New answers tagged dft
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The FFT of a strictly real input results in a conjugate mirrored result. For a sinewave, both a positive and a negative frequency peak will appear. If the input is not integer periodic over the full aperture or is zero-padded, the two response peaks will be Sinc shaped, with lots of extended ripples in the frequency domain. The positive and negative ...
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A frequency bin 'only' covers the value of its center frequency (+ potential spectral leaks to that center frequency). Is that correct?Is that correct?
No. All frequencies show up in all frequency bins EXCEPT frequencies that are an integer multiple of $f_{\Delta}=f_s/N_{FFT}$ sample rate divided by FFT length.
In other words a frequency bin is affected ...
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The frequency response of a single FFT bin (when using the default rectangular window) is that of a periodic Sinc or Dirichlet function. If using a non-rectangular window (Hann or Hamming, etc.) the frequency response of each FFT bin is that of the transform of the window applied before the FFT.
It's not the average of the signal energy within the bin width,...
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I'll start with a counter-example:
Assume your continuous-time signal is a sinusoid at freqeuncy $f_0$. You then sample that signal, and use a rectangular window to select $N$ samples. Feed those $N$ samples to your DFT, and you will see non-zero values in the whole spectrum (unless in the special cases where $f_0$ is a multiple of $f_s/N$).
So you start ...
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You are right, the DFT bins correspond to frequencies $f_k=\frac{k}{N}f_s$. In order to see this let's consider finite length signals with potentially non-zero elements in the index range $[0,N-1]$. In this case, the DFT is just a sampled version of the DTFT (discrete-time Fourier transform):
$$\textrm{DTFT:}\quad X(e^{j\omega})=\sum_{n=0}^{N-1}x[n] e^{-jn\...
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Depends a bit on the indexing convention, but in typically you would interpret the frequency interval as $[-f_s/2, f_s/2]$ and not as $[0,f_s]$. The DFT is periodic in N so you you have
$$X(N-1) = X(-1) $$
Keep in mind that that the sampling theorem requires the signal to be band limited to $f_s/2$ so assuming that you have actual independent information ...
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Could you develop on the usage of integer frequencies?
When dealing with sampled discrete signals, there can only be an integer amount of resolvable frequencies within a captured interval.
When you take an $x(t), t \in \mathbb{R}$ and apply sample-and-hold on it at some sampling frequency $Fs$, it is turned into an $x \left[n \cdot \frac{1}{Fs} \right ], n ...
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I would like to give another take on @DanielSank's answer. We first suppose that $v_{n} \sim \mathcal{CN}(0, \sigma^{2})$ and is i.i.d. Its Discrete Fourier Transform is then:
$$ V_{k} = \frac{1}{N} \sum_{n=0}^{N-1} v_{n} e^{-j 2 \pi \frac{n}{N} k}$$.
We want to calculate the distribution of $V_{k}$ To start, we note that since $v_{n}$ is white Gaussian ...
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