ch05 - F I V E Reduction of Multiple Subsystems SOLUTIONS...

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F I V E Reduction of Multiple Subsystems SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Designing a Closed-Loop Response a. Drawing the block diagram of the system: + - 10 Π i u K 150 s+150 u o 0.16 s(s+1.32) Pots Pre amp Power amp Motor, load and gears Thus, T(s) = 76.39K s 3 +151.32s 2 +198s+76.39K b. Drawing the signal flow-diagram for each subsystem and then interconnecting them yields: Π i u K u o 1 s 1 s -150 -1.32 150 0.8 1 s 0.2 Π - pot pot pre amp power amp motor and load gears x 1 2 x 3 x
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Solutions to Case Studies Challenges 127 x . 1 = x 2 x . 2 = - 1.32x 2 + 0.8x 3 x . 3 = -150x 3 +150K( 10 π (q i 0.2x 1 )) = -95.49Kx 1 - 150x 3 + 477.46K θ i θ o = 0.2x 1 In vector-matrix notation, x . = 01 0 0 -1.32 0.8 -95.49K 0 -150 x + 0 0 477.46K θ i θ o 0 . 200 x c. T 1 = 10 ( K )(150) 1 s (0.8) 1 s 1 s (0.2) = 76.39 s 3 G L 1 = 150 s ; G L 2 = 1.32 s G L 3 = ( K )(150) 1 s (0.8) 1 s 1 s (0.2) 10 = 76.39 K s 3 Nontouching loops: G L1 G L2 = 198 s 2 = 1 - [G L1 + G L2 + G L3 ] + [G L1 G L2 ] = 1 + 150 s + 1.32 s + 76.39K s 3 + 198 s 2 1 = 1 T(s) = T 1 1 = 76.39K s 3 +151.32s 2 +198s+76.39K d. The equivalent forward path transfer function is G(s) = 10 π 0.16K s(s+1.32) . Therefore, T(s) = 2.55 s 2 +1.32s+2.55 The poles are located at -0.66 ± j1.454. ω n = 2.55 = 1.597 rad/s; 2 ζω n = 1.32, therefore, ζ = 0.413.
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128 Chapter 5: Reduction of Multiple Subsystems % OS = e ζπ /1 ζ 2 x 100 = 24%; T s = 4 ζω n = 4 0.66 = 6.06 seconds; T p = π ω n 1- ζ 2 = π 1.454 = 2.16 seconds; Using Figure 4.16, the normalized rise time is 1.486. Dividing by the natural frequency, T r = 1.486 2.55 = 0.93 seconds. e. f. Since G(s) = 0.51K s(s+1.32) , T(s) = 0.51K s 2 +1.32s+0.51K . Also, ζ = - ln ( %OS 100 ) π 2 + ln 2 ( %OS 100 ) = 0.517 for 15% overshoot; ω n = 0.51K ; and 2 ζω n = 1.32. Therefore, ω n = 1.32 2 ζ = 1.32 2(0.5147) = 1.277 = 0.51K . Solving for K, K=3.2. 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. -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:
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Solutions to Case Studies Challenges 129 Ý x 1 = x 2 Ý x 2 =− 1.483 x 2 + x 3 0.125 x 4 Ý x 3 0.24897 x 2 (0.125* 0.437) x 4 Ý x 4 = 2 x 1 + 2 x 2 2 x 4 2 u In vector-matrix form: Ý x = 010 0 0 1.483 1 0.125 0 0.24897 0 0.054625 220 2 x + 0 0 0 2 u y = 10 00 [] 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) [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
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130 Chapter 5: Reduction of Multiple Subsystems 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 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.
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This note was uploaded on 04/04/2008 for the course MECH COntrol Sy taught by Professor Khurshid during the Spring '08 term at Michigan State University.

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ch05 - F I V E Reduction of Multiple Subsystems SOLUTIONS...

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