ch05 - FIVE Equivalent Systems SOLUTIONS TO CASE STUDIES...

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F I V E Equivalent Systems SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Designing a Closed-Loop Response a. Drawing the block diagram of the system: Thus, () 32 76.39 151.32 198 76.39 K Ts s ss K = ++ + b. Drawing the signal flow-diagram for each subsystem and then interconnecting them yields:
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5-2 Chapter 5: Equivalent Systems () 12 22 3 33 1 1 3 1 1.32 0.8 10 150 150 0.2 95.49 150 477.46 0.2 ii o xx x x xK q x K K x θ π = =+ ⎛⎞ =− + = + ⎜⎟ ⎝⎠ = ± ± ± In vector-matrix notation, 01 0 0 0 1.32 0.8 0 95.49 0 150 477.46 i KK ⎡⎤ ⎢⎥ + −− ⎣⎦ ± [ ] 0.2 0 0 o = x c. ( ) 1 3 10 1 1 1 76.39 150 0.8 0.2 TK s ss s ⎛ ⎞ == ⎜ ⎟ ⎝ ⎠ 3 3 150 1.32 1 1 1 10 76.39 ; ; 150 0.8 0.2 LL L K GG G K s s s s = = Nontouching loops: 2 198 s = [] [ ] 123 1 2 32 150 1.32 76.39 198 11 LL L L L K GGG G G s sss Δ= − + + + = + + + + 1 1 Δ= 76.39 151.32 198 76.39 T K Ts s K Δ Δ+ + + d. The equivalent forward path transfer function is 10 0.16 1.32 K Gs = + Therefore, 2 2.55 1.32 2.55 s s = ++
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Solutions to Problems 5-3 The poles are located at 0.66 1.454 j −± . 2.55 1.597rad/s ω == n ; 2 1.32 n ζω = , therefore, 0.413 ζ = . 2 /1 % 100 24% OS e ζπ −− = ; 44 6.06seconds 0.66 s n T = ; 2 2.16 1.454 1 p n T π ωζ = seconds; Using Figure 4.16, the normalized rise time is 1.486. Dividing by the natural frequency, 1.486 0.93 2.55 r T seconds. e. () [] 2 2 2 2 0.66 2.55 1.32 2.55 1 1 25 33 25 1.32 2.55 25 0.66 11.347 2.1144 11 25 0.66 2.1144 1 cos 1.454 0.454sin 1.454 t Cs ss s s s s s s ct e t t = ++ + =− + f. Since 0.51 1.32 K Gs = + , 2 0.51 1.32 0.51 K Ts s sK = + + . Also, ( ) 22 ln % 100 0.517 % ln 100 OS OS ⎛⎞ + ⎜⎟ ⎝⎠ for 15% overshoot; 0.51 n K = ; and 2 1.32 n = . Therefore, 1.32 1.32 1.277 0.5 2 2 0.5147 n K = = . Solving for K , 3.2 K = . UFSS Vehicle: Pitch-Angle Control Representation a. Use the observer canonical form for the vehicle dynamics so that the output yaw rate is a state variable.
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5-4 Chapter 5: Equivalent Systems -2 2 -1 -0.125 1 s 1 1 s 1 1 -1 u 1 s 1 y x 1 0.437 1 s 1 -0.24897 -1.483 x 2 x 3 x 4 b. Using the signal flow graph to write the state equations: () 12 22 3 4 32 4 41 2 4 1.483 0.125 0.24897 0.125*0.437 2222 xx x x x x x u = =− + =+−− ± ± ± ± In vector-matrix form: 010 0 0 0 1.483 1 0.125 0 0 0.24897 0 0.054625 0 220 2 2 u ⎡⎤ ⎢⎥ −− =+ ⎣⎦ ± [ ] 1000 y = x c. Program: numg1=-0.25*[1 0.437]; deng1=poly([-2 -1.29 -0.193 0]); 'G(s)' G=tf(numg1,deng1) numh1=[-1 0]; denh1=[0 1]; 'H(s)' H=tf(numh1,denh1) 'Ge(s)' Ge=feedback(G,H) 'T(s)' T=feedback(-1*Ge,1)
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Solutions to Problems 5-5 [numt,dent]=tfdata(T,'V'); [Acc,Bcc,Ccc,Dcc]=tf2ss(numt,dent) Computer response: ans = G(s) Transfer function: -0.25 s - 0.1093 -------------------------------------- s^4 + 3.483 s^3 + 3.215 s^2 + 0.4979 s ans = H(s) Transfer function: -s ans = Ge(s) Transfer function: -0.25 s - 0.1093 -------------------------------------- s^4 + 3.483 s^3 + 3.465 s^2 + 0.6072 s ans = T(s) Transfer function: 0.25 s + 0.1093 ----------------------------------------------- s^4 + 3.483 s^3 + 3.465 s^2 + 0.8572 s + 0.1093 Acc = -3.4830 -3.4650 -0.8572 -0.1093 1.0000 0 0 0 0 1.0000 0 0 0 0 1.0000 0 Bcc = 1 0 0 0 Ccc = 0 0 0.2500 0.1093 Dcc = 0
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5-6 Chapter 5: Equivalent Systems ANSWERS TO REVIEW QUESTIONS 1. Signals, systems, summing junctions, pickoff points 2. Cascade, parallel, feedback 3. Product of individual transfer functions, sum of individual transfer functions, forward gain divided by one plus the product of the forward gain times the feedback gain 4. Equivalent forms for moving blocks across summing junctions and pickoff points 5. As K is varied from 0 to , the system goes from overdamped to critically damped to underdamped. When the system is underdamped, the settling time remains constant.
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ch05 - FIVE Equivalent Systems SOLUTIONS TO CASE STUDIES...

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