# 15.12 Plancharel and parseval's theorems

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This module contains the definition of the Plancharel theorem and Parseval's theorem along with proofs and examples.

## Parseval's theorem

Continuous Time Fourier Series preserves signal energy

i.e.:

$\begin{array}{ccc}\hfill \underset{0}{\overset{T}{\int }}{|f\left(t\right)|}^{2}dt& =T\sum _{n=-\infty }^{\infty }{|{C}_{n}|}^{2}\hfill & \hfill \phantom{\rule{4.pt}{0ex}}\text{with}\phantom{\rule{4.pt}{0ex}}\text{unnormalized}\phantom{\rule{4.pt}{0ex}}\text{basis}\phantom{\rule{4.pt}{0ex}}{e}^{j\frac{2\pi }{T}nt}\\ \hfill \underset{0}{\overset{T}{\int }}{|f\left(t\right)|}^{2}dt& =\sum _{n=-\infty }^{\infty }{|{C}_{n}^{\text{'}}|}^{2}\hfill & \hfill \phantom{\rule{4.pt}{0ex}}\text{with}\phantom{\rule{4.pt}{0ex}}\text{unnormalized}\phantom{\rule{4.pt}{0ex}}\text{basis}\phantom{\rule{4.pt}{0ex}}\frac{{e}^{j\frac{2\pi }{T}nt}}{\sqrt{T}}\\ \hfill \underset{{L}^{2}\left[0,T\right)energy}{\underbrace{{||f||}_{2}^{2}}}& =\underset{{l}^{2}\left(Z\right)energy}{\underbrace{||{C}_{n}^{\text{'}}{||}_{2}^{2}}}\hfill & \end{array}$

## Prove: plancherel theorem

$\begin{array}{cc}\hfill \text{Given}\phantom{\rule{4.pt}{0ex}}f\left(t\right)& \stackrel{CTFS}{\to }{c}_{n}\hfill \\ \hfill g\left(t\right)& \stackrel{CTFS}{\to }{d}_{n}\hfill \\ \hfill \text{Then}\phantom{\rule{4.pt}{0ex}}\underset{0}{\overset{T}{\int }}f\left(t\right){g}^{*}\left(t\right)dt& =T\sum _{n=-\infty }^{\infty }{c}_{n}{d}_{n}^{*}\phantom{\rule{4.pt}{0ex}}\text{with}\phantom{\rule{4.pt}{0ex}}\text{unnormalized}\phantom{\rule{4.pt}{0ex}}\text{basis}\phantom{\rule{4.pt}{0ex}}{e}^{j\frac{2\pi }{T}nt}\hfill \\ \hfill \underset{0}{\overset{T}{\int }}f\left(t\right){g}^{*}\left(t\right)dt& =\sum _{n=-\infty }^{\infty }{c}_{n}^{\text{'}}{\left({d}_{n}^{\text{'}}\right)}^{*}\phantom{\rule{4.pt}{0ex}}\text{with}\phantom{\rule{4.pt}{0ex}}\text{normalized}\phantom{\rule{4.pt}{0ex}}\text{basis}\phantom{\rule{4.pt}{0ex}}\frac{{e}^{j\frac{2\pi }{T}nt}}{\sqrt{T}}\hfill \\ \hfill {〈f,g〉}_{{L}_{2}\left(0,T\right]}& ={〈c,d〉}_{{l}_{2}\left(\mathbb{Z}\right)}\hfill \end{array}$

## Periodic signals power

$\begin{array}{cc}\hfill \text{Energy}\phantom{\rule{4.pt}{0ex}}& ={||f||}^{2}=\underset{-\infty }{\overset{\infty }{\int }}{|f\left(t\right)|}^{2}dt=\infty \hfill \\ \hfill \text{Power}\phantom{\rule{4.pt}{0ex}}& =\underset{T\to \infty }{lim}\frac{\text{Energy}\phantom{\rule{4.pt}{0ex}}\text{in}\phantom{\rule{4.pt}{0ex}}\left[0,T\right)}{T}\hfill \\ & =\underset{T\to \infty }{lim}\frac{T{\sum }_{n}{|{c}_{n}|}^{2}}{T}\hfill \\ & =\sum _{n\in \mathbb{Z}}{|{c}_{n}|}^{2}\phantom{\rule{4.pt}{0ex}}\text{(unnormalized}\phantom{\rule{4.pt}{0ex}}\text{FS)}\hfill \end{array}$

## Fourier series of square pulse iii -- compute the energy

$\begin{array}{cc}\hfill f\left(t\right)& =\sum _{n=-\infty }^{\infty }{c}_{n}{e}^{j\frac{2\pi }{T}nt}\stackrel{\mathbb{FS}}{\to }{c}_{n}=\frac{1}{2}\frac{sin\frac{\pi }{2}n}{\frac{\pi }{2}n}\hfill \\ \hfill \text{energy}\phantom{\rule{4.pt}{0ex}}\text{in}\phantom{\rule{4.pt}{0ex}}\text{time}\phantom{\rule{4.pt}{0ex}}\text{domain:}\phantom{\rule{4.pt}{0ex}}& {||f||}_{2}^{2}=\underset{0}{\overset{T}{\int }}{|f\left(t\right)|}^{2}dt=\frac{T}{2}\hfill \\ \hfill \text{apply}\phantom{\rule{4.pt}{0ex}}\text{Parseval's}\phantom{\rule{4.pt}{0ex}}\text{Theorem:}\phantom{\rule{4.pt}{0ex}}& T\sum _{n}{|{c}_{n}|}^{2}\hfill \\ & =\frac{T}{4}\sum _{n}{\left(\frac{sin\frac{\pi }{2}n}{\frac{\pi }{2}n}\right)}^{2}\hfill \\ & =\frac{T}{4}\frac{4}{{\pi }^{2}}\sum _{n}\frac{{\left(sin,\frac{\pi }{2}n\right)}^{2}}{{n}^{2}}\hfill \\ & =\frac{T}{{\pi }^{2}}\left[\frac{{\pi }^{2}}{4},+,\underset{\frac{{\pi }^{2}}{4}}{\underbrace{{\sum }_{n\phantom{\rule{4.pt}{0ex}}\text{odd}}\frac{1}{{n}^{2}}}}\right]\hfill \\ & =\frac{T}{2}\square \hfill \end{array}$

## Plancharel theorem

The inner product of two vectors/signals is the same as the $\ell ^{2}$ inner product of their expansion coefficients.

Let $\{{b}_{i}\}$ be an orthonormal basis for a Hilbert Space $H$ . $x\in H$ , $y\in H$ $x=\sum {\alpha }_{i}{b}_{i}$ $y=\sum {\beta }_{i}{b}_{i}$ then ${x\dot y}_{H}=\sum {\alpha }_{i}\overline{{\beta }_{i}}$

Applying the Fourier Series, we can go from $f(t)$ to $\{{c}_{n}\}$ and $g(t)$ to $\{{d}_{n}\}$ $\int_{0}^{T} f(t)\overline{g(t)}\,d t=\sum_{n=()}$ c n d n inner product in time-domain = inner product of Fourier coefficients.

$x=\sum {\alpha }_{i}{b}_{i}$ $y=\sum {\beta }_{j}{b}_{j}$ ${x\dot y}_{H}=\sum {\alpha }_{i}{b}_{i}\dot \sum {\beta }_{j}{b}_{j}=\sum {\alpha }_{i}({b}_{i}\dot \sum {\beta }_{j}{b}_{j})=\sum {\alpha }_{i}\sum \overline{{\beta }_{j}}({b}_{i}\dot {b}_{j})=\sum {\alpha }_{i}\overline{{\beta }_{i}}$ by using inner product rules

${b}_{i}\dot {b}_{j}=0$ when $i\neq j$ and ${b}_{i}\dot {b}_{j}=1$ when $i=j$

If Hilbert space H has a ONB, then inner products are equivalent to inner products in $\ell ^{2}$ .

All H with ONB are somehow equivalent to $\ell ^{2}$ .

square-summable sequences are important

## Parseval's theorem

Energy of a signal = sum of squares of its expansion coefficients

Let $x\in H$ , $\{{b}_{i}\}$ ONB

$x=\sum {\alpha }_{i}{b}_{i}$ Then $(, x)^{2}=\sum \left|{\alpha }_{i}\right|^{2}$

Directly from Plancharel $(, x)^{2}={x\dot x}_{H}=\sum {\alpha }_{i}\overline{{\alpha }_{i}}=\sum \left|{\alpha }_{i}\right|^{2}$

Fourier Series $\frac{1}{\sqrt{T}}e^{i{w}_{0}nt}$ $f(t)=\frac{1}{\sqrt{T}}\sum {c}_{n}\frac{1}{\sqrt{T}}e^{i{w}_{0}nt}$ $\int_{0}^{T} \left|f(t)\right|^{2}\,d t=\sum_{n=()}$ c n 2

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