ECE382/ME482 Spring 2005 Homework 6 Solution April 17, (s/2 + 1) s(2s + 1)[(s/8) 2 + (s/20) + 1]

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1 ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 1 Solution to HW6 P8.17 We are given a system with open loop transfer function G(s) = 4(s/2 + 1) s(2s + 1)[(s/8) 2 + (s/2) + 1] (1) and unity negative feedback. We are asked to determine (a) sketch the Bode diagram and (b) find the frequency ω 1 at which the magnitude is db and the frequency ω 2 where the phase is 18 degrees. Solution: (a) In order to sketch the Bode diagram we first rewrite the open loop transfer function as G(s) = = 4(1/2)(s + 2) s(2)(s + 1/2)(1/64)[s s/5 + 64] 64(s + 2) s(s + 1/2)[s s/5 + 64] and note that for the quadratic factor in the denominator ω n = 8 rad/s and ζ = (1/2)(16/5)(1/8) =.2. For the magnitude plot we thus have corner frequencies.5, 2, and 8. We start with magnitude 2log 1 4 (we must use the gain from (1) not (2) see the form of equation (8.26) in section 8.2 of Dorf and Bishop). The slope will be 2 db/decade for frequencies less than.5. 4 db/decade for.5 < ω < 2, 2 db/decade for 2 < ω < 8 and 4 db/ decade for frequencies above 8 rad/s. From Figure 8.1 of the text, we see that we should have a peak of about 8 db corresponding to the value ζ =.2. For the phase plot, we see that the phase angle for low frequencies will be 9 degrees because of the pole at the origin, while for high frequencies the phase angle will ten to 27 degrees because the transfer function has three more poles than zeros. Because the corner frequencies are so close together it would be extremely difficult to accurately draw the phase angle by hand. The Bode diagram generated by Matlab is shown in Figure 1. (b) From the Bode diagram generated by Matlab we determine that ω rad/s and ω 2 = The margin command >> margin(tf(64*[1 2],conv([1 1/2 ],[1 16/5 64]))) gives us the values exactly as shown in Figure 2. AP8.1 We are given a mechanical system in which a mass is suspended from a fixed surface by a spring and supported by a damper. We are also given a Bode diagram. We are asked to determine the spring, damper, and mass parameters k, b, and m, respectively. Solution: (2) (3) (4)

2 ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 2 Figure 1: Bode Diagram for P8.17a 1 Bode Diagram for P A force balance on the mass yields so the transfer function of the system is ky + bẏ + mÿ = r (5) G(s) = 1 ms 2 + bs + k. (6) From the diagram we obtain the following estimates: low frequency gain 26 db, corner frequency ω n = 3 rad/s, and difference between low frequency gain and peak gain 4 db. From Figure 8.1a of the text we estimate the value of ζ to be.3. Then G() = 1/k so 2log 1 k 26 db, yielding k 2 N/m. The natural frequency ω 2 n = k/m then yields m 2/9 2.2 kg. Finally, b = 2ω n ζ 1.8 Ns/m. DP8.1 We are given a system with unity negative feedback and forward path transfer function We are asked to G c (s)g(s) = K(s + 2) s 2 (s + 12). (7)

3 ECE382/ME482 Spring 25 Homework 6 Solution April 17, Figure 2: Bode Diagram with gain and phase margins for P8.17b Bode Diagram Gm = 9.33 db (at 7.7 rad/sec), Pm = 51.2 deg (at 1.59 rad/sec) (a,b) For K = 1, sketch the Bode diagrams for the open and closed loop systems, respectively. (c) Repeat (a,b) for K = 5. (d) Choose K so that M pω 2 and the bandwidth is maximized. (e) Find the steady state error for a ramp input. Solution: The closed loop transfer function is T(s) = K(s + 2) s s 2 + K(s + 2). (8) When K = 1, the poles obtained using the roots function of Matlab are , and.349±.48j. When K = 5, the poles are and ±2.822j. To find the corner frequencies we would factor s 2 + 2ζω n s + ωn 2 to find that the real part of a complex root is ζω n and the imaginary part is ±ω n 1 ζ 2, then solve for ω n and ζ.

4 ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 4 (a) For the open loop transfer function we have corner frequencies 2 and 12 rad/s. The phase will start at 18 degrees due to the double pole at the origin. There will be an increase of 9 degrees due to the zero at 2 rad/s over the range 2/1 to 1(2) rad/s and a decrease of 9 degrees due to the pole at 12 over the range 12/1 to 1(12) rad/s. It is clear that these ranges overlap so we expect the maximum phase to be less than 9 degrees. The magnitude will decrease by 4 db/decade for frequencies less than 2 rad/s, decrease by only 2 db/dec for frequencies between the corner frequencies, and decrease by 4 db/dec for frequencies above 12 rad/s. The Bode plot is shown in Figure 3. 5 Figure 3: Bode Diagram for DP8.1a Bode plot for DP8.1 Open Loop System with K= (b) The Bode plot for the closed loop system is shown in Figure 4. (c) The Bode plot for the open loop system with K = 5 is shown in Figure 5. The Bode plot for the closed loop system with K = 5 is shown in Figure 6. (d) Since M pω is given by M pω = 1 2ζ ζ <.77, (9) 1 ζ2,

5 ECE382/ME482 Spring 25 Homework 6 Solution April 17, Figure 4: Bode Diagram for DP8.1b Bode plot for DP8.1 Closed Loop System with K= constraining its value constrains the value of ζ. Solving (9) for ζ in terms of M pω yields ζ 4 ζ M 2 p ω = (1) or 1 ± 1 1 ζ = 2 where we will choose the appropriate square roots to make the damping coefficient real and positive. For M pω = 2 we obtain ζ =.9659 or Since, in fact, we are trying to obtain bounds on ζ, we test the value of M pω resulting for ζ = 1/2 which is between these and find that the resulting M pω = so we have determined that we may safely choose any value for the damping coefficient between.26 and.96. Now by doing long division of s 3 +12s 2 +Ks+2K by s 2 +2ζω n s+ω 2 n and constraining the remainder to be zero, I was able to obtain an expression for ω n in terms of ζ and K. However, it was not a very tractable expression so instead we will do a root locus plot to find a reasonable value of K and see whether that helps. M 2 pω (11)

6 ECE382/ME482 Spring 25 Homework 6 Solution April 17, Figure 5: Open Loop Bode Diagram for DP8.1c Bode plot for DP8.1 Open Loop System with K= Checking the values of the damping coefficient as we travel along the loci corresponding to the complex conjugate pair of roots we find that K = 9 corresponds roughly to ζ =.26 but that for the default root locus plot we never obtain ζ =.96. We enlarge the region shown by specifying a set of gains in the second argument to the rlocus command, rlocus(oltf,[.1:.1:1]) and then check further along the loci. We discover that the value of ζ decreases as we increase the gain further and that around K = 38 we again have ζ =.26. Since bandwidth should increase with K, we test values of K near the large end of the range to find the largest K that meets the constraint M pω < 2. I used the following command for K = [25:1:4]; [mag,phase,w]=bode(feedback(tf(k*[1 2],[1 12 ]),1)); if max(mag)> 2 disp([ K =,num2str(k-1)]) stop

7 ECE382/ME482 Spring 25 Homework 6 Solution April 17, Figure 6: Closed Loop Bode Diagram for DP8.1c Bode plot for DP8.1 Closed Loop System with K= end; end; to determine that K = 337 is the maximum allowable gain. (This is sloppy code. There is no Matlab command stop so an error is generated when the command is encountered. This stops the loop since we have the answer we need.) Finally, we determine the resulting system bandwidth from the Bode plot shown in Figure 7 to be 26.7 rad/s. (e) Applying the Final Value Theorem, we obtain ( ) 1 e ss = lim s (1 T(s)) (12) s 1 = lim s s s 2 ( s s 2 ) + K(s + 2) K(s + 2) s s 2 + K(s + 2) (13) s 2 (s + 12) = lim s s(s s 2 =. (14) + Ks + 2K) DP8.3 We are given a system having transfer function G(s) in the forward path and transfer

8 ECE382/ME482 Spring 25 Homework 6 Solution April 17, Figure 7: Closed Loop Bode Diagram for DP8.1d Bode Diagram for DP8.1 Closed Loop System with K= System: untitled1 : 26.7 : function H(s) in the (negative) feedback path where and G(s) = K s 2 + 2s + 2 (15) H(s) = 5 s + 5. (16) We are asked to (a) choose a value for K for which the system is stable and plot the Bode diagram, and (b) adjust the gain K to obtain 2log 1 M pω = 3 db, then (c) with this gain, determine the bandwidth ω B and the steady state error e ss for a step input. Solution: The closed loop transfer function is T(s) = K(s + 5) s 3 + 7s s + (1 + 5K) (17)

9 ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 9 (a) The corresponding Routh array is given in Table 1. From the fourth row we determine that we need K > 2 for stability. The third row simplifies to the constraint K < or K < 74/5 so the range of values of K for which the system is stable is 2 < K < (18) Table 1: Routh array for DP8.3a s 3 : 1 12 s 2 : K s 1 1 : 7 (1 + 5K 84) s : 1 + 5K We choose K = 1 and obtain the Bode plot shown in Figure 8. Figure 8: Closed Loop Bode Diagram for DP8.3a Bode plot for DP8.3 Closed Loop System with K= (b) To determine the value of K within the stable range that yields M pω = 3 db I used the following Matlab commands which identified the appropriate gain to be K = 4.13.

10 ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 1 for K = [1:.1:14]; [mag,phase,w]=bode(tf(k*[1 5],[ (1+5*K)])); if 2*log1(max(mag))> 3 disp([ K =,num2str(k-.1)]) stop end; end; (c) We obtain the Bode plot shown in Figure 9 with bandwidth 3 rad/s (obtained by clicking on magnitude curve and finding the frequency at which the magnitude is 3 db). 2 Figure 9: Closed Loop Bode Diagram for DP8.3b with K=4.13 Bode plot for DP8.3 Closed Loop System with K= Applying the final value theorem we obtain ( 1 e ss = lim s s s ) (1 T(s)) = 1 T() = K (19) so for K = 4.13, e ss =.3263.

11 ECE382/ME482 Spring 25 Homework 6 Solution April 17, MP8.5 We are given a closed loop transfer function T(s) = 1 s 2 + 6s + 1 (2) and asked to (a) determine ω r, M pω, and ω B from the Bode plot, (b) estimate ζ and ω n from the Bode plot, (c) find ω n and ζ analytically, then predict ω r and M pω from them and compare the results. In part (a), we are asked to use the logspace function to generate a frequency vector with log spacing and frequency range [.1,1] rad/s. Solution: Part of the purpose of this exercise is to familiarize the reader with the units used in the bode and logspace commands. logspace takes two arguments: the exponents of the power of 1 corresponding to the minimum and maximum frequencies for the vector to be generated. When used without return variables, bodegenerates plots of magnitude in db and phase in deg. However if return variables are used, it generates a vector of magnitudes, which are not in db, and a vector of phases in deg. (a) I used the Matlab script shown below to find the requested values M pω = 4.8 db, ω r = 8.7 rad/s, and bandwidth 15.2 rad/s. % % MP8_5.m solves part of problem MP8_5 from Dorf and Bishop 1th ed. % % 17 April 5 --sk % cltf = tf([1],[1 6 1]) % (a) closed loop Bode plot figure(5) w = logspace(-1,3,1); bode(cltf,w) grid title( Bode plot for MP8.5 Closed Loop System ) print -deps mp8_5a [mag,phase,w]=bode(cltf,w); smag = squeeze(mag); % for some reason mag is 1 x 1 x 1. make smag 1 x 1. [Mpw,loc] = max(smag); MpwdB = 2*log1(Mpw) wr = w(loc) [foo,loc]=min(abs(2*log1(smag)-(-3*ones(size(smag))))) wb = w(loc) The Bode plot is shown in Figure 1. (b) From (11) above (in DP8.1), using the value of M pω (not in db) obtained above we find that ζ =.3. We reject the option ζ =.95 resulting using the positive square root because the formula (8.37) is only valid for ζ <.77. Then inserting this value

12 ECE382/ME482 Spring 25 Homework 6 Solution April 17, Figure 1: Closed Loop Bode Diagram for MP8.5a Bode plot for MP8.5 Closed Loop System for ζ and the value determined earlier for for ω r into (8.36) and solving for ω n yields ω n = Here are the Matlab commands I used. % (b) invert (8.36) and (8.37) zetap = sqrt(1/2+1/2*(sqrt(1-1/mpw^2))) zetam = sqrt(1/2-1/2*(sqrt(1-1/mpw^2))) wn = wr/sqrt(1-2*zetam^2) %%% zetap >.77 so eqn used isn t valid (c) From the denominator of T(s) we see that ω n = 1 = 1 and ζ = 6/(2ω n ) =.3. This is quite good agreement. It s possible we would have gotten even better agreement if we had use more points in our log-spaced frequency vector. (The default is 5 points. I used 1.) MP8.9 We are asked to design a filter G(s) such that the magnitude of the frequency response satisfies the design specifications given in Table 2. We are also asked to maximize the peak magnitude close to 4 rad/s. Solution: This is rather an open-ended question. We are not told how complicated we

13 ECE382/ME482 Spring 25 Homework 6 Solution April 17, Table 2: Design Specs for magnitude of frequency response of filter of MP8.9 Frequency Range Magnitude Requirement ω < 1 G(jω) < db 1 < ω < 1 G(jω) db ω > 1 G(jω) < db are allowed to make the filter. We also are not told what additional criteria we should use to decide whether a design is good or not. Obviously, the more poles and zeros we use, the sharper the cutoffs we can obtain (at the cost of some ripple in the middle of the range). Keeping in mind that simplicity is desirable, we note that we will need at least one zero to increase the magnitude so that it goes through db at 1 rad/s and at least two poles to decrease it so that it goes back through db at 1 rad/s. I played around with the gain and zero and pole locations to get the filter whose Bode plot is shown in Figure 11. The transfer function of my filter is G(s) = 1(s +.9) s s (21)

14 ECE382/ME482 Spring 25 Homework 6 Solution April 17, Figure 11: Closed Loop Bode Diagram for MP8.9 Bode Diagram

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