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

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!!!

[Jan]

We cannot know what "delta[n]" and "5π(n-2)" means.

I suggest to read the Getting Started chapters of the documentation and learn how to use Matlab syntax. Otherwise even the smallest modifications of the posted source code would be an impossible task for you.

Hint: for n=0:299, ... end

[Hin Kwan Wong]

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

5π(n-2) is 5*pi*(n-2) I suppose

[Jan]

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

Answer 1

[Hin Kwan Wong]

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)

[Lee Jae-yeon]

in fact,I haven't learnt 'filter'.... so is there another way to solve this problem???

[Jan]

@Lee Jae-yeon: If you haven't learnt it, learn it now: "help filter" and "doc filter" contain a sufficient number of information.

[Hin Kwan Wong]

You can also use the convolution command to do the same thing

conv(h,X)

But it will give you 2 more values at the end due to transient.

Also, for both methods there will be a starting transient because it is not a steady state solution.

Since X is periordic, it may be beneficial to use

Y=filter([1/3 1/3 1/3],1,repmat(X(:),2,1));

X=(length(X)/2+1:end);

Which takes the 2nd period of the result to eliminate the transient.

[Lee Jae-yeon]

But the condition is that the problem cannnot be used filter utility... Then there are other ways??

[Hin Kwan Wong]

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

[Hin Kwan Wong]

you can even use fast fourier transform

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

Answer 2

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