ECE382/ME482 Spring 2005 Homework 1 Solution February 10,

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1 ECE382/ME482 Spring 25 Homework 1 Solution February 1, 25 1 Solution to HW1 P2.33 For the system shown in Figure P2.33 on p. 119 of the text, find T(s) = Y 2 (s)/r 1 (s). Determine a relationship that will make Y 2 (s) independent of R 1 (s). Solution: We apply Mason s signal-flow gain formula (2.96), p. 74 of the text, to the signal flow graph in the figure, which has paths and loops P 121 = G 1 G 8 G 6 (1) P 122 = G 2 G 5 G 6 (2) P 123 = G 2 G 7 G 4 H 2 G 1 G 8 G 6 (3) L 1 = G 1 G 3 G 4 H 2 (4) L 2 = G 2 G 5 G 6 H 1 (5) L 3 = G 2 G 7 G 4 H 2 G 1 G 8 G 6 H 1. (6) Since all loops touch paths P 121 and P 123, the corresponding path cofactors are 121 = 123 = 1. Path P 122 does not touch loop L 1 so 122 = 1 L 1 = 1 + G 1 G 3 G 4 H 2. Using (2.97), p. 74, the determinant of the graph is = 1 L 1 L 2 L 3 + L 1 L 2 (7) = 1 + G 1 G 3 G 4 H 2 + G 2 G 5 G 6 H 1 G 1 G 2 G 4 G 6 G 7 G 8 H 1 H 2 + G 1 G 2 G 3 G 4 G 5 G 6 H 1 H 2. Substituting into the formula (2.96) we then have T 12 = P P P = G 1G 8 G 6 + ( G 2 G 5 G 6 )(1 + G 1 G 3 G 4 H 2 ) + G 2 G 7 G 4 H 2 G 1 G 8 G 6 where we have not substituted for only because there is no efficient way to simplify it and it is too long to fit in the available space. To decouple Y 2 (s) from R 1 (s), we need T 12 =, which is achieved if the numerator of the expression for T 12 is set equal to zero. For this we need G 1 G 6 G 8 = G 2 G 5 G 6 + G 1 G 2 G 3 G 4 G 5 G 6 H 2 + G 1 G 2 G 4 G 6 G 7 G 8 H 2. P2.34 Find the transfer function Y (s)/r(s) for the block diagram given in Figure P2.34, p Solution: The block diagram can be transformed by simple transformations to obtain the desired transfer function. 1. Duplicate the fuel gain block K 6 and the summer that follows it so that we can separate the loops through it. 2. Eliminate the feedback loop consisting of the forward path through the dynamics block G 3 (s) and feedback path through spark gain H 1 (s), using transformation 6. The resulting block will have transfer function G 3 (s)/(1 G 3 (s)h 1 (s)) because the feedback is positive. Call this new block G I (s). (8)

2 ECE382/ME482 Spring 25 Homework 1 Solution February 1, Eliminate the feedback loop consisting of the forward path through G I (s) and feedback path through (one copy of the) constant fuel gain K 6, using transformation 6. The resulting block will have transfer function G I (s)/(1 G I (s)k 6 ) because the feedback is positive. 4. Combine the feedforward paths through K 5 K 6 and G 2 (s) to obtain a block G II (s) = K 5 K 6 + G 2 (s) between the manifold block G 1 (s) and the block G I obtained earlier. 5. Eliminate the remaining two loops (one with feedback path through K 4 and one with feedback path through the air bypass H 2 (s) by applying transformation 6 twice more. The resulting transfer function is Y (s) R(s) = G 1 (s)g 3 (s)(k 5 K 6 + G 2 (s)) 1 G 3 (s)h 1 (s) K 6 G 3 (s) + G 1 (s)g 3 (s)(k 4 + H 2 (s))(k 5 K 6 + G 2 (s)). (9) P2.37 We are given a system whose transfer function is T(s) = Y (s) R(s) = 12 s 3 + 8s s + 12 and asked to use partial fraction expansions to determine the responses to ramp and impulse inputs, including the output values at t = 1.5 s, and to check our work with Matlab. We are also asked to use the tf and parallel commands to obtain the transfer function from the partial fraction expansion. This will be done in part (e) below. Solution: We will use L to indicate the Laplace transform operator. (a) First we must obtain the Laplace transform of the ramp input R(s) = L {r(t)} = L {t} = 1 s 2 (1) and factor the denominator of the transfer function to obtain s 3 + 8s s + 12 = (s + 1)(s + 3)(s + 4). (11) We then use residues to find the coefficients A, B, C, D, and E of the partial fraction expansion ( ) ( ) 1 12 s 2 (s + 1)(s + 3)(s + 4) = A s 2 + B s + C s D s E s + 4. (12) We obtain A = s2 T(s) s 2 = 1, (13) s= B = d ( ) s 2 T(s) ds s 2 = 12(19) s= 12 2 = 19 12, (14) C = s + 1 s 2 T(s) = = 2, (15) s= 1

3 ECE382/ME482 Spring 25 Homework 1 Solution February 1, 25 3 and similarly, D = 2/3 and E = 1/4. Thus and, inverse transforming, Y (s) = 1 s /2 s + 2 s /3 s /4 s + 4 (16) y(t) = t 19/12 + 2e t 2 3 e 3t e 4t, t (17) (b) To generate the plot of y(t) corresponding to a ramp input, we use the Matlab code below to generate the the plot. We use that fact that the ramp response of the original system is the same as the impulse response of the system obtained by multiplying the Laplace transform of the input with the transfer function. tf2_37 = tf([12],[ ]) rs2_37 = tf([1],[1 ]) ys2_37 = series(rs2_37,tf2_37) t = [:.1:1]; y = impulse(ys2_37,t); t(16) y(16) plot(t,y) title( Ramp Response of P2.37 ) xlabel( Time (s) ) ylabel( y(t) ) grid print -deps p2_27b.eps The plot obtained is shown in Figure 1. We find the value y(1.5) =.3561 using the commands above. (c) In the case of the impulse response, R(s) = 1 so we obtain a partial fraction expansion for T(s) itself, obtaining Y (s) = 2 s s s + 4 (18) and inverse transform y(t) = 2e t 6e 3t + 4e 4t (19) (d) [y,t]=impulse(t2_37,t); y(16) plot(t,y) title( Impulse Response of P2.37 ) xlabel( Time (s) ) ylabel( y(t) ) grid print -deps p2_27d.eps The plot obtained is shown in Figure 1. We find the value y(1.5) =.3895 using the commands above.

4 ECE382/ME482 Spring 25 Homework 1 Solution February 1, Figure 1: Plot of response of transfer function of P2.37 to a ramp input Ramp Response of P y(t) Time (s) (e) This procedure yields a cautionary tale, as show in the Matlab transcript below. The result of reconstructing the transfer function after obtaining the partial fraction expansion is incorrect. Somehow the numerator and denominator have both been multiplied by s. As usual, we must be careful in how we use software tools. >> s2_37 = tf([12],[ ]) s^5 + 8 s^ s^ s^2 >> [n2_37,d2_37]=tfdata(s2_37, v ) n2_37 = 12

5 ECE382/ME482 Spring 25 Homework 1 Solution February 1, Figure 2: Plot of impulse response of transfer function of P2.37 Impulse Response of P y(t) Time (s) d2_37 = >> [r,p,k]=residue(n2_37,d2_37) r =

6 ECE382/ME482 Spring 25 Homework 1 Solution February 1, 25 6 p = k = [] >> t1 = tf([r(1)],[1 -p(1)]) s + 4 >> t2 = tf([r(2)],[1 -p(2)]) s + 3 >> t3 = tf([r(3)],[1 -p(3)]) s + 1 >> t4 = tf([r(4)],[1 -p(4)]) s >> t5 = tf([r(5)],[1 -p(5) ]) 1

7 ECE382/ME482 Spring 25 Homework 1 Solution February 1, s^2 >> tf = parallel(t1,parallel(t2,parallel(t3,parallel(t4,t5)))) 1.86e-15 s^ e-15 s^ e-15 s^ e-16 s^ s s^6 + 8 s^ s^ s^3 Notice also the four coefficients in the numerator that should be zero, but aren t quite. DP2.4 We are asked to find the transfer function from input voltage V 1 to output voltage V for the operational amplifier circuit shown in Figure DP2.4 on page 126 of the textbook, assuming an ideal op amp. We are then asked to find the output voltage corresponding to an input voltage v 1 (t) = At, t. We assume that v 1 (t) =, t <. Solution: The ideal op amp has input voltages v + = v and input currents i + = i =. (a) Accordingly, when we apply Kirchoff s current law at the negative and positive terminals we obtain, respectively, in the frequency domain V 1 V = V V o R 1 R 1 (2) V 1 V + = V + R 2 1/Cs. (21) Then from (2), V = V 1 + 2V + and from (21), V + = V 1 /(R 2 Cs + 1) so ( ) 2V 1 V = V 1 + R 2 Cs + 1 = R2 Cs 1 V 1 (22) R 2 Cs + 1 and T(s) = V ( ) (s) V 1 (s) = R2 Cs 1 R 2 Cs + 1 (b) The Laplace transform of the input voltage is A/s 2 so we find the partial fraction expansion of ( ) ( ) R2 Cs 1 A V (s) = T(s)V 1 (s) = R 2 Cs + 1 s 2. (24) We let β = 1/(R 2 C) and solve for B, C, and D in B s 2 + C s + D s + β = A(s β) s 2 (s + β) to obtain B = A, C = 2A/β, and D = 2A/β so that ( 1 V (s) = A s 2 2/β s + 2/β ) s + β and, inverse transforming, v (t) = A (23) (25) (26) (t 2β + 2β e βt ), t (27)

8 ECE382/ME482 Spring 25 Homework 1 Solution February 1, 25 8 MP2.1 We are given a block diagram and asked to simplify it by hand and using Matlab. Simplification by hand is straightforward but results in a very messy expression. We are then asked to use the commands pzmap and pole and zero to identify the locations of the poles and zeros of the transfer function. Finally we are asked to plot the step response and use the final value theorem to verify that our result has the correct behavior as t becomes large. Solution: First we have three feedback loops to transform. The one with a double integrator 1/s 2 in the forward path and a gain of 5 in the backward path becomes a single block with transfer function 1/(s 2 5). The loop with a series connection of 1/(s + 1) and s/(s 2 +2) in the forward path and (4s+2)/(s+1) 2 in the backward path reduces to a single block with transfer function (s 3 +2s 2 +s)/(s 5 +3s 4 +5s 3 +11s 2 +8s+2). Simplifying the resulting loop whose forward path consists of these first two resulting blocks in series and whose backward path contains the block with transfer function (s 2 + 2)/(s ) and taking into account the constant gain of 4 in the block multiplying the input R(s) we obtain the transfer function T(S) = = Y (s) R(s) 4s 6 + 8s 5 + 4s s s s s 1 + 3s 9 45s 8 125s 7 2s s s s s s 14 (28) If you obtained that by hand, you are truly to be commended. I gave up and used Matlab about half way through as shown below. >> rfblock = tf([1],[1-5]) s^2-5 >> lfblock = feedback(tf([1 ],[ ]),tf([4 2],[1 2 1])) s^3 + 2 s^2 + s s^5 + 3 s^4 + 5 s^ s^2 + 8 s + 2 >> fbblock = tf([1 2],[1 14]) s^ s^3 + 14

9 ECE382/ME482 Spring 25 Homework 1 Solution February 1, 25 9 >> mp26tf = 4*feedback(series(lfblock,rfblock),fbblock) 4 s^6 + 8 s^5 + 4 s^ s^ s^ s s^1 + 3 s^9-45 s^8-125 s^7-2 s^ s^ s^ s^ s^ s - 14 (a) To reduce the block diagram using Matlab alone we would calculate the transfer functions of the forward path blocks using the following commands. >> rfblock = feedback(tf([1],[1 ]),tf([5],[1]),1) s^2-5 >> lfblock = feedback(series(tf([1],[1 1]),tf([1 ],[1 2])),... tf([4 2],[1 2 1])) s^3 + 2 s^2 + s s^5 + 3 s^4 + 5 s^ s^2 + 8 s + 2 Note that we use a third argument, 1 to indicate to the feedback command that the feedback is positive rather than (the default) negative. (b) The output generated by the command pzmap(mp26tf) is shown in Figure 3. (c) The poles and zeros obtained using the pole and zero commands are shown in the transcript fragment below. Note that although the pole-zero map found in the previous part of the problem appears to indicate that some poles and zeros coincide, the listings below indicate that they do not, so care must be taken in interpreting the output of the pzmap command. >> pole(mp26tf) ans = i

10 ECE382/ME482 Spring 25 Homework 1 Solution February 1, 25 1 Figure 3: Pole-Zero Map generated using the Matlab command pzmap for Problem MP Pole Zero Map for MP Imaginary Axis Real Axis i i i i i >> zero(mp26tf) ans = i i

11 ECE382/ME482 Spring 25 Homework 1 Solution February 1, We were also asked to plot the step response and determine whether the behavior for large t agreed with that predicted by the final value theorem. The final value theorem (page 5) states that lim y(t) = lim sy (s) (29) t s so long as there is not more than one pole at the origin, no poles on the imaginary axis, no poles in the right-half plane, and no repeated poles. In other words, this was a trick question. The list of poles above shows that the transfer function does have poles in the right half plane in fact it has five of them so we cannot apply the final value theorem. Not surprisingly, given the poles in the right half plane, the step response appears to increase without bound, as shown in Figure 4, which was generated using the following commands. t=[:.1:1]; [y,t]=step(mp26tf,t); plot(t,y) title( Step Response for System of MP2.6 ) xlabel( Time (s) ) ylabel( y(t) ) print -deps mp2_6s.eps If the conditions of the final value theorem had not been violated by the transfer function, we would have multiplied T(s) by R(s) = 1/s for the unit step to obtain Y (s).

12 ECE382/ME482 Spring 25 Homework 1 Solution February 1, Figure 4: Plot of step response of the system of MP2.6 Step Response for System of MP y(t) Time (s)