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I have dt signal $x[n]={[6.29, 8.11,-7.46,8.26,2.64,-8.04,-4.43,0.93,-9.29]}$

And I need to give the function value of:

1) sum of $x[n] = \sum\limits_{k=0}^{N-1} X[k] \, e^{-j 3 \pi k/5}$ from k=0 to 9

How can I find that

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  • $\begingroup$ hi. Your question is practically unreadable. Please fix that and explain why you haven't been able to simply apply the formula. $\endgroup$ – Marcus Müller Feb 21 at 23:04
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I don't know if I understand exactly what you mean, because your question is very unclear (you may need to edit it), but I will try to answer what you may need in your task.

For DFT you will need this equation:

$ X[k] = \sum\limits_{n=0}^{N-1} x[n] \, e^{-j 2 \pi nk/N} $

where $k = 0, 1, . . . , N − 1$

But I guess you will be asking for IDFT more, so first you need to understand how IDFT works. I will try to explain this in my picture:

enter image description here

So the formula will look like this:

$ x[n] = \frac{1}{N} \sum\limits_{k=0}^{N-1} X[k] \, e^{j 2 \pi nk/N} $

where $n = 0, 1, . . . , N − 1$

Note that here, x[n] will be our complex time series, and X[k] is complex frequencies.

Now, we can get your sum, thanks to a simple code:

x = [6.29 8.11 -7.46 8.26 2.64 -8.04 -4.43 0.93 -9.15 9.29]; 
XF = fft(x);
k =  0:9;
YF = exp(-i*3*pi*k/5).*XF;
output = [sum(YF)];
disp(output)

And, as we can see, the sum equals 9.3.

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