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数学代写|信息论代写Information Theory代考|BANDLIMITED CHANNELS

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数学代写|信息论代写Information Theory代考|BANDLIMITED CHANNELS

A common model for communication over a radio network or a telephone line is a bandlimited channel with white noise. This is a continuoustime channel. The output of such a channel can be described as the convolution
$$Y(t)=(X(t)+Z(t)) * h(t),$$
where $X(t)$ is the signal waveform, $Z(t)$ is the waveform of the white Gaussian noise, and $h(t)$ is the impulse response of an ideal bandpass filter, which cuts out all frequencies greater than $W$. In this section we give simplified arguments to calculate the capacity of such a channel.
We begin with a representation theorem due to Nyquist [396] and Shannon [480], which shows that sampling a bandlimited signal at a sampling rate $\frac{1}{2 W}$ is sufficient to reconstruct the signal from the samples. Intuitively, this is due to the fact that if a signal is bandlimited to $W$, it cannot change by a substantial amount in a time less than half a cycle of the maximum frequency in the signal, that is, the signal cannot change very much in time intervals less than $\frac{1}{2 W}$ seconds.
Theorem 9.3.1 Suppose that a function $f(t)$ is bandlimited to $W$, namely, the spectrum of the function is 0 for all frequencies greater than $W$. Then the function is completely determined by samples of the function spaced $\frac{1}{2 W}$ seconds apart.
Proof: Let $F(\omega)$ be the Fourier transform of $f(t)$. Then
\begin{aligned} f(t) & =\frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{i \omega t} d \omega \ & =\frac{1}{2 \pi} \int_{-2 \pi W}^{2 \pi W} F(\omega) e^{i \omega t} d \omega, \end{aligned}
since $F(\omega)$ is zero outside the band $-2 \pi W \leq \omega \leq 2 \pi W$. If we consider samples spaced $\frac{1}{2 W}$ seconds apart, the value of the signal at the sample points can be written
$$f\left(\frac{n}{2 W}\right)=\frac{1}{2 \pi} \int_{-2 \pi W}^{2 \pi W} F(\omega) e^{i \omega \frac{n}{2 W}} d \omega$$

数学代写|信息论代写Information Theory代考|PARALLEL GAUSSIAN CHANNELS

In this section we consider $k$ independent Gaussian channels in parallel with a common power constraint. The objective is to distribute the total power among the channels so as to maximize the capacity. This channel models a nonwhite additive Gaussian noise channel where each parallel component represents a different frequency.

Assume that we have a set of Gaussian channels in parallel as illustrated in Figure 9.3. The output of each channel is the sum of the input and Gaussian noise. For channel $j$,
$$Y_j=X_j+Z_j, \quad j=1,2, \ldots, k,$$
with
$$Z_j \sim \mathcal{N}\left(0, N_j\right),$$
and the noise is assumed to be independent from channel to channel. We assume that there is a common power constraint on the total power used, that is,
$$E \sum_{j=1}^k X_j^2 \leq P .$$
We wish to distribute the power among the various channels so as to maximize the total capacity.
The information capacity of the channel $C$ is
$$C=\max _{f\left(x_1, x_2, \ldots, x_k\right): \Sigma E X_i^2 \leq P} I\left(X_1, X_2, \ldots, X_k ; Y_1, Y_2, \ldots, Y_k\right) .$$
We calculate the distribution that achieves the information capacity for this channel. The fact that the information capacity is the supremum of achievable rates can be proved by methods identical to those in the proof of the capacity theorem for single Gaussian channels and will be omitted.
Since $Z_1, Z_2, \ldots, Z_k$ are independent,
\begin{aligned} I & \left(X_1, X_2, \ldots, X_k ; Y_1, Y_2, \ldots, Y_k\right) \ & =h\left(Y_1, Y_2, \ldots, Y_k\right)-h\left(Y_1, Y_2, \ldots, Y_k \mid X_1, X_2, \ldots, X_k\right) \ & =h\left(Y_1, Y_2, \ldots, Y_k\right)-h\left(Z_1, Z_2, \ldots, Z_k \mid X_1, X_2, \ldots, X_k\right) \ & =h\left(Y_1, Y_2, \ldots, Y_k\right)-h\left(Z_1, Z_2, \ldots, Z_k\right) \ & =h\left(Y_1, Y_2, \ldots, Y_k\right)-\sum_i h\left(Z_i\right) \ & \leq \sum_i h\left(Y_i\right)-h\left(Z_i\right) \ & \leq \sum_i \frac{1}{2} \log \left(1+\frac{P_i}{N_i}\right) \end{aligned}
where $P_i=E X_i^2$, and $\sum P_i=P$. Equality is achieved by
$$\left(X_1, X_2, \ldots, X_k\right) \sim \mathcal{N}\left(0,\left[\begin{array}{cccc} P_1 & 0 & \cdots & 0 \ 0 & P_2 & \cdots & 0 \ \vdots & \vdots & \ddots & \vdots \ 0 & 0 & \cdots & P_k \end{array}\right]\right)$$

数学代写|信息论代写Information Theory代考|BANDLIMITED CHANNELS

$$Y(t)=(X(t)+Z(t)) * h(t),$$

\begin{aligned} f(t) & =\frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{i \omega t} d \omega \ & =\frac{1}{2 \pi} \int_{-2 \pi W}^{2 \pi W} F(\omega) e^{i \omega t} d \omega, \end{aligned}

$$f\left(\frac{n}{2 W}\right)=\frac{1}{2 \pi} \int_{-2 \pi W}^{2 \pi W} F(\omega) e^{i \omega \frac{n}{2 W}} d \omega$$

数学代写|信息论代写Information Theory代考|PARALLEL GAUSSIAN CHANNELS

$$Y_j=X_j+Z_j, \quad j=1,2, \ldots, k,$$

$$Z_j \sim \mathcal{N}\left(0, N_j\right),$$

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

$$C=\max _{f\left(x_1, x_2, \ldots, x_k\right): \Sigma E X_i^2 \leq P} I\left(X_1, X_2, \ldots, X_k ; Y_1, Y_2, \ldots, Y_k\right) .$$

\begin{aligned} I & \left(X_1, X_2, \ldots, X_k ; Y_1, Y_2, \ldots, Y_k\right) \ & =h\left(Y_1, Y_2, \ldots, Y_k\right)-h\left(Y_1, Y_2, \ldots, Y_k \mid X_1, X_2, \ldots, X_k\right) \ & =h\left(Y_1, Y_2, \ldots, Y_k\right)-h\left(Z_1, Z_2, \ldots, Z_k \mid X_1, X_2, \ldots, X_k\right) \ & =h\left(Y_1, Y_2, \ldots, Y_k\right)-h\left(Z_1, Z_2, \ldots, Z_k\right) \ & =h\left(Y_1, Y_2, \ldots, Y_k\right)-\sum_i h\left(Z_i\right) \ & \leq \sum_i h\left(Y_i\right)-h\left(Z_i\right) \ & \leq \sum_i \frac{1}{2} \log \left(1+\frac{P_i}{N_i}\right) \end{aligned}

$$\left(X_1, X_2, \ldots, X_k\right) \sim \mathcal{N}\left(0,\left[\begin{array}{cccc} P_1 & 0 & \cdots & 0 \ 0 & P_2 & \cdots & 0 \ \vdots & \vdots & \ddots & \vdots \ 0 & 0 & \cdots & P_k \end{array}\right]\right)$$

MATLAB代写

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