# ECE382/ME482 Spring 2005 Homework 3 Solution March 7,

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1 ECE382/ME482 Spring 2005 Homework 3 Solution March 7, 2005 Solution to HW3 AP4.5 We are given a block diagram in Figure AP4.5 on page 237 of the text and asked to find steady state errors due to (a) a step disturbance and (b) a ramp input. We are then asked in part (c) to find a suitable value of the gain K, not more than 0, and plot the response to the unit step disturbance. Solution: (a) First we note that when the input R(s) is zero, the error E(s) is equal to the negative of the output. The transfer function from the disturbance input to the output is determined by noting that in this case, the block containing the transfer function /(s+2) 2 is in the feedforward path and the remaining two blocks are in the feedback path. Noting that the feedback into the first summer is negative, we thus have after simplification a block with gain K/s in the feedback path. Setting the input R(s) to zero and noting that the feedback into the second summer is positive we have transfer function from disturbance input D(s) to output of D(s) = (s+2) 2 K s(s+2) 2 = s s 3 + 4s 2 + 4s + K. () The steady state error due to a step disturbance R(s) = /s is thus, by the final value theorem, ( ) s e ss = lim s s 0 s s 3 + 4s 2 = 0. (2) + 4s + K (b) The transfer function from the input R(s) to the error E(s) is E(s) R(s) = + K = s(s+2) 2 s(s + 2)2 s(s + 2) 2 + K (3) where we have noted that all of the blocks are in the feedback path. Then the steady state error corresponding to a ramp input is ( ) s(s + 2) 2 e ss = lim s s 0 s 2 s(s + 2) 2 + K = 4 K. (4) This indicates that increasing the gain K decreases the steady state error resulting from a ramp input. (c) We don t have much information here from which to determine what a suitable value of K should be. Let s assume that since we were asked about steady state error due to a ramp input, that is important, so we should probably choose a large gain (close to our maximum of 0) so long as the system remains stable with this gain value. (We can check that in Matlab by executing the command roots([ 4 4 K]), after assigning our chosen value to the variable K, and making sure that all of the roots have negative real part. In fact they do for any value of K between 0 and 0. We will learn a method for determining this conclusively in Chapter 6.

2 ECE382/ME482 Spring 2005 Homework 3 Solution March 7, To see whether there is a tradeoff involved, let s have Matlab plot responses to step disturbances for several values of K. Using the following Matlab code we obtain the plot shown in Figure. clf;for k=[:0];[y,t]=step(tf([ 0],[ 4 4 k]));plot(t,y),hold on; end; grid xlabel( Time (s) ) ylabel( e(t) ) title( Error due to step disturbance for K=,2,...,0 ) print -deps AP4.5c Figure : Plot of Output Error Corresponding to Step Disturbance for a Range of Gain Values 0.25 Error due to step disturbance for K=,2,..., e(t) Time (s) Now by plotting the results for individual gains K we can see that the smallest error is obtained for the largest K so given the information we have, K = 0 appears to be the best choice. (That doesn t mean a smaller positive gain is unsuitable. We are not given any information to judge that.) If we had tested gains less than one, we would have seen that the maximum error increases with increasing gain. If we did not want any oscillation at all, we could see from the plot that we could achieve this by choosing a gain of, but

3 ECE382/ME482 Spring 2005 Homework 3 Solution March 7, there would be no reason to choose a gain less than one since that would increase the maximum error. AP4.7 In this problem we are given a block diagram of a system having both disturbance and sensor noise inputs. We are asked to determine the effect of (a) the disturbance and (b) the sensor noise on the output. We are then asked to select a value of the gain K in the range to 00, inclusive that minimizes the effects of step disturbances and sensor noise on the steady state error. In all parts of the problem we will assume that the error E(s) is defined to be the output of the leftmost summer. Solution: We ll apply the final value theorem to compute the steady state errors due to step disturbance and noise inputs from the transfer functions obtained in parts (a) and (b). Then we ll use this information to obtain an answer to part (c). (a) The transfer function from the disturbance to the output is D(s) = 2s s(s + 2) + 2K (5) so a step input of D(s) = A/s leads to a steady state error of ( ) A 2s e ss = lim s = 0, (6) s 0 s s(s + 2) + 2K i.e. the value of the gain K does not affect the value of the steady state error due to a step disturbance. (b) The transfer function from the noise to the output is N(s) = 2K s(s+2) 2K s(s+2) = 2K s(s + 2) + 2K (7) so a step input of N(s) = B/s leads to a steady state error of ( ) B 2K e ss = lim s = B, (8) s 0 s s(s + 2) + 2K and again the gain K has no effect on the steady state error. (c) We ve just shown that this was a trick question. There is no best value for K in terms of minimizing the effect of step noise and disturbance inputs on steady state error. As in the previous problem, we could plot the error as a function of time for different values of K and choose one that seems a reasonable compromise between the reducing the maximum error and reducing the settling time (the time for the error to become small a more precise definition will be given in Chapter 5). AP4.8 Given the block diagram in Figure AP4.8, we are asked to determine (a) the transfer function T(s) of the system, (b) the sensitivity Sb T of the transfer function T(s) to variation in the parameter b, and finally we are asked to choose a gain K [,50] so that the disturbance effects and the sensitivity are both minimized. Solution:

4 ECE382/ME482 Spring 2005 Homework 3 Solution March 7, (a) The transfer function is (b) The sensitivity is thus T(s) = R(s) = Kb s+ + Kb s+ S T b = SN b S D b = N b = Kb s + + Kb. (9) b N D b where N and D are the numerator and denominator of the transfer function T. We obtain Sb T = Kb Kb Kb s + + Kb = s + s + + Kb. () (c) The transfer function from the disturbance to the output is D(s) = b s+ + Kb s+ = b D b s + + Kb so if we define the error to be the output of the leftmost summer, thus e(t) = y(t) we have the following response to a unit step input: ( ) b e ss = lim s s 0 s s + + Kb = b + Kb. (3) We now have shown that both S T b and e ss resulting from a step disturbance decrease with increasing K so we should choose the maximum allowable K, namely K = 50. DP4.6 We are given the system whose block diagram is shown in Figure DP4.6 and asked to consider the effect of various choices of controllers on steady state error for unit step disturbances. In particular, we are asked to determine appropriate values of the controller gains for each case. It is noted that the parameter J represents the pitching moment of inertia, hence is always positive. Solution: (a) When G c (s) = K, the closed loop transfer function is T(s) = R(s) = (0) (2) K Js 2 + K. (4) Accordingly, if K is positive, the roots of the denominator are s = ±j K/J so both poles are on the jω axis and the system is marginally stable. If K is negative, the roots of the denominator are s = ± K/J so one pole is in the right-half plane and the system is unstable. (b) The steady state error to a unit step disturbance when G c (s) = K can be determined by first obtaining the transfer function from D(s) to then applying the final value theorem. The transfer function is D(s) = Js 2 + K Js 2 = Js 2 + K. (5)

5 ECE382/ME482 Spring 2005 Homework 3 Solution March 7, We can not use the final value theorem to calculate the steady state error because we have poles on the imaginary axis. (Refer to the explanation on page 50 of the conditions under which the final value theorem can be applied.) In fact, since the system will oscillate indefinitely, there is no final value. (c) Now supposing that the controller is instead G(s) = K P + K D s we obtain the closed loop system transfer function T(s) = R(s) = The poles are now K D ± KD 2 4JK P s = 2J so for stability we need to consider three cases. K p + K D s Js 2 + K D s + K P. (6) (i) The expression under the square root is negative. Then the R(s) = K D /(2J) so the system is stable if and only if K D > 0 (since J > 0). The expression under the square root is negative if 4JK P > KD 2, i.e. if K P > KD 2 /(4J), so the system is stable if both K D > 0 and K P > KD 2 /(4J) are true. (ii) The expression under the square root is negative. This occurs when KD 2 > 4JK P. First, suppose K P > 0. Then the square root will always be less than K D, so, as long as both K P and K D are positive, the system will be stable. (This is a weaker condition than the one we found above.) Next, suppose K P is negative. In that case, the square root would be greater than K D so regardless of whether K D were positive or negative, one of the roots would end up positive. Finally, suppose K P = 0. If K P = 0 then one of the roots will be zero and the system will be marginally stable. (iii) The expression under the square root is zero. Then the system is stable if and only if K D > 0. In summary, we ve found that for stability we need both gains strictly positive. (d) We assume here that K D and K P are positive. Now the transfer function from the disturbance D(s) to the output Θ(s) is Θ(s) D(s) = Js 2 + K P +K D s Js 2 = (7) Js 2 + K D s + K P (8) so the steady state error resulting from a step disturbance is ( ) e ss = lim s s 0 s Js 2 = (9) + K D s + K P K P MP4.3 We are given a closed loop transfer function T(s) = 0K s s + 0K and asked to obtain and plot step responses for K = 0, 00, and 500. We are also asked to develop a table comparing percent overshoot, settling time, and steady state error for (20)

6 ECE382/ME482 Spring 2005 Homework 3 Solution March 7, these values of K. This is a little complicated since we haven t yet defined settling time and percent overshoot, so you were allowed some leeway in your answers. Here s the answer to the problem using the definitions given in Chapter 5. Solution: First, let s look at how these values of K affect the poles (roots of the denominator) of the system transfer function. Using Matlab we find that for K = 0 we obtain unequal poles in the left half plan; for K = 00 we obtain a repeated pole of multiplicity 2 in the left half plane; and for K = 500 we obtain a complex pair of poles in the left half plane as shown in the following Matlab transcript. >> roots([ 20 0]) ans = >> roots([ 20 00]) ans = >> roots([ ]) ans = i i Accordingly, we expect different behavior in the three cases. By calculating the damping ratio and natural frequency we can tell more. The natural frequency will be ω n = K and the damping ratio will be ζ = 20/(2ω n ) = 0/ K so the system is overdamped (ζ > ) if K = 0, critically damped (ζ = ) if K = 00, and underdamped (ζ < ) if K = 500. Accordingly we expect to see overshoot only in the case K = 500. Note that our findings agree with the discussion in Section 2.4, on p. 50 of the text. There it was noted that if ζ < there are (distinct) real roots; if ζ = there are real repeated roots; and if ζ < there is a complex conjugate pair of roots. Before we plot the step responses, let s find out more about the sort of responses we expect. The steady state output for a step input will be ( ) y ss = lim s T(s) = T(0) = 0. (2) s 0 s (This means that the steady state error e ss = y ss = 9 does not depend on the value of K.)

7 ECE382/ME482 Spring 2005 Homework 3 Solution March 7, The 2% settling time given by (5.3) in the text is T s,2% = 4/(ζω n ). Since ω n = K and ζ = 20/(2ω n ) here, we would seem to have T s,2% = 4 0 independent of K. However, the plot of the step responses clearly shows that this is not the case. >> t=[0:.0:2] ; >> clf; >> for k=[0,00,500];[y,t]=step(tf([0*k],[ 20 k]),t);plot(t,y),hold on;end; >> xlabel( Time (s) ); >> ylabel( y(t) ) >> title( MP4.3 Step responses for K=0, 00, and 000 ) >> print -deps MP4.3.eps What went wrong here? The problem is that equation (5.3) for the settling time was derived assuming that the system response was a sinusoid multiplied by e ζωnt. This is only true if the second order system is underdamped. From the analysis above and the computed step responses we determine the values shown in Table for the percent overshoot, settling time and steady state error. (22) Table : Table of Response Parameters for MP4.3 Gain Percent Overshoot 2% Settling Time (s) Steady-State Error MP4.7 We are given a block diagram in Figure MP4.7 of a closed loop system with a controller G c (s). We are asked to compare the response of the closed-loop system to a unit step input for each of the two possible controllers. Solution: We will compare the steady state errors first by hand and then using the Matlab code shown in the transcript below. (a) Using the proportional controller G c (s) = K = 2 we obtain ( ) e ss = lim s 0 s + 0K = + K = 3. (23) s+0 (b) Using the proportional-integral controller G c (s) = K 0 + K /s = /2 we obtain e ss = lim s 0 ( ) s + 0K 0+K /s s+0 s 2 + 0s = lim s 0 s 2 = 0. (24) + 0( + K 0 )s + K

8 ECE382/ME482 Spring 2005 Homework 3 Solution March 7, Figure 2: Plot of Output Corresponding to Step Disturbance for a Range of Gain Values in MP4.3 4 MP4.3 Step responses for K=0, 00, and y(t) Time (s) (c) Obviously by using a more complex controller we have managed to reduce the steady state error for a step input to zero. Here are the Matlab calculations, which were, actually, superfluous except that the problem requested them. The plot of the step responses for both controllers is shown in Figure 3 >> %MP4.7 >> sysa = feedback(tf([20],[ 0]),) Transfer function: s + 30 >> sysb = feedback(tf(0*[2 20],[ 0 0]),)

9 ECE382/ME482 Spring 2005 Homework 3 Solution March 7, Transfer function: 20 s s^ s >> step(sysa, r-,sysb, b-. ) >> print -deps MP4.7.eps Figure 3: Plots of Step Responses for P and PI Controllers in MP4.7 Step Response Amplitude Time (sec)

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