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Probability distribution of square of amplitude of QPSK signal

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Jun 11, 2009
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Hi all,

I just wonder if anyone can show me how to get probability distribution (or standard deviation) of square of amplitude of M-PSK signals (QPSK for example).

I think with QPSK signal, there is two components, in-phase (X) and quadrature (Y)

X = S_I + n_I
Y = S_Q + n_Q
where n_I and n_Q present the in-phase and quadrature components of noise, being zero-mean i.i.d Gaussian random variables with variance of sigma^2.

Now, the square of amplitude can be taken as

A = X^2 + Y^2;

Assume that SNR = 30 dB, Signal Power = 0 dBW, by doing MATLAB simulation, I can get the standard deviation of A is about 0.0448 (see the code below).

My question is:
How to calculate the standard deviation theoretically? I mean how to know probability distribution function (and the standard deviation as well) of square of amplitude of the signals? Please help!!!!


M = 4;
Len = 10^6;
SNR = 30;
hMod = comm.PSKModulator(M,'BitInput',true);
data = logical(randi([0 1], Len, 1)); % Generate data
modData = step(hMod, data); % Modulate
tx = awgn(modData,SNR,0);
squareamptx = (abs(tx)).^2;

Thank you very much.


The theoretical formula for standard deviation look in google.
First find the mean, then differences between that mean and all samples amplitudes etc.

I think you may only build a histogram, then approximate its shape with one of the known formulas (or invent your own).

If you add a gaussian noise in matlab, then the distribution function should be gaussian (normal).
But I'm not sure, maybe this is for vector of real values only.
If the normally distributed noise samples are added to I and Q components independently,
so the amplitude should be distributed by raleigh law, and the amplitude square should have then chi-squared distribution.

Thank you so much, Mityan.

Actually, I am thinking if the amplitude square should have the noncentral chi-squared distribution. But i am not sure. then I will post it here so that you or other friends may have any comments.

We have
X = S_I + n_I
Y = S_Q + n_Q
where n_I and n_Q are independent normally distributed N(0, σ^2).

Thus, X is distributed normally with mean μ1 and variance σ^2, and Y is distributed normally with mean μ2 and variance σ^2. Note that because of PSK signal, μ1^2 + μ2^2 = 1.
(X/σ)^2 + (Y/σ)^2 should have the noncentral chi-squared distribution ( with k = 2 and λ = (μ1/σ)^2 + (μ2/σ)^2 = 1/σ^2.

The variance of noncentral chi-squared distribution is 2(k+2*λ) = 4(1+1/σ^2). Thus, the standard deviation is 2*sqrt(1+1/σ^2).

So, (X^2 + Y^2)/σ^2 has the noncentral chi-squared distribution with the standard deviation of 2*sqrt(1+1/σ^2).

Since noncentral chi-squared distribution is non-linear transformations (is that right?), I couldn't do further for (X^2 + Y^2)!!!!
Can you help me about this, Mityan?
Any comments would be highly appreciated!

The articles about discussed distributions are a little different in english and russian,
so I have read that
1) if x1...xk are independent normally distributed with the same known mean u and variance sigma,
then Y = sum of ((xi - u)/sigma)^2 has the distribution chi-squared of k. And
2) if k=2, chi-squared distribution match exponential distribution - lambda*exp(-lambda*x) where x is your sum.

So if you assume that your I and Q components have the same distribution parameters (it should be so I think)
then your distribution will be exponential. Your headache is gone and nobody feels the difference (if assumption is not true).

Hi Mityan,

I think in M-PSK signal, X and Y are distributed normally with a same variance σ^2 only, the means μ1 and μ2 of X and Y, respectively, are different.
For example, if the modulated data is
data = 1 + j*0;
X = 1 + noise_I
Y = 0 + noise_Q
Since noise_I and noise_Q are zero-mean i.i.d Gaussian random variables with variance of σ^2, X and Y should only have a same variance but different means (but there is always μ1^2 + μ2^2 = 1).

So, finally, I still couldn't find the standard deviation of (X^2 + Y^2)! :(

Note that because of PSK signal, μ1^2 + μ2^2 = 1.

Why you think so? For me it is not evident.
For random data source the quantity of nulls and ones for both I and Q should be equal I think.
It seems you tend to complicate.

I just think that for QPSK, after modulating, signal has only 4 possible symbols

X_00 = 1 + noise_I ---> X_00 ~ N(1, σ^2) --> μ1 = 1
Y_00 = 0 + noise_Q ---> Y_00 ~ N(0, σ^2) --> μ2 = 0

X_01 = 0 + noise_I ---> X_01 ~ N(0, σ^2) --> μ1 = 0
Y_01 = 1 + noise_Q ---> Y_01 ~ N(1, σ^2) --> μ2 = 1

X_11 = -1 + noise_I ---> X_11 ~ N(-1, σ^2) --> μ1 = -1
Y_11 = 0 + noise_Q ---> Y_11 ~ N(0, σ^2) --> μ2 = 0

X_10 = 0 + noise_I ---> X_10 ~ N(0, σ^2) --> μ1 = 0
Y_10 = -1 + noise_Q ---> Y_10 ~ N(-1, σ^2) --> μ2 = -1

(where X_00 + j*Y_00 is the modulated symbol for the pair bit 00, and so on)

In all possible four symbols above, μ1^2 + μ2^2 = 1.

What I need to find is the standard deviation of square of amplitude (it means: X^2 + Y^2)
I just wonder if the above explanation is correct?

- - - Updated - - -

Hold on!
I think there is something wrong...:x

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