I'm just beginning to study Fourier transforms and I'm having trouble getting this concept down. I know what a low/high pass filter does to a signal.

  • What is the relationship between Fourier magnitudes (dB) and frequency (Hz)?
  • How would Fourier magnitudes get affected when a low/high pass filter is passed on a signal?
  • How about an all pass filter?
  • $\begingroup$ You can try to google "Bode Diagram" and stare at some of the pictures that come up. These show the change of magnitude and phase with frequency. $\endgroup$
    – Hilmar
    Oct 29, 2013 at 12:03

2 Answers 2


(This explanation is going to be rough around the edges so that I can build your intuition. There are subtleties involved, but this should be good enough to start you off).

Forget about the dB scale for a moment. Let us take a step back:

Imagine that you have some signal that occupies a bandwidth from say, 50 Hz to 100 Hz. Its spectrum is flat across this band. That is, it has the same power at 50 Hz, as it does at 51 Hz, as it does at 52 Hz, etc etc. In this example, we say that it has a constant (flat) power-spectral-density within this band. Let us say that the value of the PSD in this band is 1 Watts / Hz. This means that across 1 Hz of frequency, we will measure 1 Watt in power. Across 50 to 100 Hz, we would measure of course, 50 Watts in power. (To compute your PSD values, they are simply the absolute value of your fourier amplitudes, squared).

So far so good.

Now, imagine a filter, that removes frequencies above 75 Hz and also removes frequencies below 50 Hz. It however keeps frequencies between 50 and 75 Hz. To "keep" a frequency, means it multiplies the power of a signal at that frequency by 1. To "remove" a frequency, means it multiplies the power of a signal at that frequency by 0.

So, you have a signal that is 1 Watt/Hz from 50 to 100 Hz. And you pass it through a filter that multiplies all frequency powers from 50 to 75 by 1, ("keeps them"), and it also multiplies all frequency powers less than 50 and above 75 Hz by 0. ("removes them").

If you pass your signal through this filter, what might you get? Well, you just removed all the energy of your signal from 75 to 100 Hz. Previously those frequencies had a power spectral density of 1 Wattage/Hz, however now they are 0. You have just filtered your signal.

If I asked you at this point, "Plot for me, the PSD of your signal after your filtering", you would draw 1's for frequencies in the 50 to 75 Hz range, and draw 0's for all other frequencies. Now if I said, "I dont like this linear scale. I want you to draw this for me on a dB scale. A log scale." This means that you now simply take the log of the PSD axis. So everywhere we had 0 watts/Hz, they are now -infinity watts/Hz dBW/Hz, (db Watts), and everywhere you had 1 Watt/Hz, we now have 0 dBW /Hz.

Taking it so dB scale doesn't change your frequency axis, only your fourier amplitude/power axis.

  • $\begingroup$ I think you should try to use more precise language here. There isn't "1 W" of power at any particular frequency. What you're probably getting at is that the power spectral density is constant across the 50 to 100 Hz band. You could say that the signal has a spectral density of 1 W/Hz over that band, but that doesn't mean that there is 1 W of power at 50 Hz, or 52 Hz, or 76.323 Hz. It means that if you integrate the spectral density over 1 Hz between 50 and 100 Hz, you would get a total power of 1 W. $\endgroup$
    – Jason R
    Oct 29, 2013 at 14:39
  • $\begingroup$ @JasonR Edited. $\endgroup$ Oct 29, 2013 at 15:00

fourier magnitudes in dB is just the logarithm of the fourier transform of the signal of interest. For example, Let's say you have an ideal low pass filter which looks like a square function centered at 0, and between -f and f in the fourier domain. (Note: f and -f are the frequencies). Now the fourier magnitudes in db will be 0 between f and -f and -$\inf$ outside that interval.

But in practice we cannot design an ideal filter so we wouldn't necessarily be having -$\inf$ outside the bounds and mainly we cannot have a square function in fourier domain.

  • $\begingroup$ Thanks for the explanation. I'm still having trouble visualizing how a filter would affect the magnitude. How would you think would each filter affect a dB vs Hz graph? $\endgroup$ Oct 29, 2013 at 7:27

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