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I'm looking at the example for convolutions here, and I can't quite understand how the output of a FIR filter can be interpreted. Obviously, it's just the convolution of the impulse response and the input, converting to the frequency domain, modifying the input, and converting back into the time domain. The part I don't get is, the input to the filter is of length $n$, and the output is of length $n + m - 1$.

Obviously, you can only output samples of the same input length, so how would you interpret this? Do you crop off part of the output sequence, do you take a sliding window weighted average, or what?

Could someone provide an example using a filter with input $\{4, -2, 2\}$ and output $\{4, 2, 8, 6, 0, 4\}$ (the example in the link above).

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    $\begingroup$ There is no reason for the output of a filter to be restricted to be the same length as the input. Consider a LPF with an impulse as input: the input lasts for an instant, and the output lasts potentially forever. $\endgroup$ – MBaz Nov 10 '16 at 23:17
  • $\begingroup$ Maybe this could help. $\endgroup$ – msm Nov 12 '16 at 13:45
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When you talk about the application of the Convolution Theorem (i.e. convolution in time is multiplication in frequency), you need to take care of discrete vs continuous transforms: Assuming an impulse response g[n] and an input x[n], the convolution theorem states

$$ y[n] = x[n]*g[n] = F^{-1}(F(x[n])F(g[n])) $$

where F stands for (any) Fourier transform. Now, if you consider $F$ to be the DTFT, you are calculating the linear convolution (the conventional convolution). Here, your output will be of length $n+m-1$, where $n$ is the length of $x[n]$ and $m$ is the length of $g[n]$

However, if F becomes the N-Point DFT (where $N=max(n,m)$), your convolution becomes circulant. Then, obviously, your output is of length N. You can have a look at Circular convolution.

Regarding your example, can you please specify, which part you dont understand?

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