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So let's say I'm given a sequence of samples of a continuous time signal and I take the $FFT$ of that at 64 points. And so I have my frequency graph which is interpolated along those 64 points. So basically that's a vector $X$ of length 64, and so applying a linear filter to that is essentially performing coordinate-wise multiplication of $X$ with my specially designed filter vector $H$ of length 64, to produce my new filtered vector $Y$ of length 64, is that correct?

And so then I take the $Inverse FFT$ of $Y$ to get my filtered signal in the time domain, and my filter coefficients will be given by the $InverseFFT$ of $H$, so there will be 64 of them, is that correct?

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    $\begingroup$ None of the assertions about which you ask "Is that correct?" are correct. $\endgroup$ – Dilip Sarwate May 5 '12 at 3:00
  • $\begingroup$ I meant to say coordinate-wise multiplication, does that change things? $\endgroup$ – Zaubertrank May 5 '12 at 3:22
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    $\begingroup$ Unfortunately, No. There seem to be serious issues in your understanding of this problem as well as many others in your recent questions. Do you have access to a good book on signal processing and the time to read it? $\endgroup$ – Dilip Sarwate May 5 '12 at 3:25
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    $\begingroup$ Yes I think that's what I'm asking, I've been under the impression that multiplication in the frequency domain is equivalent to convolution in the time domain, where you are convolving x(n) with h(n) (your filter coefficients, I thought!). Thus I thought if you have H(z) then you take the inverse transform to get h(n). $\endgroup$ – Zaubertrank May 5 '12 at 4:06
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    $\begingroup$ @ Zaubertrank : This is one of those cases where the "devil is in the details" and the many details are explained in the books you haven't yet read. $\endgroup$ – hotpaw2 May 5 '12 at 4:14
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One detail that is rendering your statements incorrect is that vector multiplication in one domain isn't linear convolution in the other domain. It's circular convolution in the other domain.

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