We know that a linear phase ramp in the frequency domain corresponds to a time shift in the time domain. Qualititavely, it seems a sawtooth phase mask in the frequency domain could produce the same time shift effect as the linear phase ramp, but I need help with the mathematical derivation to see what the actual effect is.

  • $\begingroup$ Can you explain exactly, mathematically, what you mean by a "sawtooth phase mask"? $\endgroup$ Jun 21, 2015 at 22:18
  • $\begingroup$ Since phase is effectively modulo $2\pi$ even if your linear phase ramp has value a gazillion, all linear phase ramps are actually sawtooth phase ramps. Now, if the phase "mask" you use does not match up in period with the linear phase ramp, you can get some interesting effects quite different from simple time shifts.... $\endgroup$ Jun 22, 2015 at 11:10
  • $\begingroup$ @DilipSarwate I understand the phase wrapping in linear ramps, but it's exactly because of the special case(s) that you mentioned with the sawtooth that I would like to derive a mathematical expression - to know what I should expect. I'm stuck with the derivation though. $\endgroup$ Jun 22, 2015 at 14:58

1 Answer 1



$$ \phi_0 \triangleq \phi - 2 \pi \left \lfloor \frac{\phi}{2\pi} \right \rfloor $$

where $\lfloor \cdot \rfloor$ is the floor() function, the largest integer not exceeding the argument.

it shouldn't be too hard to show that

$$ e^{j \phi_0} = e^{j \phi} $$

for all real $\phi$.

  • $\begingroup$ Doesn't this assume that the amplitude of the sawtooth is larger than 2pi? Or am I mistaken? $\endgroup$ Jun 26, 2015 at 23:02
  • $\begingroup$ $\phi_0$ is a sawtooth, if $\phi$ is increasing linearly, and $0 \le \phi_0 < 2\pi$. $\endgroup$ Jun 27, 2015 at 20:34
  • $\begingroup$ Oh OK, I was going the other way round. What if we start with a sawtooth where $0 \le \phi_0 < 1$ for example. Meaning, we're not phase-wrapping an increasingly linear phase, but rather phase-unwrapping a sawtooth phase with a small amplitude. $\endgroup$ Jun 29, 2015 at 19:07

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