My question is around down conversion of a signal (e.g. RF) to DC and the affect of negative frequencies.

If I have a signal whos spectral content is say between 2-4kHz and I mix this with an intermediate frequency of 3kHz, ignoring the upper mirror, I now have a signal whos spectral content is centered around DC +- 1kHz with a bandwidth of 2kHz.

What does the signal at -1Khz to DC represent- I believe this is mirrored onto the DC-1kHz signal but does this create artifacts if the spectral content is asymmetric?

  • $\begingroup$ Your signal's spectrum also has spectral content in the range $[-4, -2]$ kHz and this spectral content also gets demodulated down to the range $[-1, 1]$ kHz. If the spectral content is not symmetric about the $3$ kHz "carrier frequency, then this stuff coming up from $[-4, -2]$ kHz (which is asymmetric the other way) exactly cancels out the asymmetry that concerns you. That is, there are no artifacts and everything is hunkydory.. $\endgroup$ Nov 24, 2016 at 22:02
  • $\begingroup$ Thank you, that makes sense. This was the image that got me confused as it seems to suggest signal "corruption"? $\endgroup$ Nov 24, 2016 at 22:28

1 Answer 1


"If the signal spectrum is asymmetric, the original signal spectrum will be corrupted"

If your mixer is a simple multiplication with a single oscillator, then yes, that will be the case.

As Dilip explained, real (RF) signals are always symmetrical in spectrum to the 0Hz axis. That means your original doesn't only exist around 3kHz, it also exists around -3kHz, only mirrored!

I think your image captures that pretty fine:


Now, if you shift your 3kHz signal (black) to 0Hz, the higher spectral peak ends up on the positive frequency side, and the lower on the negative frequency side.

Problem is that for real-valued signals, there's no physical difference between positive and negative frequencies – and thus, your mirrored signal also gets overlayed with this. So you have the small and the big spectral peak overlay - and since they were originally two different parts of your signal, this overlay is not actually just a "shifted copy" of your original signal, but a "corrupted" version.

The only way this wouldn't happen is if the "upper" half of your signal (3-4 kHz) and the "lower" half (2-3 kHz) were already symmetrical in spectrum to its center frequency – because then, the your mirrored spectrum would be identical to the original, and nothing was lost.

But: simply read on. The fact that your text uses negative frequencies definitely works towards showing you how a complex baseband mixer works. In two or three pages, you'll understand what needs to be done to still be

  • $\begingroup$ Thanks for your answer, my signal is not symmetric around the center frequency so I believe this may have undesired effects- perhaps spectral inversion is a potential solution - I will continue reading on the subject. $\endgroup$ Nov 25, 2016 at 10:51

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