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.

Pulso rectangular normalizada a T = 1

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

Ahora, la transformada de Fourier de un pulso rectangular:

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

Sinc(t)

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

Función triangular normalizada a T = 1

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

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:

Sign(t)

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

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:

función escalón considerando u(0) = 1/2

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

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

Diagrama esquemático de la función 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.

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

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