MAE 143A: Homework 7 Solutions

Transcription

1 MAE 143A: Homework 7 Solutions Jacob Huffman and Elena Menyaylenko Date Due: February 6 th, 13 Problem 1: (i) Lets utilize the formula cos(a + B) = cos A cos B sin A sin B. (ii) u(t) = Acos(ωt + φ) u(t) = A cos φ cos ωt A sin φ sin ωt s U(s) = Acosφ s + ω Asinφ ω + ω s U(s) = A cos φ s + ω A sin φ ω s + ω U(s) = A s + ω The Y (s) is omitted here! This is because the zero-input response Y (s) has poles at the poles of the system. The zero-state response H(s)U(s) has poles at the poles of the system and at the poles of the input signal transform. Notice, the poles of the system are at s = p 1, p,..., p l, jω, jω. Partial Fraction Expansion.. (iii) y(t) = Y (s) = a 1,n 1 (s p 1 ) n1 + a 1,n 1 1 (s p 1 ) n a l,1 s p l + b s jω + b s + jω [ a1,n1 (n 1 1)! tn1 1 e p1t + a ] 1,n 1 1 (n 1 )! tn1 e p1t + + a l,a e plt + be jωt jωt + be 1(t). Since H(s) is stable, all of its poles, p i, are in the open left half-plane. So all of the terms e pit and t. Only two last two terms do not tend to zero. y(t) be jωt + be jωt as t. 1

2 (iv) (v) { } b = lim AH(s) s jω s + ω (s jω), { } = lim AH(s) (s + jω), s jω (s + jω)(s jω) { } = lim AH(s), s jω s + jω jω cos φ ω sin φ = AH(jω), jω = AH(jω) 1 [cos φ + j sin φ], = AH(jω) e jφ. You are just guided by professor s thinking. Plug in answers from above. (vi) We have from Part (iii) that y(t) be jωt + be jωt, Problem : H(jω) = H(jω) e j H(jω). = A H(jω) ej[φ+ H(jω)] e jωt + A H(jω) e j[φ+ H(jω)] e jωt, = A H(jω) 1 [ e j[ωt+φ+ H(jω)] + e j[ωt+φ+ H(jω)]], = A H(jω) cos [ωt + φ + H(jω)]. For this question, using pole() could have also been used to determine where the poles lie. Answers to questions 1)The frequency break point is also called the critical frequency. This is the frequency w at which your zero or pole breaks the bode plot, or changes it in some way- for example, poles make its magnitude fall, zeros make it rise. At a constant gain, your Bode plot has a constant magnitude, just until you stumble upon a pole or zero (at a break point/frequency). Above the critical frequency, they represent a ramp function of db per decade, with the poles giving a negative slope and the zeros giving a positive slope. These magnitudes and phases are shown to be additive in part(iv). Pole-position-wise, H1 has the pole at -1 (smallest frequency), H at -1, H3 at -1 (highest frequency). The H(s) have similar form, so all the bode plots fall off due to their respective poles at the same slope of db/decade, from H1 to H3-smallest to highest frequency. The phase change, likewise, occurs in H1 first, then H, H3.

3 The more left the pole lies, the faster this pole is (the system H3 is fastest, with a pole furthest to the left on the s-plane).the system with the pole 1 times faster, has an impulse response 1 times faster. And since the gain is also 1 times more, the starting impulse response starts at 1 times that as well. The system is stable. ) System is stable. 3) System with the smallest damping coefficient is the most resonant (highest peak). It is stable, since the real part of the poles is in the open left half plane. 4)s= pole makes the composite system unstable, and you can also see this with the step(). Note that the bode plots are additive, in both gain and phase. )All are stable, even thought we ve got a zero in the right half plane, thats ok. Matlab Code: 1 % Code for Question, Hwk 7 : MAE 143A 3 clc 4 clear all close all 6 7 % Part 1 8 H1=tf(1,[1 1]); 9 H=tf(1,[1 1]); 1 H3=tf(1,[1 1]); 11 1 figure(1) 13 subplot(1,,1) 14 hold on 1 bode(h1) 16 bode(h) 17 bode(h3) legend('h1','h','h3') subplot(1,,) 1 hold on impulse(h1); 3 impulse(h); 4 impulse(h3); axis([ ]) % Part 9 H4=tf([1 1],[.1 1]); 3 zpk(h4) 31 figure() 3 subplot(1,,1) 33 bode(h4) 34 title('lead Compensator') 3 subplot(1,,) 36 impulse(h4) % Part 3 39 k=[ ]; 4 figure(3) 41 subplot(1,,1) 4 for i=1:7; 3

4 43 H=tf(1,[1/1ˆ *k(i)/1 1]); 44 % pole(h) will show you stability via pole location. 4 hold on 46 bode(h) legendinfo{i} = ['\zeta=' numstr(k(i))]; 49 if i==7 legend(legendinfo) 1 end end 3 4 subplot(1,,) for ii=1:7; 6 H=tf(1,[1/1 *k(ii)/1 1]); 7 hold on 8 impulse(h) 9 end 6 61 % Part 4 6 H6A=tf(1,[1 ]); 63 H6B=tf(1,[1 1]); 64 H6C=tf([1.],1); 6 H6D=tf([ 1],[1 1 ]); 66 figure(4) 67 subplot(1,,1) 68 bode(h6a) 69 title('component 1') 7 subplot(1,,) 71 bode(h6b) 7 title('component ') 73 subplot(1,,3) 74 bode(h6c) 7 title('component 3') 76 subplot(1,,4) 77 bode(h6d) 78 title('composite') 79 subplot(1,,) 8 impulse(h6d) 81 zpk(h6d)% See pole at s=! 8 title('composite Impulse') % Part 86 H7A=tf(1*[1 1],[1 1]); 87 H7B=tf(1*[1-1],[1 1]); figure() 9 subplot(1,,1) 91 hold on 9 bode(h7a) 93 bode(h7b,'g--') 94 legend('h1','h') 9 hold off 96 subplot(1,,) 97 hold on 98 impulse(h7a) 99 impulse(h7b,'g--') 1 legend('h1','h') 11 hold off 4

5 Bode Diagram H1 H H Figure 1: Part 1 Plots

6 Lead Compensator Figure : Part 4 Bode Diagram ζ=. ζ=.1 ζ=. ζ=.3 ζ=. ζ=.7 ζ= Figure 3: Part 3 6

7 Component 1 Component Component 3 Composite Composite Impulse Figure 4: Part 4 Bode Diagram H1 H H1 H Figure : Part 7