15.12 Plancharel and parseval's theorems

Parseval's theorem

Continuous Time Fourier Series preserves signal energy

i.e.:

Prove: plancherel theorem

Periodic signals power

Fourier series of square pulse iii -- compute the energy

$$\begin{array}{cc}\hfill f\left(t\right)& =\sum _{n=-\infty }^{\infty }{c}_n{e}^{j\frac{2\pi }{T}nt}\stackrel{\mathbb{FS}}{\to }{c}_n=\frac{1}{2}\frac{sin\frac{\pi }{2}n}{\frac{\pi }{2}n}\hfill \\ \hfill \text{energy}\text{in}\text{time}\text{domain:}& {\left|\right|f\left|\right|}_2^2=\underset{0}{\overset{T}{\int }}{\left|f\left(t\right)\right|}^{2}dt=\frac{T}{2}\hfill \\ \hfill \text{apply}\text{Parseval's}\text{Theorem:}& T\sum _n{\left|c_n\right|}^2\hfill \\ & =\frac{T}{4}\sum _n{\left(\frac{sin\frac{\pi }{2}n}{\frac{\pi }{2}n}\right)}^2\hfill \\ & =\frac{T}{4}\frac{4}{{\pi }^2}\sum _n\frac{{\left(sin,\frac{\pi }{2}n\right)}^2}{n^2}\hfill \\ & =\frac{T}{{\pi }^2}\left[\frac{{\pi }^2}{4},+,\underset{\frac{{\pi }^2}{4}}{\underbrace{{\sum }_{n\text{odd}}\frac{1}{n^2}}}\right]\hfill \\ & =\frac{T}{2}\square \hfill \end{array}$$

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Plancharel theorem

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 $${x\cdot y}_H=\sum {\alpha }_i\overline{{\beta }_i}$$

$$x=\sum {\alpha }_i{b}_i$$ $$y=\sum {\beta }_j{b}_j$$ $${x\cdot y}_H=\sum {\alpha }_i{b}_i\cdot \sum {\beta }_j{b}_j=\sum {\alpha }_i(b_i\cdot \sum {\beta }_j{b}_j)=\sum {\alpha }_i\sum \overline{{\beta }_j}(b_i\cdot b_j)=\sum {\alpha }_i\overline{{\beta }_i}$$ by using inner product rules

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 $\ell ^2$ .

Plancharels theorem demonstration

Parseval's theorem: a different approach

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 $$(, x)^2={x\cdot x}_H=\sum {\alpha }_i\overline{{\alpha }_i}=\sum \left|{\alpha }_i\right|^2$$