6.1 Sketch the z-domain root locus and find the critical gain for the following systems K., the closed-loop characteristic equation is K + z 0.

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1 6. Sketch the z-domain root locus and find the critical gain for the following systems K (i) Gz () z 4. (ii) Gz K () ( z+ 9. )( z 9. ) (iii) Gz () Kz ( z. )( z ) (iv) Gz () Kz ( + 9. ) ( z. )( z 8. ) (i) To sketch the z-domain root locus by hand, we apply the root locus rules. The plots can more easily be obtained using the MATLAB command rlocus. For Gz K () z 4., the closed-loop characteristic equation is K + z.4 At z, we obtain the critical gain, K cr.4 (see also the root locus plot) Root Locus.8.6 Imaginary Axis System: g Gain:.4 Pole: Damping:.755 Overshoot (%): 99.8 Frequency (rad/sec): (ii) For Gz Real Axis Figure 6- Root locus for Problem 6.(i) K () ( z+ 9. )( z 9. ), the closed-loop characteristic equation isk + z.8 On the unit circle, z and we obtain K cr.8

2 Root Locus.8 System: g Gain:.8 Pole: -8.3e-7 + i Damping: -.79 Overshoot (%): Frequency (rad/sec): 57 Imaginary Axis Real Axis Figure 6- Root locus for Problem 6.(ii) Kz (iii) For G() z, the closed-loop characteristic equation is ( z. )( z ) K z + z. z +.. On the unit circle, z, K and the critical gain is K cr.4 Root Locus Imaginary Axis System: g Gain:.4 Pole: - Damping: -.94 Overshoot (%): Frequency (rad/sec): Real Axis Figure 6-3 Root locus for Problem 6.(iii) (iv) For G() z Kz ( + 9. ) ( z. )( z 8. ), the closed-loop characteristic equation is

3 z z K z +.9 K. On the unit circle, z.9 K +.6 and we obtain K cr Root Locus Imaginary Axis.5 System: g Gain:.934 Pole: i Damping:.39 Overshoot (%): 99 Frequency (rad/sec): Real Axis Figure 6-4 Root locus for Problem 6.(iv)

4 6.5 Design proportional controllers for the systems of Problem 6. to meet the following specifications where possible. If the design specification cannot be met, explain why and suggest a more appropriate controller. (a) A damping ratio of.7. (b) A steady-state error of % due to a unit step. (c) A steady-state error of % due to a unit ramp. K (i) Gz () z 4. (a) A damping ratio of.7 This is a first order system and does not have an oscillatory response. Hence, the design specification cannot be met with proportional control.

5 A PI controller would give a system with the desired damping ratio but it is probably preferable to use proportional control to obtain a good time response with the appropriate time constant unless the steady-state error is required to be zero. (b) A steady-state error of % due to a unit step The position error constant is K p K G( ) 9.4 e( )% Hence, the gain for % error due to step is K 5.4. Unfortunately, the critical gain for the system is.4 and a gain of 5.4 would make the system unstable and the design specifications cannot be met with proportional control. Using a PD controller would allow us to increase the gain without causing instability. Alternatively, we could use a PI controller to reduce the error due to a step to zero, which would meet the desired specifications. (c) A steady-state error of % due to a unit ramp. The system is type and cannot track a ramp input. Hence, the design specification cannot be met with proportional control. A PI controller is needed for the system to become type and have a finite steady-state error due to ramp. (ii) K G ( z) ( z +.9)( z.9) (a) A damping ratio of.7 The closed-loop characteristic polynomial is ζωnt z.8+ K z cos ω T e z + e ζωnt ( ) d For a gain K>.8, we have imaginary poles so that the angle of the pole is ω T ω T ζ ω T.5 π / d n The square magnitude of the poles is ζω K.8 e n T e.4599 The required gain is K n Note that we only need the value of the product ω n T to solve the problem rather than each of the two values. Hence, the result can be verified with MATLAB using a unity sampling period and the command rlocus.

6 Root Locus Imaginary Axis System: g Gain:.856 Pole: -.54e-7 +.4i Damping:.7 Overshoot (%): 4.59 Frequency.7 (rad/sec): Real Axis Figure 6.5 Root locus for Problem 6.5 (ii) and the gain for a damping ratio of.7. (b) A steady-state error of % due to a unit step. The steady-state error is given by e ( )% + K p The error constant for the system is K p 9 K G() ( +.9)(.9) We solve for the gain K The specification can be met because the gain is less than the critical gain. (c) A steady-state error of % due to a unit ramp. The system is type and cannot track a ramp input. Hence, the design specification cannot be met with proportional control. A PI controller is needed for the system to become type and have a finite steady-state error due to ramp. (iii) K z G ( z) ( z.)( z ) (a) A damping ratio of.7 The closed-loop characteristic polynomial is

7 z + ζωnt ζωnt ( K.) z +. z cos( ω T ) e z + e Equating coefficients, we have e ζωn T ω T n cos..447 ln(. ) ( ω T ) cos ω T ζ cos(.496.5). 685 d K. cos n ( ω T ) d e ζω n T d K We obtain approximately the same answer using the MALAB command rlocus. Root Locus Imaginary Axis System: g Gain:.595 Pole: i Damping:.698 Overshoot (%): 4.66 Frequency (rad/sec): Real Axis Figure 6.6 Root locus for Problem 6.5 (iii) and the gain for a damping ratio of.7. (b) A steady-state error of % due to a unit step. The steady-state error is zero since the system is type I and the specification can be met for any stable gain. The gain can be selected as in Part (a) to obtain a satisfactory transient response. (c) A steady-state error of % due to a unit ramp. The system is type and can track a ramp input. The velocity error constant is K v ( z ) G( z) T K 8T z T K (.) The sampling period is typically sufficiently small to result in a small gain that is less than the critical value and the specifications can be met.

8 Kz ( + 9. ) (iv) Gz () ( z. )( z 8. ) (a) A damping ratio of.7 cos The closed-loop characteristic polynomial is z + K z K z cos ω dt Equating coefficients, we have e ζω ω T ln n n T ζωnt ζωnt ( ) ( ) e z + e.6 +.9K (.6 +.9K ). 4 ( ω T ) cos ω nt ζ cos ln(.6.9k ) d + (.5.4) ( ln.6 +.9K.5.4) (.6. K ) ζω T ( ω T ) e cos ( ) K cos + d n 9 This is a nonlinear equation in K and must be solved numerically. By trial and error, we obtain gain value K.434. Approximately, the same value can be obtained using the MATLAB command rlocus..5 Root Locus.5 System: g Gain:.4 Pole: i Damping:.699 Overshoot (%): 4.64 Frequency (rad/sec): Imaginary Axis Real Axis Figure 6.7 Root locus for Problem 6.5 (iv) and the gain for a damping ratio of.7.

9 (b) A steady-state error of % due to a unit step. The steady-state error is given by e ( )% + K p The error constant for the system is K p 9 K( +.9) G() (.)(.8) We solve for the gain K The specification can be met because the gain is less than the critical gain. (c) A steady-state error of % due to a unit ramp. The system is type and cannot track a ramp input. Hence, the design specification cannot be met with proportional control. A PI controller is needed for the system to become type and have a finite steady-state error due to ramp.

10 6.8 Design a digital PID controller (with T.) for the plant G( s) e s + 5s by applying the Ziegler-Nichols tuning rules of Table 5.. The Ziegler-Nichols tuning rules shown in Table 5. can be applied by considering K, τ, L5. We obtain K p.4, T i, T d.5. Thus, by applying (6.3) we have. z + z.4z C z) z. z + ( 4z z

11 6. Use direct control design for the system of Problem 5.7 (with T.), namely, design a controller for the transfer function G ( s) ( s + )( s + 5) to obtain (i) zero steady-state error due to step, (ii) a settling time of less than s, and (iii) an undamped natural frequency of 5 rad/s. Obtain the discretized and the analog output. Then, apply the designed controller to the system G ( s) ( s + )( s + 5)(.s + ) and obtain the discretized and the analog output in order to verify the robustness of the control system. First, we find the discretized process transfer function: ( z ) G( s) G ZAS ( z) Z s.47( z +.889) ( z.948)( z.665) Then, the desired analog characteristic polynomial is s ζω + ωn ns T s 4 ζω n where, according to the specifications, ω n 5 and, which implies that ζ.4. Thus, by taking into account that a zero steady-state error is required, the desired closed-loop transfer function is

12 G cl ( s) s 5 + 4s + 5 Then, the desired closed-loop transfer function is obtained by using z e st, namely, G cl ( z).9 z By applying (6.43) we have z +.469z C ( z).576( z.948)( z.665)( z + ) ( z )( z +.889)( z.5694) The obtained discretized and analog closed-loop system output are shown in Figure 6-6 and Figure 6-7 while the corresponding control variable is plotted in Figure 6-8. If the same controller is applied to the system G ( s) ( s + )( s + 5)(.s + ) the process output obtained is that in Figure 6-9. It can be seen that the additional lag causes a more significant overshoot and an increment of the settling time..4. process output time [s] Figure 6-6 Discretized process output for Problem 6.

13 6.3 Design a deadbeat controller for the system of Problem 5.7 to obtain perfect tracking of a unit step in minimum finite time. Obtain the analog output for the system and compare your design to that obtained in Problem 5.7. Apply then the controller to the process G ( s) ( s + )( s + 5)(.s + ) in order to verify the robustness of the control system. We consider the same discretized process transfer function of Problem 6., but in this case we set G cl ( z) z. Thus, by applying (6.43) we have C ( z) ( z.948)( z.665) ( z )( z +.889) The corresponding discretized and analog output are shown in Figure 6-3, the control variable in Figure 6-3. We observe that the deadbeat controller causes wide intersampling oscillations and requires a much higher control effort. It results in an increased overshoot without significantly decreasing the settling time. When the system G ( s) ( s + )( s + 5)(.s + ) is considered, the process output obtained is plotted in Error! Reference source not found...5 process output time [s] Figure 6-3 Discretized and analog output for Problem 6.3

14 3 control variable time [s] Figure 6-3 Control variable for Problem 6.3. process output time [s] Figure 6-3 Analog output for Problem 6.3 with uncertainty

15 6.8 To examine the effect of the sampling period on the relative stability and transient response of a digital control system, consider the system G ( s) ( s + )( s + 5) a) Obtain the transfer function of the system, the root locus and the critical gain for T.,.5,. s. Using the MATLAB commands zpk and cd, we obtain:

16 >> gzpk([],[-,-5],) Zero/pole/gain: (s+) (s+5) >> gdcd(g,.) Zero/pole/gain: 4.93e-5 (z+.98) (z-.99) (z-.95) Sampling time:. >> gdcd(g,.5) Zero/pole/gain:.37 (z+.949) (z-.95) (z-.7788) Sampling time:.5 >> gd3cd(g,.) Zero/pole/gain:.47 (z+.889) (z-.948) (z-.665) Sampling time:. Using the command >> [Gm,PM]margin(gd) We obtain the gain margin, which is the critical gain value. The values obtain are summarized in the following table. Sampling Period Zero Pole Pole K cr e b) Obtain the step response for each system at a gain of 5.

17 .4 Step Response. Amplitude Time (sec) Figure 6-4 Step response for sampling period T. (blue), T.5 (green), and T. s (red). c) Discuss the effect of the sampling period on the transient response and relative stability of the system based on your results from (a) and (b). The transfer functions for the different sampling period with generally faster poles for slower sampling. Faster poles are poles that are closer to the origin. The systems with slower sampling does not monitor the process as closely as ones with faster sampling. As a result, the step response of the system is more oscillatory.-

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