Design of Transmit & Receive Filters

Consider the system model given below.

Question:

If we are given a c(f), how should we choose G_{T}(f) and G_{r}(f) such that the signal to ratio for the samples y_{m} is maximized?

We want

(1)
\begin{equation} G_T(f) C(f) G_R(f) = X_d(f) = X_{RC}(f) \end{equation}

so,

(2)
\begin{align} G_T(f) G_R(f) = \frac{X_{RC}(f)}{C(f)} \end{align}

How should we split $\frac{X_{RC}(f)}{C(f)}$ between G_{T}(f) and G_{R}(f) ?

Let us consider BPSK and look at,

(3)
\begin{align} y_m= x_0 I_m +\nu_m \end{align}

say we set x_{0} = 1 and make I_{m} = + d, so $y_m = I_m + \nu_m$

(4)
\begin{eqnarray} E[I_m^2] &=& d^2 \hspace{3mm} : \hspace{3mm} \mbox{Energy of the signal component at the Receiver} \\ E[\nu_m^2] &=& \sigma_{\nu}^2 \hspace{3mm} : \hspace{3mm} \mbox{Energy of the noise component at the Receiver as $X_{RC}(f)$ has unit energy} \end{eqnarray}

Transmitted energy (P_{av}T) is given by,

(5)
\begin{align} P_{av}T = d^2 \int_{-\infty}^{\infty}|G_T(f)|^2 df \end{align}

(6)
\begin{align} d^2= \frac{P_{av}T}{\int_{-\infty}^{\infty}|G_T(f)|^2 df} \end{align}

Notice that the transmitted energy depends on G_{T}(f) not just on X_{RC}(f)!!

(7)
\begin{align} E[\nu_m^2] = \int_{-\infty}^{\infty} S_{nn}{f}|G_R(f)|^2 df = \frac{N_0}{2} \int_{-\infty}^{\infty}|G_R(f)|^2 df \end{align}

SNR corresponding to the discrete samples y_{m} is calculated as,

(8)
\begin{align} SNR_0 = \frac{d^2}{E[\nu_m^2]} = \frac{P_{av} T}{\frac{N_0}{2} \int_{-\infty}^{\infty}|G_T(f)|^2 df . \int_{-\infty}^{\infty}|G_R(f)|^2 df} \end{align}

We want to express SNR_{0} (at the receiver) as the function of transmit power P_{av} or transmit energy (P_{av}T). The optimal design problem can be stated as, maximize SNR_{0} given $G_T(f)G_R(f) = \frac{X_{RC}}{C(f)}$. In this case we minimize $\int |G_T(f)|^2 df . \int |G_R(T)|^2 df$ subject to a condition that $G_T(f) G_R(f) = \frac {X_{RC}(f)}{C(f)}$.The solution can be shown to be,

(9)
\begin{align} G_T(f) = G_R^*(f) \mbox{ or } |G_T(f)| = |G_R(f)| = \frac{\sqrt{X_{RC}(f)}}{\sqrt{|C(f)|}} \mbox{ for } |f|<W \end{align}

Phase can also be equally split.

(10)
\begin{align} SNR_0 = \frac{2P_{av} T}{ N_0}. \frac{1}{\left[ \int_{-W}^{W} \frac{X_{RC}(f)}{|C(f)| df \right]^2}} \end{align}

on a ideally bandlimited channel $SNR_0 = \frac{2P_{av}T}{N_0}$, so $\left[ 10\log_{10} \frac{X_{RC}(f)}{|c(f)|} df \right]^2$ is the SNR loss for a non-ideal channel with frequency response C(f)!!

Example:

Consider a binary transmission scheme with a symbol rate of 4800 bits/sec over a channel with a frequency response given by,

(11)
\begin{align} |C(f)| = \left\{ \begin{array}{cc} \frac{1}{\sqrt{1 + \left( \frac{f}{4800} \right) ^2} } & |f|< 4800 \\ 0 & otherwise \end{array} \end{align}

Since R_{s}=4800 and Band-Width(W)=4800, excess Band-Width=100%, $\beta=1$.

(12)
\begin{align} X_{RC}(f) = \frac{T}{2} \left[ 1 + cos(\pi T |f|) \right] = T \cos ^2 \left( \frac{\pi|f|}{9600} \right) \end{align}

(13)
\begin{align} |G_T(f)| = |G_R(f)| = \frac{1}{\left( 1 + \frac{f}{4800} \right)^{\frac{1}{4}}} \cos \left( \frac{\pi |f|}{9600} \right) \hspace{3mm} |f|<4800 \end{align}