ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 1 Homework Assignment 3 Assigned September 30, 2008. Due in lecture October 7, 2008. Note that you must include all of your work to obtain full credit. Also the paper you submit must be your own work. To copy someone else s homework is cheating, which is not permitted and will result in a score of 0 for both the original and the copy. 1. Consider the close-loop transfer function T(s) = Y (s) R(s) = 5(s + 3) (s + 7)(s 2 + 8s + 36). (1) (a) If r(t) is a step input, find the steady-state error. Solution: The final value theorem indicates that Thus y ss = lim s 0 sy (s) = lim s 0 st(s)r(s). (2) ( ( ) 5(s + 3) 1 y ss = lim s s 0 (s + 7)(s 2 + 8s + 36)) s = 15/252 = 5/84. (3) The error is defined to be e(t) := y(t) r(t) so the steady state error is 79/84 (which would indicate that if we were actually trying to generate y ss = 1, we d need to change the gain). (b) Suppose that the real pole can be ignored as discussed in the section on the effects of a third pole and a zero on second order system response. Calculate the overshoot and 2% settling time for this second order system. Solution: Since 2ζω n = 8, we expect the 2% settling time to be 4/(ζω) = 1. Since ω n = 6, we find that ζ = 2/3. To calculate the percent overshoot we have to be a little careful. Using the formula in the text, the percent overshoot would be 100y ss e ζπ/ 1 ζ 2 = 6%. This does not match the observed value of 30% calculated from the actual response in the Matlab transcript below. The reason for this is the zero, which is not accounted for in the calculation in the textbook. Here s the transcript of my Matlab session. >> ps3p1 Actual System: Transfer function: 5 s + 15 ------------------------- s^3 + 15 s^2 + 92 s + 252 Max Value of y(t): 0.077143
ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 2 Steady State Value of y(t): 0.059524 Percent Overshoot: 0.29601 Max Value of y(t): 0.077143 Steady State Value of y(t): 0.059524 Reduced Order System: Transfer function: 5 s + 15 ------------------ 7 s^2 + 56 s + 252 omnz = 4 zeta = 0.6667 Percent Overshoot: 0.45629 Max Value of y2(t): 0.086684 Steady State Value of y2(t): 0.059524 2-percent settling time: 1.16 Expected settling time: 1 Expected Percent Overshoot: 6.0209 Here s the script I used. tofs = tf(5*[1 3],[1 7+8 56+36 7*36]); tofs2 = tf(5*[1 3],7*[1 8 36]); t = [0:.01:5]; y=step(tofs,t); y2=step(tofs2,t); figure(1) plot(t,y,t,y2) title( Problem 1: Step Response Comparison ) xlabel( t ) ylabel( y(t) ) legend( Original System, Reduced Order System ) grid disp( Actual System: ) tofs = tf(5*[1 3],[1 7+8 56+36 7*36]) disp([ Max Value of y(t):,num2str(max(y))]) disp([ Steady State Value of y(t):,num2str(y(max(size(y))))])
ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 3 ymax = max(y); yss = y(max(size(y))); disp([ Percent Overshoot:,num2str( (ymax-yss)/yss )]) ts2p = 0; for index = max(size(t)):-1:1 if ts2p == 0, if abs( (y(index)-yss)/yss ) > 0.02, ts2p = t(index+1); disp([ Max Value of y(t):,num2str(max(y))]) disp([ Steady State Value of y(t):,num2str(y(max(size(y))))]) disp( ) disp( Reduced Order System: ) tofs2 = tf(5*[1 3],7*[1 8 36]) omnz = 8/2 zeta = omnz/sqrt(36) y2max = max(y2); y2ss = y2(max(size(y2))); disp([ Percent Overshoot:,num2str( (y2max-y2ss)/y2ss )]) ts2p2 = 0; for index = max(size(t)):-1:1 if ts2p2 == 0, if abs( (y2(index)-y2ss)/y2ss ) > 0.02, ts2p2 = t(index+1); disp([ Max Value of y2(t):,num2str(max(y2))]) disp([ Steady State Value of y2(t):,num2str(y2(max(size(y2))))]) disp([ 2-percent settling time:,num2str(ts2p2)]) disp([ Expected settling time:,num2str(4/omnz)]) disp([ Expected Percent Overshoot:,num2str(100*exp((-pi*zeta)/sqrt(1-zeta^2)))]) print -deps ps3p1c.eps (c) Use Matlab to plot the response of the actual system and the second order system on the same (properly labelled) plot. Solution: The plot that was generated by the script above is shown in Figure 1c. (d) Now, consider T(s) = Y (s) R(s) = 5(s + 3) (s + ω)(s 2 + 8s + 36), (4) select an ω such that the criterion in the text is satisfied. Plot the response of the original and second order systems for this case.
ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 4 0.09 0.08 Problem 1: Step Response Comparison Original System Reduced Order System 0.07 0.06 0.05 y(t) 0.04 0.03 0.02 0.01 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 t
ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 5 Solution: The criterion in the text is 1/γ 10 ζω n where 1/γ is the ω that we are asked to find. Thus we need omega 40 so let s choose ω = 50. In order to get the same y ss we ll have to multiply the transfer function by 50. The step responses are compared in Figure 1d 0.014 0.012 Problem 1(d): Step Response Comparison Original System Reduced Order System 0.01 0.008 y(t) 0.006 0.004 0.002 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 t 2. Consider a unity negative feedback system with forward path (open loop) transfer function G(s) = K(s2 + 120s + 110) s 2. (5) + 8s + 7 Find the gain K that minimizes the damping ratio of the closed loop system poles and indicate the resulting damping ratio.
ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 6 Solution: The closed loop transfer function will be G(s)/(1 + G(s)) = = K(s 2 + 120s + 110) s 2 + 8s + 7 + K(s 2 + 120s + 110) (6) K(s 2 + 120s + 110) (1 + K)s 2 + (8 + 120K)s + 7 + 110K. (7) Thus the damping ratio is ( ) ( ( ) 8 + 120K 1 + K 1 8 + 120K ζ = = 1 + K 7 + 110K 2) 2. (8) (1 + K)(7 + 110K) We then must solve dζ/dk = 0 for K. After some algebra, dζ dk = 2(1535K + 93) (110K + 7) 3 (K + 1), (9) so K = 93/1535. To verify that we have minimized as opposed to maximized ζ we compute d 2 ζ/dk 2 K=K which turns out to be positive, indicating that ζ = 8 70/103 = 0.65 is the minimal damping ratio. 3. Consider the unity negative feedback system with forward path (open loop) transfer function ( ) ( ) K 15 G a (s)g d (s) =. (10) s + 7 s(s + 3) (This is a simple model of an aircraft aileron actuator and airplane dynamics.) (a) Find the closed loop transfer function. Solution: The closed loop transfer function will be G a (s)g d (s) 1 + G a (s)g d (s) = 15K s(s + 3)(s + 7) + 15K. (11) (b) Find a value of K for which the system has 3 real roots; a value for which the system has a complex conjugate pair; and a value for which the behavior of the system can be approximated using the concept of dominant roots. Solution: The characteristic equation is s(s + 3)(s + 7) + 15K = s 3 + 10s 2 + 21s + 15K. (12) Suppose that we would like (s+1) to be a factor of the characteristic equation. Then by long division (s 3 + 10s 2 + 21s + 15K)/(s + 1) = s 2 + 9s + 12 + (15K 12)/(s + 1) (13) Let s start by selecting a real root and then checking the others. If we let K = 12/15 = 4/5 we then have s = 1 as one root. The other two are s = 9 + 81 48 2 = 9 ± 33 2 (14)
ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 7 which are also real. Next, we examine the characteristic equation again and see that to get complex conjugate roots we should decrease the value of the coefficient of s 1 or increase the value of the coefficient of s 0 in (15). To do this we increase the magnitude of the real root, we try a few values before obtaining (s 3 + 10s 2 + 21s + 15K)/(s + 8) = s 2 + 2s + 19 + (15K 19)/(s + 8) (15) where 4 76 will result in a pair of complex conjugate roots. For this we need K = 19/15 and obtain the complex conjugate pair s = 4 + 4 76 2 = 4 ± j 72 2 = 2 ± j 18. (16) To find a value of K such that (s + η) is a factor of the characteristic equation and η 10 ζω n we first observe that the previous value of K doesn t achieve this because 4 10. This tells us that we need to continue increasing η to obtain an acceptable value. Let s take η = 20 to obtain (s 3 + 10s 2 + 21s + 15K)/(s + 20) = s 2 10s + 31 + (15K 31)/(s + 20) (17) which requires K = 31/15. (c) For this last value of K, calculate the expected peak time and overshoot. (d) Plot the actual response and the predicted response. (e) Find a value of K for which the peak time is reduced by 1/3 (i.e. new peak time is 2/3 of old peak time) if this is possible. 4. Write a Matlab script to generate the plot of transient response of a second order system for a range of values of ζ as is done in the text (p. 282 of 11th edition). Properly label the resulting plot. (You must turn in both the script and the plot.) Solution: The plot is shown in Figure 4. The Matlab script used to generate the plot was: figure(4) t = [0:.01:30] ; y = zeros(max(size(t)),6); zetas = [0.1,0.2,0.4,0.7,1,2] for index = 1:6 y(:,index) = step(tf([1],[1 2*zetas(index) 1]),t); plot(t,y(:,1),t,y(:,2),t,y(:,3),t,y(:,4),t,y(:,5),t,y(:,6)) title( Second Order Step Response ) xlabel( Normalized Time \omega_n t ) ylabel( y(t) ) legend( \zeta = 0.1, \zeta = 0.2, \zeta = 0.4, \zeta = 0.7,... \zeta = 1, \zeta = 2 ) grid print -depsc ps3p4.eps
ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 8 1.8 1.6 1.4 Second Order Step Response ζ = 0.1 ζ = 0.2 ζ = 0.4 ζ = 0.7 ζ = 1 ζ = 2 1.2 1 y(t) 0.8 0.6 0.4 0.2 0 0 5 10 15 20 25 30 Normalized Time ω n t
ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 9 5. For ζ = 0.2, plot (on the same graph) the step response of a second order system for the following values of ω n : 1/2, 1, 2, 4, 8 rad/sec. (You must turn in both the script and the properly labelled plot.) Solution: The plot is shown in Figure 5. The Matlab script used to generate the plot 1.6 Second Order Step Response 1.4 1.2 1 y(t) 0.8 0.6 ω n = 1/2 ω n = 1 0.4 0.2 ω n = 2 ω n = 4 ω n = 8 0 0 5 10 15 20 25 Time t was: figure(5) t = [0:.01:25] ; y = zeros(max(size(t)),6); zet = 0.2 omegas = [1/2,1,2,4,8] for index = 1:5 y(:,index) = step(tf([omegas(index)^2],[1 2*zet*omegas(index) omegas(index)^2]),t);
ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 10 plot(t,y(:,1),t,y(:,2),t,y(:,3),t,y(:,4),t,y(:,5)) title( Second Order Step Response ) xlabel( Time t ) ylabel( y(t) ) legend( \omega_n = 1/2, \omega_n = 1, \omega_n = 2, \omega_n = 4,... \omega_n = 8 ) grid print -depsc ps3p5.eps