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20105ee103_1_hw7_sols

# 20105ee103_1_hw7_sols - L Vandenberghe EE103 Homework 7...

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Unformatted text preview: L. Vandenberghe 11/18/10 EE103 Homework 7 solutions 1. Exercise 11.3. As in exercise 11.2, we can express the end state of the first vehicle as a function of the actuator input, using the recursion s ( t + 1) = Fs ( t ) + gu ( t ) , s (0) = 0 , where F = bracketleftBigg 1 1 0 0 . 95 bracketrightBigg , g = bracketleftBigg . 1 bracketrightBigg . The end state is s ( N ) = Bu where B = bracketleftBig F N − 1 g F N − 2 g ··· Fg g bracketrightBig , u = ( u (0) , u (1) , . . . , u ( N − 1)) . Similarly, if we define G = bracketleftBigg 1 1 0 0 . 8 bracketrightBigg , h = bracketleftBigg . 2 bracketrightBigg , then the end state of the second vehicle is p ( N ) = bracketleftBigg 1 bracketrightBigg + Cv where C = bracketleftBig G N − 1 h G N − 2 h ··· Gh h bracketrightBig , v = ( v (0) , v (1) , . . . , v ( N − 1)) . To formulate the problem as a least-norm problem we define a 2 N-vector of variables x = ( u (0) , u (1) , ··· , u ( N − 1) , v (0) , v (1) , ··· , v ( N − 1)) . We are asked to minimize bardbl x bardbl 2 subject to the three conditions s 1 ( N ) = p 1 ( N ) , s 2 ( N ) = 0 , p 2 ( N ) = 0 . These are three linear equations in x , and equivalent to Ax = b with A = B 21 B 22 ··· B 2 ,N − 1 B 2 N ··· ··· C 21 C 22 ··· C 2 ,N − 1 C 2 N B 11 B 12 ··· B 1 ,N − 1 B 1 N − C 11 − C 12 ··· − C 1 ,N − 1 − C 1 N b = 1 . The first equation is the condition s 2 ( N ) = 0. The second equation states p 2 ( N ) = 0. The third equation states s 1 ( N ) = p 1 ( N ). The Matlab code is as follows. 1 N = 20; F = [1 1; 0 0.95]; g = [0;0.1]; G = [1 1; 0 0.8]; h = [0;0.2]; B = zeros(2, N); C = zeros(2, N); B(:,N) = g; C(:,N) = h; for t = N-1:-1:1 B(:,t) = F*B(:,t+1); C(:,t) = G*C(:,t+1); end; A = [ B(1,:),-C(1,:); B(2,:), zeros(1,N); zeros(1,N), C(2,:)]; b = [1;0;0]; x = A’*((A*A’)\b); u = x(1:N); v = x(N+[1:N]); % simulate the two systems and plot s = zeros(2,N+1); p = zeros(2,N+1); p(1,1) = 1; for t = 1:N s(:,t+1) = F*s(:,t) + g*u(t); p(:,t+1) = G*p(:,t) + h*v(t); end; plot(0:N, s(1,:), ’-’, 0:N, p(1,:), ’--’) The solid line in the figure is s ( t ). The dashed line is p ( t )....
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• Spring '08
• VANDENBERGHE,LIEVEN
• Singular value decomposition, Triangular matrix, Linear least squares, QR factorization, Matrix decomposition

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20105ee103_1_hw7_sols - L Vandenberghe EE103 Homework 7...

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