hw07solns

# hw07solns - 1 Homework 7 Solution[1 8.3 1 From the figure...

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Unformatted text preview: 1 Homework 7 Solution [1.] 8.3 1. From the figure, we have f = Bli -f. m = x Using Laplace Transformation, F (s) = ms X(s) = H 2. From Kirchhoff's circuit law, we have iC m x esp iR R IC (s) iC vin Vin (s) H = 2 BlI(s) -F (s) Bl X(s) = - 2. I(s) ms (1) = iR + iL = IC (s) = IR (s) + IL (s) ms2 = -BliL = IL (s) = - X(s) Bl = Blx = Esp (s) = BlsX(s) diL 1 -mLs3 + B 2 l2 s = L + esp = IR (s) = (LsIL (s) + BlsX(s)) = X(s) dt R BlR -mLs3 - mRs2 + B 2 l2 s = X(s) BlR = C VC = IC (s) = CsVC (s) 1 -mLs3 + B 2 l2 s = vC + iR R = Vin (s) = IC (s) + X(s) Cs Bl 3 2 2 2 3 2 2 -mLs - mRs + B l s -mLs + B l s = X(s) + BlRCs Bl = X(s) = Vin (s) -mLs3 - mRs2 + B 2 l2 s -mLs3 + B 2 l2 s + BlRCs Bl -1 (2) (3) (4) (5) (6) (7) (8) (9) (10) [2.] 9.1 As illustrated in the figure, we have i = Vout vin H = = = C VC = I(s) = CsVC (s) 1 iR = I(s) = Vout (s) R 1 + CRs vC + vout = Vin (s) = Vout (s) CRs Vout CRs = Vin 1 + CRs (11) (12) (13) (14) 2 = 0.2, n = 2(-4 1.9596i) 2 1 Im(s) 0 -1 -2 -0.5 -0.45 -0.4 Re(s) -0.35 Im(s) 2 = 0.4, n = 2(-0.8 1.8330i) 1 0 -1 -2 -0.9 -0.85 -0.8 Re(s) -0.75 -0.7 = 0.2, n = 5(-1 4.8990i) 5 Im(s) Im(s) -1.05 = 0.05, n = 5(-0.25 4.9937i) 5 0 0 -5 -1.1 -1 Re(s) -0.95 -0.9 -5 -0.3 -0.25 Re(s) -0.2 -0.15 Figure 1: Poles changed as , n changing. [3.] 9.2 Y (s) = = = 1 R(s) sn c0 sn-1 + c1 sn-2 + c2 sn-3 + + cn-1 + sn c0 c1 c2 cn-1 ^ + 2 + 3 + + n + R(s) s s s s ^ R(s) So regardless of the input, y(t) will contain an n - 1th order polynomial in t, i.e., c2 c3 cn-1 n-1 y(t) = c0 + c1 t + t2 + t3 + + t + other terms 2 3! (n - 1)! [4.] 9.4 The poles locations are illustrated in Figure . And the corresponding step responses are illustrated as follows 3 Table 1: Step Responses. = 0.2, n = 2 2 1.5 = 0.4, n = 2 1.5 1 y(t) 1 y(t) 0.5 0 0 5 10 15 20 0 0.5 0 Time 5 Time 10 15 20 = 0.2, n = 5 2 2 = 0.05, n = 5 1.5 1.5 y(t) 1 y(t) 0 5 10 15 20 1 0.5 0.5 0 0 Time 0 5 Time 10 15 20 4 Table 2: Poles location. 4 s2 +2s+4 4 (s2 +2s+4)(s+20) 1 1 Im(s) 0 Im(s) -1.05 0 -1 -1 -1.1 -1 Re(s) -0.95 -0.9 -20 -15 -10 Re(s) -5 [5.] 9.5 Poles/Zeros 1A 1B 2A 2B 3A 3B Step Response 3A 2A 1A 1B 2B 3B Transfer Function 1 s2 +s+1.25 2 s +s+1.25 (s2 +s+1.25)(s2 +s+9.25) 1 s2 +3s+2 s+1 s2 +3s+2 s-0.5 s2 +s+1.25 s+0.5 s2 +s+1.25 [6.] They are illustrated in Table 2, 3 and 4. [7.] {-20, -5, -1, -0.1, 0.1 1} are illustrated in Table 5, 6, 7, 8, 9 and 10. [8.]{-20, -5, -1, -0.1, 0.1 1} are illustrated in Table 11, 12, 13, 14, 15 and 16. [9.] As illustrated, Y = [(R - DY )AB + (R - DY )C] E 1 Y ( + CD + ABD) E Y E(AB + C) = . R ABDE + CDE + 1 R(AB + C) = H = (15) 5 Table 3: Step response. 1.5 4 s2 +2s+4 0.06 4 (s2 +2s+4)(s+20) 0.05 1 0.04 y(t) y(t) 0.5 0 0 1 2 3 4 5 0.03 0.02 0.01 0 Time 0 1 2 Time 3 4 5 Table 4: Impulse response. 1.5 4 s2 +2s+4 0.06 4 (s2 +2s+4)(s+20) 0.05 1 0.04 y(t) y(t) 0 1 2 3 4 5 0.03 0.5 0.02 0.01 0 0 -0.5 -0.01 Time 0 1 2 Time 3 4 5 6 Table 5: z = -20, H = -0.2s+4 s2 +2s+4 Poles/Zeros 1.5 zeros poles 1 1 Step response Im(s) 0 y(t) 0.5 0 0 5 10 Re(s) 15 20 0 1 2 -1 Time 3 4 5 Table 6: z = -5, H = -0.8s+4 s2 +2s+4 Poles/Zeros 1.5 zeros poles 1 1 Step response Im(s) y(t) 0 0.5 -1 0 -0.5 -1 0 1 2 Re(s) 3 4 5 0 1 2 Time 3 4 5 7 Table 7: z = -1, H = -4s+4 s2 +2s+4 Poles/Zeros 1.5 zeros poles 1 1 Step response y(t) -1 -0.5 0 Re(s) 0.5 1 0.5 Im(s) 0 0 -1 -0.5 -1 0 1 2 Time 3 4 5 Table 8: z = -0.1, H = -40s+4 s2 +2s+4 Poles/Zeros 5 zeros poles 1 0 Step response Im(s) y(t) 0 -5 -1 -10 -15 -1 -0.5 Re(s) 0 0 1 2 Time 3 4 5 8 Table 9: z = 0.1, H = 40s+4 s2 +2s+4 Poles/Zeros 15 zeros poles 1 10 Step response Im(s) 0 y(t) -1 -0.8 -0.6 Re(s) -0.4 -0.2 5 -1 0 -5 0 1 2 Time 3 4 5 Table 10: z = 1, H = 4s+4 s2 +2s+4 Poles/Zeros 2 zeros poles 1 1.5 Step response Im(s) y(t) 0 1 -1 0.5 0 -1.1 -1.05 -1 Re(s) -0.95 -0.9 0 1 2 Time 3 4 5 9 Table 11: p = -20, H = 4 (s2 +2s+4)(-0.05s+1) Poles/Zeros 1000000 poles (no zeros) 0 1 -1000000 Step response (Unstable) Im(s) 0 y(t) 0 5 10 Re(s) 15 20 -2000000 -3000000 -1 -4000000 -5000000 0 0.2 0.4 Time 0.6 0.8 1 Table 12: p = -5, H = 4 (s2 +2s+4)(-0.02s+1) Poles/Zeros 50000000 poles (no zeros) Step response (Unstable) 1 0 Im(s) 0 -50000000 -1 -100000000 -150000000 0 10 20 Re(s) 30 40 50 0 0.1 0.2 Time 0.3 0.4 0.5 10 Table 13: p = -1, H = 4 (s2 +2s+4)(-s+1) Poles/Zeros 20 poles (no zeros) 0 1 -20 Step response (Unstable) Im(s) 0 y(t) -1 -0.5 0 Re(s) 0.5 1 -40 -60 -1 -80 -100 0 1 2 Time 3 4 5 Table 14: p = -0.1, H = 4 (s2 +2s+4)(-10s+1) Poles/Zeros 50 poles (no zeros) Step response (Unstable) 1 0 Im(s) y(t) 0 -50 -1 -100 -150 -1 -0.5 Re(s) 0 0 10 20 Time 30 40 50 11 Table 15: p = 0.1, H = 4 (s2 +2s+4)(10s+1) Poles/Zeros 1 poles (no zeros) Step response 0.8 1 y(t) -1 -0.8 -0.6 Re(s) -0.4 -0.2 0.6 Im(s) 0 0.4 -1 0.2 0 0 10 20 Time 30 40 50 60 Table 16: p = 1, H = 4 (s2 +2s+4)(s+1) Poles/Zeros 1 poles (no zeros) Step response 0.8 1 y(t) 0.6 Im(s) 0 0.4 -1 0.2 0 -1.1 -1.05 -1 Re(s) -0.95 0 1 2 Time 3 4 5 12 Poles 4 0.01 0.1 3 0.4 0.8 1 2 2 4 5 7 1 Im(s) 10 0 -1 -2 -3 -4 -3 -2.5 Re(s) Figure 2: Poles. -2 -1.5 -1 [10.] As illustrated, Y = [(R - DY )AB + CR] E 1 Y ( + ABD) E Y E(AB + C) = . R ABDE + 1 R(AB + C) = H = (16) [11.] The transfer function is calculated as Y H = = = and the poles are illustrated in Figure 2. (R - Y )kT Y kT = R 1 + kT k s2 + 4s + k + 3 (17) 13 - a. Since overshoot = e 1-2 < 10% = > 0.591 = = 2/n = 2/ k + 3 > 0.591 = k < 2 (2/0.591) - 3 8.11. Then k = {0.01, 0.1, 0.4, 0.8, 1, 2, 4, 5, 7}. b. Since tr 1.8 n 2 < 0.6 = n > 3 = k + 3 = n > 9 = k > 6, so k = {7, 10}. c. k = 7 satisfy both conditions simultaneously. ...
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