PSet5Solutions

# PSet5Solutions - Problem Set#5 16.30 Estimation and Control...

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Problem Set #5 16.30: Estimation and Control of Aerospace Systems Posted: April 11, Due: April 18 1 Problem 9.11 in VDV. The continuous-time transfer function is G ( s ) = K s ( s + a ) . Use Tustin’s method to derive the discrete-time transfer function: replace s by 2 /T (1 - z - 1 ) / (1 + z - 1 ). We obtain G ( z ) = K 2 T 1 - z - 1 1 + z - 1 2 T 1 - z - 1 1 + z - 1 + a - 1 = KT 2 2 (1 - z - 1 )(2 - 2 z - 1 + aT + aTz - 1 ) (1 + z - 1 ) 2 - 1 = KT 2 2 ( aT + 2) + ( aT - 2) z - 1 - ( aT + 2) z - 1 - ( aT - 2) z - 2 1 + 2 z - 1 + z - 2 - 1 = KT 2 2 1 + 2 z - 1 + z - 2 ( aT + 2) - 4 z - 1 - ( aT - 2) z - 2 . Given the input sequence ( u 0 , u 1 , . . . ) we can compute the output sequence ( y 0 , y 1 , . . . ), solving the following set of equations iteratively (remember that because of causality we set y i = 0 and u i = 0 for i < 0): ( aT + 2) y i - 4 y i - 1 - ( aT - 2) y i - 2 = KT 2 2 ( u i + 2 u i - 1 + u i - 2 ) . For example, we get: y 0 = KT 2 2( aT + 2) u 0 y 1 = 4 aT + 2 y 0 + KT 2 2( aT + 2) ( u 1 + 2 u 0 ) = 2 KT 2 ( aT + 2) 2 u 0 + KT 2 2( aT + 2) ( u 1 + 2 u 0 ) . . . 2 Problem 9.12 in VDV. The continuous-time transfer function is G ( s ) = K ( s + a )( s + b ) . The pulse transfer function is simply the z -transform of G ( s ), i.e., G ( z ) = Z K ( s + a )( s + b ) = K b - a Z 1 s + a - 1 s + b = K b - a z z - e - aT - z z - e - bT = K b - a ( e - aT - e - bT ) z z 2 - ( e - aT + e - bT ) z + e ( a + b ) T , which translates to the difference equation

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y i = ( e - aT + e - bT ) y i - 1 - e ( a + b ) T y i - 2 + K ( e - aT - a - bT ) b - a u i - 1 . Moreover, for the above problems, implement the algorithms in Matlab, and compare the output to a unit step input to the output from the continuous time system. Choose a = 1 , b = 5 , and T = 0 . 01 , 0 . 1 , 0 . 2 . The following commands in Matlab plot the first 100 samples in the step response for the case in problem 9.11. (Notice that since matlab does not understand negative or zero indices, I offset all indices by three, i.e., u 3 is the first input sample to be equal to 1.) >> a = 1; k=1 ; T=0.01; >> y = zeros(1,100); u = [0 0 ones(1,98)]; >> for i=3:100; y(i) = (K*T^2/2*(u(i)+2*u(i-1)+u(i-2))+4*y(i-1)+(a*T-2)*y(i-2))/(a*T+2); end >> plot(T*(0:97),y(3:100),’b.’) 3 Problem 9.22 in VDV.
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• Spring '07
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