ch12 - T W E L V E Design via State Space SOLUTION TO CASE...

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T W E L V E Design via State Space SOLUTION TO CASE STUDY CHALLENGE Antenna Control: Design of Controller and Observer a. We first draw the signal-flow diagram of the plant using the physical variables of the system as state variables. Writing the state equations for the physical variables shown in the signal-flow diagram, we obtain z . = 01 0 0 -1.32 0.8 00- 1 0 0 z + 0 0 2000 u ; y = 0.2 0 0 z The characteristic polynomial for this system is s 3 + 101.32s 2 + 132s + 0. Hence, the A and B matrices of the phase-variable form are Ax Bx 0100 0010 0 -132 -101.32 1 Writing the controllability matrices and their determinants for both systems yields
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Solution to Case Study Challenge 523 CMz Controllability Matrix of z CMx Controllability Matrix of x 0 0 1600 0 0 1 0 1600 -162112 0 1 -101.32 2000 -200000 20000000 1 -101.32 10133.7424 Det(CMz) -5.12E+09 Det(CMx) -1 where the system is controllable. Using Eq. (12.39), we find the transformation matrix and its inverse to be P Transformation Matrix z=Px PINV 1600 0 0 0.000625 0 0 0 1600 0 0 0.000625 0 0 2640 2000 0 -0.000825 0.0005 The characteristic polynomial of the phase-variable system with state feedback is s 3 + (k 3 + 101.32)s 2 + (k 2 + 132)s + (k 1 + 0) For 15% overshoot, T s = 2 seconds, and a third pole 10 times further from the imaginary axis than the dominant poles, the characteristic polynomial is (s + 20)(s 2 + 4s + 14.969) = s 3 + 24s 2 + 94.969s + 299.38 Equating coefficients, the controller for the phase-variable system is Kx Controller for x 299.38 -37.031 -77.32 Using Eq. (12.42), the controller for the original system is Kz Controller for z 0.1871125 0.04064463 -0.03866 b. Using K z , gain from θ m = - 0.1871125 (including gear train, pot, and operational amplifier); gain from tachometer = - 0.04064463; and gain from power amplifier output = 0.03866.
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524 Chapter 12: Design via State Space c. Using the original system from part (a) and its characteristic polynomial, we find the observer canonical form which has the following A and C matrices: Ax -101.32 1 0 -132 0 1 00 0 Cx 100 Writing the observability matrices and their determinants for both systems yields OMz Observability Matrix of z OMx Observability Matrix of x 0 . 200100 0 0.2 0 -101.32 1 0 0 -0.264 0.16 10133.7424 -101.32 1 Det(OMz) 0.0064 Det(OMx) 1 where the system is observable. Using Eq. (12.89), we find the transformation matrix and its inverse to be P Transformation Matrix z=Px PINV 5 0 0 0.20 0.00 0.00 -506.6 5 0 20.26 0.20 0.00 62500 -625 6.25 26.40 20.00 0.16 The characteristic polynomial of the dual phase-variable system with state feedback is
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Solution to Case Study Challenge 525 s 3 + (l 1 + 101.32)s 2 + (l 2 + 132)s + (l 3 + 0) For 10% overshoot, ω n = 10 14.969 = 38.69 rad/s, and a third pole 10 times further from the imaginary axis than the dominant observer poles, the characteristic polynomial is (s + 228.72)(s 2 + 45.743s + 1496.916) = s 3 + 274.46s 2 + 11959s + 3.4237x10 5 Equating coefficients, the observer for the observer canonical system is Lx Observer for x 173.14 11827 342370 Using Eq. (12.92), the observer for the original system is Lz Observer for z 865.7 -28577.724 5569187.5 d.
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526 Chapter 12: Design via State Space e. Program: 'Controller' A=[0 1 0;0 -1.32 0.8;0 0 -100]; B=[0;0;2000]; C=[0.2 0 0]; D=0; pos=input('Type desired %OS '); Ts=input('Type desired settling time '); z=(-log(pos/100))/(sqrt(pi^2+log(pos/100)^2)); wn=4/(z*Ts); %Calculate required natural %frequency.
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ch12 - T W E L V E Design via State Space SOLUTION TO CASE...

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