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

$\mathbf{X}$ 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.

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  • 2
    $\begingroup$ 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? $\endgroup$ – Alexander Armen Berian Oct 29 '18 at 1:20

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