The Gaussian Channel

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Definitions :: Band-limited :: Kuhn-Tucker :: Parallel :: Colored Noise

Parallel Gaussian channels

Parallel Gaussian channels are used to model bandlimited channels with a non-flat frequency response. We assume we have k Gaussian channels,

$\begin{displaymath} Y_j = X_j + Z_j,\qquad j=1,2,\ldots, k. \end{displaymath}$

where

$\begin{displaymath} Z_j \sim \Nc(0,N_j) \end{displaymath}$

and the channels are independent. The total power used is constrained:

$\begin{displaymath} E\sum_{j=1}^k X_j^2 \leq P. \end{displaymath}$

One question we might ask is: how do we distribute the power across the k channels to get maximum throughput.

We can find the maximum mutual information (the information channel capacity) as

$\begin{displaymath}\begin{aligned} I(X_1,\ldots,X_k;Y_1,\ldots,Y_k) &= h(Y_1,\ld... ...i) \\ &\leq \sum_{i} \frac{1}{2} \log(1+P_i/N_i) \end{aligned}\end{displaymath}$

Equality is obtained when the X s are independent normally distributed. We want to distribute the power available among the various channels, subject to not exceeding the power constraint:

$\begin{displaymath} J(P_1,\ldots,P_k) = \sum_i \frac{1}{2}\log(1+\frac{P_i}{N_i}) + \lambda \sum_{i=1}^k P_i \end{displaymath}$

with a side constraint (not shown) that $P_i \geq 0$ . Differential w.r.t. P _j to obtain

$\begin{displaymath} \frac{1}{P_j + N_j} + \lambda \geq 0. \end{displaymath}$

with equality only if all the constraints are inactive. After some fiddling, we obtain

$\begin{displaymath} P_j = \nu - N_j \end{displaymath}$

(since λ is a constant). However, we must also have $P_j \geq 0$ , so we must ensure that we don't violate that if $N_j > \nu$ . Thus, we let

$\begin{displaymath} P_j = (\nu - N_j)^+ \end{displaymath}$

where

$\begin{displaymath} (x)^+ = \begin{cases} x & x \geq 0 \\ 0 & x < 0 \end{cases} \end{displaymath}$

and

is chosen so that

$\begin{displaymath} \sum_{i=1}^n (\nu - N_i)^+ = P \end{displaymath}$

Draw picture; explain "water filling.''

Copyright 2008, by the Contributing Authors. Cite/attribute Resource . admin. (2006, May 17). The Gaussian Channel. Retrieved January 07, 2011, from Free Online Course Materials — USU OpenCourseWare Web site: http://ocw.usu.edu/Electrical_and_Computer_Engineering/Information_Theory/lecture11_3.htm. This work is licensed under a Creative Commons License

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Parallel Gaussian channels

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	Info The Gaussian Channel Document Actions Definitions :: Band-limited :: Kuhn-Tucker :: Parallel :: Colored Noise Parallel Gaussian channels Parallel Gaussian channels are used to model bandlimited channels with a non-flat frequency response. We assume we have k Gaussian channels, $\begin{displaymath} Y_j = X_j + Z_j,\qquad j=1,2,\ldots, k. \end{displaymath}$ where $\begin{displaymath} Z_j \sim \Nc(0,N_j) \end{displaymath}$ and the channels are independent. The total power used is constrained: $\begin{displaymath} E\sum_{j=1}^k X_j^2 \leq P. \end{displaymath}$ One question we might ask is: how do we distribute the power across the k channels to get maximum throughput. We can find the maximum mutual information (the information channel capacity) as $\begin{displaymath}\begin{aligned} I(X_1,\ldots,X_k;Y_1,\ldots,Y_k) &= h(Y_1,\ld... ...i) \\ &\leq \sum_{i} \frac{1}{2} \log(1+P_i/N_i) \end{aligned}\end{displaymath}$ Equality is obtained when the X s are independent normally distributed. We want to distribute the power available among the various channels, subject to not exceeding the power constraint: $\begin{displaymath} J(P_1,\ldots,P_k) = \sum_i \frac{1}{2}\log(1+\frac{P_i}{N_i}) + \lambda \sum_{i=1}^k P_i \end{displaymath}$ with a side constraint (not shown) that $P_i \geq 0$ . Differential w.r.t. P _j to obtain $\begin{displaymath} \frac{1}{P_j + N_j} + \lambda \geq 0. \end{displaymath}$ with equality only if all the constraints are inactive. After some fiddling, we obtain $\begin{displaymath} P_j = \nu - N_j \end{displaymath}$ (since λ is a constant). However, we must also have $P_j \geq 0$ , so we must ensure that we don't violate that if $N_j > \nu$ . Thus, we let $\begin{displaymath} P_j = (\nu - N_j)^+ \end{displaymath}$ where $\begin{displaymath} (x)^+ = \begin{cases} x & x \geq 0 \\ 0 & x < 0 \end{cases} \end{displaymath}$ and is chosen so that $\begin{displaymath} \sum_{i=1}^n (\nu - N_i)^+ = P \end{displaymath}$ Draw picture; explain "water filling.'' Copyright 2008, by the Contributing Authors. Cite/attribute Resource . admin. (2006, May 17). The Gaussian Channel. Retrieved January 07, 2011, from Free Online Course Materials — USU OpenCourseWare Web site: http://ocw.usu.edu/Electrical_and_Computer_Engineering/Information_Theory/lecture11_3.htm. This work is licensed under a Creative Commons License