# Pares clásicos de la transformada de Fourier
## Tabla de Pares clásicos de la transformada de Fourier

${\begin{aligned}&\mathbb {F} [f(t)]=F(\omega )=\int \limits _{-\infty }^{\infty }{f(t).e^{-j\omega t}dt}\\&\mathbb {F} ^{-1}[F(\omega )]=f(t)={\frac {1}{2\pi }}\int \limits _{-\infty }^{\infty }{F(\omega ).e^{+j\omega t}d\omega }\\\end{aligned}}$
| Pulso rectangular $\prod {\left({\frac {t}{T}}\right)}=\left\{{\begin{aligned}&1,\left|t\right|\leq^{T}\!\!\diagup \!\!_{2}\;\\&0,\left|t\right|>^{T}\!\!\diagup \!\!_{2}\;\\\end{aligned}}\right.$ | $\mathbb {F} \left[\prod {\left({\frac {t}{T}}\right)}\right]=2{\frac {\sin \left(\omega^{T}\!\!\diagup \!\!_{2}\;\right)}{\omega }}=T{\text{sinc}}\left(T{\frac {\omega }{2\pi }}\right)$ |
| Pulso triangular $\Lambda \left({\frac {t}{T}}\right)=\left\{{\begin{aligned}&1-{\frac {\left|t\right|}{T}},{\text{ }}\left|t\right|\leq T\\&0{\text{ }},{\text{ }}\left|t\right|>T\\\end{aligned}}\right.$ | $\mathbb {F} \left[\Lambda \left({\frac {t}{T}}\right)\right]={\frac {2(1-\cos(\omega T))}{\omega ^{2}T}}=T\cdot {\text{sinc}}^{2}\left(T{\frac {\omega }{2\pi }}\right)$ |
| $\operatorname {sign} (t)=\left\{{\begin{aligned}&1,{\text{ }}t>0\\&-1,{\text{ }}t<0\\\end{aligned}}\right.$ | $\mathbb {F} \left[\operatorname {sign} (t)\right]={\frac {2}{j\omega }}{\xrightarrow[{\omega =2\pi f}]}{\frac {1}{j\pi f}}$ |
| $u(t)=\left\{{\begin{aligned}&1,t>0\\&0,t<0\\\end{aligned}}\right.$ | $\mathbb {F} [u(t)]={\frac {1}{j\omega }}+\underbrace {\pi \delta (\omega )} _{\text{a veces se omite}}$ |
| Delta de Dirac $\delta (t)$ | $\mathbb {F} [\delta (t)]=1\leftrightarrow F[1]=2\pi \delta (-\omega )$ |
| $\cos(\omega _{0}t)$ | $\mathbb {F} \left[\cos(\omega _{0}t)\right]=\pi \delta (\omega -\omega _{0})+\pi \delta (\omega +\omega _{0})$ |
| $\sin(\omega _{0}t)$ | $\mathbb {F} \left[\sin(\omega _{0}t)\right]={\frac {\pi }{j}}\delta (\omega -\omega _{0})-{\frac {\pi }{j}}\delta (\omega +\omega _{0})$ |
| $\sum \limits _{k=-\infty }^{\infty }{\delta \left(t-kT_{s}\right)}$ | $\mathbb {F} \left[\sum \limits _{k=-\infty }^{\infty }{\delta \left(t-kT_{s}\right)}\right]=\mathbb {F} \left[{\frac {1}{T_{s}}}\sum \limits _{k=-\infty }^{\infty }{e^{j{\frac {2\pi }{T_{s}}}kt}}\right]={\frac {1}{T_{s}}}\cdot \sum \limits _{k=-\infty }^{\infty }{2\pi \delta }\left(\omega -k{\frac {2\pi }{T_{s}}}\right)$ |
Demostraciones:

## Pulso rectangular

Se representa mediante el mismo símbolo que el productorio, siendo la función una fracción: la parte de arriba (t) representa en función de que variable esta , la parte de abajo (T) representa la extensión de la función, que irá desde –T/2 a T/2.
${\begin{aligned}&\prod {\left({\frac {t}{T}}\right)}=\left\{{\begin{aligned}&1,-^{T}\!\!\diagup \!\!_{2}\;\leq t\leq^{T}\!\!\diagup \!\!_{2}\;\\&0,\left|t\right|>^{T}\!\!\diagup \!\!_{2}\;\\\end{aligned}}\right.\\&\prod {\left({\frac {t-t_{0}}{T}}\right)}=\left\{{\begin{aligned}&1,-^{T}\!\!\diagup \!\!_{2}\;\leq t-t_{0}\leq^{T}\!\!\diagup \!\!_{2}\;\\&0,\left|t-t_{0}\right|>^{T}\!\!\diagup \!\!_{2}\;\\\end{aligned}}\right.\\\end{aligned}}$
Ahora, la transformada de Fourier de un pulso rectangular:
${\begin{aligned}&\mathbb {F} \left[\prod {\left({\frac {t}{T}}\right)}\right]=\int \limits _{-\infty }^{\infty }{\prod {\left({\frac {t}{T}}\right)}\cdot e^{-j\omega t}\partial t}=\int \limits _{-^{T}\!\!\diagup \!\!_{2}\;}^{^{T}\!\!\diagup \!\!_{2}\;}{1\cdot e^{-j\omega t}\partial t}=\left.{\frac {e^{-j\omega t}}{-j\omega }}\right|_{^{-T}\!\!\diagup \!\!_{2}\;}^{^{T}\!\!\diagup \!\!_{2}\;}={\frac {1}{-j\omega }}\left(e^{-j\omega^{T}\!\!\diagup \!\!_{2}\;}-e^{+j\omega^{T}\!\!\diagup \!\!_{2}\;}\right)=\\&{\frac {2}{2}}{\frac {1}{j\omega }}\left(e^{+j\omega^{T}\!\!\diagup \!\!_{2}\;}-e^{-j\omega^{T}\!\!\diagup \!\!_{2}\;}\right){\xrightarrow[{\begin{smallmatrix}\sin x=\\{\frac {e^{jx}-e^{-jx}}{2j}}\end{smallmatrix}}]}=2{\frac {\sin \left(\omega^{T}\!\!\diagup \!\!_{2}\;\right)}{\omega }}\\\end{aligned}}$
Al ser este resultado bastante habitual, se representa muchas veces mediante la función sinc.
${\begin{aligned}&{\text{sinc}}\left(t\right)={\frac {\sin \left(\pi t\right)}{\pi t}};{\text{ sinc}}\left(Tt\right)={\frac {\sin(T\pi t)}{T\pi t}}\\&{\text{Nulos: }}\sin=n\pi \to {\text{sin}}(T\pi t)=0\to T\pi t=n\pi \to t=^{n}\!\!\diagup \!\!_{T}\;\\&{\text{sinc}}\left(Tt\right)=\left\{{\begin{aligned}&0,t=^{n}\!\!\diagup \!\!_{T}\;,\forall n\in \mathbb {Z} ,n\neq 0\\&1,t=0\\\end{aligned}}\right.,\\\end{aligned}}$
La función sinc se usa especialmente por comodidad y no tener que usar límites, 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 límites para saber el resultado, en cambio:
${\begin{aligned}&2{\frac {\sin \left(\omega^{T}\!\!\diagup \!\!_{2}\;\right)}{\omega }}=f({\text{sinc}}){\xrightarrow[{\omega =2\pi f}]}2{\frac {\sin \left(\left(2\pi f\right)^{T}\!\!\diagup \!\!_{2}\;\right)}{2\pi f}}{\xrightarrow[{^{T}\!\!\diagup \!\!_{T}\;}]}T\underbrace {\frac {\sin \left(\pi fT\right)}{\pi fT}} _{{\text{sinc}}(Tf)}=\\&T{\text{sinc}}(Tf)\to f=0\to T{\text{sinc}}(0)=T\\&\omega =2\pi f\leftrightarrow f=^{\omega }\!\!\diagup \!\!_{2\pi }\;\to T{\text{sinc}}(Tf)=T{\text{sinc}}\left(T{\frac {\omega }{2\pi }}\right)\\\end{aligned}}$

## Pulso Triangular

${\begin{aligned}&\Lambda \left({\frac {t}{T}}\right)=\left\{{\begin{aligned}&1-{\frac {\left|t\right|}{T}},{\text{ }}\left|t\right|\leq T\\&0{\text{ }},{\text{ }}\left|t\right|>T\\\end{aligned}}\right.\\&\mathbb {F} \left[\Lambda \left({\frac {t}{T}}\right)\right]=\int _{-\infty }^{\infty }{\underbrace {\Lambda \left({\frac {t}{T}}\right)} _{\text{par}}\cdot }\underbrace {e^{-j\omega t}} _{\underbrace {\cos(\omega t)} _{\text{par}}-j\underbrace {\sin(\omega t)} _{\text{impar}}}\partial t=2\int _{0}^{T}{\left(1-{\frac {t}{T}}\right)}\cos(\omega t)\partial t=\\&{\xrightarrow[{\int {u.\partial v}}]}\left\{{\begin{aligned}&u=1-{\frac {t}{T}}{\text{ }}\partial u={\frac {-1}{T}}\partial t\\&\partial v=\cos(\omega t)\partial t{\text{ }}v={\frac {{\text{sin}\omega {\text{t)}}}{\omega }}{\text{ }}\\\end{aligned}}\right\}{\xrightarrow[{u.v-\int {v\partial u}}]}2\left.\left(1-{\frac {t}{T}}\right){\frac {\sin(\omega t)}{\omega }}\right|_{0}^{T}+2\int _{0}^{T}{{\frac {\sin(\omega t)}{\omega T}}\partial t}\\&2\left.\left(1-{\frac {t}{T}}\right){\frac {\sin(\omega t)}{\omega }}\right|_{0}^{T}+\left.{\frac {-2\cos(\omega t)}{\omega ^{2}T}}\right|_{0}^{T}=2\left(\underbrace {1-{\frac {T}{T}}} _{{\hat {\ }}0}\right){\frac {\sin(\omega t)}{\omega }}+\left.{\frac {-2\cos(\omega t)}{\omega ^{2}T}}\right|_{0}^{T}=\\&\left.{\frac {-2\cos(\omega t)}{\omega ^{2}T}}\right|_{0}^{T}={\frac {-2}{\omega ^{2}T}}\left(\cos(\omega T)-1\right)={\frac {2(1-\cos(\omega T))}{\omega ^{2}T}}\\\end{aligned}}$
Sabiendo que
${\begin{aligned}&{\xrightarrow[{\sin ^{2}x={\frac {1-\cos 2x}{2}}}]}2{\frac {\left(2\sin ^{2}\left({\frac {\omega T}{2}}\right)\right)}{\omega ^{2}T}}=4{\frac {\sin ^{2}\left({\frac {\omega T}{2}}\right)}{\omega ^{2}T}}={\frac {\sin ^{2}\left({\frac {\omega T}{2}}\right)}{\frac {\omega ^{2}T}{4}}}=T{\frac {\sin ^{2}\left({\frac {\omega T}{2}}\right)}{\frac {\omega ^{2}T^{2}}{4}}}=T{\frac {\sin ^{2}\left({\frac {\omega T}{2}}\right)}{\left({\frac {\omega T}{2}}\right)^{2}}}=\\&{\xrightarrow[{{\text{sinc}}\left(T\omega \right)={\frac {\sin \left(\pi T\omega \right)}{\pi T\omega }}}]}T\cdot {\text{sinc}}^{2}\left(T{\frac {\omega }{2\pi }}\right){\underset {\begin{smallmatrix}{\text{o tambien}}\\\omega =2\pi f\end{smallmatrix}}{\longleftrightarrow }}T{\text{sinc}}^{2}\left(Tf\right)=F_{\Lambda }(f)\\&{\text{Sabiendo que }}\mathbb {F} \left[\prod {\left({\frac {t}{T}}\right)}\right]=T{\text{sinc}}\left(Tf\right)=F(f)\to \\&F^{2}(f)=T\cdot F_{\Lambda }(f)\leftrightarrow f(t)*f(t)=T\cdot f_{\Lambda }(t)\to \\&\prod {\left({\frac {t}{T}}\right)}*\prod {\left({\frac {t}{T}}\right)}=T\cdot \Lambda \left({\frac {t}{T}}\right)\Leftrightarrow \mathbb {F} \left[\prod {\left({\frac {t}{T}}\right)}*\prod {\left({\frac {t}{T}}\right)}\right]=T^{2}{\text{sinc}}^{2}\left(Tf\right)\\\end{aligned}}$
Recuérdese 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.

## Función sign(t)

La función sign(t) es una función auxiliar bastante utilizada en áreas de telecomunicación que, además, es fácilmente representable:
$\operatorname {sign} (t)=\left\{{\begin{aligned}&1,{\text{ }}t>0\\&-1,{\text{ }}t<0\\\end{aligned}}\right.$
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:
${\begin{aligned}&\operatorname {sign} (t)=\left\{{\begin{aligned}&1,{\text{ }}t>0\\&-1,{\text{ }}t<0\\\end{aligned}}\right.\\&\mathbb {F} \left[\operatorname {sign} (t)\right]=\int _{-\infty }^{\infty }{\operatorname {sign} (t)\cdot e^{-j\omega t}\partial t}\\\end{aligned}}$
Usaremos integración 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.
${\begin{aligned}&f(t)=\left\{{\begin{aligned}&1,t>0\\&0,t<0\\\end{aligned}}\right.\\&g(t)=\left\{{\begin{aligned}&0,t>0\\&-1,t<0\\\end{aligned}}\right.\to g(t)=-f(-t)\\&\operatorname {sign} (t)=f(t)-f(-t)\\&\mathbb {F} [f(t)]=\int _{-\infty }^{\infty }{f(t)\cdot e^{-j\omega t}\partial t=}\int _{0}^{\infty }{1\cdot e^{-j\omega t}\partial t}=\left.{\frac {e^{-j\omega t}}{-j\omega }}\right|_{0}^{\infty }={\frac {1}{j\omega }}\\&\mathbb {F} \left[f(t)\right]=F(\omega )\to \mathbb {F} \left[f(-t)\right]=F(-\omega )\\&\mathbb {F} \left[\operatorname {sign} (t)\right]=\mathbb {F} \left[f(t)-f(-t)\right]={\frac {1}{j\omega }}-{\frac {1}{j\left(-\omega \right)}}={\frac {2}{j\omega }}{\xrightarrow[{\omega =2\pi f}]}{\frac {1}{j\pi f}}\\\end{aligned}}$

## Función u(t)

La siguiente función, llamada Heaviside step function, o la función escalon unidad, se define:
$u(t)=\left\{{\begin{aligned}&1,t>0\\&0,t<0\\\end{aligned}}\right.$
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 aparición de la delta, también podemos obtenerla representando u(t) en función de sign(t).
${\begin{aligned}&\mathbb {F} \left[u(t)\right]=?\\&\mathbb {F} \left[\operatorname {sign} (t)\right]={\frac {2}{j\omega }}\\&u(t)={\frac {1}{2}}\cdot \left(1+\operatorname {sign} (t)\right)\to \mathbb {F} \left[u(t)\right]=\mathbb {F} \left[^{1}\!\!\diagup \!\!_{2}\;\right]+\mathbb {F} \left[^{1}\!\!\diagup \!\!_{2}\;\cdot \operatorname {sign} (t)\right]=\pi \delta (\omega )+{\frac {1}{j\omega }}\\\end{aligned}}$
Esta función resulta muy útil como función auxiliar, pues las señales solo existen a partir de un momento en el tiempo, a modo de ejemplo:
${\begin{aligned}&Ej:\\&f(t)=e^{-at}u(t)=\left\{{\begin{aligned}&e^{-at},t>0\\&0{\text{ }}{\text{,}}t<0\\\end{aligned}}\right.;a>0\\&\mathbb {F} [f(t)]=F(\omega )=\int \limits _{-\infty }^{\infty }{f(t)e^{-j\omega t}\partial t=}\int \limits _{0}^{\infty }{e^{-at}e^{-j\omega t}\partial t}=\left.{\frac {e^{-at}e^{-j\omega t}}{-a-j\omega }}\right|_{0}^{\infty }={\frac {1}{-a-j\omega }}\left(\underbrace {e^{-\infty }} _{0}e^{-j\omega \infty }-1\right)=\\&{\frac {1}{a+j\omega }}=F(\omega )\\\end{aligned}}$

## Delta de Dirac

La función delta de Dirac es una función muy especial tanto por su forma como por sus propiedades, se denota como:
$\delta (\omega )=\left\{{\begin{aligned}&\infty ,{\text{ }}\omega =0\\&0,{\text{ }}\omega \neq 0\\\end{aligned}}\right.$
Entre sus propiedades:
${\begin{aligned}&x(t)*\delta (t)=x(t)\\&x(t)*\delta (t-t_{0})=x(t-t_{0})\\&x(t)\cdot \delta (t_{0})=x(t_{0})\to x(t)\cdot \delta (0)=x(0)\\&\delta (t)=\delta (-t)\\\end{aligned}}$
Su transformada de Fourier es:
$\mathbb {F} [\delta (t)]=1\leftrightarrow F[1]=2\pi \delta (-\omega )=2\pi \delta (\omega )$
También tenemos que:
${\begin{aligned}&\delta (at)={\frac {1}{\left|a\right|}}\delta (t)\\&{\text{Debido a: }}\mathbb {F} [\delta (at)]{\xrightarrow[{\mathbb {F} [f(at)}]{=}}{\frac {1}{\left|a\right|}}F\left({\frac {\omega }{a}}\right)]{\mathbb {F} [\delta (t)]=1\neq f(\omega )}{\frac {1}{\left|a\right|}}\cdot 1\to \mathbb {F} ^{-1}\left[{\frac {1}{a}}\right]={\frac {1}{a}}\cdot \delta (t)\\\end{aligned}}$
Está relacionada con la función escalón unidad de la siguiente manera:
${\begin{aligned}&{\frac {\partial u(t)}{\partial t}}=\delta (t)\leftrightarrow u(t)=\int _{-\infty }^{t}{\delta (\tau )\partial \tau }\\&{\text{debido a:}}\\&\mathbb {F} \left[\int _{-\infty }^{t}{f(\tau )\partial \tau }\right]={\frac {F(\omega )}{j\omega }}\Rightarrow f(t)=\delta (t)\to F(\omega )=1\\&\mathbb {F} \left[\int _{-\infty }^{t}{\delta (\tau )\partial \tau }\right]=\underbrace {\frac {1}{j\omega }} _{\mathbb {F} [u(t)]}\\\end{aligned}}$
y también 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 función delta de Dirac, es relacionarlo con la función sinc.
${\begin{aligned}&\mathbb {F} \left[\prod {\left({\frac {t}{T}}\right)}\right]=T{\text{sinc}}\left(Tf\right)=T{\text{sinc}}\left(T^{\omega }\!\!\diagup \!\!_{2\pi }\;\right)\\&{\underset {T\to \infty }{\mathop {\lim } }}\,\prod {\left({\frac {t}{T}}\right)}=1\\&\mathbb {F} \left[{\underset {T\to \infty }{\mathop {\lim } }}\,\prod {\left({\frac {t}{T}}\right)}\right]=\underbrace {{\underset {T\to \infty }{\mathop {\lim } }}\,{\text{ }}T{\text{sinc}}\left(T{\frac {\omega }{2\pi }}\right)} _{\delta \left({\frac {\omega }{2\pi }}\right)}=\underbrace {{\underset {T\to \infty }{\mathop {\lim } }}\,{\text{ }}T{\text{sinc}}\left(Tf\right)} _{\delta (f)}=\left\{{\begin{aligned}&\infty ,f=0\\&0,{\text{ }}f\neq 0\\\end{aligned}}\right.\\&\delta (f)={\underset {T\to \infty }{\mathop {\lim } }}\,{\text{ }}T{\text{sinc}}\left(Tf\right)=\delta \left(^{\omega }\!\!\diagup \!\!_{2\pi }\;\right){\xrightarrow[{\delta (a\omega )={\frac {1}{\left|a\right|}}\delta (\omega )}]}2\pi \delta (\omega )\\\end{aligned}}$

## Sinθ y Cosθ

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

## Tren de deltas

$\sum \limits _{k=-\infty }^{\infty }{\delta \left(t-kT_{s}\right)}$
Primeramente, apreciamos que un sumatorio de deltas (llamado comúnmente tren de deltas) es una señal periódica, por lo que puede ser representada como suma de senos y cosenos según la serie de Fourier: Llamemos al periodo de la señal Ts.
${\begin{aligned}&f(t)=\sum \limits _{k=-\infty }^{\infty }{\delta \left(t-kT_{s}\right)}\\&T=T_{s}\to f_{s}={\frac {1}{T_{s}}}\to \omega _{s}=2\pi f_{s}={\frac {2\pi }{T_{s}}}\\&f(t)=\sum \limits _{k=-\infty }^{\infty }{C_{k}\cdot e^{j{\frac {2\pi }{T}}kt}}=\sum \limits _{k=-\infty }^{\infty }{C_{k}\cdot e^{j{\frac {2\pi }{T_{s}}}kt}}\\&C_{k}={\frac {1}{T}}\int \limits _{^{-T}\!\!\diagup \!\!_{2}\;}^{^{T}\!\!\diagup \!\!_{2}\;}{f(t)\cdot e^{-j{\frac {2\pi }{T_{s}}}kt}\partial t}={\frac {1}{T_{s}}}\int \limits _{^{-T_{s}}\!\!\diagup \!\!_{2}\;}^{^{T_{s}}\!\!\diagup \!\!_{2}\;}{\delta (t)\cdot e^{-j{\frac {2\pi }{T_{s}}}kt}\partial t}={\frac {1}{T_{s}}}\int \limits _{^{-T_{s}}\!\!\diagup \!\!_{2}\;}^{^{T_{s}}\!\!\diagup \!\!_{2}\;}{\delta (t)\cdot \left.e^{-j{\frac {2\pi }{T_{s}}}kt}\right|_{t=0}\partial t}={\frac {1}{T_{s}}}\int \limits _{^{-T_{s}}\!\!\diagup \!\!_{2}\;}^{^{T_{s}}\!\!\diagup \!\!_{2}\;}{\delta (t)\partial t}\\&\to \int \limits _{-\infty }^{\infty }{\delta (t)\partial t}=1\to C_{k}={\frac {1}{T_{s}}}{\text{ }}\forall k\to \\&f(t)=\sum \limits _{k=-\infty }^{\infty }{{\frac {1}{T_{s}}}e^{j{\frac {2\pi }{T_{s}}}kt}}={\frac {1}{T_{s}}}\sum \limits _{k=-\infty }^{\infty }{e^{j{\frac {2\pi }{T_{s}}}kt}}\\&\sum \limits _{k=-\infty }^{\infty }{\delta \left(t-kT_{s}\right)}={\frac {1}{T_{s}}}\sum \limits _{k=-\infty }^{\infty }{e^{j{\frac {2\pi }{T_{s}}}kt}}\\\end{aligned}}$
${\begin{aligned}&{\text{Recordemos las propiedades: }}\mathbb {F} [e^{+j\omega _{0}t}f(t)]=F(\omega -\omega _{0})\\&\omega _{0}={\frac {2\pi }{T_{s}}},\mathbb {F} [1]=2\pi \delta (\omega )\to \mathbb {F} [e^{+j\omega _{0}t}]=2\pi \delta (\omega -\omega _{0})\\&\mathbb {F} \left[{\frac {1}{T_{s}}}\sum \limits _{k=-\infty }^{\infty }{e^{j{\frac {2\pi }{T_{s}}}kt}}\right]={\frac {1}{T_{s}}}\cdot \mathbb {F} \left[1+e^{j{\frac {2\pi }{T_{s}}}t}+e^{-j{\frac {2\pi }{T_{s}}}t}+e^{j{\frac {2\pi }{T_{s}}}2t}+e^{-j{\frac {2\pi }{T_{s}}}2t}+e^{j{\frac {2\pi }{T_{s}}}3t}+...\right]=\\&{\frac {1}{T_{s}}}\cdot \left[2\pi \delta (\omega )+2\pi \delta \left(\omega -{\frac {2\pi }{T_{s}}}\right)+2\pi \delta \left(\omega +{\frac {2\pi }{T_{s}}}\right)+2\pi \delta \left(\omega -2\cdot {\frac {2\pi }{T_{s}}}\right)+...\right]\\&{\frac {1}{T_{s}}}\cdot \sum \limits _{k=-\infty }^{\infty }{2\pi \delta }\left(w-k{\frac {2\pi }{T_{s}}}\right)\\&\mathbb {F} \left[\sum \limits _{k=-\infty }^{\infty }{\delta \left(t-kT_{s}\right)}\right]=\mathbb {F} \left[{\frac {1}{T_{s}}}\sum \limits _{k=-\infty }^{\infty }{e^{j{\frac {2\pi }{T_{s}}}kt}}\right]={\frac {1}{T_{s}}}\cdot \sum \limits _{k=-\infty }^{\infty }{2\pi \delta }\left(\omega -k{\frac {2\pi }{T_{s}}}\right)\\\end{aligned}}$