# Distribution of the output of AWGN channel

$$\mathbf{Y}=\mathbf{X}+\mathbf{W}$$

$\mathbf{X}$ is input, an $n$-dimension equiprobable vector in a finite subset of complex space $\mathcal{A}^n \subset \mathbb{C}^n$ with norm $\sqrt{\rho n}$ (constant power), e.g. BPSK, QPSK, 8-PSK, etc.

$\mathbf{W}$ is circularly complex Gaussian vector $\mathcal{CN}(0, \mathbf{I}_n)$.

I want to know the closed form of $P_{\mathbf{Y}}(\mathbf{y})$. And if possible, the $P_{\mathbf{Y}}(\mathbf{y})$ induced by the QAM modulation of $\mathbf{X}$ (identical average power).

I can numerically calculate $$P_{\mathbf{Y}}(\mathbf{y}) = \sum_\mathbf{x}P_{\mathbf{Y}|\mathbf{X}}(\mathbf{y}|\mathbf{x})P_{\mathbf{X}}(\mathbf{x}) = \sum_\mathbf{x}P_{\mathbf{Y}|\mathbf{X}}(\mathbf{y}|\mathbf{x})\frac{1}{|\mathcal{A}|^n} = \frac{1}{|\mathcal{A}|^n} \sum_{\mathbf{x} \in \mathcal{A}^n} \frac{1}{\pi^n}e^{-\left\Vert\mathbf{y-x}\right\Vert ^2} \tag{1}$$

where $|\mathcal{A}|$ is the cardinality of set $\mathcal{A}$. But I want to avoid $\sum_{\mathbf{x} \in \mathcal{A}^n}$ because it is cumbersome.

If $\mathcal{A} = \mathbb{C}$ I can use the spherical symmetry trick that $$P_{\mathbf{Y}}(\mathbf{y}) = \frac{1}{S} \times P_{\left\Vert\mathbf{Y}\right\Vert^2}(\left\Vert\mathbf{y}\right\Vert^2) \tag{2}$$

where $S$ is the surface of n-dimension complex sphere with radius $\sqrt{\rho n}$ and $P_{\left\Vert\mathbf{Y}\right\Vert^2}(\left\Vert\mathbf{y}\right\Vert^2)$ is noncentral $\chi$-squared distribution with $n$ degrees of freedom and noncentrality parameter $n \rho$. Thus I have a closed form of the distribution of $\mathbf{Y}$.

I think this must be a classical problem, it would be great if someone gives me a hint or points out a paper.

• Since QAM modulation has a finite number of constellation points, I think you just expand the sum into M terms (M is # of constellation points), giving a closed form expression? Oct 29 '18 at 1:20
• Are the elements in each vector iid? Because in the coding schemes you mentioned (BPSK, QPSK, 8-PSK), each element of the vectors belong to the alphabet, and not the entire vector itself. If the vector elements are iid, then it would simplify things. Aug 14 '20 at 10:04

## 1 Answer

I see this question is open for a long time, I am not sure if I understand it correctly.

## PDF of the symbols

The first probability is a $$P_{\mathbf{Y}}: \mathbb{C}^n \to \mathbb{R}$$, you are not considering phase noise, the probability of the $$n$$ symbols are identically distributed, let's say with $$P_{\mathbf{Y}_i}: \mathbb{C}^n \to \mathbb{R}$$, since the joint probability of independent variables are the product of the individual variables this distribution can be written as

$$P_{\mathbf{Y}}(y) = \prod_{i=1}^{N}P_{\mathbf{Y}_i}(y_i)$$

Where $$P_{\mathbf{Y}_i}(y_i) = \sum_{x\in \mathbf{X}_i}p_{\mathbf{X}}(x_i) e^{-||y_i - x_i||^2}/\pi$$ which leads to

$$P_{\mathbf{Y}}(y) = \frac{1}{\pi^N}\prod_{i=1}^{N} \sum_{x\in \mathbf{X}_i} p_{\mathbf{X}}(x_i) e^{-||y_i - x_i||^2}$$

The term $$p_{\mathbb{X}}(x_i)$$ can be used to model probabilistic shaping, but for the original quesiton it will probably be a the inverse of the number of symbols in the constellation $$1/c = 1/|\mathbf{X}_i|$$, that by it's turn is the same for all symbols, thus

$$P_{\mathbf{Y}}(y) = \frac{1}{c^N\pi^N}\prod_{i=1}^{N} \sum_{x\in \mathbf{X}_i} p_{\mathbf{X}}(x_i) e^{-||y_i - x_i||^2}$$

The complex of evaluating this will be $$c N$$ exponentials. The number of additions and multiplications will be linear in $$cN$$ as well, differently from the initial formula that would compute $$c^N$$ terms.

For some applications you may want the logarithm of that expression, in such situations it is preferable to use the sum of logarithms

$$log\left(P_{\mathbf{Y}}(y)\right) = \sum_{i=1}^{N}\log\left(\frac{1}{c \pi} \sum_{x\in \mathbf{X}_i} p_{\mathbf{X}}(x_i) e^{-||y_i - x_i||^2}\right)$$

All of this is valid for QAM modulation as well, and don't have the exponential number of terms in the initial formula.

## PDF of the energy

In the second part you mention the PDF of the energy of the symbols

That is the probability distribution of

$$\sum_{i=0}^{N} \sum_{x_i \in \mathbf{X}_i} p_{\mathbf{X}_i}(x_i) ||x_i + w_i||^2$$

Assuming $$w$$ having a variance $$\sigma$$ for both real and imaginary parts, since the norm of a complex number $$||a + 1j b||^2 = a^2 + b^2$$, assuming uniform probability for the symbols in the constelation, we can divide the whole expression the symbols $$\sigma$$ (since the the formula is given for a covariance matrix being the identity). And the distribution of the energy will be of the form $$\sum_{k=1}^{2N} X_{i}^2$$ as assumed for the noncentral $$\chi^2$$ distribution with $$2N$$ degrees of freedom.

The noncentrality parameter given by

$$\lambda =\frac{N}{c \sigma^2} \sum_{x \in \mathbf{X}} ||x||^2$$

The number of degrees of freedom $$k=2N$$, because we separated real and imaginary parts.

Let $$z$$ be the energy of one sample, taking in to account that the $$\chi^2$$ distribution was normalized by dividing by the variance, the PDF of $$z$$ will be

$$p_{||\mathbf{Y}||^2}(z) = \frac{1}{2\sigma} e^{\left(\frac{z}{2\sigma} + \frac{\lambda}{2}\right)} \left(\frac{z}{\lambda \sigma}\right)^{(N-1)/2} I_{N-1}\left( \sqrt{\frac{\lambda z}{\sigma}}\right)$$

Note: In spite of being a closed formula it depends on the computation of $$I_{\alpha}(x)$$, a modified Bessel function, that for positive integer $$\alpha$$ and positive $$x$$ reduces to a series with rational coefficients in $$x$$, thus

$$I_{N-1}\left( \sqrt{\frac{\lambda z}{\sigma}}\right) = \sum_{m=0}^{\infty} \frac{\left(\sqrt{\frac{\lambda z}{\sigma}}\right)^{2m+N-1}}{m!(m+N)!}$$