SOLUTIONS TO CASE STUDIES CHALLENGES

Transcription

1 F O U R Time Response SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Open-Loop Response The forward transfer function for angular velocity is, G(s) = ω 0(s) V P (s) = 24 (s+50)(s+.32) a. ω 0 (t) = A + Be -50t + Ce -.32t 24 b. G(s) = s s+98. Therefore, 2ζω n =5.32, ω n = 4.07, and ζ = c. ω 0 (s) = s(s 2 = +5.32s+98) Therefore, ω 0 (t) = e -50t e -.32t. d. Using G(s), Defining, Thus, the state equations are, ω ω ω 0 = 24v p (t) x x 2 = x 2 x = ω 0 x 2 = ω 0 = 98x 5.32x v p (t) In vector-matrix form, y = x x 0 = x v p (t); y = [ 0]x

2 Solutions to Case Studies Challenges 75 e. Program: 'Case Study Challenge (e)' num=24; den=poly([ ]); G=tf(num,den) step(g) Computer response: ans = Case Study Challenge (e) Transfer function: s^ s + 98 Ship at Sea: Open-Loop Response a. Assuming a second-order approximation: ω 2 n = 2.25, 2ζω n = 0.5. Therefore ζ = 0.67, ω n =.5. 4 π T s = = 6; T ζω P = n ω n -ζ 2 = 2.2 ; %OS = e -ζπ / - ζ2 x 00 = 58.8%; ω n T r =.69 therefore, T r = b. θ s = 2.25 s s 2 = s s s s s s = s s

3 76 Chapter 4: Time Response = s s s Taking the inverse Laplace transform, θ(t) = - e -0.25t ( cos.479t sin.479t) c. Program: 'Case Study 2 Challenge (C)' '(a)' numg=2.25; deng=[ ]; G=tf(numg,deng) omegan=sqrt(deng(3)) zeta=deng(2)/(2*omegan) Ts=4/(zeta*omegan) Tp=pi/(omegan*sqrt(-zeta^2)) pos=exp(-zeta*pi/sqrt(-zeta^2))*00 t=0:.:2; [y,t]=step(g,t); Tlow=interp(y,t,.); Thi=interp(y,t,.9); Tr=Thi-Tlow '(b)' numc=2.25*[ 2]; denc=conv(poly([0-3.57]),[ ]); [K,p,k]=residue(numc,denc) '(c)' [y,t]=step(g); plot(t,y) title('roll Angle Response') xlabel('time(seconds)') ylabel('roll Angle(radians)') Computer response: ans = Case Study 2 Challenge (C) ans = (a) Transfer function: s^ s omegan =.5000 zeta = Ts = 6

4 Solutions to Case Studies Challenges 77 Tp = 2.24 pos = Tr = ans = (b) K = i i p = i i 0 k = [] ans = (c)

5 78 Chapter 4: Time Response ANSWERS TO REVIEW QUESTIONS.Time constant 2. The time for the step response to reach 67% of its final value 3. The input pole 4. The system poles 5. The radian frequency of a sinusoidal response 6. The time constant of an exponential response 7. Natural frequency is the frequency of the system with all damping removed; the damped frequency of oscillation is the frequency of oscillation with damping in the system. 8. Their damped frequency of oscillation will be the same. 9. They will all exist under the same exponential decay envelop. 0. They will all have the same percent overshoot and the same shape although differently scaled in time.. ζ, ω n, T P, %OS, T s 2. Only two since a second-order system is completely defined by two component parameters 3. () Complex, (2) Real, (3) Multiple real 4. Pole's real part is large compared to the dominant poles, (2) Pole is near a zero 5. If the residue at that pole is much smaller than the residues at other poles 6. No; one must then use the output equation 7. The Laplace transform of the state transition matrix is (si -A) - 8. Computer simulation 9. Pole-zero concepts give one an intuitive feel for the problem. 20. State equations, output equations, and initial value for the state-vector 2. Det(sI-A) = 0 SOLUTIONS TO PROBLEMS. a. Overdamped Case: Expanding into partial fractions, 9 C(s) = s(s 2 + 9s + 9) Taking the inverse Laplace transform, c(t) = e t -.7 e -.46t

6 Solutions to Problems 79 b. Underdamped Case: K 2 and K 3 can be found by clearing fractions with K replaced by its value. Thus, or Hence K 2 = - and K 3 = -3. Thus, 9 = (s 2 + 3s + 9) + (K 2 s + K 3 )s 9 = s 2 + 3s +9 + K 2 s 2 + K 3 s c(t) = e-3t/2 cos( 27 4 t - φ) = -.55 e -.5t cos (2.598t - φ) where φ = arctan ( 3 27 ) = 30o c. Oscillatory Case:

7 80 Chapter 4: Time Response The evaluation of the constants in the numerator are found the same way as they were for the underdamped case. The results are K 2 = - and K 3 = 0. Hence, Therefore, d. Critically Damped c(t) = - cos 3t The constants are then evaluated as Now, the transform of the response is 2. c(t) = - 3t e -3t - e -3t 5 a. C(s) = s(s+5) = s - s+5. Therefore, c(t) = - e-5t. Also, T = 5, T r = 2.2 a = = 0.44, T s = 4 a = 4 5 = b. C(s) = s(s+20) = s - s+20. Therefore, c(t) = - e-20t. Also, T = 20, T r = 2.2 a = = 0., T s = 4 a = 4 20 = 0.2.

8 Solutions to Problems 8 3. Program: '(a)' num=5; den=[ 5]; Ga=tf(num,den) subplot(,2,) step(ga) title('(a)') '(b)' num=20; den=[ 20]; Gb=tf(num,den) subplot(,2,2) step(gb) title('(b)') Computer response: ans = (a) Transfer function: s + 5 ans = (b) Transfer function: s + 20

9 82 Chapter 4: Time Response Using voltage division, V C(s) V i (s) Therefore v C (t) = 5-5e -2t. Also, T = 2, T r = 2.2 a Program: clf num=2; den=[ 2]; G=tf(num,den) step(5*g) Computer response: Transfer function: s + 2 Cs 2 = (R+ = Cs ) (s+2). Since V i(s) = 5 s, V 0 C(s) = s(s+2) = =., T s = 4 a = 4 2 = 2. = 5 s - 5 s Writing the equation of motion, Thus, the transfer function is, (Ms 2 + 8s)X(s) = F(s)

10 Solutions to Problems 83 X(s) F(s) = Ms 2 + 8s Differentiating to yield the transfer function in terms of velocity, sx(s) F(s) = Ms +8 = / M s + 8 M Thus, the settling time, T s, and the rise time, T r, are given by T s = 4 8/ M = 2 M; T r = 2.2 8/M = 0.275M 7. Program: Clf M= num=/m; den=[ 8/M]; G=tf(num,den) step(g) pause M=2 num=/m; den=[ 8/M]; G=tf(num,den) step(g) Computer response: Transfer function: M = Transfer function: s + 8 M = 2 Transfer function: s + 4

11 84 Chapter 4: Time Response From plot, time constant = 0.25 s. From plot, time constant = 0.25 s. 8. a. Pole: -2; c(t) = A + Be -2t ; first-order response. b. Poles: -3, -6; c(t) = A + Be -3t + Ce -6t ; overdamped response. c. Poles: -0, -20; Zero: -7; c(t) = A + Be -0t + Ce -20t ; overdamped response.

12 Solutions to Problems 85 d. Poles: (-3+j3 5 ), (-3-j3 5 ) ; c(t) = A + Be -3t cos (3 5 t + φ); underdamped. e. Poles: j3, -j3; Zero: -2; c(t) = A + B cos (3t + φ); undamped. f. Poles: -0, -0; Zero: -5; c(t) = A + Be -0t + Cte -0t ; critically damped. 9. Program: p=roots([ ]) Computer response: p = i i G(s) = C (si-a) - B 8 4 A = ; B = 3 ; C = [ ] (s 2 + 7s 8) (4s + 29) (s 2) (si A) = (3s + 27) (s 2 + s 77) 3 s 3 s 2 9s s 3 7s 76 (s 2 0s + 4) Therefore, G(s ) = 5s2 +36s 777 s 3 s 2 9s Factoring the denominator, or using det(si-a), we find the poles to be 9.683, , Program: A=[8-4 ;-3 2 0;5 7-9] B=[;3;7] C=[2 8-3] D=0 [numg,deng]=ss2tf(a,b,c,d,); G=tf(numg,deng) poles=roots(deng) Computer response: A = B =

13 86 Chapter 4: Time Response C = D = Transfer function: 5 s^ s s^3 - s^2-9 s + 67 poles = Writing the node equation at the capacitor, V C (s) ( R 2 + Ls + Cs) + V C(s) - V(s) R = 0. Hence, V C(s) V(s) R 0s = R + R + = 2 Ls + Cs s 2 +20s+500. The step response is 0 s 2 +20s+500.The poles 3. are at -0 ± j20. Therefore, v C (t) = Ae -0t cos (20t + φ). Program: num=[0 0]; den=[ ]; G=tf(num,den) step(g) Computer response: Transfer function: 0 s s^ s + 500

14 Solutions to Problems The equation of motion is: (Ms 2 +f v s+k s )X(s) = F(s). Hence, X(s) F(s) = Ms 2 +f v s+k s = The step response is now evaluated: X(s) = 5 (s+ 2 ) (s+ 2 ) s(s 2 +s+5) = /5 s Taking the inverse Laplace transform, x(t) = 5-5 e-0.5t ( cos = e-0.5t 9 cos ( 2 t o ). ω 2 n C(s) = s(s 2 +2ζω n s+ω 2 n ) - 5 s + 5 (s+ 2 ) t + = 9 sin 9 2 = s - s + 2ζω n s 2 +2ζω n s+ω n 2 = s - s + 2ζω n (s+ζω n ) 2 + ω n 2 - ζ 2 ω n 2 ζω n (s+ζω n ) + = s - (s + ζω n ) + ζω n (s+ζω n ) 2 + (ω n - ζ 2 ) 2 = ω n - ζ 2 ω n - ζ 2 s - (s+ζω n ) 2 + (ω n - ζ 2 ) 2 = - e -ζω n t cos ω n - ζ 2 ζ t + - ζ sin ω Hence, c(t) 2 n - ζ 2 t s 2 +s+5 t ).

15 88 Chapter 4: Time Response = - e -ζω n t + ζ2 -ζ 2 cos (ω n - ζ 2 t - φ) = - e -ζω n t where φ = tan - ζ - ζ 2 -ζ 2 cos (ω n - ζ 2 t - φ), %OS = e -ζπ / ln ( %OS 00 ) = - ζπ negative square root, ζ = a. - ζ2 x 00. Dividing by 00 and taking the natural log of both sides, ln 2 ( %OS 00 ) - ζ 2. Squaring both sides and solving for ζ2, ζ 2 = - ln ( %OS 00 ) π 2 + ln 2 ( %OS. 00 ) π 2 + ln 2 ( %OS 00 ). Taking the b. c. d.

16 Solutions to Problems 89 e. f. 8. a. N/A b. s 2 +9s+8, ω 2 n = 8, 2ζω n = 9, Therefore ζ =.06, ω n = 4.24, overdamped. c. s 2 +30s+200, ω 2 n = 200, 2ζω n = 30, Therefore ζ =.06, ω n = 4.4, overdamped. d. s 2 +6s+44, ω 2 n = 44, 2ζω n = 6, Therefore ζ = 0.25, ω n = 2, underdamped. e. s 2 +9, ω 2 n = 9, 2ζω n = 0, Therefore ζ = 0, ω n = 3, undamped. f. s 2 +20s+00, ω 2 n = 00, 2ζω n = 20, Therefore ζ =, ω n = 0, critically damped X(s) = 2 s(s 2 +00s+00 2 ) = s - s+00 (s+50) = s Therefore, x(t) = - e -50t 50 (cos 7500 t = e-50t cos (50 3 t - tan - 3 ) 50 (s+50) (s+50) = (s+50) + s (s+50) sin 7500 t) 20. a. ω 2 4 n = 6 r/s, 2ζω n = 3. Therefore ζ = 0.375, ω n = 4. T s = = s; T ζω P = n ω n -ζ 2 π = s; %OS = e -ζπ / - ζ2 x 00 = %; ω n T r = (.76ζ ζ ζ + ) =.4238; therefore, T r = s.

17 90 Chapter 4: Time Response b. ω 2 4 n = 0.04 r/s, 2ζω n = Therefore ζ = 0.05, ω n = 0.2. T s = = 400 s; T ζω P = n ω n -ζ 2 π = 5.73 s; %OS = e -ζπ / - ζ2 x 00 = %; ω n T r = (.76ζ ζ ζ + ); therefore, T r = 5.26 s. c. ω 2 n =.05 x 0 7 r/s, 2ζω n =.6 x Therefore ζ = 0.247, ω n = T s = = s; T ζω P = n π ω n -ζ 2 = 0.00 s; %OS = e -ζπ / - ζ2 x 00 = %; ω n T r = (.76ζ ζ ζ + 2. ); therefore, T r = 3.88x0-4 s. Program: '(a)' clf numa=6; dena=[ 3 6]; Ta=tf(numa,dena) omegana=sqrt(dena(3)) zetaa=dena(2)/(2*omegana) Tsa=4/(zetaa*omegana) Tpa=pi/(omegana*sqrt(-zetaa^2)) Tra=(.76*zetaa^3-0.47*zetaa^ *zetaa + )/omegana percenta=exp(-zetaa*pi/sqrt(-zetaa^2))*00 subplot(22) step(ta) title('(a)') '(b)' numb=0.04; denb=[ ]; Tb=tf(numb,denb) omeganb=sqrt(denb(3)) zetab=denb(2)/(2*omeganb) Tsb=4/(zetab*omeganb) Tpb=pi/(omeganb*sqrt(-zetab^2)) Trb=(.76*zetab^3-0.47*zetab^ *zetab + )/omeganb percentb=exp(-zetab*pi/sqrt(-zetab^2))*00 subplot(222) step(tb) title('(b)') '(c)' numc=.05e7; denc=[.6e3.05e7]; Tc=tf(numc,denc) omeganc=sqrt(denc(3)) zetac=denc(2)/(2*omeganc) Tsc=4/(zetac*omeganc) Tpc=pi/(omeganc*sqrt(-zetac^2)) Trc=(.76*zetac^3-0.47*zetac^ *zetac + )/omeganc percentc=exp(-zetac*pi/sqrt(-zetac^2))*00 subplot(223) step(tc) title('(c)') Computer response: ans = (a)

18 Solutions to Problems 9 Transfer function: s^2 + 3 s + 6 omegana = 4 zetaa = Tsa = Tpa = Tra = percenta = ans = (b) Transfer function: s^ s omeganb = zetab = Tsb = 400 Tpb =

19 92 Chapter 4: Time Response Trb = percentb = ans = (c) Transfer function:.05e s^ s +.05e007 omeganc = e+003 zetac = Tsc = Tpc = Trc = 3.880e-004 percentc =

20 Solutions to Problems Program: T=tf(6,[ 3 6]) T2=tf(0.04,[ ]) T3=tf(.05e7,[.6e3.05e7]) ltiview Computer response: Transfer function: s^2 + 3 s + 6 Transfer function: s^ s Transfer function:.05e s^ s +.05e007

21 94 Chapter 4: Time Response

22 Solutions to Problems a. ζ = - ln ( %OS 00 ) 4 π 2 + ln 2 ( %OS = 0.56, ω n = =.92. Therefore, poles = -ζω 00 ) ζt n ± jω n -ζ 2 s = ± j9.88. b. ζ = - ln ( %OS 00 ) π π 2 + ln 2 ( %OS = 0.59, ω n = 00 ) T P -ζ 2 = Therefore, poles = -ζω n ± jω n -ζ 2 = ± j c. ζω n = 4 T s = 0.57, ω n -ζ 2 = π T p =.047. Therefore, poles = ± j Re = 4 T s = 4; ζ = -ln(2.3/00) π 2 + ln 2 (2.3/00) = Re =ζω n = ω n = 4; ω n = 7.2 Im = ω n ζ 2 = 6 G(s) = ω n 2 s 2 + 2ζω n s + ω n 2 = 5.96 s 2 + 8s a. Writing the equation of motion yields, (3s 2 + 5s + 33)X(s) = F(s)

23 96 Chapter 4: Time Response Solving for the transfer function, X(s) F(s) = / 3 s 2 + 5s + b. ω 2 4 n = r/s, 2ζω n = 5. Therefore ζ = 0.754, ω n = T s = =.6 s; T ζω P = n ω n -ζ 2 π =.44 s; %OS = e -ζπ / - ζ2 x 00 = 2.7 %; ω n T r = (.76ζ ζ ζ + ); therefore, T r = 0.69 s. 26. Writing the loop equations, s 2 + s θ s s θ 2 s = T s s θ s + s + θ 2 s = 0 Solving for θ 2 (s), θ 2 s = s 2 + s T ( s ) s 0 s 2 + s s s s + = T s s 2 + s + Forming the transfer function, θ 2 T s s = s 2 + s + Thus ω n =, 2ζω n =. Thus, ζ = 0.5. From Eq. (4.38), %OS = 6.3%. From Eq. (4.42), T s = 8 seconds. From Eq. (4.34), T p = 3.63 seconds a. s(s 2 + 4s ) = s - s + 4 (s + 2) = (s + 2) + s (s + 2) Thus c(t) = - e -2t (cos4.532t+0.44 sin 4.532t) = -.09e -2t cos(4.532t ). b.

24 Solutions to Problems 97 Therefore, c(t) = e -0t - e -2t (0.7 cos 4.532t sin 4.532t) = e -0t -.89 cos(4.532t o ). c. Therefore, c(t) = -.4e -3t + e -2t (0.4 cos 4.532t sin 4.532t) = -.4e -3t cos(4.532t o ). 28. Since the third pole is more than five times the real part of the dominant pole, s s determines the transient response. Since 2ζω n =.204, and ω n = = ω n =.682, ζ = 0.358, %OS = e ζπ / ζ 2 x00 = 30%, T s = therefore, T r = π = 6.64 sec, T ζω p = n ω n -ζ 2 = 2 sec; ω nt r =.4, 29. a. Measuring the time constant from the graph, T = seconds. 3 Response Time(seconds) T = seconds

25 98 Chapter 4: Time Response Estimating a first-order system, G(s) = K s+a. But, a = /T = , and K a = 2. Hence, K = Thus, G(s) = s b. Measuring the percent overshoot and settling time from the graph: %OS = ( )/.03 = 25.3%, Response 5 0 c max = 3.82 c final = Time(seconds) T s = 2.62 seconds and T s = 2.62 seconds. Estimating a second-order system, we use Eq. (4.39) to find ζ = 0.4, and Eq. K (4.42) to find ω n = Thus, G(s) = s 2 +2ζω n s +ω 2. Since C final =.03, K n ω 2 =.03. Hence, n K = Substituting all values, G(s) = s s+4.59 c. From the graph, %OS = 40%. Using Eq. (4.39), ζ = Also from the graph, π T p = ω n ζ = 4. Substituting ζ = 0.28, we find ω 2 n = Thus, K G(s) = s 2 +2ζω n s +ω 2 = n s s

26 Solutions to Problems a. Since the amplitude of the sinusoids are of the same order of magnitude as the residue of the pole at - 2, pole-zero cancellation cannot be assumed. b. Since the amplitude of the sinusoids are of the same order of magnitude as the residue of the pole at - 2, pole-zero cancellation cannot be assumed. c. Since the amplitude of the sinusoids are of two orders of magnitude larger than the residue of the pole at -2, pole-zero cancellation can be assumed. Since 2ζω n =, and ω n = 5 = 2.236, ζ = 0.224, %OS = e ζπ / ζ 2 4 π x00 = 48.64%, T s = = 8 sec, T ζω p = n ω n -ζ 2 =.44 sec; ω nt r =.23, therefore, T r = d. Since the amplitude of the sinusoids are of two orders of magnitude larger than the residue of the pole at -2, pole-zero cancellation can be assumed. Since 2ζω n = 5, and ω n = 20 = 4.472, ζ = 0.559,

27 00 Chapter 4: Time Response %OS = e ζπ / ζ 2 x00 = 2.03%, T s =.852, therefore, T r = π =.6 sec, T ζω p = n ω n -ζ 2 = sec; ω nt r = 3. Program: %Form sc(s) to get transfer function clf num=[ 3]; den=conv([ 3 0],[ 2]); T=tf(num,den) step(t) Computer response: Transfer function: s s^3 + 5 s^2 + 6 s %OS = ( ) 0.5 = 8.67% Only part c can be approximated as a second-order system. From the exponentially decaying cosine the poles are located at s,2 = 2 ± j Thus,

28 Solutions to Problems T s = 4 Re = 4 2 = 2 s; T = π p Im = π = s Also, ω n = = 0 and ζω n = Re = 2. Hence, ζ = 0.2, yielding percent overshoot. a. () C a s = = s s = = s s s Taking the inverse Laplace transform C a (t) = e -.5t sin t (2) C a 2 s s = 2 = 8 s + 6 s s s + 36 s s 2 = + 3 s s s s = s Taking the inverse Laplace transform s s C a2 (t) = e -.5t ( cos 5.809t sin 5.809t) The total response is found as follows: C at (t) = C a (t) + C a2 (t) = e -.5t ( cos 5.809t sin 5.809t) Plotting the total response: b. () Same as () from part (a), or C b (t) = C a (t) (2) Same as the negative of (2) of part (a), or C b2 (t) = - C a2 (t)

29 02 Chapter 4: Time Response The total response is C bt (t) = C b (t) + C b2 (t) = C a (t)- C a2 (t) = e -.5t ( cos 5.809t sin 5.809t) Notice the nonminimum phase behavior for C bt (t).

30 Solutions to Problems Unit Step 0 Gain s 2 +3s+0 Transfer Fcn Scope Unit Step 0 Gain Saturation 0.25 volts s 2 +3s+0 Transfer Fcn Scope Unit Step2 0 Gain2 Saturation volts s 2 +3s+0 Transfer Fcn2 Backlash Deadzone 0.02 Scope2

31 04 Chapter 4: Time Response 35. si A = s (s + 2) 5 = 3 (s + 5) si A = s 2 + 7s + 7 Factoring yields poles at and.2087.

32 Solutions to Problems a s 2 3 si A = s = 0 (s 6) (s 2) si A = s 3 8s 2 s + 8 b. Factoring yields poles at 9., , and x = (si - A ) - (x 0 + B u ) 38. x = (si - A ) - (x 0 + B u ) 39. x = (si - A ) - (x 0 + B u )

33 06 Chapter 4: Time Response 40. x = (si A) (x 0 + Bu) x = s s 0 x = s(s + 3)(s + 5) s(s + 5) s(s + 5) 5 6 e 3t + x(t) = 5 5 e 5t 5 5 e 5t 0 e 5t 4. Program: A=[-3 0;0-6 ;0 0-5]; B=[0;;]; C=[0 ]; D=0; S=ss(A,B,C,D) step(s) y(t) = [ 0 ]x = e 5t Computer response: a = x x2 x3 x -3 0 x2 0-6 x

34 Solutions to Problems 07 b = u x 0 x2 x3 c = x x2 x3 y 0 d = u y 0 Continuous-time model. 42. Program: syms s %Construct symbolic object for %frequency variable 's'. 'a' %Display label A=[-3 0;0-6 ;0 0-5] %Create matrix A. B=[0;;]; %Create vector B. C=[0 ]; %Create C vector X0=[;;0] %Create initial condition vector,x(0). U=/s; %Create U(s). I=[ 0 0;0 0;0 0 ]; %Create identity matrix. X=((s*I-A)^-)*(X0+B*U); %Find Laplace transform of state vector. x=ilaplace(x()) %Solve for X(t). x2=ilaplace(x(2)) %Solve for X2(t). x3=ilaplace(x(3)) %Solve for X3(t). y=c*[x;x2;x3] %Solve for output, y(t). y=simplify(y) %Simplify y(t). 'y(t)' %Display label. pretty(y) %Pretty print y(t). Computer response: ans = a

35 08 Chapter 4: Time Response A = X0 = x = 0 7/6*exp(-3*t)-/3*exp(-6*t)+/5+/0*exp(-5*t) x2 = exp(-6*t)+/5-/5*exp(-5*t) x3 = /5-/5*exp(-5*t) y = 2/5+exp(-6*t)-2/5*exp(-5*t) y = 2/5+exp(-6*t)-2/5*exp(-5*t) ans = y(t) 2/5 + exp(-6 t) - 2/5 exp(-5 t) 43. λi - A = λ 2 + 5λ + λi - A = (λ ) (λ ) Therefore,

36 Solutions to Problems 09 Solving for A i 's two at a time, and substituting into the state-transition matrix To find x(t), To find the output, 44. λi - A = λ 2 + Solving for the A i 's and substituting into the state-transition matrix, To find the state vector,

37 0 Chapter 4: Time Response 45. Let the state-transition matrix be λi - A = (λ + 2) (λ i) (λ i) Since φ(0) = I, Φ. (0) = A, and φ.. (0) = A 2, we can evaluate the coefficients, A i 's. Thus, Solving for the A i 's taking three equations at a time,

38 Solutions to Problems t Using x (t) = φ (t)x (0) + φ (t-τ )B u(τ )dτ, and y = 0 0 x (t), Program: syms s t tau = 2-2 e-2t 'a' A=[-2 0;0 0 ;0-6 -] %Create matrix A. B=[;0;0] %Create vector B. C=[ 0 0] %Create vector C. %Construct symbolic object for %frequency variable 's', 't', and 'tau. %Display label. X0=[;;0] %Create initial condition vector,x(0). I=[ 0 0;0 0;0 0 ]; %Create identity matrix. 'E=(s*I-A)^-' %Display label. E=((s*I-A)^-) %Find Laplace transform of state %transition matrix, (si-a)^-. Fi=ilaplace(E(,)); %Take inverse Laplace transform Fi2=ilaplace(E(,2)); %of each element Fi3=ilaplace(E(,3)); Fi2=ilaplace(E(2,)); Fi22=ilaplace(E(2,2)); Fi23=ilaplace(E(2,3)); Fi3=ilaplace(E(3,)); Fi32=ilaplace(E(3,2)); %to find state transition matrix. Fi33=ilaplace(E(3,3)); %of (si-a)^-. 'Fi(t)' %Display label. Fi=[Fi Fi2 Fi3 %Form Fi(t). Fi2 Fi22 Fi23 Fi3 Fi32 Fi33]; pretty(fi) %Pretty print state transition matrix, Fi. Fitmtau=subs(Fi,t,t-tau); %Form Fi(t-tau). 'Fi(t-tau)' %Display label. pretty(fitmtau) %Pretty print Fi(t-tau). x=fi*x0+int(fitmtau*b*,tau,0,t); %Solve for x(t). x=simple(x); %Collect terms. x=simplify(x); %Simplify x(t). x=vpa(x,3); 'x(t)' %Display label. pretty(x) %Pretty print x(t). y=c*x; %Find y(t) y=simplify(y); y=vpa(simple(y),3); y=collect(y); 'y(t)' pretty(y) %Pretty print y(t).

39 2 Chapter 4: Time Response ans = Computer response: a A = B = 0 0 C = 0 0 X0 = ans = 0 E=(s*I-A)^- E = [ /(s+2), (s+)/(s+2)/(s^2+s+6), /(s+2)/(s^2+s+6)] [ 0, (s+)/(s^2+s+6), /(s^2+s+6)] [ 0, -6/(s^2+s+6), s/(s^2+s+6)] ans = Fi(t) [ 3 [exp(-2 t), - /8 exp(-2 t) + /8 % %2, [ 84 ] /8 exp(-2 t) - /8 % + 3/84 %2] ] [ [0, /23 %2 + %, - /23 /2 /2 /2 (-23) (exp((-/2 + /2 (-23) ) t) - exp((-/2 - /2 (-23) ) t)) ] ] [ [0, 6/23 /2 /2 /2 (-23) (exp((-/2 + /2 (-23) ) t) - exp((-/2 - /2 (-23) ) t))

40 Solutions to Problems 3 ], - /23 %2 + %] /2 % := exp(- /2 t) cos(/2 23 t) /2 /2 %2 := exp(- /2 t) 23 sin(/2 23 t) ans = Fi(t-tau) [ [exp(-2 t + 2 tau), [ 3 /2 - /8 exp(-2 t + 2 tau) + /8 %2 cos(%) %2 23 sin(%), 84 /2 ] /8 exp(-2 t + 2 tau) - /8 %2 cos(%) + 3/84 %2 23 sin(%)] ] [ /2 /2 [0, /23 %2 23 sin(%) + %2 cos(%), - /23 (-23) ( /2 exp((-/2 + /2 (-23) ) (t - tau)) /2 ] - exp((-/2 - /2 (-23) ) (t - tau)))] [ /2 /2 [0, 6/23 (-23) (exp((-/2 + /2 (-23) ) (t - tau)) /2 - exp((-/2 - /2 (-23) ) (t - tau))), /2 ] - /23 %2 23 sin(%) + %2 cos(%)] /2 % := /2 23 (t - tau) %2 := exp(- /2 t + /2 tau) ans = x(t) ans = y(t) [.375 exp(-2. t) +.25 exp(-.500 t) cos(2.40 t) exp(-.500 t) sin(2.40 t) +.500] [.209 exp(-.500 t) sin(2.40 t) + exp(-.500 t) cos(2.40 t)] [.25 i (exp(( i) t) -. exp(( i) t))].375 exp(-2. t) +.25 exp(-.500 t) cos(2.40 t)

41 4 Chapter 4: Time Response exp(-.500 t) sin(2.40 t) The state-space representation used to obtain the plot is, 0 ẋ = x Using the Step Response software, 0 u(t); y(t) = 0 x Calculating % overshoot, settling time, and peak time, 2ζω n = 0.8, ω n =, ζ = 0.4. Therefore, %OS = e ζπ / ζ 2 x00 = 25.38%, T s = π T p = = 3.43 sec. ω n -ζ2 4 ζω n = 0 sec,

42 Solutions to Problems a. P(s) = s+0.5 s(s+2)(s+5) = /20 s + /4 s+2-3/0 s+5. Therefore, p(t) = e-2t e-5t. b. To represent the system in state space, draw the following block diagram.

43 6 Chapter 4: Time Response V(s) Y(s) P(s) s 2 s s+0 For the first block,.. y + 7ẏ +0y = v(t) Let x = y, and x 2 = ẏ. Therefore, ẋ = x 2 Also, Thus, ẋ = c. Program: A=[0 ;-0-7]; B=[0;]; C=[.5 ]; D=0; S=ss(A,B,C,D) step(s) ẋ 2 = -0x - 7x 2 + v(t) p(t) = 0.5y + ẏ = 0.5x + x 2 0 x + 0 ; p(t) = 0.5 x -0-7 Computer response: a = x x2 x 0 x b = u x 0 x2 c = x x2 y 0.5 d = u y 0 Continuous-time model.

44 Solutions to Problems a. ω n = 0 = 3.6; 2ζω n = 4. Therefore ζ = %OS = e ξπ / ξ2 *00 = 7.69%. π T s = 4 = 2 seconds. T p = =.28 seconds. From Figure 4.6, T r ω n =.93. ξω ω n ξ 2 n Thus, T r = 0.6 second. To justify second-order assumption, we see that the dominant poles are at 2 ± j The third pole is at -0, or 5 times further. The second-order approximation is valid. K b. G e (s) = (s+0)(s 2 +4s+0) = K s 3 +4s 2. Representing the system in phase-variable form: +50s A = 0 0 ; B = 0 ; C = [ 0 0] K c. Program: numg=00; deng=conv([ 0],[ 4 0]); G=tf(numg,deng) step(g) Computer response: Transfer function: s^3 + 4 s^ s + 00

45 8 Chapter 4: Time Response %OS = (.08-) * 00 = 8% 5. a. ω n = 0.28 = 0.529; 2ζω n =.5. Therefore ζ = x0-2 b. P(s) = U(s) s 2 +.5s+0.28, where U(s) = s. Expanding by partial fractions, P(s) = s + natural response terms. Thus percent paralysis = 54.5%. 7.63x0-2 c. P(s) = s(s 2 +.5s+0.28) = s s s+0.8. Hence, p(t) = e -0.35t e -0.8t. Plotting,

46 Solutions to Problems 9 Fractional paralysis for % isoflurane d. P(s) = K s * for K, K = 3.67%. 7.63x0-2 s 2 +.5s+0.28 = s + natural response terms. Therefore, 7.63x0-2 K 0.28 =. Solving 52. a. Writing the differential equation, dc(t) dt = -k 0 c(t) + i(t) V d Taking the Laplace transform and rearranging, from which the transfer function is found to be For a step input, I(s) = I 0 s (s+k 0 )C(s) = I(s) V d C(s) I(s). Thus the response is I 0 V d C(s) = s(s+k 0 ) V d = s+k 0 I 0 = k 0 V ( d s - s+k ) 0 Taking the inverse Laplace transform, I c(t) = 0 ( e k t 0 ) k 0 V d where the steady-state value, C D, is I 0 C D = k 0 V d Solving for I R = I 0, I R = C D k 0 V d b. T r = 2.2 k ; T s = 4 0 k 0 c. I R = C D k 0 V d = 2 µg ml x 0.07 hr- x 0.6 liters = mg h d. Using the equations of part b, where k 0 = 0.07, T r = 3.43 hrs, and T s = 57.4 hrs.

47 20 Chapter 4: Time Response SOLUTIONS TO DESIGN PROBLEMS Writing the equation of motion, ( f v s + 2)X(s) = F(s). Thus, the transfer function is X(s) F(s) = / f v s + 2. Hence, T s = 4 a = 4 T 2 = 2 f v, or f v = s 2. f v f v / M The transfer function is, F(s) = s 2 + M s + K M. Now, T s = 2 = 4 Re = 4 2M M =. Substituting the value of M in the denominator of the transfer function yields, 4 = 8M. Thus, s 2 + 4s + 4K. Identify the roots s,2 = 2 ± j2 K. Using the imaginary part and π, from which 2 K substituting into the peak time equation yields T p = = π Im = K = Writing the equation of motion, (Ms 2 + f v s +)X(s) = F(s). Thus, the transfer function is X(s) F(s) = / M s 2 + f v M s +. Since T s =0 = 4, ζω ζω n = 0.4. But, f v n M = 2ζω = n 0.8.Also, M from Eq. (4.39) 30% overshoot implies ζ = Hence, ω n =.7. Now, /M = ω 2 n =.248. Therefore, M = Since f v M = 2ζω n = 0.8, f v = Writing the equation of motion: (Js 2 +s+k)θ(s) = T(s). Therefore the transfer function is ζ = θ(s) T(s) = J s 2 + J s+k J - ln ( %OS 00 ) π 2 + ln 2 ( %OS 00 ) = T s = = 4 ζω n 2J = 8J = Therefore J = 2. Also, T 4 4 s = 4 = =. Hence, ω ζω n (0.358)ω n = Now, K n J = ω n 2 = Finally, K = Writing the equation of motion

48 Solutions to Design Problems 2 The transfer function is [s 2 +D(5) 2 s+ 4 (0) 2 ]θ(s) = T(s) θ (s) T(s) = s 2 +25Ds+25 Also, and ζ = - ln ( %OS 00 ) π 2 + ln 2 ( %OS 00 ) = ζω n = 2(0.358)(5) = 25D Therefore D = The equivalent circuit is: where J eq = +( N N ) 2 ; D eq = ( N 2 N ) 2 ; K eq = ( N 2 N ) 2. Thus, 2 θ (s) T(s) = J eq s 2. Letting N +D eq s+k eq N = n and substituting the above values into the transfer 2 function, θ (s) T(s) = +n 2 s 2 + n2 +n 2 s + n2. Therefore, ζω n = +n 2 n 2 2(+n 2 ). Finally, T 4 s = = 8(+n2 ) ζω n n 2 = 6. Thus n =. 59. Let the rotation of the shaft with gear N 2 be θ L (s). Assuming that all rotating load has been reflected to the N 2 shaft, ( J eql s 2 + D eql s + K)θ L (s) + F(s)r = T eq (s), where F(s) is the force from the translational system, r = 2 is the radius of the rotational member, JeqL is the equivalent inertia at the N 2 shaft, and D eql is the equivalent damping at the N 2 shaft. Since J eql = (2) 2 + = 5 and D eql =

49 22 Chapter 4: Time Response (2) 2 = 4, the equation of motion becomes, ( 5s 2 + 4s + K)θ L (s) + 2F(s) = T eq (s). For the translational system (Ms 2 + s)x(s) = F(s). Substituting F(s) into the rotational equation of motion, ( 5s 2 + 4s + K)θ L (s) + ( Ms 2 + s)2x(s) = T eq (s). But,θ L (s) = X(s) r = X(s) 2 and T eq (s) = 2T(s). Substituting these quantities in the equation above yields ((5 + 4M)s 2 + 8s + K) X(s) = T(s). Thus, the transfer function is 4 X(s) T(s) = 4/(5+ 4M) 8 s 2 + (5 + 4M) s + K (5 + 4M). Now, T s =0 = 4 Re = 4 8 = (5 + 4M). 2(5 + 4M) Hence, M = 5/4. For 0% overshoot, ζ = from Eq. (4.39). Hence, 2ζω n = 8 (5 + 4M) = 0.8. Solving for ω n yields ω n = But, ω n = K (5 + 4M) = K 0 = Thus, K = The transfer function for the capacitor voltage is V C(s) V(s) For 20% overshoot, ζ = 92Ω. - ln ( %OS 00 ) Cs 0 = R+Ls+ = 6 s 2 +Rs+0 6. Cs π 2 + ln 2 ( %OS 00 ) = Therefore, 2ζω n = R = 2(0.456)(03 ) = Solving for the capacitor voltage using voltage division, V C (s) = V i (s) /(CS). Thus, the R + LS + CS transfer function is V C(s) V i (s) = /(LC) s 2 + R L s +. Since T s = 4 Re =0 3, Re = R = Thus 2L LC R = 8 KΩ. Also, since 20% overshoot implies a damping ratio of 0.46 and 2ζω n = 8000, ω n = =. Hence, C = 0.03 µf. LC 62. Using voltage division the transfer function is, V C (s) V i (s) = Cs R + Ls + Cs = LC s 2 + R L s + LC

50 Solutions to Design Problems 23 Also,T s = 2x0 3 = 4 Re = 4 R 2L = 8L R. Thus, R L ζ = But, 2ζω n = R/L. Thus, ω n = 3869 = R = Ω. = Using Eq. (4.39) with 5% overshoot, LC = L(0 5. Therefore, L = 6.7 mh and ) 63. For the circuit shown below L = i (t) R = i 2 (t) o write the loop equations as R + L s I s R I 2 s = V i s R I s + R + R 2 + I 2 s = 0 C s Solving for I 2 (s) I 2 s = R + L s V i ( s ) R 0 R + L s R But, V o s = C s I 2 s. Thus, V o V i Substituting component values, V o (s) V i (s) = s s = R R + R 2 + C s R R 2 + R C L s 2 + C R 2 R + L s + R (R )C s 2 + (000000CR 2 + ) (R )C s (R )C 4 For 5% overshoot, ζ = For T s = 0.00, ζω n = 0.00 = Hence, ω n = Thus, or, R C =

51 24 Chapter 4: Time Response Also, C = () R C R 2 + R C = 8000 (2) Solving () and (2) simultaneously, R 2 = Ω, and C =.6573 x 0-2 µf si A = s 0 0 s ( K ) c 0.255x x = s ( K 9 c ) 0.255x x0 s si A = s 2 + ( x0 5 K c )s + (5520K c ) (2ζω n ) 2 = [2*0.9] 2 *(5520K c ) = ( x0 5 K c ) 2 or K 2 c 8.87x0 4 K c 2.022x0 0 = 0 Solving for K c, K c = 8.89x0 4 a. The transfer function from Chapter 2 is, Y h (s) Y cat (s) F up (s) = (s ) (s s )(s s ) The dominant poles come from s s Using this polynomial, 2ζω n = 8.9, and ω n 2 = Thus, ω n = 9.4 and ζ = Using Eq. (4.38), %OS = 5.05%. Also,T s = 4 π = s, and T ζω p = = 0.66 s. To find rise time, use 2 n ω n ζ Figure 4.6. Thus,ω n T r =.236 or T r = s. b. The other poles have a real part of 5.47/2 = Dominant poles have a real part of 8.9/2 = Thus, 7.735/4.06 =.9. This is not at least 5 times. c. Program: syms s numg=0.7883*(s+53.85); deng=(s^2+5.47*s+9283)*(s^2+8.9*s+376.3); 'G(s) transfer function' G=vpa(numg/deng,3); pretty(g) numg=sym2poly(numg); deng=sym2poly(deng); G=tf(numg,deng)

52 Solutions to Design Problems 25 step(g) Computer response: ans = G(s) transfer function.788 s (s s ) (s s ) Transfer function: s s^ s^ s^ e004 s e006 The time response shows 58 percent overshoot, T s = 0.86 s, T p = 0.3 s, T r = 0.05 s.