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Lee Jae-yeon on 30 Nov 2011
Answered: insat code on 30 Oct 2018
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
Please solve this problem, matlab EXPERT!!!
Jan on 1 Dec 2011
@Hin Kwan Wong: The backward shift operator - good guess!

Hin Kwan Wong on 30 Nov 2011
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)
Hin Kwan Wong on 1 Dec 2011
you can even use fast fourier transform
ifft(fft([1/3;1/3;1/3;zeros(length(X)-3,1)]).*fft(X))

insat code on 30 Oct 2018
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