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# 15.13: Teoremas de Plancharel y Parseval

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## Teorema de Parseval

La serie de tiempo continuo de Fourier conserva la energía de la señal

es decir:

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

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

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

### Demostrar: Teorema de Plancherel

\ [\ begin {alineado}
\ text {dado} f (t) &\ stackrel {\ text {CTFS}} {\ longrightarrow} c_ {n}\\
g (t) &\ stackrel {\ texto {CTFS}} {\ larightarrow} d_ {n}
\ end {alineado}\ nonumber\]

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

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

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

### Potencia de señales periódicas

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

\ [\ begin {alineado}
\ text {Poder} &=\ lim _ {T\ fila derecha\ infty}\ frac {\ text {Energía en} [0, T)} {T}\\
&=\ lim _ {T\ fila derecha\ infty}\ frac {T\ sum_ {n}\ izquierda|c_ {n}\ derecha|^ {2}} {T}\\
&=\ sum_ {n\ in\ mathbb {Z}}\ izquierda|c_ {n}\ derecha|^ {2} (\ texto {no normalizado}\ mathrm {FS})
\ end {alineado}\ nonumber\]

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

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

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

Aplicar el Teorema de Parseval:

$\quad T \sum_{n}\left|c_{n}\right|^{2} \nonumber$
\ [\ begin {array} {l}
=\ frac {T} {4}\ sum_ {n}\ izquierda (\ frac {\ sin\ frac {\ pi} {2} n} {\ frac {\ pi} {2} n}\ derecha) ^ {2}\\
=\ frac {T} {4}\ frac {4} {\ pi^ {2}} sum_ {n}\ frac {\ left (\ sin\ frac {\ pi} {2} n\ derecha) ^ {2}} {n^ {2}}\\
=\ frac {T} {\ pi^ {2}}\ izquierda [\ frac {\ pi^ {2}} { 4} +\ underbrackets {\ suma_ {n}\ nombreoperador {impar}\ frac {1} {n^ {2}}} _ {\ frac {\ pi^ {2}} {4}}\ derecha]\\
=\ frac {T} {2}\ cuadrado
\ final {array}\ nonumber\]

## Teorema de Plancharel

Teorema$$\PageIndex{1}$$: Plancharel Theorem

El producto interno de dos vectores/señales es el mismo que el producto$$\ell^2$$ interno de sus coeficientes de expansión.

Dejar$$\left\{b_{i}\right\}$$ ser una base ortonormal para un Espacio Hilbert$$H$$. $$x \in H$$,$$y \in H$$

\ [\ begin {array} {l}
x=\ suma_ {i}\ alpha_ {i} b_ {i}\\
y=\ suma_ {i}\ beta_ {i} b_ {i}
\ end {array}\ nonumber\]

entonces

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

Ejemplo$$\PageIndex{2}$$

Aplicando la Serie de Fourier, podemos ir de$$f(t)$$ a$$\left\{c_{n}\right\}$$ y$$g(t)$$ a$$\left\{d_{n}\right\}$$

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

producto interno en el dominio del tiempo = producto interno de los coeficientes de Fourier.

Prueba:

\ [\ begin {array} {l}
x=\ suma_ {i}\ alpha_ {i} b_ {i}\\
y=\ suma_ {j}\ beta_ {j} b_ {j}
\ end {array}\ nonumber\]

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

mediante el uso de reglas internas del producto (Sección 15.4)

Nota

$$\left\langle b_{i}, b_{j}\right\rangle=0$$cuándo$$i \neq j$$ y$$\left\langle b_{i}, b_{j}\right\rangle=1$$ cuándo$$i=j$$

Si Hilbert space H tiene un ONB, entonces los productos internos son equivalentes a los productos internos en$$\ell^2$$.

Todos los H con ONB son de alguna manera equivalentes a$$\ell^2$$.

Punto de Interés

Las secuencias sumables al cuadrado son importantes

## Teorema de Parseval: un enfoque diferente

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

Energía de una señal = suma de cuadrados de sus coeficientes de expansión

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

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

Entonces

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

Prueba:

Directamente desde Plancharel

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

Ejemplo$$\PageIndex{3}$$

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

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

This page titled 15.13: Teoremas de Plancharel y Parseval is shared under a CC BY license and was authored, remixed, and/or curated by Richard Baraniuk et al..