MIMO 模型信道容量公式详细推导
考虑MIMO信道的一般表达式:
$$Y= HX +N \qquad (1)$$
其中
$$X-H_t \times 1 $$
$$Y-H_r \times 1 $$
$$ X,N- i.i.d Gaussian Distribution$$
同时定义
$$ R_{xx}= E[XX^H] $$
$$ R_{nn}= E[NN^H] = \sigma ^2 I_{Nr}$$
$$ R_{yy}= E[YY^H]= HR_{xx}H^H+ \sigma ^2 I_{Nr}$$
那么 $P=tr(R_{xx})$ 为发射信号总功率,$H(Y|X)= H(HX+N|X) = H(N|X)= H(N)$ 并且
$$C= \arg max \quad I(X,Y)= H(Y)-H(N)$$
当信道AWGN信道时,信源X服从高斯分布时$I(X,Y)$取得最大值,此时接收信号Y也服从高斯分布。
(1) 考虑一维连续高斯分布,此时需要计算连续信息熵 (微分熵), 假设 $X \sim \phi (0, \sigma) = (1/\sqrt {2 \pi \sigma ^2}) e^{-x^2/2 \sigma ^2}$
可计算得到微分熵
$$ h(X) = - \int x \ln x dx $$
$$\qquad = - \int \phi [- \frac {-x^2}{2\sigma ^2}-ln \sqrt {2\pi \sigma ^2}]dx $$
$$\qquad = \frac {E[X^2]}{2 \sigma ^2} + \frac {1}{2} \ln 2\pi \sigma ^2 $$
$$\qquad = \frac {1}{2} + \frac {1}{2}\ln 2 \pi \sigma ^2$$
$$\qquad = \frac {1}{2} \ln \text {e} + \frac {1}{2} 2 \pi \sigma ^2$$
$$\qquad = \frac {1}{2} \ln 2 \pi \text {e} \sigma ^2$$
(2) 对于N维高斯信道,其中概率密度函数表达式如下
$$ f_X(x_1,,,x_N) = \frac {1}{\sqrt {(2\pi)^N|\sum|}} \text {exp} [-\frac {1}{2}(X-\mu )^T \Sigma ^{-1}(X-\mu)] \qquad (2)$$
假设$x_1,x_2,,,x_N$ 为$i.i.d$分布,即 $\sum = \sigma_X ^2 I_{N}$ 并且 $|\sum| = \prod _{i=1}^{N} \sigma ^2 $. (2)式可进一步简化为
$$f_X(x_1,,,x_N) = \prod _{i=1}^{N} \frac {1}{\sqrt {2\pi \sigma ^2}}\text {exp} ^{-\frac {(x_i-\mu _i)^2}{2\sigma ^2}} = \prod _{i=1}^{N} f(x_i)$$
重新计算微分熵可得
$$ h(X)= - \int f_X(x_1,,,x_N) \ln ^{f_X(x_1,,,x_N)} dx_1...dx_N$$
$$ \qquad = - \int f_X(x_1,,,x_N) [-\ln ^{\sqrt {(2\pi)^N|\Sigma|}}- \frac {1}{2}(X-\mu )^T \Sigma ^{-1}(X-\mu)]dx_1...dx_N$$
$$ \qquad = \frac {1}{2}\ln ^{(2\pi)^N|\Sigma|} + \frac {1}{2} \int _{x_1}f(x_1)[\int _{x_2}f(x_2)[.....[\int _{x_N}f(x_N) \sum _{i=1}^{N} \frac {(x_i-\mu _{i})^2}{\sigma ^2}dx_N]dx_{N-1}]....]dx_1 $$
$$\qquad = \frac {1}{2}\ln ^{(2\pi)^N|\Sigma|}+ \frac {N}{2} $$
$$\qquad = \frac {1}{2} \ln ^{(2\pi)^N|\Sigma|\text{e}^N} $$
$$\qquad = \frac {1}{2} \ln ^{|\Sigma|(2\pi \text{e})^N}$$
将上述结果代入 $C= \arg max \quad I(X,Y)= H(Y)-H(N)$ 中可得
$$ C = 1/2 \log _{2} {(2\pi \text{e})^{N}|\Sigma _Y|} -1/2 \log _{2} {(2\pi \text{e})^{N}|\Sigma _N|} $$
$$ \qquad =1/2 \log _2 {\frac {|HR_{xx}H^T+\sigma ^{2}I_{N_r}|}{|\sigma ^{2}I_{N_r}|}}$$
$$\qquad = 1/2 \log _2 {| I_{N_r} + \frac {HR_{xx}H^T}{\sigma ^2}|}$$
证明完毕