This is taken from Analog-to-Digital Conversion by Marcel Pelgrom. enter image description here

enter image description here

Okay, I understand that with 1-bit quantization, the output is block-wave (Square) and that of course when you do the Fourier Series, has harmonics. That makes sense.

What I don't understand is, how come when we go to higher quantization bit levels (2, 3, 4, 5), the harmonic power decreases? I mean there is still square/block waves asssociated with those, just with smaller jumps.

TLDR: Why does increasing number of quantization bits, decrease the harmonic power?


1 Answer 1


The total quantization noise is proportional to the difference between the fundamental or original waveform and the quantized one. The more steps, the likelier one can get a closer fit or approximation to the waveform. So the jumps are still there and will still produce harmonics, but there is less energy available for the jumps to put into the spectrum which those jumps produce, as more of the energy is in the better approximated fundamental. Conservation of energy, or Parseval’s theorem limits the total power.

  • \$\begingroup\$ Hmm. I see, so because there's a closer fit to the real data, more energy appears in the fundamental and less energy is available for the harmonics? I've never looked at it like that. \$\endgroup\$ Commented Dec 26, 2019 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.