15.13: Plancharel and Parseval's Theorems

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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

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

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

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

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 \(\frac{1}{\sqrt{T}} e^{j w_{0} n t}\)

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