15.13: Teoremas de Plancharel y Parseval
- Page ID
- 86575
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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

Figura\(\PageIndex{1}\)
\[f(t)=\sum_{n=-\infty}^{\infty} c_{n} e^{j \frac{2 \pi}{T} n t} \stackrel{\mathbb{F}\mathbb{S}}{\rightarrow} c_{n}=\frac{1}{2} \frac{\sin \frac{\pi}{2} n}{\frac{\pi}{2} n} \nonumber \]
\[\text { energy in time domain: }\|f\|_{2}^{2}=\int_{0}^{T}|f(t)|^{2} d t=\frac{T}{2} \nonumber \]
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\]
\[\langle x, y\rangle_{H}=\sum_{i} \alpha_{i} \overline{\beta_{i}} \nonumber \]
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
Demostración del Teorema de Plancharels

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\]