I'm a bit of a noob here, looking for a simple formula or just a table which can tell me the absolute minimum phase shift angle possible for different variations of QAM. Particularly QAM 16, 64, 256 and 1024. If the answer is zero (is it?), then what is the minimum phase shift possible in each case other than 0?

Additionally, what is the limiting factor which governs the rate at QAM can switch between different phases? In other words, what is the maximum rate at which QAM can switch between different phases, and how is that determined for a particular device? Is there a relation to frequency of transmission (say a 1 GHz signal), or are frequency of transmission and phase-switching rate completely independent?

  • $\begingroup$ hm, the questions you ask sound like you want to do something that you really don't understand – what's the purpose of getting that table? (and yes, since in any QAM>4, there's multiple constellation points on the same line through the origin, there's multiply constellation points with the same phase) A simple pencil drawing on a piece of paper would've allowed you to answer this question yourself! $\endgroup$ Commented May 11, 2019 at 16:17
  • $\begingroup$ and your last question (is there a relationship between QAM and carrier frequency) points to you ignoring the very basic idea of equivalent baseband being identical regardless of carrier frequency, so: what are you doing? $\endgroup$ Commented May 11, 2019 at 16:19
  • $\begingroup$ To me it doesn't matter if multiple constellation points have the same phase -- more so curious what the absolute minimum phase shift is possible. Basically I'm trying to simulate very small phase shifts using existing techniques available in commodity hardware. I understand that smaller angles are possible with higher order QAM, but I would like some hard data on what those minimum possible shifts actually are in practice. $\endgroup$
    – Jordan
    Commented May 11, 2019 at 16:21
  • $\begingroup$ To me it doesn't matter if multiple constellation points have the same phase -- more so curious what the absolute minimum phase shift is possible Um, what is $x-y$ if $x = y$?!? If two constellation points have the same phase, the phase shift is zero. it can't be smaller than that, so for all these constellations, the minimum phase shift is 0. I don't understand the question? $\endgroup$ Commented May 11, 2019 at 16:22
  • 1
    $\begingroup$ I'm by the way pretty sure that you can come up with the geometric construction of $M$-QAM: you take a grid of $\sqrt M\times\sqrt M$ points. You center that on the coordinate grid origin. Done! Scaling doesn't matter to the angles; only that the distance between neighbors is constant (i.e. that it is a square grid). $\endgroup$ Commented May 11, 2019 at 17:09

1 Answer 1


The minimum phase shift in 16-QAM, 64-QAM, 256-QAM etc is 0 (same value regardless of whether you are using degrees or radians!) and occurs when the symbol transition is from one constellation point on some radius vector to another constellation point on the same radius vector. If we take the constellation points as having coordinates $(i,j)$ where $(i,j) \in \{-3, -1, +1, +3\}$ in 16-QAM, then $(1,1)$ and $(3,3)$ lie on the same radius vector (line of slope 1 through the origin) and so the RF phase remains unchanged in the transition from one constellation point to the other. Obviously, the same example will also work for 64-QAM, 256-QAM, and 1024-QAM.

What is the minimum non-zero phase shift?

Well, for 16-QAM, the constellation points $(1,3)$ and $(3,3)$ are distance 2 apart, and together with the origin, form an obtuse triangle with sides of lengths $2$, $\sqrt{10}$, and $\sqrt{18}$. From the standard formula $$c^2 = a^2 + b^2 - 2ab\cos C$$ relating the sides $a$, $b$, $c$ of a triangle opposite to angles $A$, $B$ and $C$, we get that $$2^2 = 10 + 18 - 2\sqrt{180} \cos C$$ giving $\arccos \frac{2}{\sqrt{5}}$ as the minimum nonzero phase shift in 16-QAM. Similar calculations for 64-QAM, 256-QAM and 1024-QAM as left as an exercise for the reader.


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.