# 4.4 Demodulation

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## Demodulation

Convert the continuous time received signal into a vector without loss of information (or performance).

${r}_{t}={s}_{m}(t)+{N}_{t}$
${r}_{t}=\sum_{n=1}^{N} {s}_{mn}{\psi }_{n}(t)+\sum_{n=1}^{N} {\eta }_{n}{\psi }_{n}(t)+\stackrel{˜}{{N}_{t}}$
${r}_{t}=\sum_{n=1}^{N} ({s}_{mn}+{\eta }_{n}){\psi }_{n}(t)+\stackrel{˜}{{N}_{t}}$
${r}_{t}=\sum_{n=1}^{N} {r}_{n}{\psi }_{n}(t)+\stackrel{˜}{{N}_{t}}$

The noise projection coefficients ${\eta }_{n}$ 's are zero mean, Gaussian random variables and are mutually independent if ${N}_{t}$ is a white Gaussian process.

${\mu }_{\eta }(n)=({\eta }_{n})=(\int_{0}^{T} {N}_{t}{\psi }_{n}(t)\,d t)$
${\mu }_{\eta }(n)=\int_{0}^{T} ({N}_{t}){\psi }_{n}(t)\,d t=0$
$({\eta }_{k}\overline{{\eta }_{n}})=(\int_{0}^{T} {N}_{t}{\psi }_{k}(t)\,d t\int_{0}^{T} \overline{{N}_{{t}^{\prime }}}\overline{{\psi }_{k}({t}^{\prime })}\,d {t}^{\prime })=\int_{0}^{T} \int_{0}^{T} \overline{{N}_{t}{N}_{{t}^{\prime }}}{\psi }_{k}(t){\psi }_{n}({t}^{\prime })\,d t\,d {t}^{\prime }$
$({\eta }_{k}\overline{{\eta }_{n}})=\int_{0}^{T} \int_{0}^{T} {R}_{N}(t-{t}^{\prime }){\psi }_{k}(t)\overline{{\psi }_{n}()}\,d t\,d {t}^{\prime }$
$({\eta }_{k}\overline{{\eta }_{n}})=\frac{{N}_{0}}{2}\int_{0}^{T} \int_{0}^{T} \delta (t-{t}^{\prime }){\psi }_{k}(t)\overline{{\psi }_{n}({t}^{\prime })}\,d t\,d {t}^{\prime }$
$({\eta }_{k}\overline{{\eta }_{n}})=\frac{{N}_{0}}{2}\int_{0}^{T} {\psi }_{k}(t)\overline{{\psi }_{n}(t)}\,d t=\frac{{N}_{0}}{2}{\delta }_{kn}=\begin{cases}\frac{{N}_{0}}{2} & \text{if k=n}\\ 0 & \text{if k\neq n}\end{cases}$
${\eta }_{k}$ 's are uncorrelated and since they are Gaussian they are also independent. Therefore, ${\eta }_{k}\approx \mathrm{Gaussian}(0, \frac{{N}_{0}}{2})$ and ${R}_{\eta }(k, n)=\frac{{N}_{0}}{2}{\delta }_{kn}$

The ${r}_{n}$ 's, the projection of the received signal ${r}_{t}$ onto the orthonormal bases ${\psi }_{n}(t)$ 's, are independent from the residual noise process $\stackrel{˜}{{N}_{t}}$ .

The residual noise $\stackrel{˜}{{N}_{t}}$ is irrelevant to the decision process on ${r}_{t}$ .

Recall ${r}_{n}={s}_{mn}+{\eta }_{n}$ , given ${s}_{m}(t)$ was transmitted. Therefore,

${\mu }_{r}(n)=({s}_{mn}+{\eta }_{n})={s}_{mn}$
$\mathrm{Var}({r}_{n})=\mathrm{Var}({\eta }_{n})=\frac{{N}_{0}}{2}$
The correlation between ${r}_{n}$ and $\stackrel{˜}{{N}_{t}}$
$(\stackrel{˜}{{N}_{t}}\overline{{r}_{n}})=(({N}_{t}-\sum_{k=1}^{N} {\eta }_{k}{\psi }_{k}(t))\overline{{s}_{mn}+{\eta }_{n}})$
$(\stackrel{˜}{{N}_{t}}\overline{{r}_{n}})=({N}_{t}-\sum_{k=1}^{N} {\eta }_{k}{\psi }_{k}(t)){s}_{mn}+({\eta }_{k}\overline{{\eta }_{n}})-\sum_{k=1}^{N} ({\eta }_{k}\overline{{\eta }_{n}}){\psi }_{k}(t)$
$(\stackrel{˜}{{N}_{t}}\overline{{r}_{n}})=({N}_{t}\int_{0}^{T} \overline{{N}_{{t}^{\prime }}}\overline{{\psi }_{n}({t}^{\prime })}\,d {t}^{\prime })-\sum_{k=1}^{N} \frac{{N}_{0}}{2}{\delta }_{kn}{\psi }_{k}(t)$
$(\stackrel{˜}{{N}_{t}}\overline{{r}_{n}})=\int_{0}^{T} \frac{{N}_{0}}{2}\delta (t-{t}^{\prime }){\psi }_{n}({t}^{\prime })\,d {t}^{\prime }-\frac{{N}_{0}}{2}{\psi }_{n}(t)$
$(\stackrel{˜}{{N}_{t}}\overline{{r}_{n}})=\frac{{N}_{0}}{2}{\psi }_{n}(t)-\frac{{N}_{0}}{2}{\psi }_{n}(t)=0$
Since both $\stackrel{˜}{{N}_{t}}$ and ${r}_{n}$ are Gaussian then $\stackrel{˜}{{N}_{t}}$ and ${r}_{n}$ are also independent.

The conjecture is to ignore $\stackrel{˜}{{N}_{t}}$ and extract information from $\left(\begin{array}{c}{r}_{1}\\ {r}_{2}\\ \dots \\ {r}_{N}\end{array}\right)$ . Knowing the vector $r$ we can reconstruct the relevant part of random process ${r}_{t}$ for $0\le t\le T$

${r}_{t}={s}_{m}(t)+{N}_{t}=\sum_{n=1}^{N} {r}_{n}{\psi }_{n}(t)+\stackrel{˜}{{N}_{t}}$

Once the received signal has been converted to a vector, the correct transmitted signal must be detected based uponobservations of the input vector. Detection is covered elsewhere .

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