- CIn(100 r:100 N-1 CIn(100 r-1:100 N-2 CIn(100 r-2:100 N-3 CIn(100 r-3:100 N-4 b = CFout1(100 r 1:100 N prmtrs = A\b A least squares solution is

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% part a) in order to obtain a least squares solution obtain the matrix A. % From the simulink document WienerHammersteinConvolutionOfTwoFilters we % obtain data. Before tha % both filters. This is the question. It seems that changing the cascade % row does not change the row of the roots!!!!! %clear all N = 300; %# of training data r = 9; CFout1 = sim_out1.signals.values; % output of convolution filter 1 and filter 2 CFout2 = sim_out2.signals.values; % output of convolution filter 2 and filter 1 CIn = sim_in.signals.values; % input to the both systems. A = [-CFout1(100+r:100+N-1) -CFout1(100+r-1:100+N-2) -CFout1(100+r-2:100+N-3). .. -CFout1(100+r-3:100+N-4) -CFout1(100+r-4:100+N-5) -CFout1(100+r-5:100+N-6). .. -CFout1(100+r-6:100+N-7) -CFout1(100+r-7:100+N-8) CIn(100+r+1:100+N)
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Unformatted text preview: ... CIn(100+r:100+N-1) CIn(100+r-1:100+N-2) CIn(100+r-2:100+N-3) ... CIn(100+r-3:100+N-4) ]; b = CFout1(100+r+1:100+N); prmtrs = A\b % A least squares solution is obtained. prmtr_a = prmtrs(1:r-1), prmtr_b = prmtrs(r:end) p % for the convolution of filter 2 and filter 1 A2 = [-CFout2(100+r:100+N-1) -CFout2(100+r-1:100+N-2) -CFout2(100+r-2:100+N-3). ..-CFout2(100+r-3:100+N-4) -CFout2(100+r-4:100+N-5) -CFout2(100+r-5:100+N-6). ..-CFout2(100+r-6:100+N-7) -CFout2(100+r-7:100+N-8) CIn(100+r+1:100+N) ... CIn(100+r:100+N-1) CIn(100+r-1:100+N-2) CIn(100+r-2:100+N-3) ... CIn(100+r-3:100+N-4) ]; bp2 = CFout2(100+r+1:100+N); prmtrs2 = A2\bp2 % A least squares solution is obtained. prmtr2_a = prmtrs2(1:r-1), prmtr2_b = prmtrs2(r:end)...
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This note was uploaded on 07/04/2011 for the course ECE 501 taught by Professor Deniz during the Spring '11 term at Istanbul Universitesi.

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