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Lee Jae-yeon el 30 de Nov. de 2011
Respondida: insat code el 30 de Oct. de 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!!!
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Hin Kwan Wong el 30 de Nov. de 2011
most probably, delta[n] is the backward shift operator
5π(n-2) is 5*pi*(n-2) I suppose
Jan el 1 de Dic. de 2011
@Hin Kwan Wong: The backward shift operator - good guess!

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Hin Kwan Wong el 30 de Nov. de 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)
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Hin Kwan Wong el 1 de Dic. de 2011
I have already posted other ways, including the use of the convolution function conv.
Hin Kwan Wong el 1 de Dic. de 2011
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 el 30 de Oct. de 2018
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