could you please make a complete code about the following question???

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Lee Jae-yeon 2011-11-30

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回答： insat code 2018-10-30

采纳的回答： Hin Kwan Wong

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this question is related to Convoution in Signal...

X[n]*h[n] = [20cos(5πn+ π/3)+cos(200πn)] * 1/3 ( delta [n-2]+ delta [n-1]+ delta [n] )
= 20/3cos[5π(n-2)+ π/3]+ 20/3cos[5π(n-1)+ π/3]+ 20/3cos[5π(n)+ π/3]+
1/3cos[200π(n-2)]+ 1/3cos[200π(n-1)]+ 1/3cos[200π(n)]
//////////////  But, n should be from 0 to 299

Unfortunately, I cannot do creat matlab code about this question...

Please solve this problem, matlab EXPERT!!!

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Hin Kwan Wong 2011-11-30

most probably, delta[n] is the backward shift operator

5π(n-2) is 5*pi*(n-2) I suppose

Jan 2011-12-1

@Hin Kwan Wong: The backward shift operator - good guess!

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Hin Kwan Wong 2011-11-30

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It's not hard of a question if I understand what Lee means as it stands From the first part: X[n]*h[n]=[20cos(5πn+ π/3)+cos(200πn)] * 1/3 ( delta [n-2]+ delta [n-1]+ delta [n] )

Clearly implies h the impulse response is 1/3(z^-2 + z^-1+1) , which is just a moving average filter.

X is a signal 20cos(5πn+ π/3)+cos(200πn)]

you can find the result by

N=2,step=0.01
h = [1/3 1/3 1/3];
t=0:step:N;
X=20*cos(5*pi*t+pi/3)+cos(200*pi*t)
Y=filter([1/3 1/3 1/3],1,X)

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Hin Kwan Wong 2011-12-1

I have already posted other ways, including the use of the convolution function conv.

Hin Kwan Wong 2011-12-1

you can even use fast fourier transform

ifft(fft([1/3;1/3;1/3;zeros(length(X)-3,1)]).*fft(X))

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insat code 2018-10-30

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