15.13: Plancharel and Parseval's Theorems

Parseval's Theorem

Continuous Time Fourier Series preserves signal energy

i.e.:

$$\int_{0}^{T}|f(t)|^{2} d t=T \sum_{n=-\infty}^{\infty}\left|C_{n}\right|^{2} \quad \text { with unnormalized basis } e^{j \frac{2 \pi}{T} n t} \nonumber$$

$$\int_{0}^{T}|f(t)|^{2} d t=\sum_{n=-\infty}^{\infty}\left|C_{n}\right|^{2} \quad \text { with unnormalized basis } \frac{e^{j \frac{2 \pi}{T} n t}}{\sqrt{T}} \nonumber$$

$$\underbrace{\|f\|_{2}^{2}}_{L^{2}[0, T) e n e r g y}=\underbrace{\left\|C_{n}^{\prime}\right\|_{2}^{2}}_{l^{2}(Z) e n e r g y} \nonumber$$

Prove: Plancherel theorem

$$\begin{aligned} \text { Given } f(t) & \stackrel{\text { CTFS }}{\longrightarrow} c_{n} \\ g(t) & \stackrel{\text { CTFS }}{\longrightarrow} d_{n} \end{aligned} \nonumber$$

$$\text { Then } \int_{0}^{T} f(t) g^{*}(t) d t=T \sum_{n=-\infty}^{\infty} c_{n} d_{n}^{*} \text { with unnormalized basis } e^{j \frac{2 \pi}{T} n t} \nonumber$$

$$\int_{0}^{T} f(t) g^{*}(t) d t=\sum_{n=-\infty}^{\infty} c_{n}^{\prime}\left(d_{n}^{\prime}\right)^{*} \text { with normalized basis } \frac{e^{j \frac{2 \pi}{T} n t}}{\sqrt{T}} \nonumber$$

$$\langle f, g\rangle_{L_{2}(0, T]}=\langle c, d\rangle_{l_{2}(\mathbb{Z})} \nonumber$$

Periodic Signals Power

$$\text { Energy }=\|f\|^{2}=\int_{-\infty}^{\infty}|f(t)|^{2} d t=\infty \nonumber$$

$$\begin{aligned} \text { Power } &=\lim _{T \rightarrow \infty} \frac{\text { Energy in }[0, T)}{T} \\ &=\lim _{T \rightarrow \infty} \frac{T \sum_{n}\left|c_{n}\right|^{2}}{T} \\ &=\sum_{n \in \mathbb{Z}}\left|c_{n}\right|^{2}(\text { unnormalized } \mathrm{FS}) \end{aligned} \nonumber$$

Example $\PageIndex{1}$: Fourier Series of Square Pulse III - Compute the Energy

Figure $\PageIndex{1}$

$$f(t)=\sum_{n=-\infty}^{\infty} c_{n} e^{j \frac{2 \pi}{T} n t} \stackrel{\mathbb{F}\mathbb{S}}{\rightarrow} c_{n}=\frac{1}{2} \frac{\sin \frac{\pi}{2} n}{\frac{\pi}{2} n} \nonumber$$

$$\text { energy in time domain: }\|f\|_{2}^{2}=\int_{0}^{T}|f(t)|^{2} d t=\frac{T}{2} \nonumber$$

Apply Parseval's Theorem:

$$\quad T \sum_{n}\left|c_{n}\right|^{2} \nonumber$$
$$\begin{array}{l} =\frac{T}{4} \sum_{n}\left(\frac{\sin \frac{\pi}{2} n}{\frac{\pi}{2} n}\right)^{2} \\ =\frac{T}{4} \frac{4}{\pi^{2}} \sum_{n} \frac{\left(\sin \frac{\pi}{2} n\right)^{2}}{n^{2}} \\ =\frac{T}{\pi^{2}} \left[ \frac{\pi^{2}}{4}+\underbrace{\sum_{n} \operatorname{odd} \frac{1}{n^{2}}}_{\frac{\pi^{2}}{4}} \right] \\ =\frac{T}{2} \square \end{array} \nonumber$$

Plancharel Theorem

Theorem $\PageIndex{1}$: Plancharel Theorem

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

Let $\left\{b_{i}\right\}$ be an orthonormal basis for a Hilbert Space $H$. $x \in H$, $y \in H$

$$\begin{array}{l} x=\sum_{i} \alpha_{i} b_{i} \\ y=\sum_{i} \beta_{i} b_{i} \end{array} \nonumber$$

then

$$\langle x, y\rangle_{H}=\sum_{i} \alpha_{i} \overline{\beta_{i}} \nonumber$$

Example $\PageIndex{2}$

Applying the Fourier Series, we can go from $f(t)$ to $\left\{c_{n}\right\}$ and $g(t)$ to $\left\{d_{n}\right\}$

$$\int_{0}^{T} f(t) \overline{g(t)} \mathrm{d} t=\sum_{n=-\infty}^{\infty} c_{n} \overline{d_{n}} \nonumber$$

inner product in time-domain = inner product of Fourier coefficients.

Proof:

$$\begin{array}{l} x=\sum_{i} \alpha_{i} b_{i} \\ y=\sum_{j} \beta_{j} b_{j} \end{array} \nonumber$$

$$\langle x, y\rangle_{H}=\left\langle\sum_{i} \alpha_{i} b_{i}, \sum_{j} \beta_{j} b_{j}\right\rangle=\sum_{i} \alpha_{i}\left\langle\left(b_{i}, \sum_{j} \beta_{j} b_{j}\right)\right\rangle=\sum_{i} \alpha_{i} \sum_{j} \bar{\beta}_{j}\left\langle\left(b_{i}, b_{j}\right)\right\rangle=\sum_{i} \alpha_{i} \bar{\beta}_{i} \nonumber$$

by using inner product rules (Section 15.4)

Note

$\left\langle b_{i}, b_{j}\right\rangle=0$ when $i \neq j$ and $\left\langle b_{i}, b_{j}\right\rangle=1$ 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 $\ell^2$.

Point of Interest

Square-summable sequences are important

Plancharels Theorem Demonstration

Figure $\PageIndex{2}$: Interact (when online) with a Mathematica CDF demonstrating Plancharels Theorem visually. To Download, right-click and save target as .cdf.

Parseval's Theorem: a different approach

Theorem $\PageIndex{2}$: Parseval's Theorem

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

Let $x \in H$, $\left\{b_{i}\right\}$ ONB

$$x=\sum_{i} \alpha_{i} b_{i} \nonumber$$

Then

$$\left(\|x\|_{H}\right)^{2}=\sum_{i}\left(\left|\alpha_{i}\right|\right)^{2} \nonumber$$

Proof:

Directly from Plancharel

$$\left(\|x\|_{H}\right)^{2}=\langle x, x\rangle_{H}=\sum_{i} \alpha_{i} \overline{\alpha_{i}}=\sum_{i}\left(\left|\alpha_{i}\right|\right)^{2} \nonumber$$

Example $\PageIndex{3}$

Fourier Series $\frac{1}{\sqrt{T}} e^{j w_{0} n t}$

$$\begin{array}{c} f(t)=\frac{1}{\sqrt{T}} \sum_{n} c_{n} \frac{1}{\sqrt{T}} e^{j w_{0} n t} \\ \int_{0}^{T}(|f(t)|)^{2} d t=\sum_{n=-\infty}^{\infty}\left(\left|c_{n}\right|\right)^{2} \end{array} \nonumber$$