Chapter 6- Baseband Transmission of Binary Data

Some Signal Definitions

Antipodal Signals: s₀(t)=-s₁(t)

Orthogonal Signals: $\int s_0(t) s_1(t) dt =0$

The Integrate-and-Dump Receiver

Input for AGN channel is Y(t)=s_i(t)+X(t)Output is

$$Z=\int_{0}^{T} Y(t) dt$$

Understand the analysis of this receiver, pages 6-8 to 6-12.

Energy in a Signal

$$\begin{eqnarray*}E_i&=&\int_{-\infty}^{\infty} \left[ s_i(t) \right]^2 dt\\ &=&... ...t[ s_i(t) \right]^2 dt \mbox{ for time-limited signal on [0,T]} \end{eqnarray*}$$

Noise Variance from Integrator for AWGN

$$\begin{eqnarray*}E\left\{ \left[ \int_{0}^{T} X(t) dt \right]^2 \right\} &=& E\l... ...s \\ &=& \int_{0}^{T} \frac{N_0}{2} dt \\ &=& \frac{N_0 T}{2} \end{eqnarray*}$$

Signal-to-Noise Ratio in dB

$$\begin{eqnarray*}\left(\frac{E}{N_0}\right)_{dB}= 10 \log_{10} \frac{E}{N_0} \\ \frac{E}{N_0}= 10^{ \left(E/N_0\right)_{dB}/10} \end{eqnarray*}$$

General Model for Linear Receiver

Study section beginning on p. 6-20

$$\begin{eqnarray*}Z(t)=\hat{s}(t)+\hat{X}(t) \\ \hat{s}=s \ast h\\ \hat{X}=X \ast h\\ R_{\hat{X}} = \tilde{h} \ast h \ast R_X \end{eqnarray*}$$

For AWGN channel,

$$\begin{eqnarray*}R_{\hat{X}}(\tau) = \frac{N_0}{2} f(\tau)=\frac{N_0}{2}\int_{-\... ...N_0}{2} \int_{-\infty}^{\infty} \left\vert H(f) \right\vert^2 df \end{eqnarray*}$$

For AGN channel, the decision statistic Z(T₀) is a Gaussian with mean $\mu(T_0)$ and variance $\sigma^2$. By our convention, we choose s₀ if Z(T₀) >0 and choose s₀ otherwise. Thus,

$$\begin{eqnarray*}P_{e,0}&=&P\left( Z_0(T_0) < \gamma \right)\\ &=&\mbox{Q}\left(\frac{\mu_0(T_0) - \gamma}{\sigma}\right). \end{eqnarray*}$$

Similarly

$$P_{e,1}=\mbox{Q}\left(\frac{\gamma-\mu_1(T_0)}{\sigma}\right).$$

Minimax Criterion

Minimize $P_{e,m}(\gamma) = \max\left\{ P_{e,0}(\gamma), P_{e,1}(\gamma) \right\}$ with respect to $\gamma$. In other words,

$$\gamma_m = \underset{\gamma}{\mbox{argmin}}\mbox{ }\max\left\{ P_{e,0}(\gamma), P_{e,1}(\gamma) \right\}$$

For the AGN channel, $P_{e,0}(\gamma_m)=P_{e,1}(\gamma_m)$, and

$$\begin{eqnarray*}\gamma_m = \frac{\mu_0(T_0) +\mu_1(T_1)}{2} = \mbox{ midpoint between two means} \end{eqnarray*}$$

The error probability is

$$P_{e,m}^{*}=\mbox{Q}\left(\frac{\mu_0(T_0) -\mu_1(T_1)}{2 \sigma} \right)$$

Choice of sampling time: For a general filter h(t), choose T₀ that maximizes $\mu_0(T_0) -\mu_1(T_1)$. The minimax threshold for antipodal signals is $\gamma_m =0$.

Bayes criterion

Minimizes $\pi_0 P_{e,0} + \pi_1 P_{e,1} =\overline{P_e}$. The constants $\pi_0$ and $\pi_1$ are costs for choosing s₀ or s₁. For our purposes, we take $\pi_0$ and $\pi_1$ to be the a priori probabilities for s₀ and s₁; that is; $\pi_0$ represents the probability that signal s₀ is sent, etc. Thus $0 < \pi_i < 1$ and $\pi_0+\pi_1=1$. The Bayes threshold minimizes $\overline{P_e}$, which is the average error probability. The threshold for this criterion is

$$\overline{\gamma} = \gamma_m + \frac{\sigma^2 \ln\left(\frac{\pi_1}{\pi_0}\right) }{\mu_0(T_0)-\mu_1(T_0)}$$

If $\pi_0=\pi_1$, then $\overline{\gamma}=\gamma_m$: the Bayes threshold equals the minimax threshold.

Section 6.4 of the Book is Not on the Test

The Matched Filter

We define a signal-to-noise ratio term called SNR by

$$\mbox{SNR}=\frac{\mu_0(T_0)-\mu_1(T_0)}{2\sigma}$$

Signal Space Math

The norm of a function is denoted $\left\vert\left\vert f\right\vert\right\vert$ and is given by

$$\left\vert \left\vert f \right\vert \right\vert = \sqrt{\int_{-\infty}^{\infty} \left[ f(t)\right]^2 dt }$$

Inner product of two function is denoted (f,g) and is given by

$$(f,g)= \int_{-\infty}^{\infty} f(t)g(t)dt$$

Matched Filter

Let

$$s(t)=\frac{s_0(t)-s_1(t)}{2}$$

The matched filter is

$$h_M(t)=\lambda s(T_0-t)=c\left[ s_0(T_0-t) - s_1(T_0-t)\right],$$

where $\lambda,c$ are arbitrary constants. The matched filter compensates for sampling time T₀. The minimum sampling time for causal filtering of a time-limited signal on [0,T] is T. The SNR for the matched filter is

$$\mbox{SNR}=\frac{\left\vert\left\vert s_{T_0}\right\vert\rig... ...qrt{N_0/2}}=\sqrt{\frac{\overline{E} \left(1-r\right)}{N_0}},$$

where

$$\overline{E}=\frac{E_0+E_1}{2}=\frac{\left\vert\left\vert s_0... ...ht\vert^2+\left\vert\left\vert s_1\right\vert\right\vert^2}{2}$$

and

$$r=\frac{\rho}{\overline{E}}=\frac{\left(s_0,s_1\right)}{\overline{E}} \mbox{ note that } -1 \le r \le 1$$

For antipodal signals, r=-1, which implies

$$P_E=\mbox{Q}\left(\mbox{SNR}\right)=\mbox{Q}\left(\sqrt{\frac{2 E}{N_0}}\right)$$

Freq. Domain Interpretation of the Matched Filter

Let $V_i(\omega)={\mathcal F}\left\{ s_i(t) \right\}$ Then

$$h_M(\omega)=c e^{-j \omega T_0} \left[ V_0^{\ast}(\omega) - V_1^{\ast} (\omega) \right]$$

Correlation Receiver

Note that

$$Z=\int_{-\infty}^{\infty}Y(t)s(t)dt= \int_{0}^{T}Y(t)s(t)dt$$

produces the same decision statistic as the matched filter sampled at time T₀.

Optimum Thresholds for Matched Filter

$$\begin{array}{ll} \mbox{Minimax: }&\gamma_m=\frac{E_0-E_1}{2}... ...\frac{N_0}{2}c \ln\left(\frac{\pi_1}{\pi_0}\right), \end{array}$$

where c is gain of matched filter in form $h(t)=c\left[s_0(T_0-t)-s_1(T_0-t) \right]$.

Signal Set Design

Want signals that minimize r, $-1 \le r\le 1$. Antipodal signals give r=-1.

$$\Rightarrow \mbox{Q}\left( \sqrt{\frac{2E}{N_0}} \right).$$

Optimum Filter for AGN Channel

Whitening Filter $H_W(\omega)$ such that

$$\left\vert H_W(\omega)\right\vert^2 = \frac{1}{S_X(\omega)}$$

Optimum filter is whitening filter plus matched filter, and is given by

$$H_{\mbox{opt}}(\omega)=\frac{H_m(\omega)}{S_X(\omega)},$$

where $H_m(\omega)$ is matched to the original signal set (s₀,s₁)on AWGN channel.

Signal Space Concepts

Know and understand Gram-Schmidt procedure to find signal space representation of signals over some orthonormal basis functions; find $\psi_0, \psi_1$ and $(\alpha_0, \alpha_1)$, $(\beta_0,\beta_1)$. Distance between s₀ and s₁ is $d=\left\vert \left\vert s_0 - s_1 \right\vert \right\vert= \sqrt{(\alpha_0-\beta_0)^2+(\alpha_1-\beta_1)^2}$ and the probability of error is

$$P_e=\mbox{Q}\left(\frac{d}{\sqrt{2 N_0}}\right).$$