Pares clasicos de la transformada de Fourier

De Wikiversidad

Contenido

[editar] Tabla de Pares clasicos de la transformada de Fourier

\begin{align} & \mathbb{F}[f(t)]=F(\omega )=\int\limits_{-\infty }^{\infty }{f(t).e^{-j\omega t}\partial t} \\ & \mathbb{F}^{-1}[F(\omega )]=f(t)=\frac{1}{2\pi }\int\limits_{-\infty }^{\infty }{F(\omega ).e^{+j\omega t}\partial \omega } \\ \end{align}


Pulso rectangular

\prod{\left( \frac{t}{T} \right)}=\left\{ \begin{align} & 1,\left| t \right|\le {}^{T}\!\!\diagup\!\!{}_{2}\; \\ & 0,\left| t \right|>{}^{T}\!\!\diagup\!\!{}_{2}\; \\ \end{align} \right.

\mathbb{F}\left[ \prod{\left( \frac{t}{T} \right)} \right]=2\frac{\sin \left( \omega {}^{T}\!\!\diagup\!\!{}_{2}\; \right)}{\omega }=T\text{sinc}\left( T\frac{\omega }{2\pi } \right)
Pulso triangular la funcion abuelito

\Lambda \left( \frac{t}{T} \right)=\left\{ \begin{align} & 1-\frac{\left| t \right|}{T},\text{ }\left| t \right|\le T \\ & 0\text{ },\text{ }\left| t \right|>T \\ \end{align} \right.

\mathbb{F}\left[ \Lambda \left( \frac{t}{T} \right) \right]=\frac{2(1-\cos (\omega T))}{\omega ^{2}T}=T\cdot \text{sinc}^{2}\left( T\frac{\omega }{2\pi } \right)
\operatorname{sign}(t)=\left\{ \begin{align} & 1,\text{ }t>0 \\ & -1,\text{ }t<0 \\ \end{align} \right.

\mathbb{F}\left[ \operatorname{sign}(t) \right]=\frac{2}{j\omega }\xrightarrow[\omega =2\pi f]{}\frac{1}{j\pi f}
u(t)=\left\{ \begin{align} & 1,t>0 \\ & 0,t<0 \\ \end{align} \right. \mathbb{F}[u(t)]=\frac{1}{j\omega }+\underbrace{\pi \delta (\omega )}_{\text{a veces se omite}}
Delta de Dirac δ(t) \mathbb{F}[\delta (t)]=1\leftrightarrow F[1]=2\pi \delta (-\omega )
cos(ω0t) \mathbb{F}\left[ \cos (\omega _{0}t) \right]=\pi \delta (\omega -\omega _{0})+\pi \delta (\omega +\omega _{0})
sin(ω0t) \mathbb{F}\left[ \sin (\omega _{0}t) \right]=\frac{\pi }{j}\delta (\omega -\omega _{0})-\frac{\pi }{j}\delta (\omega +\omega _{0})
\sum\limits_{k=-\infty }^{\infty }{\delta \left( t-kT_{s} \right)}

\mathbb{F}\left[ \sum\limits_{k=-\infty }^{\infty }{\delta \left( t-kT_{s} \right)} \right]=\mathbb{F}\left[ \frac{1}{T_{s}}\sum\limits_{k=-\infty }^{\infty }{e^{j\frac{2\pi }{T_{s}}kt}} \right]=\frac{1}{T_{s}}\cdot \sum\limits_{k=-\infty }^{\infty }{2\pi \delta }\left( \omega -k\frac{2\pi }{T_{s}} \right)

Demostraciones:

[editar] Pulso rectangular

Se representa mediante el mismo simbolo que el productorio, siendo la funcion una fraccion: la parte de arriba (t) representa en funcion de que variable esta , la parte de abajo (T) representa la extension de la funcion, que irá desde –T/2 a T/2.


Pulso rectangular normalizada a T = 1

\begin{align} & \prod{\left( \frac{t}{T} \right)}=\left\{ \begin{align} & 1,-{}^{T}\!\!\diagup\!\!{}_{2}\;\le t\le {}^{T}\!\!\diagup\!\!{}_{2}\; \\ & 0,\left| t \right|>{}^{T}\!\!\diagup\!\!{}_{2}\; \\ \end{align} \right. \\ & \prod{\left( \frac{t-t_{0}}{T} \right)}=\left\{ \begin{align} & 1,-{}^{T}\!\!\diagup\!\!{}_{2}\;\le t-t_{0}\le {}^{T}\!\!\diagup\!\!{}_{2}\; \\ & 0,\left| t-t_{0} \right|>{}^{T}\!\!\diagup\!\!{}_{2}\; \\ \end{align} \right. \\ \end{align}


Ahora, la transformada de Fourier de un pulso rectangular:

\begin{align} & \mathbb{F}\left[ \prod{\left( \frac{t}{T} \right)} \right]=\int\limits_{-\infty }^{\infty }{\prod{\left( \frac{t}{T} \right)}\cdot e^{-j\omega t}\partial t}=\int\limits_{-{}^{T}\!\!\diagup\!\!{}_{2}\;}^{{}^{T}\!\!\diagup\!\!{}_{2}\;}{1\cdot e^{-j\omega t}\partial t}=\left. \frac{e^{-j\omega t}}{-j\omega } \right|_{{}^{-T}\!\!\diagup\!\!{}_{2}\;}^{{}^{T}\!\!\diagup\!\!{}_{2}\;}=\frac{1}{-j\omega }\left( e^{-j\omega {}^{T}\!\!\diagup\!\!{}_{2}\;}-e^{+j\omega {}^{T}\!\!\diagup\!\!{}_{2}\;} \right)= \\ & \frac{2}{2}\frac{1}{j\omega }\left( e^{+j\omega {}^{T}\!\!\diagup\!\!{}_{2}\;}-e^{-j\omega {}^{T}\!\!\diagup\!\!{}_{2}\;} \right)\xrightarrow[\begin{smallmatrix} \sin x= \\ \frac{e^{jx}-e^{-jx}}{2j} \end{smallmatrix}]{}=2\frac{\sin \left( \omega {}^{T}\!\!\diagup\!\!{}_{2}\; \right)}{\omega } \\ \end{align}

Sinc(t)

Al ser este resultado bastante habitual, se representa muchas veces mediante la funcion sinc().

\begin{align} & \text{sinc}\left( t \right)=\frac{\sin \left( \pi t \right)}{\pi t};\text{ sinc}\left( Tt \right)=\frac{\sin (T\pi t)}{T\pi t} \\ & \text{Nulos: }\sin ()=n\pi \to \text{sin}(T\pi t)=0\to T\pi t=n\pi \to t={}^{n}\!\!\diagup\!\!{}_{T}\; \\ & \text{sinc}\left( Tt \right)=\left\{ \begin{align} & 0,t={}^{n}\!\!\diagup\!\!{}_{T}\;,\forall n\in \mathbb{Z},n\ne 0 \\ & 1,t=0 \\ \end{align} \right., \\ \end{align}

La funcion sinc() se usa especialmente por comodidad y no tener que usar limites, pues:

\left. 2\frac{\sin \left( \omega {}^{T}\!\!\diagup\!\!{}_{2}\; \right)}{\omega } \right|_{\omega =0}=\frac{0}{0}\to \underset{\omega \to 0}{\mathop{\lim }}\,2\frac{\sin \left( \omega {}^{T}\!\!\diagup\!\!{}_{2}\; \right)}{\omega }\xrightarrow[\sin x\approx x]{}2\frac{\omega {}^{T}\!\!\diagup\!\!{}_{2}\;}{\omega }=T


Nos obliga a utilizar limites para saber el resultado, en cambio:

\begin{align} & 2\frac{\sin \left( \omega {}^{T}\!\!\diagup\!\!{}_{2}\; \right)}{\omega }=f(\text{sinc()})\xrightarrow[\omega =2\pi f]{}2\frac{\sin \left( \left( 2\pi f \right){}^{T}\!\!\diagup\!\!{}_{2}\; \right)}{2\pi f}\xrightarrow[{}^{T}\!\!\diagup\!\!{}_{T}\;]{}T\underbrace{\frac{\sin \left( \pi fT \right)}{\pi fT}}_{\text{sinc}(Tf)}= \\ & T\text{sinc}(Tf)\to f=0\to T\text{sinc}(0)=T \\ & \omega =2\pi f\leftrightarrow f={}^{\omega }\!\!\diagup\!\!{}_{2\pi }\;\to T\text{sinc}(Tf)=T\text{sinc}\left( T\frac{\omega }{2\pi } \right) \\ \end{align}

[editar] Pulso Triangular

Funcion triangular normalizada a T = 1


\begin{align} & \Lambda \left( \frac{t}{T} \right)=\left\{ \begin{align} & 1-\frac{\left| t \right|}{T},\text{ }\left| t \right|\le T \\ & 0\text{ },\text{ }\left| t \right|>T \\ \end{align} \right. \\ & \mathbb{F}\left[ \Lambda \left( \frac{t}{T} \right) \right]=\int_{-\infty }^{\infty }{\underbrace{\Lambda \left( \frac{t}{T} \right)}_{\text{par}}\cdot }\underbrace{e^{-j\omega t}}_{\underbrace{\cos (\omega t)}_{\text{par}}-j\underbrace{\sin (\omega t)}_{\text{impar}}}\partial t=2\int_{0}^{T}{\left( 1-\frac{t}{T} \right)}\cos (\omega t)\partial t= \\ & \xrightarrow[\int{u.\partial v}]{}\left\{ \begin{align} & u=1-\frac{t}{T}\text{ }\partial u=\frac{-1}{T}\partial t \\ & \partial v=\cos (\omega t)\partial t\text{ }v=\frac{\text{sin(}\omega \text{t)}}{\omega }\text{ } \\ \end{align} \right\}\xrightarrow[u.v-\int{v\partial u}]{}2\left. \left( 1-\frac{t}{T} \right)\frac{\sin (\omega t)}{\omega } \right|_{0}^{T}+2\int_{0}^{T}{\frac{\sin (\omega t)}{\omega T}\partial t} \\ & 2\left. \left( 1-\frac{t}{T} \right)\frac{\sin (\omega t)}{\omega } \right|_{0}^{T}+\left. \frac{-2\cos (\omega t)}{\omega ^{2}T} \right|_{0}^{T}=2\left( \underbrace{1-\frac{T}{T}}_{\hat{\ }0} \right)\frac{\sin (\omega t)}{\omega }+\left. \frac{-2\cos (\omega t)}{\omega ^{2}T} \right|_{0}^{T}= \\ & \left. \frac{-2\cos (\omega t)}{\omega ^{2}T} \right|_{0}^{T}=\frac{-2}{\omega ^{2}T}\left( \cos (\omega T)-1 \right)=\frac{2(1-\cos (\omega T))}{\omega ^{2}T} \\ \end{align}

Sabiendo que


\begin{align} & \xrightarrow[\sin ^{2}x=\frac{1-\cos 2x}{2}]{}2\frac{\left( 2\sin ^{2}\left( \frac{\omega T}{2} \right) \right)}{\omega ^{2}T}=4\frac{\sin ^{2}\left( \frac{\omega T}{2} \right)}{\omega ^{2}T}=\frac{\sin ^{2}\left( \frac{\omega T}{2} \right)}{\frac{\omega ^{2}T}{4}}=T\frac{\sin ^{2}\left( \frac{\omega T}{2} \right)}{\frac{\omega ^{2}T^{2}}{4}}=T\frac{\sin ^{2}\left( \frac{\omega T}{2} \right)}{\left( \frac{\omega T}{2} \right)^{2}}= \\ & \xrightarrow[\text{sinc}\left( T\omega \right)=\frac{\sin \left( \pi T\omega \right)}{\pi T\omega }]{}T\cdot \text{sinc}^{2}\left( T\frac{\omega }{2\pi } \right)\underset{\begin{smallmatrix} \text{o tambien} \\ \omega =2\pi f \end{smallmatrix}}{\longleftrightarrow}T\text{sinc}^{2}\left( Tf \right)=F_{\Lambda }(f) \\ & \text{Sabiendo que }\mathbb{F}\left[ \prod{\left( \frac{t}{T} \right)} \right]=T\text{sinc}\left( Tf \right)=F(f)\to \\ & F^{2}(f)=T\cdot F_{\Lambda }(f)\leftrightarrow f(t)*f(t)=T\cdot f_{\Lambda }(t)\to \\ & \prod{\left( \frac{t}{T} \right)}*\prod{\left( \frac{t}{T} \right)}=T\cdot \Lambda \left( \frac{t}{T} \right)\Leftrightarrow \mathbb{F}\left[ \prod{\left( \frac{t}{T} \right)}*\prod{\left( \frac{t}{T} \right)} \right]=T\text{sinc}^{2}\left( Tf \right) \\ \end{align}

Recordar, que mientras que el pulso de rectangular \prod{\left( \frac{t}{T} \right)} tiene una anchura de T (llega de –T/2 hasta T/2), el pulso triangular \Lambda \left( \frac{t}{T} \right) tiene una anchura de 2T (desde –T a T). Es conveniente recordarlo, pues suele ser un error habitual.

[editar] Funcion sign(t)

La funcion sign(t) es una funcion auxiliar bastante utilizada en areas de telecomunicacion que, ademas, es facilmente representable:

Sign(t)

\operatorname{sign}(t)=\left\{ \begin{align} & 1,\text{ }t>0 \\ & -1,\text{ }t<0 \\ \end{align} \right.

El valor de sign(t) cuando t=0 es:

\ell =\frac{\underset{t\to 0^{+}}{\mathop{\lim }}\,\text{ }\operatorname{sign}(t)+\underset{t\to 0^{-}}{\mathop{\lim }}\,\text{ sign}(t)}{2}=\frac{1+\left( -1 \right)}{2}=0

Veamos ahora, su transformada de Fourier:

\begin{align} & \operatorname{sign}(t)=\left\{ \begin{align} & 1,\text{ }t>0 \\ & -1,\text{ }t<0 \\ \end{align} \right. \\ & \mathbb{F}\left[ \operatorname{sign}(t) \right]=\int_{-\infty }^{\infty }{\operatorname{sign}(t)\cdot e^{-j\omega t}\partial t} \\ \end{align}

Usaremos integracion por partes y las propiedades de Fourier para sacar su transformada. Realizaremos la transformada de la parte positiva, usando funciones auxiliares y cambios de variables.

\begin{align} & f(t)=\left\{ \begin{align} & 1,t>0 \\ & 0,t<0 \\ \end{align} \right. \\ & g(t)=\left\{ \begin{align} & 0,t>0 \\ & -1,t<0 \\ \end{align} \right.\to g(t)=-f(-t) \\ & \operatorname{sign}(t)=f(t)-f(-t) \\ & \mathbb{F}[f(t)]=\int_{-\infty }^{\infty }{f(t)\cdot e^{-j\omega t}\partial t=}\int_{0}^{\infty }{1\cdot e^{-j\omega t}\partial t}=\left. \frac{e^{-j\omega t}}{-j\omega } \right|_{0}^{\infty }=\frac{1}{j\omega } \\ & \mathbb{F}\left[ f(t) \right]=F(\omega )\to \mathbb{F}\left[ f(-t) \right]=F(-\omega ) \\ & \mathbb{F}\left[ \operatorname{sign}(t) \right]=\mathbb{F}\left[ f(t)-f(-t) \right]=\frac{1}{j\omega }-\frac{1}{j\left( -\omega \right)}=\frac{2}{j\omega }\xrightarrow[\omega =2\pi f]{}\frac{1}{j\pi f} \\ \end{align}


== Esto esta mal, no tiene coherencia con la transformada de sign(t) ==

[editar] Funcion u(t)

La siguiente funcion, llamada Heaviside step function, o la funcion escalon unidad, se define:

función escalón considerando u(0) = 1/2

u(t)=\left\{ \begin{align} & 1,t>0 \\ & 0,t<0 \\ \end{align} \right.

Para solucionar el valor de u(t) cuando t=0, se usa:

\ell =\frac{\underset{t\to t^{+}}{\mathop{\lim }}\,\text{ }u(t)+\underset{t\to t^{-}}{\mathop{\lim }}\,\text{ }u(t)}{2}=\frac{1+0}{2}=\frac{1}{2}

Ahora, la transformada de Fourier:

\mathbb{F}[u(t)]=\int_{-\infty }^{\infty }{u(t)\cdot e^{-j\omega t}\partial t=}\int_{0}^{\infty }{1\cdot e^{-j\omega t}\partial t}=\left. \frac{e^{-j\omega t}}{-j\omega } \right|_{0}^{\infty }=\frac{1}{j\omega }+\underbrace{\pi \delta (\omega )}_{F[{}^{1}\!\!\diagup\!\!{}_{2}\;]}


Para clarificar la aparicion de la delta, tambien podemos obtenerla representando u(t) en funcion de sign(t).

\begin{align} & \mathbb{F}\left[ u(t) \right]=? \\ & \mathbb{F}\left[ \operatorname{sign}(t) \right]=\frac{2}{j\omega } \\ & u(t)=\frac{1}{2}\cdot \left( 1+\operatorname{sign}(t) \right)\to \mathbb{F}\left[ u(t) \right]=\mathbb{F}\left[ {}^{1}\!\!\diagup\!\!{}_{2}\; \right]+\mathbb{F}\left[ {}^{1}\!\!\diagup\!\!{}_{2}\;\cdot \operatorname{sign}(t) \right]=\pi \delta (\omega )+\frac{1}{j\omega } \\ \end{align}

Esta funcion resulta muy util como funcion auxiliar, pues la señales solo existen a partir de un momento en el tiempo, a modo de ejemplo:

\begin{align} & Ej: \\ & f(t)=e^{-at}u(t)=\left\{ \begin{align} & e^{-at},t>0 \\ & 0\text{ }\text{,}t<0 \\ \end{align} \right.;a>0 \\ & \mathbb{F}[f(t)]=F(\omega )=\int\limits_{-\infty }^{\infty }{f(t)e^{-j\omega t}\partial t=}\int\limits_{0}^{\infty }{e^{-at}e^{-j\omega t}\partial t}=\left. \frac{e^{-at}e^{-j\omega t}}{-a-j\omega } \right|_{0}^{\infty }=\frac{1}{-a-j\omega }\left( \underbrace{e^{-\infty }}_{0}e^{-j\omega \infty }-1 \right)= \\ & \frac{1}{a+j\omega }=F(\omega ) \\ \end{align}

[editar] Delta de Dirac

Diagrama esquemático de la función delta de Dirac

La funcion delta de Dirac es una funcion muy especial tanto por su forma como por sus propiedades, se denota como:

\delta (\omega )=\left\{ \begin{align} & \infty ,\text{ }\omega =0 \\ & 0,\text{ }\omega \ne 0 \\ \end{align} \right.

Entre sus propiedades:

\begin{align} & x(t)*\delta (t)=x(t) \\ & x(t)*\delta (t-t_{0})=x(t-t_{0}) \\ & x(t)\cdot \delta (t_{0})=x(t_{0})\to x(t)\cdot \delta (0)=x(0) \\ & \delta (t)=\delta (-t) \\ \end{align}

Su transformada de Fourier es:

\mathbb{F}[\delta (t)]=1\leftrightarrow F[1]=2\pi \delta (-\omega )=2\pi \delta (\omega )

Tambien tenemos que:

\begin{align} & \delta (at)=\frac{1}{\left| a \right|}\delta (t) \\ & \text{Debido a: }\mathbb{F}[\delta (at)]\xrightarrow[\mathbb{F}[f(at)]=\frac{1}{\left| a \right|}F\left( \frac{\omega }{a} \right)]{\mathbb{F}[\delta (t)]=1\ne f(\omega )}\frac{1}{\left| a \right|}\cdot 1\to \mathbb{F}^{-1}\left[ \frac{1}{a} \right]=\frac{1}{a}\cdot \delta (t) \\ \end{align}

Esta relacionada con la funcion escalon unidad de la siguiente manera:

\begin{align} & \frac{\partial u(t)}{\partial t}=\delta (t)\leftrightarrow u(t)=\int_{-\infty }^{t}{\delta (\tau )\partial \tau } \\ & \text{debido a:} \\ & \mathbb{F}\left[ \int_{-\infty }^{t}{f(\tau )\partial \tau } \right]=\frac{F(\omega )}{j\omega }\Rightarrow f(t)=\delta (t)\to F(\omega )=1 \\ & \mathbb{F}\left[ \int_{-\infty }^{t}{\delta (\tau )\partial \tau } \right]=\underbrace{\frac{1}{j\omega }}_{\mathbb{F}[u(t)]} \\ \end{align}

y tambien tenemos: \int_{-\infty }^{\infty }{\delta (t)\partial t=1}\text{ debido a: }\mathbb{F}\text{ }\!\![\!\!\text{ }\delta \text{(t) }\!\!]\!\!\text{ =1}\to \left. \text{1} \right|_{\omega =0}=\int_{-\infty }^{\infty }{f(t)\partial t}

La mejor de entender la funcion delta de Dirac, es relacionarlo con la funcion sinc().

\begin{align} & \mathbb{F}\left[ \prod{\left( \frac{t}{T} \right)} \right]=T\text{sinc}\left( Tf \right)=T\text{sinc}\left( T{}^{\omega }\!\!\diagup\!\!{}_{2\pi }\; \right) \\ & \underset{T\to \infty }{\mathop{\lim }}\,\prod{\left( \frac{t}{T} \right)}=1 \\ & \mathbb{F}\left[ \underset{T\to \infty }{\mathop{\lim }}\,\prod{\left( \frac{t}{T} \right)} \right]=\underbrace{\underset{T\to \infty }{\mathop{\lim }}\,\text{ }T\text{sinc}\left( T\frac{\omega }{2\pi } \right)}_{\delta \left( \frac{\omega }{2\pi } \right)}=\underbrace{\underset{T\to \infty }{\mathop{\lim }}\,\text{ }T\text{sinc}\left( Tf \right)}_{\delta (f)}=\left\{ \begin{align} & \infty ,f=0 \\ & 0,\text{ }f\ne 0 \\ \end{align} \right. \\ & \delta (f)=\underset{T\to \infty }{\mathop{\lim }}\,\text{ }T\text{sinc}\left( Tf \right)=\delta \left( {}^{\omega }\!\!\diagup\!\!{}_{2\pi }\; \right)\xrightarrow[\delta (a\omega )=\frac{1}{\left| a \right|}\delta (\omega )]{}2\pi \delta (\omega ) \\ \end{align}

[editar] Sin()y Cos()

\begin{align} & \mathbb{F}\left[ \cos (\omega _{0}t) \right]=\pi \delta (\omega -\omega _{0})+\pi \delta (\omega +\omega _{0}) \\ & \mathbb{F}\left[ \sin (\omega _{0}t) \right]=\frac{\pi }{j}\delta (\omega -\omega _{0})-\frac{\pi }{j}\delta (\omega +\omega _{0}) \\ \end{align}

Demostracion:

\begin{align} & \mathbb{F}[1]=2\pi \delta (\omega )\text{ y }\mathbb{F}[f(t)\cdot e^{+j\omega _{0}t}]=F(\omega -\omega _{0}) \\ & \mathbb{F}[1\cdot e^{+j\omega _{0}t}]=2\pi \delta (\omega -\omega _{0}) \\ & \mathbb{F}\left[ \cos (\omega _{0}t) \right]=\mathbb{F}\left[ \frac{e^{j\omega _{0}t}+e^{-j\omega _{0}t}}{2} \right]=\frac{1}{2}\left[ 2\pi \delta (\omega -\omega _{0})+2\pi \delta (\omega +\omega _{0}) \right] \\ \end{align}

Analogamente:

\begin{align} & \mathbb{F}[e^{+j\omega _{0}t}]=2\pi \delta (\omega -\omega _{0}) \\ & \mathbb{F}\left[ \sin (\omega _{0}t) \right]=\mathbb{F}\left[ \frac{e^{j\omega _{0}t}-e^{-j\omega _{0}t}}{2j} \right]=\frac{1}{2j}\left[ 2\pi \delta (\omega -\omega _{0})-2\pi \delta (\omega +\omega _{0}) \right] \\ \end{align}

[editar] Tren de deltas

\sum\limits_{k=-\infty }^{\infty }{\delta \left( t-kT_{s} \right)}

Primeramente, apreciamos que un sumatorio de deltas (llamado comunmente tren de deltas) es una señal periodica, por lo que puede ser representada como suma de senos y cosenos segun la serie de Fourier: Llamemos al periodo de la señal Ts.

\begin{align} & f(t)=\sum\limits_{k=-\infty }^{\infty }{\delta \left( t-kT_{s} \right)} \\ & T=T_{s}\to f_{s}=\frac{1}{T_{s}}\to \omega _{s}=2\pi f_{s}=\frac{2\pi }{T_{s}} \\ & f(t)=\sum\limits_{k=-\infty }^{\infty }{C_{k}\cdot e^{j\frac{2\pi }{T}kt}}=\sum\limits_{k=-\infty }^{\infty }{C_{k}\cdot e^{j\frac{2\pi }{T_{s}}kt}} \\ & C_{k}=\frac{1}{T}\int\limits_{{}^{-T}\!\!\diagup\!\!{}_{2}\;}^{{}^{T}\!\!\diagup\!\!{}_{2}\;}{f(t)\cdot e^{-j\frac{2\pi }{T_{s}}kt}\partial t}=\frac{1}{T_{s}}\int\limits_{{}^{-T_{s}}\!\!\diagup\!\!{}_{2}\;}^{{}^{T_{s}}\!\!\diagup\!\!{}_{2}\;}{\delta (t)\cdot e^{-j\frac{2\pi }{T_{s}}kt}\partial t}=\frac{1}{T_{s}}\int\limits_{{}^{-T_{s}}\!\!\diagup\!\!{}_{2}\;}^{{}^{T_{s}}\!\!\diagup\!\!{}_{2}\;}{\delta (t)\cdot \left. e^{-j\frac{2\pi }{T_{s}}kt} \right|_{t=0}\partial t}=\frac{1}{T_{s}}\int\limits_{{}^{-T_{s}}\!\!\diagup\!\!{}_{2}\;}^{{}^{T_{s}}\!\!\diagup\!\!{}_{2}\;}{\delta (t)\partial t} \\ & \to \int\limits_{-\infty }^{\infty }{\delta (t)\partial t}=1\to C_{k}=\frac{1}{T_{s}}\text{ }\forall k\to \\ & f(t)=\sum\limits_{k=-\infty }^{\infty }{\frac{1}{T_{s}}e^{j\frac{2\pi }{T_{s}}kt}}=\frac{1}{T_{s}}\sum\limits_{k=-\infty }^{\infty }{e^{j\frac{2\pi }{T_{s}}kt}} \\ & \sum\limits_{k=-\infty }^{\infty }{\delta \left( t-kT_{s} \right)}=\frac{1}{T_{s}}\sum\limits_{k=-\infty }^{\infty }{e^{j\frac{2\pi }{T_{s}}kt}} \\ \end{align}


\begin{align} & \text{Recordemos las propiedades: }\mathbb{F}[e^{+j\omega _{0}t}f(t)]=F(\omega -\omega _{0}) \\ & \omega _{0}=\frac{2\pi }{T_{s}},\mathbb{F}[1]=2\pi \delta (\omega )\to \mathbb{F}[e^{+j\omega _{0}t}]=2\pi \delta (\omega -\omega _{0}) \\ & \mathbb{F}\left[ \frac{1}{T_{s}}\sum\limits_{k=-\infty }^{\infty }{e^{j\frac{2\pi }{T_{s}}kt}} \right]=\frac{1}{T_{s}}\cdot \mathbb{F}\left[ 1+e^{j\frac{2\pi }{T_{s}}t}+e^{-j\frac{2\pi }{T_{s}}t}+e^{j\frac{2\pi }{T_{s}}2t}+e^{-j\frac{2\pi }{T_{s}}2t}+e^{j\frac{2\pi }{T_{s}}3t}+... \right]= \\ & \frac{1}{T_{s}}\cdot \left[ 2\pi \delta (\omega )+2\pi \delta \left( \omega -\frac{2\pi }{T_{s}} \right)+2\pi \delta \left( \omega +\frac{2\pi }{T_{s}} \right)+2\pi \delta \left( \omega -2\cdot \frac{2\pi }{T_{s}} \right)+... \right] \\ & \frac{1}{T_{s}}\cdot \sum\limits_{k=-\infty }^{\infty }{2\pi \delta }\left( w-k\frac{2\pi }{T_{s}} \right) \\ & \mathbb{F}\left[ \sum\limits_{k=-\infty }^{\infty }{\delta \left( t-kT_{s} \right)} \right]=\mathbb{F}\left[ \frac{1}{T_{s}}\sum\limits_{k=-\infty }^{\infty }{e^{j\frac{2\pi }{T_{s}}kt}} \right]=\frac{1}{T_{s}}\cdot \sum\limits_{k=-\infty }^{\infty }{2\pi \delta }\left( \omega -k\frac{2\pi }{T_{s}} \right) \\ \end{align}

Obtenido de "http://es.wikiversity.org/wiki/Pares_clasicos_de_la_transformada_de_Fourier"
Herramientas personales