# 4.2 Geometric representation of modulation signals

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Geometric representation of signals provides a compact, alternative characterization of signals.

Geometric representation of signals can provide a compact characterization of signals and can simplify analysis of theirperformance as modulation signals.

Orthonormal bases are essential in geometry. Let $\{{s}_{1}(t), {s}_{2}(t), \dots , {s}_{M}(t)\}$ be a set of signals.

Define ${\psi }_{1}(t)=\frac{{s}_{1}(t)}{\sqrt{{E}_{1}}}$ where ${E}_{1}=\int_{0}^{T} {s}_{1}(t)^{2}\,d t$ .

Define ${s}_{21}={s}_{2}\dot {\psi }_{1}=\int_{0}^{T} {s}_{2}(t)\overline{{\psi }_{1}(t)}\,d t$ and ${\psi }_{2}(t)=\frac{1}{\sqrt{\stackrel{^}{{E}_{2}}}}({s}_{2}(t)-{s}_{21}{\psi }_{1}())$ where $\stackrel{^}{{E}_{2}}=\int_{0}^{T} ({s}_{2}(t)-{s}_{21}{\psi }_{1}(t))^{2}\,d t$

In general

${\psi }_{k}(t)=\frac{1}{\sqrt{\stackrel{^}{{E}_{k}}}}({s}_{k}(t)-\sum_{j=1}^{k-1} {s}_{\mathrm{kj}}{\psi }_{j}(t))$
where $\stackrel{^}{{E}_{k}}=\int_{0}^{T} ({s}_{k}(t)-\sum_{j=1}^{k-1} {s}_{\mathrm{kj}}{\psi }_{j}(t))^{2}\,d t$ .

The process continues until all of the $M$ signals are exhausted. The results are $N$ orthogonal signals with unit energy, $\{{\psi }_{1}(t), {\psi }_{2}(t), \dots , {\psi }_{N}(t)\}$ where $N\le M$ . If the signals $\{{s}_{1}(t), \dots , {s}_{M}(t)\}$ are linearly independent, then $N=M$ .

The $M$ signals can be represented as

${s}_{m}(t)=\sum_{n=1}^{N} {s}_{\mathrm{mn}}{\psi }_{n}(t)$
with $m\in \{1, 2, \dots , M\}$ where ${s}_{\mathrm{mn}}={s}_{m}\dot {\psi }_{n}$ and ${E}_{m}=\sum_{n=1}^{N} {s}_{\mathrm{mn}}^{2}$ . The signals can be represented by ${s}_{m}=\left(\begin{array}{c}{s}_{m1}\\ {s}_{m2}\\ ⋮\\ {s}_{\mathrm{mN}}\end{array}\right)$

${\psi }_{1}(t)=\frac{{s}_{1}(t)}{\sqrt{A^{2}T}}$
${s}_{11}=A\sqrt{T}$
${s}_{21}=-(A\sqrt{T})$
${\psi }_{2}(t)=({s}_{2}(t)-{s}_{21}{\psi }_{1}(t))\frac{1}{\sqrt{\stackrel{^}{{E}_{2}}}}=(-A+\frac{A\sqrt{T}}{\sqrt{T}})\frac{1}{\sqrt{\stackrel{^}{{E}_{2}}}}=0$

Dimension of the signal set is 1 with ${E}_{1}={s}_{11}^{2}$ and ${E}_{2}={s}_{21}^{2}$ .

${\psi }_{m}(t)=\frac{{s}_{m}(t)}{\sqrt{{E}_{s}}}$ where ${E}_{s}=\int_{0}^{T} {s}_{m}(t)^{2}\,d t=\frac{A^{2}T}{4}$

${s}_{1}=\left(\begin{array}{c}\sqrt{{E}_{s}}\\ 0\\ 0\\ 0\end{array}\right)$ , ${s}_{2}=\left(\begin{array}{c}0\\ \sqrt{{E}_{s}}\\ 0\\ 0\end{array}\right)$ , ${s}_{3}=\left(\begin{array}{c}0\\ 0\\ \sqrt{{E}_{s}}\\ 0\end{array}\right)$ , and ${s}_{4}=\left(\begin{array}{c}0\\ 0\\ 0\\ \sqrt{{E}_{s}}\end{array}\right)$

$\forall mn\colon {d}_{\mathrm{mn}}=\left|{s}_{m}-{s}_{n}\right|=\sqrt{\sum_{j=1}^{N} ({s}_{\mathrm{mj}}-{s}_{\mathrm{nj}})^{2}}=\sqrt{2{E}_{s}}$
is the Euclidean distance between signals.

Set of 4 equal energy biorthogonal signals. ${s}_{1}(t)=s(t)$ , ${s}_{2}(t)={s}^{\perp }(t)$ , ${s}_{3}(t)=-s(t)$ , ${s}_{4}(t)=-{s}^{\perp }(t)$ .

The orthonormal basis ${\psi }_{1}(t)=\frac{s(t)}{\sqrt{{E}_{s}}}$ , ${\psi }_{2}(t)=\frac{{s}^{\perp }(t)}{\sqrt{{E}_{s}}}$ where ${E}_{s}=\int_{0}^{T} {s}_{m}(t)^{2}\,d t$

${s}_{1}=\left(\begin{array}{c}\sqrt{{E}_{s}}\\ 0\end{array}\right)$ , ${s}_{2}=\left(\begin{array}{c}0\\ \sqrt{{E}_{s}}\end{array}\right)$ , ${s}_{3}=\left(\begin{array}{c}-\sqrt{{E}_{s}}\\ 0\end{array}\right)$ , ${s}_{4}=\left(\begin{array}{c}0\\ -\sqrt{{E}_{s}}\end{array}\right)$ . The four signals can be geometrically represented using the 4-vector of projection coefficients ${s}_{1}$ , ${s}_{2}$ , ${s}_{3}$ , and ${s}_{4}$ as a set of constellation points.

${d}_{21}=\left|{s}_{2}-{s}_{1}\right|=\sqrt{2{E}_{s}}$
${d}_{12}={d}_{23}={d}_{34}={d}_{14}$
${d}_{13}=\left|{s}_{1}-{s}_{3}\right|=2\sqrt{{E}_{s}}$
${d}_{13}={d}_{24}$
Minimum distance ${d}_{\mathrm{min}}=\sqrt{2{E}_{s}}$

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