ECE535_10S_hw7soln

ECE535_10S_hw7soln - ECE 535 DISCRETE TIME SYSTEMS HOMEWORK...

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ECE 535 DISCRETE TIME SYSTEMS SPRING 2010 HOMEWORK ASSIGNMENT #7 Solutions 1. (a) We have (10 s 2 + s ) Y ( s ) = U ( s ) ¨ y ( t ) + 1 10 ˙ y ( t ) = u ( t ) ˙ x 2 ( t ) + 1 10 ˙ x 2 ( t ) = u ( t ) so that ˙ x 2 ( t ) = - 1 10 ˙ x 2 ( t ) + u ( t ) . Also, ˙ x 1 ( t ) = x 2 ( t ) . Therefore ˙ x ( t ) = 0 1 0 - 1 10 x ( t ) + 0 1 10 u ( t ) y ( t ) = £ 1 0 / x ( t ) . (b) The Matlab code F = [0 1 ; 0 -0.1] ; G = [0 ; 0.1] ; H = [1 0 ]; J = [0]; [A,B,C,D] = c2dm(F,G,H,J,1); cl_poles = exp([-0.5+j*sqrt(3)/2 -0.5-j*sqrt(3)/2]); K = place(A,B,cl_poles) [y,t] = step(ss(A-B*K,B,C-D*K,D,1)); plot(t,y,’-’,t,y,’o’) xlabel(’time (samples)’),ylabel(’y(k)’) determines K = £ 6 . 1157 8 . 6494 / and generates the step response shown in figure 1. (c) The Matlab command L = ( place(A,C,[0 0]) )’ determines L = £ 1 . 9048 0 . 8603 / T .
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0 2 4 6 8 10 12 14 16 18 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 time (samples) y(k) Figure 1: Step response for problem 1(b). The estimated state then evolves according to
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ECE535_10S_hw7soln - ECE 535 DISCRETE TIME SYSTEMS HOMEWORK...

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