Re: DFT point = decimated filter output?

On Apr 9, 9:47 am, "NewLine" <umts_remove_this_and_t...@xxxxxxxxx>
wrote:
Hi,

I am struggling with some FFT interpretation.

To me it looks that each sample of an FFT can be considered as the output of
a process consiting of downconversion, fitlering ,decimating.
e.g .the DC bin output is calculated by:
- downconverting the signal by 0 (hence doing nothing)
- applying a very simple filter that has N taps equal to 1 (for an FFT of
size N).
- decimating the filtered signal by N (hence aliasing some content in the
due to far from perfect filter)

Is this an OK way of thinking? This makes things like leakage, windowing
somewhat easier to grab for me.


The best way to look at the DFT is to consider it as a projection.
It's right there in the formula:

X_k = \sum_{n=0}^{N-1} x[n] \exp(-j 2\pi n k / \Omega).

Write it out as inner product X_k = < x[n], f_k[n] >. Now,
what is the correct f_k[n] ?

Remember that the DFT is for periodic, discrete-time signals,
so taking a finite segment of a signal and taking the DFT
of that segment can be a two-edged sword!

Julius

.

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