4
$\begingroup$

I wonder why the phase range in DSP applications is between $-\pi$ and $+\pi$. For example, why is that not between $0$ and $2\pi$?

This question popped when I was reading about synchronization in digital communications.

It said instead of $\arg( Im(x) / Re(x) )$ we can use just $Im(x)$ because they are of the same sign because phase (here $x$) is always between $-\pi$ and $+\pi$ !!!

$\endgroup$

4 Answers 4

5
$\begingroup$

It is just a convention, but it is useful in some cases. For example, the phase of the DFT of a real discrete-time signal is odd only if the angles are expressed in the range $[-\pi, \pi)$. Sometimes you just have to adapt to the convention used by your tools -- for example, MATLAB functions like angle and atan2 return angles in $[-\pi, \pi)$.

Note that the frequency of a discrete-time signal, measured in radians per sample, is in the range $(-\pi, \pi]$, but in this case it has a different interpretation.

$\endgroup$
2
  • 2
    $\begingroup$ (+1) Afaik for frequencies Opp & Sch uses the range $[-\pi, \pi)$ (in addition to $[0, 2\pi)$) where the lower limit is included, (upper limit discluded)... $\endgroup$
    – Fat32
    Apr 24, 2021 at 17:30
  • $\begingroup$ Good point. And again a matter of preference :-) $\endgroup$
    – MBaz
    Apr 24, 2021 at 18:17
5
$\begingroup$

First, when you're talking angles, in DSP pretty much all angles are $\mod 2\pi$. So $2\pi \equiv 0$. Usually it's more convenient to keep angles on the interval $\left [-\pi, \pi \right )$, because we're usually most interested in angles around $0$. You don't have to do this, however -- if your problem at hand is easier to solve if your angle lies on $[-2\pi, 0)$, or $[0, 2\pi)$, or any such interval, by all means use that. Just be careful to point it out along the way, in case any innocent bystanders are trying to understand your thinking.

It's a judgement call, though -- sometimes if you're working with actual physically rotating objects, or you're otherwise dealing with a sequence of angular steps, you may want to treat a sequence of angles as continuous -- which is common enough that most math packages have an "unwrap" function hidden within them someplace.

This can even extend to phase-locked loops, where you may want a phase detector that, instead of reporting an angular error that jumps at the $-\pi$ to $\pi$ boundary, smoothly transitions from $\pi - \epsilon$ to $\pi + \epsilon$ (and the equivalent in the negative direction) and from $-2\pi$ or $2\pi$ to 0 -- such a detector is called a "phase-frequency" detector and nicely extends a loop's lock range.

$\endgroup$
1
$\begingroup$

I have thought about some points which could help find the answer:

1- I think there might be something related to $\operatorname{arctan}(x)$ which is continuous in $(-\pi/2 \ \ \pi/2 )$ but I am not sure how.

2- We almost always work with phase DIFFERENCE rather than the absolute phase itself. Phase difference could be both positive and negative. So, it might be better to consider a signed range than an unsigned one.

3- Because of the nonlinearity of the phase argument in the applicable functions (e.g. $\sin(\cdot)$, $\cos(\cdot)$, etc.), it would be better to work with the functions instead of the phase itself. Among these functions, sinusoidal-like functions show the advantage of having the same sign as their argument (phase). Plus, the range of the argument to keep this is $[-\pi, \pi)$.

$\endgroup$
1
  • $\begingroup$ I think your #2 is the most salient. Most often we are dealing with an angle increment, that angle can also decrement which is a negative increment. so a a bipolar value is, i think, more useful in general. $\endgroup$ May 2, 2021 at 13:20
0
$\begingroup$

It has to do with the unit circle in the i vs Re plane - instead of going counterclockwise by 360 degrees, we could equivalently go +/- 180 degrees.

In my experience with audio, this thought process allows one to minimize phase delay.

For example, let's say I've got a L and R audio signal arriving at some listening point in space, where (arbitrarily) the L signal arrives with a phase delay of 270 degrees relative to the R signal.

In this case, I could either (1) delay the R signal by 270 degrees, or (2) delay the L signal by 90 degrees.

As far as relative phase is concerned (not absolute phase), both methods create perfect phase coherence (0 and 360 phase difference respectively). However, delaying the L signal by 90 degrees induces less phase delay than a 270 degree phase delay. (The audible differences between 0 vs 360 degree phase difference is an entirely different discussion in psychoacoustics).

$\endgroup$

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.