locus for the parameter \(\beta\).
where \(k\) is an integer. The locus of roots follows a zero-degree locus in contrast with the \(180^{\circ}\) locus considered previously. However, the root locus rules of Section 7.3 may be altered to account for the zero-degree phase angle requirement, and then the root locus may be obtained as in the preceding sections. Therefore, to obtain the effect of reducing \(\beta\), we determine the zero-degree locus in contrast to the \(180^{\circ}\) locus, as shown by a dotted locus in Figure 7.22. To find the effect of a \(20\%\) change of the parameter \(\beta\), we evaluate the new roots for \(\Delta\beta = \pm 0.20\), as shown in Figure 7.22. The root sensitivity is readily evaluated graphically and, for a positive change in \(\beta\), is
The root sensitivity for a negative change in \(\beta\) is
As the percentage change \(\Delta\beta/\beta\) decreases, the sensitivity measures \(S_{\beta +}^{n_{1}}\) and \(S_{\beta -}^{\eta_{-}}\) will approach equality in magnitude and a difference in angle of \(180^{\circ}\). Thus, for small changes when \(\Delta\beta/\beta \leq 0.10\), the sensitivity measures are related as
and
Often, the desired root sensitivity measure is desired for small changes in the parameter. When the relative change in the parameter is of the order \(\Delta\beta/\beta = 0.10\), we can estimate the increment in the root change by approximating the root locus with the line at the angle of departure \(\theta_{d}\). This approximation is shown in Figure 7.22 and is accurate for only relatively small changes in \(\Delta\beta\). However, the use of this approximation allows the analyst to avoid sketching the complete root locus diagram. Therefore, for Figure 7.22, the root sensitivity may be evaluated for \(\Delta\beta/\beta = 0.10\) along the departure line, and we obtain
The root sensitivity measure for a parameter variation is useful for comparing the sensitivity for various design parameters and at different root locations. Comparing Equation (7.85) for \(\beta\) with Equation (7.83) for \(\alpha\), we find (a) that the sensitivity for \(\beta\) is greater in magnitude by approximately \(50\%\) and (b) that the angle for \(S_{\beta -}^{r_{1}}\) indicates that the approach of the root toward the \(j\omega\)-axis is more sensitive for changes in \(\beta\). Therefore, the tolerance requirements for \(\beta\) would be more stringent than for \(\alpha\). This information provides the designer with a comparative measure of the required tolerances for each parameter. To utilize the root sensitivity measure for the analysis and design of control systems, a series of calculations must be performed; they will determine the various selections of possible root configurations and the zeros and poles of the loop transfer function. Therefore, the root sensitivity measure as a design technique is somewhat limited by the relatively large number of calculations required and the lack of an obvious direction for adjusting the parameters in order to provide a minimized or reduced sensitivity. However, the root sensitivity measure can be utilized as an analysis measure, which permits the designer to compare the sensitivity for several system designs based on a suitable method of design. The root sensitivity measure is a useful index of the system sensitivity to parameter variations expressed in the \(s\)-plane. The weakness of the sensitivity measure is that it relies on the ability of the root locations to represent the performance of the system. The root locations represent the performance quite adequately for many systems, but due consideration must be given to the location of the zeros of the closed-loop transfer function and the dominant roots. The root sensitivity measure is a suitable measure of system performance sensitivity and can be used reliably for system analysis and design.
244.1. PID CONTROLLERS
One form of controller widely used in industrial process control is the three-term, PID controller \(\lbrack 4,10\rbrack\). This controller has a transfer function
The equation for the output in the time domain is
The three-term controller is called a PID controller because it contains a proportional, an integral, and a derivative term represented by \(K_{p},K_{I}\), and \(K_{D}\), respectively. The transfer function of the derivative term is actually
but \(\tau_{d}\) is usually much smaller than the time constants of the process itself, so it is neglected.
If we set \(K_{D} = 0\), then we have the proportional plus integral (PI) controller
When \(K_{I} = 0\), we have
which is called a proportional plus derivative (PD) controller. The PID controller can also be viewed as a cascade of the PI and the PD controllers. Consider the PI controller
and the PD controller
where \({\widehat{K}}_{P}\) and \({\widehat{K}}_{I}\) are the PI controller gains and \({\bar{K}}_{P}\) and \({\bar{K}}_{D}\) are the PD controller gains. Cascading the two controllers (that is, placing them in series) yields
where we have the following relationships between the PI and PD controller gains and the PID controller gains
Consider the PID controller
where \(a = K_{P}/K_{D}\) and \(b = K_{I}/K_{D}\). A PID controller introduces a transfer function with one pole at the origin and two zeros that can be located anywhere in the \(s\)-plane.
Consider the system shown in Figure 7.23 where we use a PID controller with complex zeros \(- z_{1}\) and \(- z_{2}\), where \(- z_{1} = - 3 + j1\) and \(- z_{2} = - {\widehat{z}}_{1}\). We plot the root
FIGURE 7.24
Root locus for plant with a PID controller with complex zeros.
locus as shown in Figure 7.24. As the gain, \(K_{D}\), of the controller is increased, the complex roots approach the zeros. The closed-loop transfer function is
The percent overshoot to a step will be P.O. \(\leq 2\%\), and the steady-state error for a step input will be \(e_{SS} = 0\). The settling time will be approximately \(T_{s} = 1\text{ }s\). If a shorter settling time is desired, then we select \(z_{1}\) and \(z_{2}\) to lie further left in the left-hand \(s\)-plane and set \(K_{D}\) to drive the roots near the complex zeros.
The popularity of PID controllers can be attributed partly to their good performance over a wide range of operating conditions and partly to their functional simplicity that allows engineers to operate them in a simple, straightforward manner. To implement the PID controller, three parameters must be determined, the proportional gain, denoted by \(K_{P}\), integral gain, denoted by \(K_{I}\), and derivative gain denoted by \(K_{D}\lbrack 10\rbrack\).
There are many methods available to determine acceptable values of the PID gains. The process of determining the gains is often called PID tuning. A common approach to tuning is to use manual PID tuning methods, whereby the PID control gains are obtained by trial-and-error with minimal analytic analysis using step responses obtained via simulation, or in some cases, actual testing on the system and deciding on the gains based on observations and experience. A more analytic method is known as the Ziegler-Nichols tuning method. The Ziegler-Nichols tuning method actually has several variations. We discuss in this section a Ziegler-Nichols tuning method based on open-loop responses to a step input and a related a Ziegler-Nichols tuning method based on closed-loop response to a step input. Table 7.4 Effect of Increasing the PID Gains \(K_{p},K_{D}\), and \(K_{\text{I}\text{~}}\) on the Step Response
| PID Gain | Percent | Steady-State | |
| Increasing \(K_{P}\) | Overshoot | Settling Time | Error |
| Increasing \(K_{I}\) | Increases | Minimal impact | Decreases |
| Increasing \(K_{D}\) | Decreases | Increases | Zero steady-state error |
One approach to manual tuning is to first set \(K_{I} = 0\) and \(K_{D} = 0\). This is followed by slowly increasing the gain \(K_{P}\) until the output of the closed-loop system oscillates just on the edge of instability. This can be done either in simulation or on the actual system if it cannot be taken off-line. Once the value of \(K_{P}\) (with \(K_{I} = 0\) and \(K_{D} = 0\) ) is found that brings the closed-loop system to the edge of stability, you reduce the value of gain \(K_{P}\) to achieve what is known as the quarter amplitude decay. That is, the amplitude of the closed-loop response is reduced approximately to one-fourth of the maximum value in one oscillatory period. A rule-of-thumb is to start by reducing the proportional gain \(K_{P}\) by onehalf. The next step of the design process is to increase \(K_{I}\) and \(K_{D}\) manually to achieve a desired step response. Table 7.4 describes in general terms the effect of increasing \(K_{I}\) and \(K_{D}\).
245. EXAMPLE 7.7 Manual PID tuning
Consider the closed-loop system in Figure 7.25, where \(b = 10,\zeta = 0.707\), and \(\omega_{n} = 4\). To begin the manual tuning process, set \(K_{I} = 0\) and \(K_{D} = 0\) and increase \(K_{P}\) until the closed-loop system has sustained oscillations. As can be seen in Figure 7.26(a), when \(K_{P} = 885.5\), we have a sustained oscillation of magnitude \(A = 1.9\) and period \(P = 0.83\) s. The root locus shown in Figure 7.26(b) corresponds to the characteristic equation
The root locus shown in Figure 7.26(b) illustrates that when \(K_{P} = 885.5\), we have closed-loop poles at \(s = \pm 7.5j\) leading to the oscillatory behavior in the step response in Figure 7.26(a).
FIGURE 7.25
Unity feedback control system with PID controller.
FIGURE 7.26
(a) Step response with \(K_{P} = 885.5\), \(K_{D} = 0\), and \(K_{I} = 0\). (b) Root locus showing \(K_{P} = 885.5\) results in marginal stability with \(s = \pm 7.5j\).
(a)
(b)
Reduce \(K_{P} = 885.5\) by half as a first step to achieving a step response with approximately a quarter amplitude decay. You may have to iterate on the value \(K_{P} = 442.75\). The step response is shown in Figure 7.27 where we note that the peak amplitude is reduced to one-fourth of the maximum value in one period, as desired. To accomplish this reduction, we refined the value of \(K_{P}\) by slowly reducing the value from \(K_{P} = 442.75\) to \(K_{P} = 370\).
The root locus for \(K_{P} = 370,K_{I} = 0\), and \(0 \leq K_{D} < \infty\) is shown in Figure 7.28. In this case, the characteristic equation is
FIGURE 7.27
Step response with \(K_{P} = 370\) showing the quarter amplitude decay. FIGURE 7.28
Root locus for \(K_{P} = 370\),
\(K_{I} = 0\), and \(0 \leq K_{D} < \infty\).
FIGURE 7.29
Percent overshoot and settling time with \(K_{P} = 370\), \(K_{I} = 0\), and \(5 \leq K_{D} < 75\).
We see in Figure 7.28 that as \(K_{D}\) increases, the root locus shows that the closed-loop complex poles move left, and in doing so, increases the associated damping ratio and thereby decreases the percent overshoot. The movement of the complex poles to the left also increases the associated \(\zeta\omega_{n}\), thereby reducing the settling time. These effects of varying \(K_{D}\) are consistent with information provided in Table 7.4. As \(K_{D}\) increases (when \(K_{D} > 75\) ), the real root begins to dominant the response and the trends described in Table 7.4 become less accurate. The percent overshoot and settling time as a function of \(K_{D}\) are shown in Figure 7.29.
FIGURE 7.30
Root locus for \(K_{P} = 370\), \(K_{D} = 0\), and \(0 \leq K_{l} < \infty\).
The root locus for \(K_{P} = 370,K_{D} = 0\), and \(0 \leq K_{I} < \infty\) is shown in Figure 7.30. The characteristic equation is
We see in Figure 7.30 that as \(K_{I}\) increases, the root locus shows that the closedloop complex pair poles move right. This decreases the associated damping ratio and thereby increasing the percent overshoot. In fact, when \(K_{I} = 778.2\), the system is marginally stable with closed-loop poles at \(s = \pm 4.86j\). The movement of the complex poles to the right also decreases the associated \(\zeta\omega_{n}\), thereby increasing the settling time. The percent overshoot and settling time as a function of \(K_{I}\) are shown in Figure 7.31. The trends in Figure 7.31 are consistent with Table 7.4.
To meet the percent overshoot and settling time specifications, we can select \(K_{P} = 370,K_{D} = 60\), and \(K_{I} = 100\). The step response shown in Figure 7.32 indicates a \(T_{S} = 2.4\text{ }s\) and P.O. \(= 12.8\%\) meeting the specifications.
Two important PID controller gain tuning methods were published in 1942 by John G. Ziegler and Nathaniel B. Nichols intended to achieve a fast closed-loop step response without excessive oscillations and excellent disturbance rejection. The two approaches are classified under the general heading of Ziegler-Nichols tuning methods. The first approach is based on closed-loop concepts requiring the computation of the ultimate gain and ultimate period. The second approach is based on open-loop concepts relying on reaction curves. The Ziegler-Nichols tuning methods are based on assumed forms of the models of the process, but the models do not have to be precisely known. This makes the tuning approach very practical in FIGURE 7.31
Percent overshoot and settling time with \(K_{P} = 370\), \(K_{D} = 0\), and \(50 \leq K_{l} < 600\).
FIGURE 7.32
Percent overshoot and settling time with final design \(K_{p} = 370\),
\(K_{D} = 60\), and \(K_{l} = 100\).
process control applications. Our suggestion is to consider the Ziegler-Nichols rules to obtain initial controller designs followed by design iteration and refinement. Remember that the Ziegler-Nichols rules will not work with all plants or processes.
The closed-loop Ziegler-Nichols tuning method considers the closed-loop system response to a step input (or step disturbance) with the PID controller in the loop. Initially the derivative and integral gains, \(K_{D}\) and \(K_{I}\), respectively, are set to zero. The proportional gain \(K_{P}\) is increased (in simulation or on the actual system) until the closed-loop system reaches the boundary of instability. The gain on the border of instability, denoted by \(K_{U}\), is called the ultimate gain. The period of the sustained oscillations, denoted by \(P_{U}\), is called the ultimate period. Once \(K_{U}\) and Table 7.5 Ziegler-Nichols PID Tuning Using Ultimate Gain, \(K_{U}\), and Oscillation Period, \(P_{U}\)
Ziegler-Nichols PID Controller Gain Tuning Using Closed-Loop Concepts
\(P_{U}\) are determined, the PID gains are computed using the relationships in Table 7.5 according to the Ziegler-Nichols tuning method.
246. EXAMPLE 7.8 Closed-loop Ziegler-Nichols PID tuning
Re-consider the system in Example 7.7. The gains \(K_{P},K_{D}\), and \(K_{I}\) are computed using the formulas in Table 7.5. We found in Example 7.7 that \(K_{U} = 885.5\) and \(T_{U} = 0.83\text{ }s\). By using the Ziegler-Nichols formulas we obtain
Comparing the step response in Figures 7.33 and 7.34 we note that the settling time is approximately the same for the manually tuned and the Ziegler-Nichols tuned PID controllers. However, the percent overshoot of the manually tuned controller is less than that of the Ziegler-Nichols tuning. This is due to the fact that the ZieglerNichols tuning is designed to provide the best disturbance rejection performance rather than the best input response performance.
FIGURE 7.33
Time response for the Ziegler-Nichols PID tuning with
\(K_{P} = 531.3\),
\(K_{I} = 1280.2\), and \(K_{D} = 55.1\).
FIGURE 7.34
Disturbance response for the Ziegler-Nichols PID tuning versus the manual tuning.
FIGURE 7.35
Reaction curve illustrating parameters \(R\) and \(\Delta T\) required for the Ziegler-Nichols open-loop tuning method.
In Figure 7.34, we see that the step disturbance performance of the ZieglerNichols PID controller is indeed better than the manually tuned controller. While Ziegler-Nichols approach provides a structured procedure for obtaining the PID controller gains, the appropriateness of the Ziegler-Nichols tuning depends on the requirements of the problem under investigation.
The open-loop Ziegler-Nichols tuning method utilizes a reaction curve obtained by taking the controller off-line (that is, out of the loop) and introducing a step input (or step disturbance). This approach is very commonly used in process control applications. The measured output is the reaction curve and is assumed to have the general shape shown in Figure 7.35. The response in Figure 7.35 implies that the process is a first-order system with a transport delay. If the actual system does not match the assumed form, then another approach to PID tuning should be
considered. However, if the underlying system is linear and lethargic (or sluggish and characterized by delay), the assumed model may suffice to obtain a reasonable PID gain selection using the open-loop Ziegler-Nichols tuning method.
The reaction curve is characterized by the transport delay, \(\Delta T\), and the reaction rate, \(R\). Generally, the reaction curve is recorded and numerical analysis is performed to obtain estimates of the parameters \(\Delta T\) and \(R\). A system possessing the reaction curve shown in Figure 7.35 can be approximated by a first-order system with a transport delay as
where \(M\) is the magnitude of the response at steady-state, \(\Delta T\) is the transport delay, and \(p\) is related to the slope of the reaction curve. The parameters \(M,\tau\), and \(\Delta T\) can be estimated from the open-loop step response and then utilized to compute \(R = M/\tau\). Once that is accomplished, the PID gains are computed as shown in Table 7.6. You can also use the Ziegler-Nichols open-loop tuning method to design a proportional controller or a proportional-plus-integral controller.
247. EXAMPLE 7.9 Open-loop Ziegler-Nichols PI controller tuning
Consider the reaction curve shown in Figure 7.36. We estimate the transport lag to be \(\Delta T = 0.1\text{ }s\) and the reaction rate \(R = 0.8\).
Using the Ziegler-Nichols tuning for the PI controller gains we have
The closed-loop system step response (assuming unity feedback) is shown in Figure 7.37. The settling time is \(T_{S} = 1.28\text{ }s\) and the percent overshoot is \(P.O. = 78\%\). Since we are using a PI controller, the steady-state error is zero, as expected.
The manual tuning method and the two Ziegler-Nichols tuning approaches presented here will not always lead to the desired closed-loop performance. The three
248. Table 7.6 Ziegler-Nichols PID Tuning Using Reaction Curve Characterized by Time Delay,
, and Reaction Rate,
| Controller Type | $$K_{P}$$ | $$K_{l}$$ | $$K_{D}$$ |
| Proportional \((P)\) | 1 | ||
| $$G_{c}(s) = K_{P}$$ | $$R\Delta T$$ | - | - |
| Proportional-plus-integral (PI) | |||
| $$G_{(c)}\ K_{I}$$ | 0.9 | 0.27 | - |
| $$G_{c}(s) = K_{P} + \frac{- 1}{s}$$ | $$R\Delta T$$ | $$R\Delta T^{2}$$ | |
| Proportional-plus-integral-plus-derivative (PID) | |||
| $$G_{(s)} - K_{D} + K_{I} + K_{D}$$ | 1.2 | 0.6 | 0.6 |
| $$U_{c}(s) = \Lambda_{P} + \frac{\bar{s}}{s} + \Lambda_{D}$$ | $$R\Delta T$$ | $$\overline{R\Delta T^{2}}$$ | $$R$$ |
Ziegler-Nichols PID Controller Gain Tuning Using Open-Loop Concepts FIGURE 7.36
Reaction curve with \(T_{d} = 0.1\text{ }s\) and \(R = 0.8\).
FIGURE 7.37
Time response for the Ziegler-Nichols PI tuning with \(K_{P} = 11.25\) and \(K_{I} = 33.75\).
methods do provide structured design steps leading to candidate PID gains and should be viewed as first steps in the design iteration. Since the PID (and the related PD and PI) controllers are in wide use today in a variety of applications, it is important to become familiar with various design approaches. We will use the PD controller later in this chapter to control the hard disk drive sequential design problem (see Section 7.10).
248.1. NEGATIVE GAIN ROOT LOCUS
As discussed in Section 7.2, the dynamic performance of a closed-loop control system is described by the closed-loop transfer function, that is, by the poles and zeros of the closed-loop system. The root locus is a graphical illustration of the variation of the roots of the characteristic equation as a single parameter of interest varies. We know that the roots of the characteristic equation and the closed-loop poles are one in the same. In the case of the single-loop negative unity feedback system shown in Figure 7.1, the characteristic equation is
where \(K\) is the parameter of interest. The orderly seven-step procedure for sketching the root locus described in Section 7.3 and summarized in Table 7.2 is valid for the case where \(0 \leq K < \infty\). Sometimes the situation arises where we are interested in the root locus for negative values of the parameter of interest where \(- \infty < K \leq 0\). We refer to this as the negative gain root locus. Our objective here is to develop an orderly procedure for sketching the negative gain root locus using familiar concepts from root locus sketching as described in Section 7.2.
Rearranging Equation (7.86) yields
Since \(K\) is negative, it follows that
where \(k = 0, \pm 1, \pm 2, \pm 3,\ldots\). The magnitude and phase conditions in Equation (7.87) must both be satisfied for all points on the negative gain root locus. Note that the phase condition in Equation (7.87) is different from the phase condition in Equation (7.4). As we will show, the new phase condition leads to several key modifications in the root locus sketching steps from those summarized in Table 7.2.
249. EXAMPLE 7.10 Negative gain root locus
Consider the system shown in Figure 7.38. The loop transfer function is
and the characteristic equation is
Sketching the root locus yields the plot shown in Figure 7.39(a) where it can be seen that the closed-loop system is not stable for any \(0 \leq K < \infty\). The negative gain root locus is shown in Figure 7.39(b). Using the negative gain root locus in Figure 7.39(b) we find that the stability is \(- 5.0 < K < - 2.5\). The system in Figure 7.38 can thus be stabilized with only negative gain, \(K\).
To locate the roots of the characteristic equation in a graphical manner on the \(s\)-plane for negative values of the parameter of interest, we will re-visit the seven steps summarized in Table 7.2 to obtain a similar orderly procedure to facilitate the rapid sketching of the locus. FIGURE 7.38
(a) Signal flow graph and (b) block diagram of unity feedback system with controller gain, \(K\).
(a)
(b)
(a)
(b)
FIGURE 7.39
(a) Root locus for \(0 \leq K < \infty\)
(b) Negative gain root locus for \(- \infty < K \leq 0\). Step 1: Prepare the root locus sketch. As before, you begin by writing the characteristic equation and rearranging, if necessary, so that the parameter of interest, \(K\), appears as the multiplying factor in the form,
For the negative gain root locus, we are interested in determining the locus of roots of the characteristic equation in Equation (7.88) for \(- \infty < K \leq 0\). As in Equation (7.24), factor \(P(s)\) in Equation (7.88) in the form of poles and zeros and locate the poles and zeros on the \(s\)-plane with "x" to denote poles and "o" to denote zeros.
When \(K = 0\), the roots of the characteristic equation are the poles of \(P(s)\), and when \(K \rightarrow - \infty\) the roots of the characteristic equation are the zeros of \(P(s)\). Therefore, the locus of the roots of the characteristic equation begins at the poles of \(P(s)\) when \(K = 0\) and ends at the zeros of \(P(s)\) as \(K \rightarrow - \infty\). If \(P(s)\) has \(n\) poles and \(M\) zeros and \(n > M\), we have \(n - M\) branches of the root locus approaching the zeros at infinity and the number of separate loci is equal to the number of poles. The root loci are symmetrical with respect to the horizontal real axis because the complex roots must appear as pairs of complex conjugate roots.
Step 2: Locate the segments of the real axis that are root loci. The root locus on the real axis always lies in a section of the real axis to the left of an even number of poles and zeros. This follows from the angle criterion of Equation (7.87).
Step 3: When \(n > M\), we have \(n - M\) branches heading to the zeros at infinity as \(K \rightarrow - \infty\) along asymptotes centered at \(\sigma_{A}\) and with angles \(\phi_{A}\). The linear asymptotes are centered at a point on the real axis given by
The angle of the asymptotes with respect to the real axis is
where \(k\) is an integer index.
Step 4: Determine where the locus crosses the imaginary axis (if it does so), using the Routh-Hurwitz criterion.
Step 5: Determine the breakaway point on the real axis (if any). In general, due to the phase criterion, the tangents to the loci at the breakaway point are equally spaced over \(360^{\circ}\). The breakaway point on the real axis can be evaluated graphically or analytically. The breakaway point can be computed by rearranging the characteristic equation
as
where \(p(s) = - d(s)/n(s)\) and finding the values of \(s\) that maximize \(p(s)\). This is accomplished by solving the equation
Equation (7.91) yields a polynomial equation in \(s\) of degree \(n + M - 1\), where \(n\) is the number of poles and \(M\) is the number of zeros. Hence the number of solutions is \(n + M - 1\). The solutions that exist on the root locus are the breakaway points.
Step 6: Determine the angle of departure of the locus from a pole and the angle of arrival of the locus at a zero using the phase angle criterion. The angle of locus departure from a pole or angle of arrival at a zero is the difference between the net angle due to all other poles and zeros and the criterion angle of \(\pm k360^{\circ}\).
Step 7: The final step is to complete the sketch by drawing in all sections of the locus not covered in the previous six steps.
The seven steps for sketching a negative gain root locus are summarized in Table 7.7.
250. Table 7.7 Seven Steps for Sketching a Negative Gain Root Locus (color text denotes changes from root locus steps in Table 7.2)
- Prepare the root locus sketch.
(a) Write the characteristic equation so that the parameter of interest, \(K\), appears as a multiplier.
(b) Factor \(P(s)\) in terms of \(n\) poles and \(M\) zeros
(c) Locate the open-loop poles and zeros of \(P(s)\) in the s-plane with selected symbols.
(d) Determine the number of separate loci, \(SL\).
(e) The root loci are symmetrical with respect to the horizontal real axis.
-
Locate the segments of the real axis that are root loci.
-
The loci proceed to the zeros at infinity along asymptotes centered at \(\sigma_{A}\) and with angles \(\phi_{A}\). (a) \(1 + KP(s) = 0\)
(b) \(1 + K\frac{\prod_{i = 1}^{n}\mspace{2mu}\left( s + z_{i} \right)}{\prod_{j = 1}^{n}\mspace{2mu}\left( s + p_{j} \right)} = 0\)
(c) \(\times =\) poles, \(O =\) zeros
(d) Locus begins at a pole and ends at a zero.
\(SL = n\) when \(n \geq M;n =\) number of finite, \(M =\) number of finite zeros .
Locus lies to the left of an even number of poles and zeros.
251. Table 7.7 (continued)
Step
-
Determine the points at which the locus crosses the imaginary axis (if it does so).
-
Determine the breakaway point on the real axis (if any).
-
Determine the angle of locus departure from complex at or poles and the angle of locus arrival at complex zeros using the phase criterion.
252. Related Equation or Rule
Use Routh-Hurwitz criterion.
a) Set \(K = p(s)\)
b) Determine roots of \(dp(s)/ds = 0\) or use graphical method to find maximum of \(p(s)\).
\(\angle P(s) = \pm k360^{\circ}\) at \(s = - p_{j}\) or \(- z_{i}\)
- Complete the negative gain root locus sketch.
252.1. DESIGN EXAMPLES
In this section we present two illustrative examples. The first example is a wind turbine control system. The feedback control system uses a PI controller to achieve a fast settling time and rise time while limiting the percent overshoot to a step input. In the second example, the automatic control of the velocity of an automobile is considered. The root locus method is extended from one parameter to three parameters as the three gains of a PID controller are determined. The design process is emphasized, including considering the control goals and associated variables to be controlled, the design specifications, and the PID controller design using root locus methods.
253. EXAMPLE 7.11 Wind turbine speed control
Wind energy conversion to electric power is achieved by wind energy turbines connected to electric generators. Of particular interest are wind turbines, as shown in Figure 7.40, that are located offshore [33]. The new concept is to allow the wind turbine to float rather than positioning the structure on a tower tied deep into the
FIGURE 7.40
Wind turbine placed offshore can help alleviate energy needs. (IS-200501/ Cultura RM/Alamy Stock Photo)
ocean floor. This allows the wind turbine structure to be placed in deeper waters up to 100 miles offshore far enough not to burden the landscape with unsightly structures [34]. Moreover, the wind is generally stronger on the open ocean, potentially leading to the production of \(5MW\) versus the more typical \(1.5MW\) for wind turbines onshore. However, the irregular character of wind direction and power results in the need for reliable, steady electric energy by using control systems for the wind turbines. The goal of these control devices is to reduce the effects of wind intermittency and of wind direction change. The rotor and generator speed control can be achieved by adjusting the pitch angle of the blades.
A basic model of the generator speed control system is shown in Figure 7.41 [35]. A linearized model from the collective pitch to the generator speed is given by \(\ ^{1}\)
The model corresponds to a \(600KW\) turbine with hub height \(= 36.6\text{ }m\), rotor diameter \(= 40\text{ }m\), rated rotor speed \(= 41.7rpm\), rated generator speed \(= 1800rpm\), and maximum pitch rate \(= 18.7deg/s\). Note that the linearized model in Equation (7.92) has zeros in the right half-plane at \(s_{1} = 827.1\) and \(s_{2,3} = 0.0274 \pm 0.1367j\) making this a nonminimum phase system.
A simplified version of the model in Equation (7.92) is given by the transfer function
where \(\tau = 5\) s and \(K = - 7200\). We will design a PI controller to control the speed of the turbine generator using the simplified first-order model in Equation (7.93) and confirm that the design specifications are satisfied for both the first-order model and the third-order model in Equation (7.92). The PI controller, denoted by \(G_{c}(s)\), is given by
where \(\tau_{c} = K_{I}/K_{P}\) and the gains \(K_{P}\) and \(K_{I}\) are to be determined. A stability analysis indicates that negative gains \(K_{I} < 0\) and \(K_{P} < 0\) will stabilize the system.
FIGURE 7.41
Wind turbine generator speed control system.
\(\ ^{1}\) Provided by Dr. Lucy Pao and Jason Laks in private correspondence. FIGURE 7.42
Wind turbine generator speed control root locus with a PI controller.
The main design specification is to have a settling time \(T_{S} < 4\) s to a unit step input. We also desire a limited percent overshoot \((\) P.O. \(< 25\%)\) and a short rise time \(\left( T_{r} < 1\text{ }s \right)\) while meeting the settling time specification. To this end, we will target the damping ratio of the dominant roots to be \(\zeta > 0.4\) and the natural frequency \(\omega_{n} > 2.5rad/s\).
The root locus is shown in Figure 7.42 for the characteristic equation
where \(\tau_{c} = 2\) and \({\widehat{K}}_{P} = - K_{p} > 0\). The placement of the controller zero at \(s = - \tau_{c} = - 2\) is a design parameter. We select the value of \({\widehat{K}}_{P}\) such that the damping ratio of the closed-loop complex poles is \(\zeta = 0.707\). Selecting \({\widehat{K}}_{P} = 0.0025\) yields \(K_{P} = - 0.0025\) and \(K_{I} = - 0.005\). The PI controller is
The step response is shown in Figure 7.43 using the simplified first-order model in Equation (7.93). The step response has \(T_{s} = 1.8\text{ }s,T_{r} = 0.34\text{ }s\), and \(\zeta = 0.707\) which translates to P.O. \(= 19\%\). The PI controller is able to meet all the control specifications. The step response using the third-order model in Equation (7.92) is shown in Figure 7.44 where we see the effect of the neglected components in the design as small oscillations in the speed response. The closed-loop impulse disturbance response in Figure 7.45 shows fast and accurate rejection of the disturbance in less than 3 seconds due to a \(1^{\circ}\) pitch angle change. FIGURE 7.43
Step response of the wind turbine generator speed control system using the first-order model in Equation (7.93) with the designed PI controller showing all specifications are satisfied with P.O. \(= 19\%\), \(T_{s} = 1.8\text{ }s\), and \(T_{r} = 0.34\text{ }s\).
FIGURE 7.44
Step response of the third-order model in Equation (7.92) with the PI controller showing that all specifications are satisfied with P.O. \(= 25\%\), \(T_{s} = 1.7\text{ }s\), and \(T_{r} = 0.3\text{ }s\).
254. EXAMPLE 7.12 Automobile velocity control
The automotive electronics market is expected to surpass \(\$ 300\) billion. It is predicted that there will be an annual growth rate of over \(7\%\) in electronic braking, steering, and driver information. Much of the additional computing power will FIGURE 7.45
Disturbance response of the wind turbine generator speed control system with a PI controller shows excellent disturbance rejection characteristics.
be used for new technology for smart cars and smart roads, such as IVHS (intelligent vehicle/highway systems) \(\lbrack 14,30,31\rbrack\). New systems on-board the automobile will support semi-autonomous automobiles, safety enhancements, emission reduction, and other features including intelligent cruise control, and brake by wire systems eliminating the hydraulics [32].
The term IVHS refers to a varied assortment of electronics that provides real-time information on accidents, congestion, and roadside services to drivers and traffic controllers. IVHS also encompasses devices that make vehicles more autonomous: collision-avoidance systems and lane-tracking technology that alert drivers to impending disasters and allow a car to drive itself.
An example of an automated highway system is shown in Figure 7.46. A velocity control system for maintaining the velocity between vehicles is shown in Figure 7.47. The output \(Y(s)\) is the relative velocity of the two automobiles; the input \(R(s)\) is the desired relative velocity between the two vehicles. Our design goal is to develop a controller that can maintain the prescribed velocity between the vehicles and maneuver the active vehicle (in this case the rearward automobile) as commanded. The elements of the design process emphasized in this example are depicted in Figure 7.48.
The control goal is
255. Control Goal
Maintain the prescribed velocity between the two vehicles, and maneuver the active vehicle as commanded. FIGURE 7.46
Automated highway system.
FIGURE 7.47 Vehicle velocity control system.
The variable to be controlled is the relative velocity between the two vehicles:
256. Variable to Be Controlled
The relative velocity between vehicles, denoted by \(y(t)\).
The design specifications are
257. Design Specifications
DS1 Zero steady-state error to a step input.
DS2 Steady-state error due to a ramp input of \(e_{ss} \leq 25\%\) of the input magnitude.
DS3 Percent overshoot of P.O. \(\leq 5\%\) to a step input.
DS4 Settling time of \(T_{s} \leq 1.5\text{ }s\) to a step input (using a \(2\%\) criterion to establish settling time).
From the design specifications and knowledge of the open-loop system, we find that we need a type 1 system to guarantee a zero steady-state error to a step input. The open-loop system transfer function is a type 0 system; therefore, the controller needs to increase the system type by at least 1 . A type 1 controller (that is, a controller
FIGURE 7.48 Elements of the control system design process emphasized in the automobile velocity control example.
with one integrator) satisfies DS1. To meet DS2 we need to have the velocity error constant
where
and \(G_{c}(s)\) is the controller (yet to be specified).
The percent overshoot specification DS3 allows us to define a target damping ratio
Similarly from the settling time specification DS4 we have
Solving for \(\zeta\omega_{n}\) yields \(\zeta\omega_{n} \geq 2.6\).
The desired region for the poles of the closed-loop transfer function is shown in Figure 7.49. Using a proportional controller \(G_{c}(s) = K_{P}\), is not reasonable, because DS2 cannot be satisfied. We need at least one pole at the origin to track a ramp input. Consider the PI controller
The question is where to place the zero at \(s = - K_{I}/K_{P}\).
We ask for what values of \(K_{P}\) and \(K_{I}\) is the system stable. The closed-loop transfer function is
The corresponding Routh array is
The first requirement for stability (from column one, row four) is
From the first column, third row, we have the inequality
FIGURE 7.49
Desired region in the complex plane for locating the dominant system poles.
It follows from DS2 that
Therefore, the integral gain must satisfy
If we select \(K_{I} > 64\), then the inequality in Equation (7.97) is satisfied. The valid region for \(K_{P}\) is then given by Equation (7.98), where \(K_{I} > 64\).
We need to consider DS4. Here we want to have the dominant poles to the left of the \(s = - 2.6\) line. We know from our experience sketching the root locus that since we have three poles (at \(s = 0, - 2\), and -8 ) and one zero (at \(s = - K_{I}/K_{P}\) ), we expect two branches of the loci to go to infinity along two asymptotes at \(\phi = - 90^{\circ}\) and \(+ 90^{\circ}\) centered at
where \(n_{p} = 3\) and \(n_{z} = 1\). In our case
We want to have \(\alpha < - 2.6\) so that the two branches will bend into the desired regions. Therefore,
or
So as a first design, we can select \(K_{P}\) and \(K_{I}\) such that
Suppose we choose \(K_{I}/K_{P} = 2.5\). Then the closed-loop characteristic equation is
The root locus is shown in Figure 7.50. To meet the \(\zeta = 0.69\) (which evolved from DS3), we need to select \(K_{P} < 30\). We selected the value at the boundary of the performance region (see Figure 7.50) as carefully as possible. FIGURE 7.50
Root locus for \(K_{l}/K_{P} = 2.5\).
Selecting \(K_{P} = 26\), we have \(K_{I}/K_{P} = 2.5\) which implies \(K_{I} = 65\). This satisfies the steady-state tracking error specification (DS2) since \(K_{I} = 65 > 64\).
The resulting PI controller is
The step response is shown in Figure 7.51.
The percent overshoot is P.O. \(= 8\%\), and the settling time is \(T_{s} = 1.45\text{ }s\). The percent overshoot specification is not precisely satisfied, but the controller in Equation (7.101) represents a very good first design. We can iteratively refine it. Even though the closed-loop poles lie in the desired region, the response does not exactly meet the specifications because the controller zero influences the response. The closed-loop system is a third-order system and does not have the performance of a second-order system. We might consider moving the zero to \(s = - 2\) (by choosing \(K_{I}/K_{P} = 2\) ) so that the pole at \(s = - 2\) is cancelled and the resulting system is a second-order system.
257.1. THE ROOT LOCUS USING CONTROL DESIGN SOFTWARE
An approximate root locus sketch can be obtained by applying the orderly procedure summarized in Table 7.2. Alternatively, we can use control design software to obtain an accurate root locus plot. However, we should not be tempted to rely solely on the computer for obtaining root locus plots while neglecting the manual steps in developing an approximate root locus. The fundamental concepts behind the root locus FIGURE 7.51
Automobile velocity control using the \(PI\) controller in Equation (7.101).
method are embedded in the manual steps, and it is essential to understand their application fully.
The section begins with a discussion on obtaining a computer-generated root locus plot. This is followed by a discussion of the connections between the partial fraction expansion, dominant poles, and the closed-loop system response. Root sensitivity is covered in the final paragraphs.
The functions covered in this section are rlocus, rlocfind, and residue. The functions rlocus and rlocfind are used to obtain root locus plots, and the residue function is utilized for partial fraction expansions of rational functions.
Obtaining a Root Locus Plot. Consider the closed-loop control system in Figure 7.10. The closed-loop transfer function is
The characteristic equation can be written as
The form of the characteristic equation in Equation (7.102) is necessary to use the rlocus function for generating root locus plots. The general form of the characteristic equation necessary for application of the rlocus function is
FIGURE 7.52
The rlocus function.
where \(K\) is the parameter of interest to be varied from \(0 \leq K < \infty\). The rlocus function is shown in Figure 7.52, where we define the transfer function object sys \(= G(s)\). The steps to obtaining the root locus plot associated with Equation (7.102), along with the associated root locus plot, are shown in Figure 7.53. Invoking the rlocus function without left-hand arguments results in an automatic generation of the root locus plot. When invoked with left-hand arguments, the rlocus function returns a matrix of root locations and the associated gain vector.
The steps to obtain a computer-generated root locus plot are as follows:
-
Obtain the characteristic equation in the form given in Equation (7.103), where \(K\) is the parameter of interest.
-
Use the rlocus function to generate the plots.
Referring to Figure 7.53, we can see that as \(K\) increases, two branches of the root locus break away from the real axis. This means that, for some values of \(K\), the closedloop system characteristic equation will have two complex roots. Suppose we want to find the value of \(K\) corresponding to a pair of complex roots. We can use the rlocfind
Obtaining root locations \(r\) associated with various values of the gain \(K\). function to do this, but only after a root locus has been obtained with the rlocus function. Executing the rlocfind function will result in a cross-hair marker appearing on the root locus plot. We move the cross-hair marker to the location on the locus of interest and hit the enter key. The value of the parameter \(K\) and the value of the selected point will then be displayed in the command display. The use of the rlocfind function is illustrated in Figure 7.54.
Control design software packages may respond differently when interacting with plots, such as with the rlocfind function on the root locus. The response of rlocfind in Figure 7.54 corresponds to MATLAB. Refer to the companion website for more information on other control design software applications.
Continuing our third-order root locus example, we find that when \(K = 20.5775\), the closed-loop transfer function has three poles and two zeros, at
Considering the closed-loop pole locations only, we would expect that the real pole at \(s = - 0.8989\) would be the dominant pole. To verify this, we can study the closed-loop system response to a step input, \(R(s) = 1/s\). For a step input, we have
FIGURE 7.54 Using the rlocfind function.
FIGURE 7.55
Partial fraction expansion of Equation (7.104).
Generally, the first step in computing \(y(t)\) is to expand Equation (7.104) in a partial fraction expansion. The residue function can be used to expand Equation (7.104), as shown in Figure 7.55. The residue function is described in Figure 7.56.
The partial fraction expansion of Equation (7.104) is
Comparing the residues, we see that the coefficient of the term corresponding to the pole at \(s = - 0.8989\) is considerably smaller than the coefficient of the terms corresponding to the complex-conjugate poles at \(s = - 2.0505 \pm j4.3227\). From this, we expect that the influence of the pole at \(s = - 0.8989\) on the output response \(y(t)\) is not dominant. The settling time (to within \(2\%\) of the final value) is then predicted by considering the complex-conjugate poles. The poles at \(s = - 2.0505 \pm j4.3227\) correspond to a damping of \(\zeta = 0.4286\) and a natural frequency of \(\omega_{n} = 4.7844\). Thus, the settling time is predicted to be
FIGURE 7.56 The residue function.
Using the step function, as shown in Figure 7.57, we find that \(T_{s} = 1.6\text{ }s\). Hence, our approximation of settling time \(T_{s} \simeq 1.95\text{ }s\) is a fairly good approximation. The percent overshoot is predicted (considering the zero of \(T(s)\) at \(s = - 3\) ) to be P.O. \(= 60\%\). As can be seen in Figure 7.57, the actual overshoot is P.O. \(= 50\%\).
When using the step function, we can right-click on the figure to access the pulldown menu, which allows us to determine the step response settling time and peak response, as illustrated in Figure 7.57. On the pull-down menu select "Characteristics" and select "Settling Time." A dot will appear on the figure at the settling point. Place the cursor over the dot to determine the settling time.
In this example, the role of the system zeros on the transient response is illustrated. The proximity of the zero at \(s = - 1\) to the pole at \(s = - 0.8989\) reduces the impact of that pole on the transient response. The main contributors to the transient
for the closedloop system in Figure 7.10 with \(K = 20.5775\). FIGURE 7.58
Converting a partial fraction expansion back to a rational function.
response are the complex-conjugate poles at \(s = - 2.0505 \pm j4.3228\) and the zero at \(s = - 3\).
There is one final point regarding the residue function: We can convert the partial fraction expansion back to the polynomials num/den, given the residues \(r\), the pole locations \(p\), and the direct terms \(k\), with the command shown in Figure 7.58.
Sensitivity and the Root Locus. The roots of the characteristic equation play an important role in defining the closed-loop system transient response. The effect of parameter variations on the roots of the characteristic equation is a useful measure of sensitivity. The root sensitivity is defined in Equation (7.75). We can use Equation (7.75) to investigate the sensitivity of the roots of the characteristic equation to variations in the parameter \(K\). If we change \(K\) by a small finite amount \(\Delta K\), and evaluate the modified root \(r_{i} + \Delta r_{i}\), it follows that \(S_{K}^{r_{i}}\) is given in Equation (7.79).
The quantity \(S_{K}^{r_{i}}\) is a complex number. Referring back to the third-order example of Figure 7.10 (Equation 7.102), if we change \(K\) by a factor of \(5\%\), we find that the dominant complex-conjugate pole at \(s = - 2.0505 + j4.3228\) changes by
when \(K\) changes from \(K = 20.5775\) to \(K = 21.6064\). From Equation (7.79), it follows that
The sensitivity \(S_{K}^{r_{i}}\) can also be written in the form
The magnitude and direction of \(S_{K}^{r_{i}}\) provides a measure of the root sensitivity. The script used to perform these sensitivity calculations is shown in Figure 7.59.
The root sensitivity measure may be useful for comparing the sensitivity for various system parameters at different root locations.
257.2. SEQUENTIAL DESIGN EXAMPLE: DISK DRIVE READ SYSTEM
In this chapter, we will use the PID controller to obtain a desirable response using velocity feedback. We will proceed with our model and then select a controller. Finally, we will optimize the parameters and analyze the performance. In this chapter, we will use the root locus method in the selection of the controller parameters. FIGURE 7.59
Sensitivity calculations for the root locus for a \(5\%\) change in \(K = 20.5775\).
We use the root locus to select the controller gains. The PID controller introduced in this chapter is
Since the process model \(G_{1}(s)\) already possesses an integration, we set \(K_{I} = 0\). Then we have the PD controller
and our goal is to select \(K_{P}\) and \(K_{D}\) in order to meet the specifications. The system is shown in Figure 7.60. The closed-loop transfer function of the system is
In order to obtain the root locus as a function of a parameter, we write \(G_{c}(s)G_{1}(s)G_{2}(s)\) as
where \(z = K_{P}/K_{D}\). We use \(K_{P}\) to select the location of the zero \(z\) and then sketch the locus as a function of \(K_{D}\). We select \(z = 1\) so that
FIGURE 7.60
Disk drive control system with a PD controller.
Disturbance
FIGURE 7.61
Sketch of the root locus.
Table 7.8 Disk Drive Control System Specifications and Actual Design Performance
| Performance Measure | Desired Value | Actual Response |
| Percent overshoot | Less than \(5\%\) | $$0%$$ |
| Settling time | Less than \(250\text{ }ms\) | $$20\text{ }ms$$ |
| Maximum response to a unit disturbance | Less than \(5 \times 10^{- 3}\) | $$2 \times 10^{- 3}$$ |
The number of poles minus the number of zeros is 2 , and we expect asymptotes at \(\phi_{A} = \pm 90^{\circ}\) with a centroid
as shown in Figure 7.61. We can quickly sketch the root locus, as shown in Figure 7.61. We use the computer-generated root locus to determine the root values for various values of \(K_{D}\). When \(K_{D} = 91.3\), we obtain the roots shown in Figure 7.61. Then, obtaining the system response, we achieve the actual response measures as listed in Table 7.8. As designed, the system meets all the specifications. It takes the system a settling time of \(20\text{ }ms\) to "practically" reach the final value. In reality, the system drifts very slowly toward the final value after quickly achieving \(97\%\) of the final value.
257.3. SUMMARY
The relative stability and the transient response performance of a closed-loop control system are directly related to the location of the closed-loop roots of the characteristic equation. We investigated the movement of the characteristic roots on the \(s\)-plane as key system parameters (such as controller gains) are varied. The root locus and the negative gain root locus are graphical representations of the variation of the system closedloop poles as one parameter varies. The plots can be sketched using a given set of rules in order to analyze the initial design of a system and determine suitable alterations of the system structure and the parameter values. A computer is then commonly used to obtain the accurate root locus for use in the final design and analysis. A summary of fifteen typical root locus diagrams is shown in Table 7.9. Section 7.11 Summary \(\mathbf{511}\)
512 Chapter 7 The Root Locus Method
Furthermore, we extended the root locus method for the design of several parameters for a closed-loop control system. Then the sensitivity of the characteristic roots was investigated for undesired parameter variations by defining a root sensitivity measure. It is clear that the root locus method is a powerful and useful approach for the analysis and design of modern control systems and will continue to be one of the most important procedures of control engineering.
258. 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 7.62 as specified in the various problem statements.
FIGURE 7.62 Block diagram for the Skills Check.
In the following True or False and Multiple Choice problems, circle the correct answer.
- The root locus is the path the roots of the characteristic equation (given by \(1 + KG(s) = 0)\) trace out on the \(s\)-plane as the system parameter \(0 \leq K < \infty\) varies.
True or False
- On the root locus plot, the number of separate loci is equal to the number of poles of \(G(s)\).
True or False
- The root locus always starts at the zeros and ends at the poles of \(G(s)\).
True or False
- The root locus provides the control system designer with a measure of the sensitivity of the poles of the system to variations of a parameter of interest.
True or False
- The root locus provides valuable insight into the response of a system to various test inputs.
True or False
- Consider the control system in Figure 7.62, where the loop transfer function is
Using the root locus method, determine the value of \(K\) such that the dominant roots have a damping ratio \(\zeta = 0.5\).
a. \(K = 1.2\)
b. \(K = 4.5\)
c. \(K = 9.7\)
d. \(K = 37.4\) In Problems 7 and 8, consider the unity feedback system in Figure 7.62 with
-
The approximate angles of departure of the root locus from the complex poles are
a. \(\phi_{d} = \pm 180^{\circ}\)
b. \(\phi_{d} = \pm 115^{\circ}\)
c. \(\phi_{d} = \pm 205^{\circ}\)
d. None of the above -
The root locus of this system is given by which of the following:
(a)
(c)
(b)
(d)
- A unity feedback system has the closed-loop transfer function given by
Using the root locus method, determine the value of the gain \(K\) so that the closed-loop system has a damping ratio \(\zeta = \sqrt{2}/2\).
a. \(K = 25\)
b. \(K = 1250\)
c. \(K = 2025\)
d. \(K = 10500\) 10. Consider the unity feedback control system in Figure 7.62 where
Using the root locus method, determine that maximum value of \(z\) for closed-loop stability.
a. \(z = 7.2\)
b. \(z = 12.8\)
c. Unstable for all \(z > 0\)
d. Stable for all \(z > 0\)
In Problems 11 and 12, consider the control system in Figure 7.62 where the model of the process is
- Suppose that the controller is
Using the root locus method, determine the maximum value of the gain \(K\) for closedloop stability.
a. \(K = 2.13\)
b. \(K = 3.88\)
c. \(K = 14.49\)
d. Stable for all \(K > 0\)
-
Suppose that a simple proportional controller is utilized, that is, \(G_{c}(s) = K\). Using the root locus method, determine the maximum controller gain \(K\) for closed-loop stability.
a. \(K = 0.50\)
b. \(K = 1.49\)
c. \(K = 4.48\)
d. Unstable for \(K > 0\) -
Consider the unity feedback system in Figure 7.62 where
Determine the breakaway point on the real axis and the respective gain, \(K\).
a. \(s = - 1.8,K = 58.75\)
b. \(s = - 2.5,K = 4.59\)
c. \(s = 1.4,K = 58.75\)
d. None of the above
In Problems 14 and 15, consider the feedback system in Figure 7.62, where
- Which of the following is the associated root locus?
(a)
(c)
(b)
(d)
- The departure angles from the complex poles and the arrival angles at the complex zeros are:
a. \(\phi_{D} = \pm 180^{\circ},\phi_{A} = 0^{\circ}\)
b. \(\phi_{D} = \pm {116.6}^{\circ},\phi_{A} = \pm {198.4}^{\circ}\)
c. \(\phi_{D} = \pm {45.8}^{\circ},\phi_{A} = \pm {116.6}^{\circ}\)
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. Parameter design
The amplitude of the closed-loop response is reduced approximately to one-fourth of the maximum value in one oscillatory period.
b. Root sensitivity The path the root locus follows as the parameter becomes very large and approaches \(\infty\).
c. Root locus The center of the linear asymptotes, \(\sigma_{A}\).
d. Root locus segments The process of determining the PID controller gains on the real axis using one of several analytic methods based on openloop and closed-loop responses to step inputs. e. Root locus method A method of selecting one or two parameters using the root locus method.
f. Asymptote centroid The root locus lying in a section of the real axis to the left of an odd number of poles and zeros.
g. Breakaway point The root locus for negative values of the parameter of interest where \(- \infty < K \leq 0\).
h. Locus
The angle at which a locus leaves a complex pole in the \(s\)-plane.
i. Angle of departure A path or trajectory that is traced out as a parameter is changed.
j. Number of separate The locus or path of the roots traced out on the loci \(s\)-plane as a parameter is changed.
k. Asymptote
The sensitivity of the roots as a parameter changes from its normal value.
l. Negative gain root locus
The method for determining the locus of roots of the characteristic equation \(1 + KG(s) = 0\) as \(0 \leq K < \infty\).
m. PID tuning The process of determining the PID controller gains.
n. Quarter amplitude decay
The point on the real axis where the locus departs from the real axis of the \(s\)-plane.
o. Ziegler-Nichols PID Equal to the number of poles of the transfer function, tuning method assuming that the number of poles is greater than or equal to the number of zeros of the transfer function.
259. EXERCISES
E7.1 Consider a device that consists of a ball rolling on the inside rim of a hoop [11]. This model is similar to the problem of liquid fuel sloshing in a rocket. The hoop is free to rotate about its horizontal principal axis as shown in Figure E7.1. The angular position of the hoop may be controlled via the torque \(T(t)\) applied to the hoop from a torque motor attached to the hoop drive shaft. If negative feedback is used, the system characteristic equation is
(a) Sketch the root locus. (b) Find the gain when the roots are both equal. (c) Find these two equal roots.
FIGURE E7.1 Hoop rotated by motor. (d) Find the settling time of the system when the roots are equal.
E7.2 A tape recorder with a unity feedback speed control system has the loop transfer function
a. Sketch a root locus for \(K\), and show that the dominant roots are \(s = - 0.41 \pm j0.384\), when \(K = 8\).
b. For the dominant roots of part (a), calculate the settling time and overshoot for a step input.
E7.3 A unity feedback control system for an automobile suspension tester has the loop transfer function [12]
We desire the dominant roots to have the maximum imaginary part value. Using the root locus, show that \(K = 8.67\) is required, and the dominant roots are \(s = - 1.5 \pm j1.65\).
E7.4 Consider a unity feedback system with the loop transfer function
(a) Find the angle of departure of the root locus from the complex poles. (b) Find the entry point for the root locus as it enters the real axis.
Answers: \({123.5}^{\circ}, - 9.8\)
E7.5 Consider a unity feedback system with a loop transfer function
(a) Find the breakaway point on the real axis. (b) Find the asymptote centroid. (c) Find the value of \(K\) at the breakaway point.
E7.6 One version of a space station is shown in Figure E7.6 [28]. It is critical to keep this station in the proper orientation toward the Sun and the Earth for generating power and communications. The orientation controller may be represented by a unity feedback system with an actuator and controller, such as
Sketch the root locus of the system as \(K\) increases. Find the value of \(K\) that results in an unstable system.
Answers: \(K = 40\)
FIGURE E7.6 Space station.
E7.7 The elevator in a modern office building can travel at a speed of 25 feet per second and still stop within one-eighth of an inch of the floor outside. The loop transfer function of the unity feedback elevator position control is
Determine the gain \(K\) when the complex roots have an \(\zeta\) equal to 0.7 .
E7.8 Sketch the root locus for a unity feedback system with
(a) Find the gain when all three roots are real and equal. (b) Find the roots when all the roots are equal as in part (a).
Answers: \(K = 108;s = - 7\)
E7.9 The primary mirror of a large telescope can have a diameter of \(10\text{ }m\) and a mosaic of 36 hexagonal segments with the orientation of each segment actively controlled. Suppose this unity feedback system for the mirror segments has the loop transfer function
a. Find the asymptotes and sketch them in the \(s\)-plane.
b. Find the angle of departure from the complex poles.
c. Determine the gain when two roots lie on the imaginary axis.
d. Sketch the root locus.
E7.10 A unity feedback system has the loop transfer function
a. Find the breakaway and entry points on the real axis.
b. Find the gain and the roots when the real part of the complex roots is located at -3 .
c. Sketch the root locus.
Answers: (a) \(- 0.59, - 3.41\); (b) \(K = 3,s = - 2 \pm j\sqrt{2}\)
E7.11 A robot force control system with unity feedback has a loop transfer function [6]
a. Find the gain \(K\) that results in dominant roots with a damping ratio of 0.707 . Sketch the root locus.
b. Find the actual percent overshoot and peak time for the gain \(K\) of part (a).
E7.12 A unity feedback system has a loop transfer function
(a) Sketch the root locus for \(K > 0\). (b) Find the roots when \(K = 10\) and 20. (c) Compute the rise time, percent overshoot, and settling time (with a \(2\%\) criterion) of the system for a unit step input when \(K = 10\) and 20 . E7.13 A unity feedback system has a loop transfer function
(a) Draw the root locus as \(z\) varies from 0 to 100 .
(b) Using the root locus, estimate the percent overshoot and settling time (with a \(2\%\) criterion) of the system at \(z = 1,2\), and 3 for a step input. (c) Determine the actual overshoot and settling time at \(z = 1,2\), and 3 .
E7.14 A unity feedback system has the loop transfer function
(a) Determine the breakaway and entry points of the root locus, and sketch the root locus for \(K > 0\).
(b) Determine the gain \(K\) when the two characteristic roots have a \(\zeta\) of \(1/\sqrt{2}\). (c) Calculate the roots.
E7.15 (a) Plot the root locus for a unity feedback system with loop transfer function
(b) Calculate the range of \(K\) for which the system is stable. (c) Predict the steady-state error of the system for a ramp input.
Answers: (a) \(K > 1.67\); (b) \(e_{ss} = 0\)
E7.16 A negative unity feedback system has a loop transfer function
where \(T = 0.1\text{ }s\). Show that an approximation for the time delay is
Using
obtain the root locus for the system for \(K > 0\). Determine the range of \(K\) for which the system is stable.
E7.17 A control system, as shown in Figure E7.17, has a process
FIGURE E7.17 Feedback system.
(a) When \(G_{c}(s) = K\), show that the system is always unstable by sketching the root locus. (b) When
sketch the root locus, and determine the range of \(K\) for which the system is stable. Determine the value of \(K\) and the complex roots when two roots lie on the \(j\omega\)-axis.
E7.18 A closed-loop negative unity feedback system is used to control the yaw of an aircraft. When the loop transfer function is
determine (a) the root locus breakaway point and (b) the value of the roots on the \(j\omega\)-axis and the gain required for those roots. Sketch the root locus.
Answers: (a) Breakaway: \(s = - 2.29\); (b) \(j\omega\)-axis: \(s = \pm j1.09,K = 8\)
E7.19 A unity feedback system has a loop transfer function
(a) Determine the angle of departure of the root locus at the complex poles. (b) Sketch the root locus.
(c) Determine the gain \(K\) when the roots are on the \(j\omega\)-axis and determine the location of these roots.
E7.20 A unity feedback system has a loop transfer function
(a) Determine the range of \(K\) for stability. (b) Sketch the root locus. (c) Determine the maximum \(\zeta\) of the stable complex roots.
Answers: (a) \(K > 16\); (b) \(\zeta = 0.25\)
E7.21 A unity feedback system has a loop transfer function
Sketch the root locus. Determine the gain \(K\) when the complex roots of the characteristic equation have a damping ratio \(\zeta\) approximately equal to 0.66 . E7.22 A high-performance missile for launching a satel- E7.25 A closed-loop feedback system is shown in Figure
lite has a unity feedback system with a loop transfer function
Sketch the root locus as \(K\) varies from \(0 < K < \infty\).
E7.23 A unity feedback system has a loop transfer function
Sketch the root locus for \(0 \leq a < \infty\).
E7.24 Consider the system represented in state variable form
where
Determine the characteristic equation and then sketch the root locus as \(0 < k < \infty\).
E7.25. For what range of \(K\) is the system stable? Sketch the root locus as \(0 < K < \infty\).
E7.26 Consider the single-input, single-output system is described by
where
Compute the characteristic polynomial and plot the root locus as \(0 \leq K < \infty\). For what values of \(K\) is the system stable?
E7.27 Consider the unity feedback system in Figure E7.27. Sketch the root locus as \(0 \leq p < \infty\). For what values of \(p\) is the closed loop system stable?
E7.28 Consider the feedback system in Figure E7.28. Obtain the negative gain root locus as \(- \infty < K \leq 0\). For what values of \(K\) is the system stable?
FIGURE E7.25
Nonunity feedback system with parameter \(K\).
FIGURE E7.27 Unity feedback system with parameter \(p\).
FIGURE E7.28 Feedback system for negative gain root locus.
260. PROBLEMS
P7.1 Sketch the root locus for the following loop transfer functions of the system shown in Figure P7.1 when \(0 \leq K < \infty\) :
a. \(L(s) = G_{c}(s)G(s) = \frac{K}{s(s + 5)(s + 20)}\)
b. \(L(s) = G_{c}(s)G(s) = \frac{K}{\left( s^{2} + 2s + 2 \right)(s + 2)}\)
c. \(\ L(s) = G_{c}(s)G(s) = \frac{K(s + 10)}{s(s + 1)(s + 20)}\)
d. \(L(s) = G_{c}(s)G(s) = \frac{K\left( s^{2} + 4s + 8 \right)}{s^{2}(s + 1)}\)
P7.2 Consider the loop transfer function of a phase-lock loop system
Sketch the root locus as a function of the gain \(K_{v} = K_{a}K\). Determine the value of \(K_{v}\) attained if the complex roots have a damping ratio equal to 0.60 [13].
P7.3 A unity feedback system has the loop transfer function
Find (a) the breakaway point on the real axis and the gain \(K\) for this point, (b) the gain and the roots when two roots lie on the imaginary axis, and (c) the roots when \(K = 10\). (d) Sketch the root locus.
P7.4 Suppose that the loop transfer function of a large antenna is given by
where \(\tau = 0.2,\zeta = 0.707\), and \(\omega_{n} = 1rad/s\). Sketch the root locus of the system as \(0 \leq k_{a} < \infty\). Determine the maximum allowable gain of \(k_{a}\) for a stable system.
P7.5 Automatic control of helicopters is necessary because, unlike fixed-wing aircraft which possess a fair degree of inherent stability, the helicopter is quite unstable. A helicopter control system that utilizes an automatic control loop plus a pilot stick control is shown in Figure P7.5. When the pilot is not using the control stick, the switch may be considered to be open. The dynamics of the helicopter are represented by the transfer function
(a) With the pilot control loop open (hands-off control), sketch the root locus for the automatic stabilization loop. Determine the gain \(K_{2}\) that results in a damping for the complex roots equal to \(\zeta = 0.707\). (b) For the gain \(K_{2}\) obtained in part (a), determine the steady-state error due to a wind gust \(T_{d}(s) = 1/s\). (c) With the pilot loop added, draw the root locus as \(K_{1}\) varies from zero to \(\infty\) when \(K_{2}\) is set at the value calculated in part (a). (d) Recalculate the steady-state
FIGURE P7.1
FIGURE P7.5
Helicopter control.
FIGURE P7.6
Satellite attitude control.
error of part (b) when \(K_{1}\) is equal to a suitable value based on the root locus.
P7.6 An attitude control system for a satellite vehicle within the earth's atmosphere is shown in Figure P7.6.
(a) Draw the root locus of the system as \(K\) varies from \(0 \leq K < \infty\). (b) Determine the range of \(K\) for closedloop stability. (c) Determine the gain \(K\) that results in a system with a settling time (with a \(2\%\) criterion) of \(T_{s} \leq 12\text{ }s\) and a percent overshoot P.O. \(\leq 25\%\).
P7.7 The speed control system for an isolated power system is shown in Figure P7.7. The valve controls the steam flow input to the turbine in order to account for load changes \(\Delta L(s)\) within the power distribution network. The equilibrium speed desired results in a generator frequency equal to \(60cps\). The effective rotary inertia \(J\) is equal to 4000 and the friction constant \(b\) is equal to 0.75 . The steady-state speed regulation factor \(R\) is represented by the equation \(R \approx \left( \omega_{0} - \omega_{r} \right)/\Delta L\), where \(\omega_{r}\) equals the speed at rated load and \(\omega_{0}\) equals the speed at no load. We want to obtain a very small \(R\), usually less than 0.10 . (a) Using root locus techniques, determine the regulation \(R\) attainable when the damping ratio of the roots of the system must be greater than 0.60. (b) Verify that the steady-state speed deviation for a load torque change \(\Delta L(s) = \Delta L/s\) is, in fact, approximately equal to \(R\Delta L\) when \(R \leq 0.1\).
P7.8 Consider again the power control system of Problem P7.7 when the steam turbine is replaced by a hydroturbine. For hydroturbines, the large inertia of the water used as a source of energy causes a considerably larger time constant. The transfer function of a hydroturbine may be approximated by
where \(\tau = 1\text{ }s\). With the rest of the system remaining as given in Problem P7.7, repeat parts (a) and (b) of Problem P7.7.
P7.9 The achievement of safe, efficient control of the spacing of automatically controlled guided vehicles is an important part of the future use of the vehicles in a manufacturing plant \(\lbrack 14,15\rbrack\). It is important that the system eliminates the effects of disturbances (such as oil on the floor) as well as maintains accurate spacing between vehicles on a guide way. The system can be
FIGURE P7.7 Power system control.
FIGURE P7.9 Guided vehicle control.
represented by the block diagram of Figure P7.9. The vehicle dynamics can be represented by
(a) Sketch the root locus of the system. (b) Determine all the roots when the loop gain \(K = K_{1}K_{2}\) is equal to 2500 .
P7.10 New concepts in passenger airliner design will have the range to cross the Pacific in a single flight and the efficiency to make it economical \(\lbrack 16,29\rbrack\). These new designs will require the use of temperature-resistant, lightweight materials and advanced control systems. Noise control is an important issue in modern aircraft designs since most airports have strict noise level requirements. An advanced airliner is depicted in Figure P7.10(a). It would seat 200 passengers and cruise at just below the speed of sound. The flight control system must provide good handling characteristics and comfortable flying conditions. An automatic control system can be designed for the next generation passenger aircraft.
The desired characteristics of the dominant roots of the control system shown in Figure P7.10(b) have a \(\zeta = 0.707\). The characteristics of the aircraft are \(\omega_{n} = 2.5,\zeta = 0.30\), and \(\tau = 0.1\). The gain factor \(K_{1}\), however, will vary over the range 0.02 at medium-weight cruise conditions to 0.20 at lightweight descent conditions. (a) Sketch the root locus as a function of the loop gain \(K_{1}K_{2}\). (b) Determine the gain \(K_{2}\) necessary to yield roots with \(\zeta = 0.707\) when the aircraft is in the medium-cruise condition. (c) With the gain \(K_{2}\) as found in part (b), determine the \(\zeta\) of the roots when the gain \(K_{1}\) results from the condition of light descent.
P7.11 A computer system requires a high-performance magnetic tape transport system [17]. The environmental conditions imposed on the system result in a severe test of control engineering design. A direct-drive DC motor system for the magnetic tape reel system is shown in Figure P7.11, where \(r\) equals the reel radius, and \(J\) equals the reel and rotor inertia. A complete reversal of the tape reel direction is required in \(6\text{ }ms\), and the tape reel must follow a step command in \(3\text{ }ms\) or less. The tape is normally operating at a speed of \(100in/s\).
FIGURE P7.10
(a) A passenger jet aircraft of the future. (Muratart/ Shutterstock.)
(b) Control system.
(a)
(b) FIGURE P7.11
(a) Tape control system. (b) Block diagram.
(a)
(b)
The motor and components selected for this system possess the following characteristics:
\(K_{2}\) is adjustable.
The inertia of the reel and motor rotor is \(2.5 \times 10^{- 3}\) when the reel is empty, and \(5.0 \times 10^{- 3}\) when the reel is full. A series of photocells is used as an error-sensing device. The time constant of the motor is \(L/R = 0.5\text{ }ms\). (a) Sketch the root locus for the system when \(K_{2} = 10\) and \(J = 5.0 \times 10^{- 3},0 < K_{a} < \infty\). (b) Determine the gain \(K_{a}\) that results in a welldamped system so that the \(\zeta\) of all the roots is greater than or equal to 0.60 . (c) With the \(K_{a}\) determined from part (b), sketch a root locus for \(0 < K_{2} < \infty\).
P7.12 A precision speed control system (Figure P7.12) is required for a platform used in gyroscope and inertial system testing where a variety of closely controlled speeds is necessary. A direct-drive DC torque motor system was utilized to provide (1) a speed range of \(0.01\%\) s to \(600\%\), and (2) \(0.1\%\) steady-state error maximum for a step input. The direct-drive DC torque motor avoids the use of a gear train with its attendant backlash and friction. Also, the direct-drive motor has a high-torque capability, high efficiency, and low motor time constants. The motor gain constant is nominally \(K_{m} = 1.8\), but is subject to variations up to \(50\%\). The amplifier gain \(K_{a}\) is normally greater than 10 and subject to a variation of \(10\%\). (a) Determine the minimum loop gain necessary to satisfy the steady-state error requirement. (b) Determine the limiting value of gain for stability. (c) Sketch the root locus as \(K_{a}\) varies from 0 to \(\infty\). (d) Determine the roots when \(K_{a} = 40\), and estimate the response to a step input.
P7.13 A unity feedback system has the loop transfer function
FIGURE P7.12
Speed control.
(a) Find the breakaway point on the real axis and the gain for this point. (b) Find the gain to provide two complex roots nearest the \(j\omega\)-axis with a damping ratio \(\zeta = 0.6\). (c) Are the two roots of part (b) dominant? (d) Determine the settling time (with a \(2\%\) criterion) of the system when the gain of part (b) is used.
P7.14 The loop transfer function of a unity feedback system is
This system is called conditionally stable because it is stable only for a range of the gain \(K\) such that \(k_{1} < K < k_{2}\). Using the Routh-Hurwitz criteria and the root locus method, determine the range of the gain for which the system is stable. Sketch the root locus for \(0 < K < \infty\).
P7.15 Suppose that the dynamics of a transport vehicle can be represented by the loop transfer function
\(G_{c}(s)G(s) = \frac{K\left( s^{2} + 40s + 800 \right)(s + 40)}{s\left( s^{2} + 100s + 1000 \right)\left( s^{2} + 250s + 4500 \right)}\).
Sketch the root locus for the system. Determine the damping ratio of the dominant roots when \(K = 1000\).
P7.16 Control systems for maintaining constant tension on strip steel in a hot strip finishing mill are called "loopers." A typical system is shown in Figure P7.16. The looper is an arm 2 to 3 feet long with a roller on the end; it is raised and pressed against the strip by a motor [18]. The typical speed of the strip passing the looper is \(2000ft/min\). A voltage proportional to the looper position is compared with a reference voltage and integrated where it is assumed that a change in looper position is proportional to a change in the steel
(a)
FIGURE P7.16
Steel mill control system. (b) strip tension. The time constant \(\tau\) of the filter is negligible relative to the other time constants in the system. (a) Sketch the root locus of the control system for \(0 < K_{a} < \infty\). (b) Determine the gain \(K_{a}\) that results in a system whose roots have a damping ratio of \(\zeta = 0.707\) or greater. (c) Determine the effect of \(\tau\) as \(\tau\) increases from a negligible quantity.
P7.17 Consider the vibration absorber in Figure P7.17. Using the root locus method, determine the effect of the parameters \(M_{2}\) and \(k_{12}\). Determine the specific values of the parameters \(M_{2}\) and \(k_{12}\) so that the mass \(M_{1}\) does not vibrate when \(F(t) = asin\left( \omega_{0}t \right)\). Assume that \(M_{1} = 1,k_{1} = 1\), and \(b = 1\). Also assume that \(k_{12} < 1\) and that the term \(k_{12}\ ^{1}\) may be neglected.
P7.18 A feedback control system is shown in Figure P7.18. The filter \(G_{c}(s)\) is often called a compensator, and the design problem involves selecting the parameters \(\alpha\) and \(\beta\). Using the root locus method, determine the effect of varying the parameters. Select a suitable filter so that the time to settle (to within \(2\%\) of the final value) is \(T_{s} \leq 4\text{ }s\), and the damping ratio of the dominant roots is \(\zeta > 0.6\).
P7.19 In recent years, many automatic control systems for guided vehicles in factories have been installed. One system uses a magnetic tape applied to the floor to guide the vehicle along the desired lane \(\lbrack 10,15\rbrack\). Using transponder tags on the floor, the automatically
FIGURE P7.18 Filter design.
guided vehicles can be tasked (for example, to speed up or slow down) at key locations. An example of a guided vehicle in a factory is shown in Figure P7.19(a).
FIGURE P7.17
Vibration absorber.
FIGURE P7.19
(a) An automatically guided vehicle. (Photo courtesy of the Vanit Janthra/ Shutterstock.)
(b) Block diagram.
(a)
(b) Sketch a root locus and determine a suitable gain P7.23 Repeat Problem P7.22 for the loop transfer func\(K_{a}\) so that the damping ratio of the complex roots is \(\zeta > 0.707\).
P7.20 Determine the root sensitivity for the dominant roots of the design for Problem P7.18 for the gain \(K = 4\alpha/\beta\) and the pole \(s = - 2\).
P7.21 Determine the root sensitivity of the dominant roots of the power system of Problem P7.7. Evaluate the sensitivity for variations of (a) the poles at \(s = - 4\), and (b) the feedback gain, \(1/R\).
P7.22 Determine the root sensitivity of the dominant roots of Problem P7.1(a) when \(K\) is set so that the damping ratio of the unperturbed roots is \(\zeta = 0.707\). Evaluate and compare the sensitivity as a function of the poles and zeros of the loop transfer function \(L(s) = G_{c}(s)G(s)\). tion \(L(s) = G_{c}(s)G(s)\) of Problem P7.1(c).
P7.24 For systems of relatively high degree, the form of the root locus can often assume an unexpected pattern. The root loci of four different feedback systems of third order or higher are shown in Figure P7.24. The open-loop poles and zeros of \(KG(s)\) are shown, and the form of the root loci as \(K\) varies from zero to infinity is presented. Verify the diagrams of Figure P7.24 by constructing the root loci.
P7.25 Solid-state integrated electronic circuits are composed of distributed \(R\) and \(C\) elements. Therefore, feedback electronic circuits in integrated circuit form must be investigated by obtaining the transfer function of the distributed \(RC\) networks. It has been shown that the slope of the attenuation curve of a distributed \(RC\) network is \(10n\text{ }dB/\) decade, where \(n\) is the order of
(a)
(b)
FIGURE P7.24
Root loci of four systems.
(c)
(d) the \(RC\) filter [13]. This attenuation is in contrast with the normal \(20n\text{ }dB/\) decade for the lumped parameter circuits. An interesting case arises when the distributed \(RC\) network occurs in a series-to-shunt feedback path of a transistor amplifier. Then the loop transfer function may be written as
(a) Using the root locus method, determine the locus of roots as \(K\) varies from zero to infinity. (b) Calculate the gain at borderline stability and the frequency of oscillation for this gain.
P7.26 A unity feedback system has a loop transfer function
(a) Sketch the root locus for \(0 \leq K < \infty\). (b) Determine the range of the gain \(K\) for which the system is stable. (c) For what value of \(K\) in the range \(K \geq 0\) do purely imaginary roots exist? What are the values of these roots? (d) Would the use of the dominant roots approximation for an estimate of settling time be justified in this case for a large magnitude of gain \((K = 100)\) ?
P7.27 A unity feedback system has a loop transfer function
Sketch the root locus as a function of \(K\). Determine the values of \(K\) where the root locus enters and leaves the real axis.
P7.28 To meet current U.S. emissions standards for automobiles, hydrocarbon (HC) and carbon monoxide (CO) emissions are usually controlled by a catalytic converter in the automobile exhaust. Federal standards for nitrogen oxides \(\left( {NO}_{x} \right)\) emissions are met mainly by exhaust-gas recirculation (EGR) techniques.
Although many schemes are under investigation for meeting the emissions standards for all three emissions, one of the most promising employs a three- way catalyst - for \(HC,CO\), and \({NO}_{x}\) emissions in conjunction with a closed-loop engine-control system. The approach is to use a closed-loop engine control, as shown in Figure P7.28 [19, 23]. The exhaust-gas sensor gives an indication of a rich or lean exhaust and compares it to a reference. The difference signal is processed by the controller, and the output of the controller modulates the vacuum level in the carburetor to achieve the best air-fuel ratio for proper operation of the catalytic converter. The loop transfer function is represented by
Calculate the root locus as a function of \(K\). Calculate where the segments of the locus enter and leave the real axis. Determine the roots when \(K = 2\). Predict the step response of the system when \(K = 2\).
P7.29 A unity feedback control system has the loop transfer function
We desire the dominant roots to have a damping ratio \(\zeta = 0.707\). Find the gain \(K\) when this condition is satisfied. Show that the complex roots area \(s = - 11.1 \pm j11.1\) at this gain.
P7.30 An \(RLC\) network is shown in Figure P7.30. The nominal values (normalized) of the network elements are \(L - C = 1\) and \(R = 2.5\). Show that the root sensitivity of the two roots of the input impedance \(Z(s)\) to a change in \(R\) is different by a factor of 4 .
FIGURE P7.30 RLC network.
FIGURE P7.28
Auto engine control. P7.31 The development of high-speed aircraft and missiles requires information about aerodynamic parameters prevailing at very high speeds. Wind tunnels are used to test these parameters. These wind tunnels are constructed by compressing air to very high pressures and releasing it through a valve to create a wind. Since the air pressure drops as the air escapes, it is necessary to open the valve wider to maintain a constant wind speed. Thus, a control system is needed to adjust the valve to maintain a constant wind speed. The loop transfer function for a unity feedback system is
Sketch the root locus and show the location of the roots for \(K = 1391\).
P7.32 A mobile robot suitable for nighttime guard duty is available. This guard never sleeps and can tirelessly patrol large warehouses and outdoor yards. The steering control system for the mobile robot has a unity feedback with the loop transfer function
(a) Find \(K\) for all breakaway and entry points on the real axis. (b) Find \(K\) when the damping ratio of the complex roots is 0.707 . (c) Find the minimum value of the damping ratio for the complex roots and the associated gain \(K\). (d) Find the overshoot and the time to settle (to within \(2\%\) of the final value) for a unit step input for the gain, \(K\), determined in parts (b) and (c).
P7.33 The Bell-Boeing V-22 Osprey Tiltrotor is both an airplane and a helicopter. Its advantage is the ability to rotate its engines to \(90^{\circ}\) from a vertical position for takeoffs and landings as shown in Figure P7.33(a), and then to switch the engines to a horizontal position for cruising as an airplane [20]. The altitude control system in the helicopter mode is shown in Figure P7.33(b). (a) Determine the root locus as \(K\) varies and determine the range of \(K\) for a stable system. (b) For \(K = 280\), find the actual \(y(t)\) for a unit step input \(r(t)\) and the percentage overshoot and settling time (with a \(2\%\) criterion). (c) When \(K = 280\) and \(r(t) = 0\), find \(y(t)\) for a unit step disturbance, \(T_{d}(s) = 1/s\).
P7.34 The fuel control for an automobile uses a diesel pump that is subject to parameter variations. A unity negative feedback has a loop transfer function
(a) Sketch the root locus as \(K\) varies from 0 to 2000 . (b) Find the roots for \(K\) equal to 400, 500, and 600 . (c) Predict how the percent overshoot to a step will vary for the gain \(K\), assuming dominant roots. (d) Find the actual time response for a step input for all three gains and compare the actual overshoot with the predicted overshoot.
P7.35 A powerful electrohydraulic forklift can be used to lift pallets weighing several tons on top of 35-foot scaffolds at a construction site. The unity feedback system has a loop transfer function
(a)
FIGURE P7.33
(a) Osprey Tiltrotor aircraft. (b) Its control system.
(b) (a) Sketch the root locus for \(K > 0\). (b) Find the gain \(K\) when two complex roots have a \(\zeta = 0.707\), and calculate all three roots. (c) Find the entry point of the root locus at the real axis. (d) Estimate the expected percent overshoot to a step input, and compare it with the actual percent overshoot.
P7.36 A microrobot with a high-performance manipulator has been designed for testing very small particles, such as simple living cells [6]. The unity feedback system has a loop transfer function
(a) Sketch the root locus for \(K > 0\). (b) Find the gain and roots when the characteristic equation has two imaginary roots. (c) Determine the characteristic roots when \(K = 50\) and \(K = 100\). (d) For \(K = 50\), estimate the percent overshoot to a step input, and compare the estimate to the actual percent overshoot.
P7.37 Identify the parameters \(K,a\), and \(b\) of the system shown in Figure P7.37. The system is subject to a unit step input, and the output response has a percent overshoot but ultimately attains the final value of 1 . When the closed-loop system is subjected to a ramp input, the output response follows the ramp input with a finite steady-state error. When the gain is doubled to \(2K\), the output response to an impulse input is a pure sinusoid with a period of 0.314 second. Determine \(K\), \(a\), and \(b\).
P7.38 A unity feedback system has the loop transfer function
FIGURE P7.37 Feedback system.
(a) Determine the range of \(K\) so that the closedloop system is stable. (b) Sketch the root locus. (c) Determine the roots for \(K = 100\). (d) For \(K = 100\), predict the percent overshoot for a step input. (e) Determine the actual percent overshoot.
P7.39 High-speed trains for U.S. railroad tracks must traverse twists and turns. In conventional trains, the axles are fixed in steel frames called trucks. The trucks pivot as the train goes into a curve, but the fixed axles stay parallel to each other, even though the front axle tends to go in a different direction from the rear axle [24]. If the train is going fast, it may jump the tracks. One solution uses axles that pivot independently. To counterbalance the strong centrifugal forces in a curve, the train also has a computerized hydraulic system that tilts each car as it rounds a turn. On-board sensors calculate the train's speed and the sharpness of the curve and feed this information to hydraulic pumps under the floor of each car. The pumps tilt the car up to eight degrees, causing it to lean into the curve like a race car on a banked track.
The tilt control system is shown in Figure P7.39. Sketch the root locus, and determine the value of \(K\) when the complex roots have maximum damping. Predict the response of this system to a step input \(R(s)\).
FIGURE P7.39
Tilt control for a high-speed train.
261. ADVANCED PROBLEMS
AP7.1 The top view of a high-performance jet aircraft is shown in Figure AP7.1(a) [20]. Using the block diagram in Figure AP7.1(b), sketch the root locus and determine the gain \(K\) so that the damping ratio of the complex poles near the \(j\omega\)-axis is the maximum achievable. Evaluate the roots at this \(K\), and predict the response to a step input. Determine the actual response, and compare it to the predicted response.
AP7.2 A magnetically levitated high-speed train "flies" on an air gap above its rail system, as shown in Figure AP7.2(a) [24]. The air gap control system has a unity feedback system with a loop transfer function
FIGURE AP7.1
(a) Highperformance aircraft. (b) Pitch control system. (a)
(b)
The feedback control system is illustrated in Figure AP7.2(b). The goal is to select \(K\) so that the response for a unit step input is reasonably damped. Sketch the root locus, and select \(K\) so that the \(T_{s} \leq 3\text{ }s\) and P.O. \(\leq 20\%\). Determine the actual response for the selected \(K\) and the percent overshoot.
AP7.3 A compact disc player for portable use requires a good rejection of disturbances and an accurate position of the optical reader sensor. The position control system uses unity feedback and a loop transfer function
(a)
(b)
(b)
FIGURE AP7.2
(a) Magnetically levitated highspeed train.
(b) Feedback control system. The parameter \(p\) can be chosen by selecting the appropriate DC motor. Sketch the root locus as a function of \(p\). Select \(p\) so that the damping ratio of the complex roots of the characteristic equation is approximately \(\zeta = 1/\sqrt{2}\).
AP7.4 A remote manipulator control system has unity feedback and a loop transfer function
We want the percent overshoot for a step input to be less than or equal to \(30\%\). Sketch the root locus as a function of the parameter \(\alpha\). Determine the range of \(\alpha\) required for the desired percent overshoot. Locate the roots for the allowable value of \(\alpha\) to achieve the required maximum overshoot. Check the actual step responses of the system for various \(\alpha\) values.
AP7.5 A unity feedback system has a loop transfer function
a. Sketch the root locus and determine \(K\) for a stable system with complex roots with \(\zeta = 1/\sqrt{2}\).
b. Determine the root sensitivity of the complex roots of part (a).
c. Determine the percent change in \(K\) (increase or decrease) so that the roots lie on the \(j\omega\)-axis.
AP7.6 A unity feedback system has a loop transfer function
Sketch the root locus for \(K > 0\), and select a value for \(K\) that will maximize the damping ratio of the complex roots.
AP7.7 A feedback system with positive feedback is shown in Figure AP7.7. The root locus for \(K > 0\) must meet the condition
Sketch the root locus for \(0 < K < \infty\).
FIGURE AP7.7 A closed-loop system with positive feedback.
AP7.8 A position control system for a DC motor is shown in Figure AP7.8. Obtain the root locus for the velocity feedback constant \(K\), and select \(K\) so that all the roots of the characteristic equation are real (two are equal and real). Estimate the step response of the system for the \(K\) selected. Compare the estimate with the actual response.
AP7.9 A control system is shown in Figure AP7.9. Sketch the root loci for the following transfer functions \(G_{c}(s)\) :
a. \(G_{c}(s) = K\)
b. \(G_{c}(s) = K(s + 3)\)
FIGURE AP7.8
A position control system with velocity feedback.
FIGURE AP7.9
A unity feedback control system.
c. \(G_{c}(s) = \frac{K(s + 1)}{s + 20}\)
d. \(G_{c}(s) = \frac{K(s + 1)(s + 4)}{s + 10}\)
AP7.10 A feedback system is shown in Figure AP7.10. Sketch the root locus as \(K\) varies when \(K \geq 0\). Determine a value for \(K\) that will provide a step response with a percent overshoot of P.O. \(\leq 5\%\) and a settling time (with a \(2\%\) criterion) of \(T_{s} \leq 2.5\text{ }s\).
AP7.11 A control system is shown in Figure AP7.11. Sketch the root locus, and select a gain \(K\) so that the step response of the system has a percent overshoot of P.O. \(\leq 5\%\), and the settling time (with a \(2\%\) criterion) is \(T_{s} \leq 10\text{ }s\).
AP7.12 A control system with PI control is shown in Figure AP7.12. (a) Let \(K_{I}/K_{P} = 0.2\) and determine \(K_{P}\) so that the complex roots have maximum damping ratio. (b) Predict the step response of the system with \(K_{P}\) set to the value determined in part (a).
AP7.13 The feedback system shown in Figure AP7.13 has two unknown parameters \(K_{1}\) and \(K_{2}\). The process transfer function is unstable. Sketch the root locus for \(0 \leq K_{1},K_{2} < \infty\). What is the fastest settling time that you would expect of the closed-loop system in response to a unit step input \(R(s) = 1/s\) ? Explain.
FIGURE AP7.10
A nonunity feedback control system.
FIGURE AP7.11
A control system with parameter \(K\).
FIGURE AP7.12 A control system with a PI controller.
FIGURE AP7.13
An unstable plant with two parameters \(K_{1}\) and \(K_{2}\).
(a)
(b) AP7.14 Consider the unity feedback control system shown in Figure AP7.14. Design a PID controller using Ziegler-Nichols methods. Determine the unit step response and the unit disturbance response. What is the maximum percent overshoot and settling time for the unit step input?
FIGURE AP7.14
Unity feedback loop with PID controller.
262. DESIGN PROBLEMS
CDP7.1 The drive motor and slide system uses the output of a tachometer mounted on the shaft of the motor as shown in Figure CDP4.1 (switch-closed option). The output voltage of the tachometer is \(v_{T} = K_{1}\theta\). Use the velocity feedback with the adjustable gain \(K_{1}\). Select the best values for the gain \(K_{1}\) and the amplifier gain \(K_{a}\) so that the transient response to a step input has a percent overshoot of P.O. \(\leq 5\%\) and a settling time (to within \(2\%\) of the final value) of \(T_{s} \leq 300\text{ }ms\).
DP7.1 A high-performance aircraft, shown in Figure DP7.1(a), uses the ailerons, rudder, and elevator to steer through a three-dimensional flight path [20]. The pitch rate control system for a fighter aircraft at 10,000 \(m\) and Mach 0.9 can be represented by the system in Figure DP7.1(b). (a) Sketch the root locus when the controller is a gain, so that \(G_{c}(s) = K\), and determine \(K\) when \(\zeta\) for the roots with \(\omega_{n} > 2\) is \(\zeta \geq 0.15\) (seek a maximum \(\zeta\) ). (b) Plot the response \(q(t)\) for a step input \(r(t)\) with \(K\) as in (a). (c) A designer suggests an anticipatory controller with \(G_{c}(s) = K_{1} + K_{2}s = K(s + 2)\). Sketch the root locus for this system as \(K\) varies and determine a \(K\) so that the damping ratio of all the closedloop roots is \(\zeta > 0.8\). (d) Plot the response \(q(t)\) for a step input \(r(t)\) with \(K\) as in (c).
DP7.2 A large helicopter uses two tandem rotors rotating in opposite directions, as shown in Figure P7.33(a). The controller adjusts the tilt angle of the main rotor and thus the forward motion as shown in Figure DP7.2.
(a)
263. FIGURE DP7.1
(a) Highperformance aircraft. (b) Pitch rate control system.
(b) FIGURE DP7.2
Two-rotor helicopter velocity control.
(a) Sketch the root locus of the system, and determine \(K\) when \(\zeta\) of the complex roots is equal to 0.707. (b) Plot the response of the system to a step input \(r(t)\), and find the settling time (with a \(2\%\) criterion) and percent overshoot for the system of part (a). (c) Repeat parts (a) and (b) when the damping ratio of the complex roots is \(\zeta = 0.3\). Compare the results with those obtained in parts (a) and (b).
DP7.3 A rover vehicle has been designed for maneuvering at \(0.25mph\) over Martian terrain. Because Mars is 189 million miles from Earth, and it would take up to 40 minutes each way to communicate with Earth \(\lbrack 22,27\rbrack\), the rover must act independently and reliably. Resembling a cross between a small flatbed truck and an elevated jeep, the rover is constructed of three articulated sections, each with its own two independent, axle-bearing, one-meter conical wheels. A pair of sampling arms - one for chipping and drilling, the other for manipulating fine objects-extend from its front end like pincers. The control of the arms can be represented by the system shown in Figure DP7.3. (a) Sketch the root locus for \(K\), and identify the roots for \(K = 2\) and 25. (b) Determine the gain \(K\) that results in a percent overshoot to a step of \(P.O. = 1\%\). (c) Determine the gain that minimizes the settling time (with a \(2\%\) criterion) while maintaining a percent overshoot of P.O. \(\leq 1\%\).
DP7.4 A welding torch is remotely controlled to achieve high accuracy while operating in changing and hazardous environments [21]. A model of the welding arm position control is shown in Figure DP7.4, with the disturbance representing the environmental changes. (a) With \(T_{d}(s) = 0\), select \(K_{1}\) and \(K\) to provide highquality performance of the position control system. Select a set of performance criteria, and examine the results of your design. (b) For the system in part (a), let \(R(s) = 0\) and determine the effect of a unit step \(T_{d}(s) = 1/s\) by obtaining \(y(t)\).
DP7.5 A high-performance jet aircraft with an autopilot control system has a unity feedback and control system, as shown in Figure DP7.5. Sketch the root locus and select a gain \(K\) that leads to dominant poles. With this gain \(K\), predict the step response of the system. Determine the actual response of the system, and compare it to the predicted response.
FIGURE DP7.3
Mars vehicle robot control system.
FIGURE DP7.4 Remotely controlled welder. FIGURE DP7.5
High-performance jet aircraft.
FIGURE DP7.6
Automatic control of walking motion.
DP7.6 A system to aid and control the walk of a partially disabled person could use automatic control of the walking motion [25]. One model of a system is shown in Figure DP7.6. Using the root locus, select \(K\) for the maximum achievable damping ratio of the complex roots. Predict the step response of the system, and compare it with the actual step response.
DP7.7 A mobile robot using a vision system as the measurement device is shown in Figure DP7.7(a) [36]. The control system is shown in Figure DP7.7(b). Design the controller so that (a) the percent overshoot for a step input is \(P.O. \leq 5\%\); (b) the settling time (with a \(2\%\) criterion) is \(T_{s} \leq 6\text{ }s\); (c) the system velocity error constant \(K_{v} > 0.9\); and (d) the peak time, \(T_{P}\), for a step input is minimized.
DP7.8 Most commercial op-amps are designed to be unity-gain stable [26]. That is, they are stable when used in a unity-gain configuration. To achieve higher bandwidth, some op-amps relax the requirement to be unity-gain stable. One such amplifier has a DC gain of \(10^{5}\) and a
FIGURE DP7.7
(a) A robot and vision system.
(b) Feedback control system.
(a)
(b) FIGURE DP7.8
(a) Op-amp circuit.
(b) Control system.
(a)
(b) bandwidth of \(10kHz\). The amplifier, \(G(s)\), is connected in the feedback circuit shown in Figure DP7.8(a). The amplifier is represented by the model shown in Figure DP7.8(b), where \(K_{a} = 10^{5}\). Sketch the root locus of the system for \(K\). Determine the minimum value of the DC gain of the closed-loop amplifier for stability. Select a DC gain and the resistors \(R_{1}\) and \(R_{2}\).
DP7.9 A robotic arm actuated at the elbow joint is shown in Figure DP7.9(a), and the control system for the actuator is shown in Figure DP7.9(b). Plot the root locus for \(K \geq 0\). Select \(G_{p}(s)\) so that the steady-state error for a step input is equal to zero. Using the \(G_{p}(s)\) selected, plot \(y(t)\) for \(K\) equal to 1,1.75, and 3.0. Record the rise time, settling time (with a \(2\%\) criterion), and percent overshoot for the three gains. We wish to limit the overshoot to P.O. \(\leq 6\%\) while achieving the shortest rise time possible. Select the best system for \(1 \leq K \leq 3.0\).
DP7.10 The four-wheel-steering automobile has several benefits. The system gives the driver a greater degree of control over the automobile. The driver gets a more forgiving vehicle over a wide variety of conditions. The system enables the driver to make sharp, smooth
(a)
FIGURE DP7.9
(a) A robotic arm actuated at the joint elbow. (b) Its control system.
(b) lane transitions. It also prevents yaw, which is the swaying of the rear end during sudden movements. Furthermore, the four-wheel-steering system gives a car increased maneuverability. This enables the driver to park the car in extremely tight quarters. With additional closed-loop computer operating systems, a car could be prevented from sliding out of control in abnormal icy or wet road conditions.
The system works by moving the rear wheels relative to the front-wheel-steering angle. The control system takes information about the front wheels' steering angle and passes it to the actuator in the back. This actuator then moves the rear wheels appropriately.
When the rear wheels are given a steering angle relative to the front ones, the vehicle can vary its lateral acceleration response according to the loop transfer function
\(L(s) = G_{c}(s)G(s) = K\frac{1 + (1 + \lambda)T_{1}s + (1 + \lambda)T_{2}s^{2}}{s\left\lbrack 1 + \left( 2\zeta/\omega_{n} \right)s + \left( 1/\omega_{n}\ ^{1} \right)s^{2} \right\rbrack}\),
where \(\lambda = 2q/(1 - q)\), and \(q\) is the ratio of rear wheel angle to front wheel steering angle [14]. We will assume that \(T_{1} = T_{2} = 1\) second and \(\omega_{n} = 4\). Design a unity feedback system, selecting an appropriate set of parameters \((\lambda,K,\zeta)\) so that the steering control response is rapid and yet will yield modest overshoot characteristics. In addition, \(q\) must be between 0 and 1 .
DP7.11 A pilot crane control is shown in Figure DP7.11(a). The trolley is moved by an input \(F(t)\) in order to control \(x(t)\) and \(\phi(t)\) [13]. The model of the pilot crane control is shown in Figure DP7.11(b). Design a controller that will achieve zero steady-state error for ramp inputs, and maximize the closed-loop system damping when \(G_{c}(s) = K\left( 1 + \frac{0.25}{s} \right)\).
DP7.12 A rover vehicle designed for use on other planets and moons is shown in Figure DP7.12(a) [21]. The block diagram of the steering control is shown in Figure DP7.12(b). (a) Sketch the root locus as \(K\) varies from 0 to 10000 . Find the roots for \(K\) equal to 1000 , 1500, and 2500. Predict the overshoot, settling time (with a \(2\%\) criterion), and steady-state error for a step input, assuming dominant roots. (c) Determine the actual time response for a step input for the three values of the gain \(K\), and compare the actual results with the predicted results.
DP7.13 The automatic control of an airplane is one example that requires multiple-variable feedback methods. In this system, the attitude of an aircraft is controlled by three sets of surfaces: elevators, a rudder, and ailerons, as shown in Figure DP7.13(a). By manipulating these surfaces, a pilot can set the aircraft on a desired flight path [20].
An autopilot, which will be considered here, is an automatic control system that controls the roll angle \(\phi(t)\) by adjusting aileron surfaces. The deflection of the aileron surfaces by an angle \(\theta(t)\) generates a torque due to air pressure on these surfaces. This causes a rolling motion of the aircraft. The aileron surfaces are controlled by a hydraulic actuator with a transfer function \(1/s\).
(a)
FIGURE DP7.11
(a) Pilot crane control system.
(b) Block diagram.
(b) FIGURE DP7.12
(a) Planetary rover vehicle. (b) Steering control system.
FIGURE DP7.13
(a) An airplane with a set of ailerons.
(b) The block diagram for controlling the roll rate of the airplane.
(a)
(b)
(a)
(b)
The actual roll angle \(\phi(t)\) is measured and compared with the input. The difference between the desired roll angle \(\phi_{d}(t)\) and the actual angle \(\phi(t)\) will drive the hydraulic actuator, which in turn adjusts the deflection of the aileron surface.
A simplified model where the rolling motion can be considered independent of other motions is assumed, and its block diagram is shown in Figure DP7.13(b). Assume that the roll rate \(\overset{˙}{\phi}(t)\) is fed back using a rate gyro. We desire a zero steady-state tracking error to a unit step. The step response desired has a percent overshoot P.O. \(\leq 15\%\) and a settling time (with a \(2\%\) criterion) of \(T_{s} \leq 25\text{ }s\). Select the parameters \(K_{1}\) and \(K_{2}\).
DP7.14 Consider the feedback system shown in Figure DP7.14. The process transfer function is marginally stable. The controller is the proportional-derivative (PD) controller.
a. Determine the characteristic equation of the closed-loop system.
b. Let \(\tau = K_{P}/K_{D}\). Write the characteristic equation in the form
c. Plot the root locus for \(0 \leq K_{D} < \propto\) when \(\tau = 6\).
d. What is the effect on the root locus when \(0 < \tau < \sqrt{10}\) ?
e. Design the PD controller to meet the following specifications:
(i) \(P.O. \leq 5\%\)
(ii) \(T_{s} \leq 1\text{ }s\)
FIGURE DP7.14
A marginally stable plant with a PD controller in the loop.
264. COMPUTER PROBLEMS
CP7.1 Using the rlocus function, obtain the root locus for the following transfer functions of the system shown in Figure CP7.1 when \(0 \leq K < \infty\) :
a. \(G(s) = \frac{25}{s^{3} + 10s^{2} + 40s + 25}\)
b. \(G(s) = \frac{s + 10}{s^{2} + 2s + 10}\)
c. \(G(s) = \frac{s^{2} + 2s + 4}{s\left( s^{2} + 5s + 10 \right)}\)
c. \(G(s) = \frac{s^{5} + 6s^{4} + 6s^{3} + 12s^{2} + 6s + 4}{s^{6} + 4s^{5} + 5s^{4} + s^{3} + s^{2} + 12s + 1}\)
265. FIGURE CP7.1
A single-loop feedback system with parameter \(K\).
CP7.2 A unity negative feedback system has the loop transfer function
Develop an m-file to plot the root locus, and show with the rlocfind function that the maximum value of \(K\) for a stable system is \(K = 92.7\).
CP7.3 Compute the partial fraction expansion of
and verify the result using the residue function.
CP7.4 A unity negative feedback system has the loop transfer function
Develop an m-file to obtain the root locus as \(p\) varies; \(0 < p < \infty\). For what values of \(p\) is the closed-loop stable?
CP7.5 Consider the unity feedback system with the loop transfer function
For what value of \(K\) is the step response to a unit step such that the percent overshoot, P.O. \(< 5\%\) ? Show the step response and confirm the perfomance specification is satisfied.
CP7.6 A large antenna, as shown in Figure CP7.6(a), is used to receive satellite signals and must accurately track the satellite as it moves across the sky. The control system uses an armature-controlled motor and a controller to be selected, as shown in Figure CP7.6(b). The system specifications require a zero steady-state error for a ramp input. We also seek a FIGURE CP7.6
Antenna position control.
FIGURE CP7.7
A single-loop feedback control system with controller \(G_{c}(s)\).
(a)
(b)
percent overshoot to a step input of P.O. \(\leq 3\%\) with a settling time (with a \(2\%\) criterion) of \(T_{s} \leq 1\text{ }s\). Using root locus methods, create an m-file to assist in designing the controller. Plot the resulting unit step response and compute the percent overshoot and the settling time and label the plot accordingly. Determine the effect of the disturbance \(T_{d}(s) = Q/s\) (where \(Q\) is a constant) on the output \(Y(s)\). Draw the system output \(y(t)\) when \(Q = 1\).
CP7.7 Consider the feedback control system in Figure CP7.7. We have three potential controllers for our system:
-
\(G_{c}(s) = K\) (proportional controller)
-
\(G_{c}(s) = K/s\) (integral controller)
-
\(G_{c}(s) = K(1 + 1/s)\) (proportional, integral (PI) controller).
The design specifications are \(T_{s} \leq 10\text{ }s\) and P.O. \(\leq 10\%\) for a unit step input. a. For the proportional controller, develop an m-file to sketch the root locus for \(0 < K < \infty\), and determine the value of \(K\) so that the design specifications are satisfied.
b. Repeat part (a) for the integral controller.
c. Repeat part (a) for the PI controller.
d. Co-plot the unit step responses for the closedloop systems with each controller designed in parts (a)-(c).
e. Compare and contrast the three controllers obtained in parts (a)-(c), concentrating on the steady-state errors and transient performance.
CP7.8 Consider the spacecraft single-axis attitude control system shown in Figure CP7.8. The controller is known as a proportional-derivative (PD) controller. Suppose that we require the ratio of \(K_{p}/K_{D} = 12\). Then, develop an m-file using root locus methods find the values of \(K_{D}/J\) and \(K_{p}/J\) so that the settling time \(T_{s}\) is \(T_{s} \leq 2\text{ }s\), and the peak overshoot is FIGURE CP7.8
A spacecraft attitude control system with a proportionalderivative controller.
\(\leq 5\%\) for a unit step input. Use a \(2\%\) criterion to determine the settling time.
CP7.9 Consider the feedback control system in Figure CP7.9. Develop an \(m\)-file to plot the root locus for \(0 < K < \infty\). Find the value of \(K\) resulting in a damping ratio of the closed-loop poles equal to \(\zeta = 0.707\).
CP7.10 Consider the system represented in state variable form
where
(a) Determine the characteristic equation. (b) Using the Routh-Hurwitz criterion, determine the values of \(k\) for which the system is stable. (c) Develop an m-file to plot the root locus, and compare the results to those obtained in (b).
FIGURE CP7.9
Unity feedback
system with
parameter \(K\).
266. ANSWERS TO SKILLS CHECK
True or False: (1) True; (2) True; (3) False; (4) True; (5) True
Multiple Choice: (6) b; (7) c; (8) a; (9) c; (10) a; (11) b; (12) c; (13) a; (14) c; (15) b
Word Match (in order, top to bottom): n, k, f, o, a, d, l, \(i,h,c,b,e,m,g,j\)
267. TERMS AND CONCEPTS
Angle of departure The angle at which a locus leaves a complex pole in the \(s\)-plane.
Angle of the asymptotes The angle \(\phi_{A}\) that the asymptote makes with respect to the real axis.
Asymptote The path the root locus follows as the parameter becomes very large and approaches infinity. The number of asymptotes is equal to the number of poles minus the number of zeros.
Asymptote centroid The center \(\sigma_{A}\) of the linear asymptotes.
Breakaway point The point on the real axis where the locus departs from the real axis of the \(s\)-plane.
Dominant roots The roots of the characteristic equation that represent or dominate the closed-loop transient response.
Locus A path or trajectory that is traced out as a parameter is changed.
Logarithmic sensitivity A measure of the sensitivity of the system performance to specific parameter changes, given by \(S_{K}^{T}(s) = \frac{\partial T(s)/T(s)}{\partial K/K}\), where \(T(s)\) is the system transfer function and \(K\) is the parameter of interest.
Manual PID tuning methods The process of determining the PID controller gains by trial-and-error with minimal analytic analysis.
Negative gain root locus The root locus for negative values of the parameter of interest, where \(- \infty < K \leq 0\).
Number of separate loci Equal to the number of poles of the transfer function, assuming that the number of poles is greater than or equal to the number of zeros of the transfer function.
Parameter design A method of selecting one or two parameters using the root locus method.
PID controller A widely used controller used in industry of the form \(G_{c}(s) = K_{p} + \frac{K_{I}}{s} + K_{D}s\), where \(K_{p}\) is the proportional gain, \(K_{I}\) is the integral gain, and \(K_{D}\) is the derivative gain.
PID tuning The process of determining the PID controller gains.
Proportional plus derivative (PD) controller A twoterm controller of the form \(G_{c}(s) = K_{p} + K_{D}s\), where \(K_{p}\) is the proportional gain and \(K_{D}\) is the derivative gain.
Proportional plus integral (PI) controller A two-term controller of the form \(G_{c}(s) = K_{p} + \frac{K_{I}}{s}\), where \(K_{p}\) is the proportional gain and \(K_{I}\) is the integral gain.
Quarter amplitude decay The amplitude of the closedloop response is reduced approximately to onefourth of the maximum value in one oscillatory period.
Reaction curve The response obtained by taking the controller off-line and introducing a step input. The underlying process is assumed to be a first-order system with a transport delay.
Root contours The family of loci that depict the effect of varying two parameters on the roots of the characteristic equation.
Root locus The locus or path of the roots traced out on the \(s\)-plane as a parameter is changed.
Root locus method The method for determining the locus of roots of the characteristic equation \(1 + KP(s) = 0\) as \(K\) varies from 0 to infinity.
Root locus segments on the real axis The root locus lying in a section of the real axis to the left of an odd number of poles and zeros.
Root sensitivity The sensitivity of the roots as a parameter changes from its normal value. The root sensitivity is given by \(S_{K}^{r} = \frac{\partial r}{\partial K/K}\), the incremental change in the root divided by the proportional change of the parameter.
Ultimate gain The \(PD\) controller proportional gain, \(K_{p}\), on the border of instability when \(K_{D} = 0\) and \(K_{I} = 0\).
Ultimate period The period of the sustained oscillations when \(K_{p}\) is the ultimate gain and \(K_{D} = 0\) and \(K_{I} = 0\).
Ziegler-Nichols PID tuning method The process of determining the PID controller gains using one of several analytic methods based on open-loop and closedloop responses to step inputs.
Frequency Response Methods
8.1 Introduction 546
8.2 Frequency Response Plots 548
8.3 Frequency Response Measurements 569
8.4 Performance Specifications in the Frequency Domain 571
8.5 Log-Magnitude and Phase Diagrams 574
8.6 Design Examples 575
8.7 Frequency Response Methods Using Control Design Software 584
8.8 Sequential Design Example: Disk Drive Read System 589
8.9 Summary 591
268. PREVIEW
In this chapter, we consider the steady-state response of a system to a sinusoidal input test signal. We will see that the response of a linear constant coefficient system to a sinusoidal input signal is an output sinusoidal signal at the same frequency as the input. However, the magnitude and phase of the output signal differ from those of the input sinusoidal signal, and the amount of difference is a function of the input frequency. Thus, we will be investigating the steady-state response of the system to a sinusoidal input as the frequency varies.
We will examine the transfer function \(G(s)\) when \(s = j\omega\) and develop methods for graphically displaying the complex number \(G(j\omega)\) as \(\omega\) varies. The Bode plot is one of the most powerful graphical tools for analyzing and designing control systems, and we will cover that subject in this chapter. We will also consider polar plots and log-magnitude and phase diagrams. We will develop several time-domain performance measures in terms of the frequency response of the system, as well as introduce the concept of system bandwidth. The chapter concludes with a frequency response analysis of the Sequential Design Example: Disk Drive Read System.
269. DESIRED OUTCOMES
Upon completion of Chapter 8, students should be able to:
$\square\ $ Explain the concept of frequency response and its role in control system design.
$\square\ $ Sketch a Bode plot and also how to obtain a computer-generated Bode plot.
$\square\ $ Describe log-magnitude and phase diagrams.
$\square\ $ Identify performance specifications in the frequency domain and relative stability based on gain and phase margins.
$\square\ $ Design a controller to meet desired specifications using frequency response methods.
269.1. INTRODUCTION
A very practical and important approach to the analysis and design of a system is the frequency response method.
The frequency response of a system is defined as the steady-state response of the system to a sinusoidal input signal. The sinusoid is a unique input signal, and the resulting output signal for a linear system is sinusoidal in the steady state; it differs from the input only in amplitude and phase angle.
For example, consider the system \(Y(s) = T(s)R(s)\) with \(r(t) = Asin\omega t\). We have
and
where \(- p_{i}\) are assumed to be distinct poles. Then, in partial fraction form, we have
Taking the inverse Laplace transform yields
where \(\alpha\) and \(\beta\) are constants which are problem dependent. If the system is stable, then all \(p_{i}\) have positive real parts and
since each exponential term \(k_{i}e^{- p_{i}t}\) decays to zero as \(t \rightarrow \infty\).
In the limit for \(y(t)\), it can be shown, for \(t \rightarrow \infty\) (the steady state),
where \(\phi = /T(j\omega)\).
Thus, the steady-state output signal depends only on the magnitude and phase of \(T(j\omega)\) at a specific frequency \(\omega\). The steady-state response, as described in Equation (8.1), is true only for stable systems, \(T(s)\). One advantage of the frequency response method is the ready availability of sinusoid test signals for various ranges of frequencies and amplitudes. Thus, the experimental determination of the system frequency response is easily accomplished. The unknown transfer function of a system can often be deduced from the experimentally determined frequency response of a system [1,2]. Furthermore, the design of a system in the frequency domain provides the designer with control of the bandwidth of a system, as well as some measure of the response of the system to undesired noise and disturbances.
A second advantage of the frequency response method is that the transfer function describing the sinusoidal steady-state behavior of a system can be obtained by replacing \(s\) with \(j\omega\) in the system transfer function \(T(s)\). The transfer function representing the sinusoidal steady-state behavior of a system is then a function of the complex variable \(j\omega\) and is itself a complex function \(T(j\omega)\) that possesses a magnitude and phase angle. The magnitude and phase angle of \(T(j\omega)\) are readily represented by graphical plots that provide significant insight into the analysis and design of control systems.
The basic disadvantage of the frequency response method for analysis and design is the indirect link between the frequency and the time domain. Direct correlations between the frequency response and the corresponding transient response characteristics are somewhat tenuous, and in practice the frequency response characteristic is adjusted by using various design criteria that will normally result in a satisfactory transient response.
The Laplace transform pair is
and
where the complex variable \(s = \sigma + j\omega\). Similarly, the Fourier transform pair is
and
The Fourier transform exists for \(f(t)\) when
The Fourier and Laplace transforms are closely related, as we can see by examining Equations (8.2) and (8.4). When the function \(f(t)\) is defined only for \(t \geq 0\), as is often the case, the lower limits on the integrals are the same. Then we note that the two equations differ only in the complex variable. Thus, if the Laplace transform of a function \(f_{1}(t)\) is known to be \(F_{1}(s)\), we can obtain the Fourier transform of this same time function by setting \(s = j\omega\) in \(F_{1}(s)\) [3].
Again we might ask, Since the Fourier and Laplace transforms are so closely related, why not use the Laplace transform? Why use the Fourier transform at all? The Laplace transform enables us to investigate the \(s\)-plane location of the poles and zeros of a transfer function \(T(s)\). However, the frequency response method allows us to consider the transfer function \(T(j\omega)\) and to concern ourselves with the amplitude and phase characteristics of the system. This ability to investigate and represent the character of a system by amplitude, phase equations, and curves is an advantage for the analysis and design of control systems.
If we consider the frequency response of the closed-loop system, we might have an input that \(r(t)\) has a Fourier transform in the frequency domain as follows:
Then the output frequency response of a unity feedback control system can be obtained by substituting \(s = j\omega\) in the closed-loop system relationship, \(Y(s) = T(s)R(s)\), so that we have
Using the inverse Fourier transform, the output transient response would be
However, it is usually difficult to evaluate this inverse transform integral for all but the simplest systems, and a graphical integration may be used. Alternatively, as we will note in succeeding sections, several measures of the transient response can be related to the frequency characteristics and utilized for design purposes.
269.2. FREQUENCY RESPONSE PLOTS
The transfer function of a system \(G(s)\) can be described in the frequency domain by the relation
FIGURE 8.1
The polar plane.
