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The frequency spectrum of a time-domain signal is a representation of that signal in the frequency domain.

2
votes
I don't know that I've heard that term before, but I would interpret the term "moment" as having an analogous meaning to the physical concepts of center of mass and first and second moments of area: …
answered Feb 7 '12 by Jason R
1
vote
This seems to be more a question of style than anything really based in theory if I read this correctly. It's not clear exactly what you want. What information would you hope that your user could extr …
answered Oct 21 '12 by Jason R
9
votes
You're probably referring to the zoom FFT. It's essentially a technique that allows for complexity reduction in the case where you have a small portion of a larger band that you'd like to analyze at h …
answered Mar 4 '13 by Jason R
5
votes
One approach is simply to calculate the frequency vector for the unshifted DFT output (i.e. what you would get directly from MATLAB's fft() function, without doing an fftshift()), then remap the frequ …
answered Jul 26 '12 by Jason R
8
votes
For a real signal, content at the negative frequencies generated by using the DFT is redundant. This is due to the well-known property of real signals with respect to the Fourier transform family: the …
answered Oct 17 '11 by Jason R
6
votes
They do "distort" the signal, as you noted. Multiplying by a window function does not "retain" the spectrum of the original signal. Recall the multiplication property of the Fourier transform: $$ x(t …
answered Mar 17 '13 by Jason R
4
votes
The Goertzel algorithm (which is really just an efficient way of calculating what amounts to a single DFT bin at an arbitrary location) is defined for complex input, just like the DFT. A real input si …
answered Aug 16 '12 by Jason R
3
votes
Strictly speaking, according to the sampling theorem, you would need to sample at a rate that is at least twice the bandwidth of your signal (which is around 960 Hz in your example). The low end of yo …
answered Oct 7 '13 by Jason R
3
votes
It is common for low-cost direct-conversion (zero-IF) RF receivers to have a DC offset in their A/D hardware, which would correspond to a large spike at the frequency that you're tuned to. That's prob …
answered Jan 29 '16 by Jason R
17
votes
Your understanding is correct. If you sample at rate $f_s$, then with real samples only, you can unambiguously represent frequency content in the region $[0, \frac{f_s}{2})$ (although the caveat that …
answered Nov 20 '11 by Jason R
1
vote
At a quick glance, it looks like you're right. The impulse with an area of $1$ at zero frequency respresents a DC offset of 1. The $0.25 - j0.43$ complex value represents the amplitude and phase of th …
answered Nov 29 '16 by Jason R
3
votes
One advantage of a polyphase filterbank approach is, as you guessed, that you can control the frequency response of each channel. When using a DFT alone, you have limited control over the frequency ba …
answered Jun 16 '15 by Jason R
5
votes
The "timestep" $dt$ (using your nomenclature) between DFT windows can really be anything you choose. There are a few different cases to consider: $dt < N$: In this case, there is overlap between suc …
answered Apr 5 '13 by Jason R
6
votes
You will see lines in the spectrum of a phase-modulated signal if there is some periodicity in the modulating signal. If the modulating signal is not periodic, then the power spectrum of the modulated …
answered Nov 1 '11 by Jason R
0
votes
As you noted in your comment, the normalized frequency range [0, 1) cycles/sample corresponds to the range [0, 2*pi) radians/sample. Therefore, given a frequency in the [0, 1) range, simply multiply b …
answered Aug 30 '16 by Jason R

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