Condiciones para que exista la transformada de Fourier cuando:
Es absolutamente integrable
∫
−
∞
∞
|
f
(
t
)
|
d
t
<
∞
(Converge)
{\displaystyle \int \limits _{-\infty }^{\infty }{\left|f\left(t\right)\right|dt<\infty {\text{ (Converge)}}}}
f(t) continua por intervalos [a,b] finito
lim
t
→
t
0
±
f
(
t
)
finito
,
∀
t
0
{\displaystyle {\underset {t\to t_{0}^{\pm }}{\mathop {\lim } }}\,{\text{ }}f(t){\text{ finito}},\forall t_{0}}
lim
f
(
t
)
=
f
(
t
+
)
+
f
(
t
−
)
2
{\displaystyle {\underset {}{\mathop {\lim } }}\,f(t)={\frac {f(t^{+})+f(t^{-})}{2}}}
Para una f(t) real o compleja, con variable real t, se define la transformada de Fourier como:
F
[
f
(
t
)
]
=
F
(
ω
)
=
∫
−
∞
∞
f
(
t
)
.
e
−
j
ω
t
∂
t
{\displaystyle \mathbb {F} [f(t)]=F(\omega )=\int \limits _{-\infty }^{\infty }{f(t).e^{-j\omega t}\partial t}}
Igualmente, tenemos la función inversa de Fourier:
F
−
1
[
F
(
ω
)
]
=
f
(
t
)
=
1
2
π
∫
−
∞
∞
F
(
ω
)
.
e
+
j
ω
t
∂
ω
{\displaystyle \mathbb {F} ^{-1}[F(\omega )]=f(t)={\frac {1}{2\pi }}\int \limits _{-\infty }^{\infty }{F(\omega ).e^{+j\omega t}\partial \omega }}
De forma que se cumple
F
−
1
[
F
[
f
(
t
)
]
⏟
F
(
ω
)
]
=
f
(
t
)
F
[
F
−
1
[
F
(
ω
)
]
⏟
f
(
t
)
]
=
F
(
ω
)
{\displaystyle {\begin{aligned}&\mathbb {F} ^{-1}[\underbrace {\mathbb {F} [f(t)]} _{F(\omega )}]=f(t)\\&\mathbb {F} [\underbrace {\mathbb {F} ^{-1}[F(\omega )]} _{f(t)}]=F(\omega )\\\end{aligned}}}
f
(
t
)
↔
F
(
ω
)
,
g
(
t
)
↔
G
(
ω
)
,
x
(
t
)
↔
X
(
ω
)
{\displaystyle f(t)\leftrightarrow F(\omega ),g(t)\leftrightarrow G(\omega ),x(t)\leftrightarrow X(\omega )}
Alguna propiedad:
Integración:
F
(
ω
=
0
)
=
∫
−
∞
∞
f
(
t
)
∂
t
f
(
t
=
0
)
=
1
2
π
∫
−
∞
∞
F
(
ω
)
∂
ω
{\displaystyle {\begin{aligned}&F(\omega =0)=\int _{-\infty }^{\infty }{f(t)\partial t}\\&f(t=0)={\frac {1}{2\pi }}\int _{-\infty }^{\infty }{F(\omega )\partial \omega }\\\end{aligned}}}
La demostración es sencilla:
F
[
f
(
t
)
]
=
F
(
ω
)
=
∫
−
∞
∞
f
(
t
)
.
e
−
j
ω
t
∂
t
F
(
ω
=
0
)
=
∫
−
∞
∞
f
(
t
)
.
e
−
j
ω
t
∂
t
|
ω
=
0
=
∫
−
∞
∞
f
(
t
)
.
e
−
j
0
t
∂
t
=
∫
−
∞
∞
f
(
t
)
∂
t
{\displaystyle {\begin{aligned}&\mathbb {F} [f(t)]=F(\omega )=\int \limits _{-\infty }^{\infty }{f(t).e^{-j\omega t}\partial t}\\&F(\omega =0)=\left.\int \limits _{-\infty }^{\infty }{f(t).e^{-j\omega t}\partial t}\right|_{\omega =0}=\int \limits _{-\infty }^{\infty }{f(t).e^{-j0t}\partial t}=\int \limits _{-\infty }^{\infty }{f(t)\partial t}\\\end{aligned}}}
Análogamente:
F
−
1
[
F
(
ω
)
]
=
f
(
t
)
=
1
2
π
∫
−
∞
∞
F
(
ω
)
.
e
+
j
ω
t
∂
ω
f
(
t
=
0
)
=
1
2
π
∫
−
∞
∞
F
(
ω
)
.
e
+
j
ω
t
=
0
∂
ω
=
1
2
π
∫
−
∞
∞
F
(
ω
)
∂
ω
{\displaystyle {\begin{aligned}&\mathbb {F} ^{-1}[F(\omega )]=f(t)={\frac {1}{2\pi }}\int \limits _{-\infty }^{\infty }{F(\omega ).e^{+j\omega t}\partial \omega }\\&f(t=0)={\frac {1}{2\pi }}\int \limits _{-\infty }^{\infty }{F(\omega ).e^{+j\omega t=0}\partial \omega }={\frac {1}{2\pi }}\int \limits _{-\infty }^{\infty }{F(\omega )\partial \omega }\\\end{aligned}}}
Comentar que en matemáticas es usual dejar el resultado de la transformada de Fourier en función de ω, mientras que en ingeniería es más habitual dejarlo en f, debido a que se mantiene la simetría.
Ambas están relacionadas directamente.
ω
=
2
π
f
F
[
f
(
t
)
]
=
F
(
ω
)
=
∫
−
∞
∞
f
(
t
)
.
e
−
j
ω
t
∂
t
→
F
[
f
(
t
)
]
=
F
(
f
)
=
∫
−
∞
∞
f
(
t
)
.
e
−
j
2
π
f
t
∂
t
F
−
1
[
F
(
ω
)
]
=
f
(
t
)
=
1
2
π
∫
−
∞
∞
F
(
ω
)
.
e
+
j
ω
t
∂
ω
=
1
2
π
∫
−
∞
∞
F
(
f
)
.
e
+
j
2
π
f
t
2
π
∂
f
→
F
−
1
[
F
(
f
)
]
=
f
(
t
)
=
∫
−
∞
∞
F
(
f
)
.
e
+
j
2
π
f
t
∂
f
{\displaystyle {\begin{aligned}&\omega =2\pi f\\&\mathbb {F} [f(t)]=F(\omega )=\int \limits _{-\infty }^{\infty }{f(t).e^{-j\omega t}\partial t}\to \mathbb {F} [f(t)]=F(f)=\int \limits _{-\infty }^{\infty }{f(t).e^{-j2\pi ft}\partial t}{\text{ }}\\&\mathbb {F} ^{-1}[F(\omega )]=f(t)={\frac {1}{2\pi }}\int \limits _{-\infty }^{\infty }{F(\omega ).e^{+j\omega t}\partial \omega }={\frac {1}{2\pi }}\int \limits _{-\infty }^{\infty }{F(f).e^{+j2\pi ft}2\pi \partial f\to }\\&\mathbb {F} ^{-1}[F(f)]=f(t)=\int \limits _{-\infty }^{\infty }{F(f).e^{+j2\pi ft}\partial f}\\\end{aligned}}}
Con la ventaja que en función de f, no tenemos ese divisor de 2π, por lo que mantiene la simetría, que es más comodo al realizar cálculos (sin embargo, no olvidar que la fase de la exponencial esta invertida).
F
[
f
(
t
)
]
=
F
(
ω
)
=
∫
−
∞
∞
f
(
t
)
.
e
−
j
ω
t
∂
t
F
−
1
[
F
(
ω
)
]
=
f
(
t
)
=
1
2
π
∫
−
∞
∞
F
(
ω
)
.
e
+
j
ω
t
∂
ω
{\displaystyle {\begin{aligned}&\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{aligned}}}
Pulso rectangular
∏
(
t
T
)
=
{
1
,
|
t
|
≤
T
╱
2
0
,
|
t
|
>
T
╱
2
{\displaystyle \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.}
F
[
∏
(
t
T
)
]
=
2
sin
(
ω
T
╱
2
)
ω
=
T
sinc
(
T
ω
2
)
{\displaystyle \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}}\right)}
Pulso triangular
Λ
(
t
T
)
=
{
1
−
|
t
|
T
,
|
t
|
≤
T
0
,
|
t
|
>
T
{\displaystyle \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.}
F
[
Λ
(
t
T
)
]
=
2
(
1
−
cos
(
ω
T
)
)
ω
2
T
=
T
⋅
sinc
2
(
T
ω
2
π
)
{\displaystyle \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)}
sign
(
t
)
=
{
1
,
t
>
0
−
1
,
t
<
0
{\displaystyle \operatorname {sign} (t)=\left\{{\begin{aligned}&1,{\text{ }}t>0\\&-1,{\text{ }}t<0\\\end{aligned}}\right.}
F
[
sign
(
t
)
]
=
2
j
ω
→
ω
=
2
π
f
1
j
π
f
{\displaystyle \mathbb {F} \left[\operatorname {sign} (t)\right]={\frac {2}{j\omega }}{\xrightarrow[{\omega =2\pi f}]{}}{\frac {1}{j\pi f}}}
u
(
t
)
=
{
1
,
t
>
0
0
,
t
<
0
{\displaystyle u(t)=\left\{{\begin{aligned}&1,t>0\\&0,t<0\\\end{aligned}}\right.}
F
[
u
(
t
)
]
=
1
j
ω
+
π
δ
(
ω
)
⏟
a veces se omite
{\displaystyle \mathbb {F} [u(t)]={\frac {1}{j\omega }}+\underbrace {\pi \delta (\omega )} _{\text{a veces se omite}}}
Delta de Dirac
δ
(
t
)
{\displaystyle \delta (t)}
F
[
δ
(
t
)
]
=
1
↔
F
[
1
]
=
2
π
δ
(
−
ω
)
{\displaystyle \mathbb {F} [\delta (t)]=1\leftrightarrow F[1]=2\pi \delta (-\omega )}
cos
(
ω
0
t
)
{\displaystyle \cos(\omega _{0}t)}
F
[
cos
(
ω
0
t
)
]
=
π
δ
(
ω
−
ω
0
)
+
π
δ
(
ω
+
ω
0
)
{\displaystyle \mathbb {F} \left[\cos(\omega _{0}t)\right]=\pi \delta (\omega -\omega _{0})+\pi \delta (\omega +\omega _{0})}
sin
(
ω
0
t
)
{\displaystyle \sin(\omega _{0}t)}
F
[
sin
(
ω
0
t
)
]
=
π
j
δ
(
ω
−
ω
0
)
−
π
j
δ
(
ω
+
ω
0
)
{\displaystyle \mathbb {F} \left[\sin(\omega _{0}t)\right]={\frac {\pi }{j}}\delta (\omega -\omega _{0})-{\frac {\pi }{j}}\delta (\omega +\omega _{0})}
∑
k
=
−
∞
∞
δ
(
t
−
k
T
s
)
{\displaystyle \sum \limits _{k=-\infty }^{\infty }{\delta \left(t-kT_{s}\right)}}
F
[
∑
k
=
−
∞
∞
δ
(
t
−
k
T
s
)
]
=
F
[
1
T
s
∑
k
=
−
∞
∞
e
j
2
π
T
s
k
t
]
=
1
T
s
⋅
∑
k
=
−
∞
∞
2
π
δ
(
ω
−
k
2
π
T
s
)
{\displaystyle \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)}
¿Cuál es la transformada de fourier de una señal periódica? Para saberlo usaremos el tren de deltas utilizado anteriormente, como función auxiliar.
x
(
t
)
=
x
(
t
+
T
s
)
x
0
(
t
)
:
función fundamental
x
(
t
)
=
∑
k
=
−
∞
∞
x
0
(
t
−
k
T
s
)
=
x
0
(
t
)
∗
∑
k
=
−
∞
∞
δ
(
t
−
k
T
s
)
∑
k
=
−
∞
∞
δ
(
t
−
k
T
s
)
=
1
T
s
∑
k
=
−
∞
∞
e
j
2
π
T
s
k
t
→
C
k
=
1
T
s
∀
k
(Propiedades de Trans
. Fourier)
Recordemos las propiedades:
{
F
[
e
+
j
ω
0
t
f
(
t
)
]
=
F
(
ω
−
ω
0
)
F
[
1
]
=
2
π
δ
(
ω
)
→
F
[
e
+
j
ω
0
t
]
=
2
π
δ
(
ω
−
ω
0
)
F
[
f
(
t
)
∗
g
(
t
)
]
=
F
(
ω
)
⋅
G
(
ω
)
}
F
[
x
(
t
)
]
=
F
[
x
0
(
t
)
∗
∑
k
=
−
∞
∞
δ
(
t
−
k
T
s
)
]
=
F
[
x
0
(
t
)
∗
1
T
s
∑
k
=
−
∞
∞
e
j
2
π
T
s
k
t
]
X
(
ω
)
=
X
0
(
ω
)
.
1
T
s
∑
k
=
−
∞
∞
2
π
δ
(
ω
−
k
2
π
T
s
)
=
2
π
T
s
∑
k
=
−
∞
∞
X
0
(
k
2
π
T
s
)
⋅
δ
(
ω
−
k
2
π
T
s
)
{\displaystyle {\begin{aligned}&x(t)=x(t+T_{s})\\&x_{0}(t):{\text{ función fundamental}}\\&x(t)=\sum \limits _{k=-\infty }^{\infty }{x_{0}\left(t-kT_{s}\right)}=x_{0}(t)*\sum \limits _{k=-\infty }^{\infty }{\delta \left(t-kT_{s}\right)}\\&\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}}\to C_{k}={\frac {1}{T_{s}}}{\text{ }}\forall k{\text{ (Propiedades de Trans}}{\text{. Fourier)}}\\&{\text{Recordemos las propiedades: }}\\&\left\{{\begin{aligned}&\mathbb {F} [e^{+j\omega _{0}t}f(t)]=F(\omega -\omega _{0})\\&\mathbb {F} [1]=2\pi \delta (\omega )\to \mathbb {F} [e^{+j\omega _{0}t}]=2\pi \delta (\omega -\omega _{0})\\&\mathbb {F} \left[f(t)*g(t)\right]=F(\omega )\cdot G(\omega )\\\end{aligned}}\right\}\\&\mathbb {F} \left[x(t)\right]=\mathbb {F} \left[x_{0}(t)*\sum \limits _{k=-\infty }^{\infty }{\delta \left(t-kT_{s}\right)}\right]=\mathbb {F} \left[x_{0}(t)*{\frac {1}{T_{s}}}\sum \limits _{k=-\infty }^{\infty }{e^{j{\frac {2\pi }{T_{s}}}kt}}\right]\\&X(\omega )=X_{0}(\omega ).{\frac {1}{T_{s}}}\sum \limits _{k=-\infty }^{\infty }{2\pi \delta \left(\omega -k{\frac {2\pi }{T_{s}}}\right)}={\frac {2\pi }{T_{s}}}\sum \limits _{k=-\infty }^{\infty }{X_{0}\left({\frac {k2\pi }{T_{s}}}\right)\cdot \delta \left(\omega -k{\frac {2\pi }{T_{s}}}\right)}\\\end{aligned}}}
X
(
ω
)
=
2
π
T
s
∑
k
=
−
∞
∞
X
0
(
k
2
π
T
s
)
⋅
δ
(
ω
−
k
2
π
T
s
)
{\displaystyle X(\omega )={\frac {2\pi }{T_{s}}}\sum \limits _{k=-\infty }^{\infty }{X_{0}\left({\frac {k2\pi }{T_{s}}}\right)\cdot \delta \left(\omega -k{\frac {2\pi }{T_{s}}}\right)}}
![{\displaystyle \mathbb {F} [f(t)]=F(\omega )=\int \limits _{-\infty }^{\infty }{f(t).e^{-j\omega t}\partial t}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba96a9496b1aea2594934fe40575798ba96170bd)
![{\displaystyle \mathbb {F} ^{-1}[F(\omega )]=f(t)={\frac {1}{2\pi }}\int \limits _{-\infty }^{\infty }{F(\omega ).e^{+j\omega t}\partial \omega }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/199e191d2157d6a1e3e5157945bcd5c3186f6b0e)
![{\displaystyle {\begin{aligned}&\mathbb {F} ^{-1}[\underbrace {\mathbb {F} [f(t)]} _{F(\omega )}]=f(t)\\&\mathbb {F} [\underbrace {\mathbb {F} ^{-1}[F(\omega )]} _{f(t)}]=F(\omega )\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ade01ba8becc7f2e2ae94ce60721ac36ca52925)
![{\displaystyle {\begin{aligned}&\mathbb {F} [f(t)]=F(\omega )=\int \limits _{-\infty }^{\infty }{f(t).e^{-j\omega t}\partial t}\\&F(\omega =0)=\left.\int \limits _{-\infty }^{\infty }{f(t).e^{-j\omega t}\partial t}\right|_{\omega =0}=\int \limits _{-\infty }^{\infty }{f(t).e^{-j0t}\partial t}=\int \limits _{-\infty }^{\infty }{f(t)\partial t}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b6636959f8cc99661d323355575b54c480cd9714)
![{\displaystyle {\begin{aligned}&\mathbb {F} ^{-1}[F(\omega )]=f(t)={\frac {1}{2\pi }}\int \limits _{-\infty }^{\infty }{F(\omega ).e^{+j\omega t}\partial \omega }\\&f(t=0)={\frac {1}{2\pi }}\int \limits _{-\infty }^{\infty }{F(\omega ).e^{+j\omega t=0}\partial \omega }={\frac {1}{2\pi }}\int \limits _{-\infty }^{\infty }{F(\omega )\partial \omega }\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9406f1c0b1f3f3858ad34f98dfd0d256fbc787c0)
![{\displaystyle {\begin{aligned}&\omega =2\pi f\\&\mathbb {F} [f(t)]=F(\omega )=\int \limits _{-\infty }^{\infty }{f(t).e^{-j\omega t}\partial t}\to \mathbb {F} [f(t)]=F(f)=\int \limits _{-\infty }^{\infty }{f(t).e^{-j2\pi ft}\partial t}{\text{ }}\\&\mathbb {F} ^{-1}[F(\omega )]=f(t)={\frac {1}{2\pi }}\int \limits _{-\infty }^{\infty }{F(\omega ).e^{+j\omega t}\partial \omega }={\frac {1}{2\pi }}\int \limits _{-\infty }^{\infty }{F(f).e^{+j2\pi ft}2\pi \partial f\to }\\&\mathbb {F} ^{-1}[F(f)]=f(t)=\int \limits _{-\infty }^{\infty }{F(f).e^{+j2\pi ft}\partial f}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea089067ae7322d9824699a8cbed484c083a5964)
![{\displaystyle {\begin{aligned}&\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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c872ec41ee946d1c97f912449138ee036ca1e8d)
![{\displaystyle \mathbb {F} \left[\alpha f(t)+\beta g(t)\right]=\alpha F(\omega )+\beta G(\omega )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8f0998043dbf9220346ff90591ad9b15ec63d7c)
![{\displaystyle \mathbb {F} [f(t)]=F(\omega )\to \mathbb {F} [F(t)]=2\pi f(-\omega )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0972c56194a230c9088ac84fc05f58833ce71d51)
![{\displaystyle \mathbb {F} [f(at)]={\frac {1}{\left|a\right|}}F\left({\frac {\omega }{a}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5e2586557c11fc03e9fcac0b3b03df650663b1c3)
![{\displaystyle \mathbb {F} [f^{*}(t)]=F^{*}(-\omega )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d9ce54b5f0cbaa786f8df5de5c0aaaab46350312)
![{\displaystyle \mathbb {F} [f(t-t_{0})]=e^{-j\omega t_{0}}F(\omega )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bed93d917324c18714dfc699cebe172cc5ababa2)
![{\displaystyle \mathbb {F} [e^{+j\omega _{0}t}f(t)]=F(\omega -\omega _{0})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed5c265298a123bfe049f32e7d3a21a99dd100b1)
![{\displaystyle \mathbb {F} \left[{\frac {\partial ^{n}f(t)}{\partial t^{n}}}\right]=\left(j\omega \right)^{n}F(\omega )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3263e0534d9cae8075e5b936cf0ba9cfe8d94ca7)
![{\displaystyle \mathbb {F} \left[\left(-jt\right)^{n}f(t)\right]={\frac {\partial ^{n}F(\omega )}{\partial \omega ^{n}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c5b9126ef5daa78b9e1f5fb7f290d19aed34fd1e)
![{\displaystyle \mathbb {F} \left[\int \limits _{-\infty }^{t}{f(\tau )\partial \tau }\right]={\frac {F(\omega )}{j\omega }}+\pi F(0)\delta (\omega )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf1bb67f15941008dad5d704d0d56774dbc6917c)
![{\displaystyle {\begin{aligned}&\mathbb {F} \left[f(t)*g(t)\right]=\\&\mathbb {F} \left[\int \limits _{-\infty }^{\infty }{f(\tau )g(t-\tau )\partial \tau }\right]=F(\omega )G(\omega )\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1546dc9f547eea9d824f4e953c0b321d1594eddc)
![{\displaystyle \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}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5571c490836376f840f05a1cc42782d846759df)
![{\displaystyle \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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/626ce2c4e481969ca8869d0f9cadf3f324f49cb7)
![{\displaystyle \mathbb {F} \left[\operatorname {sign} (t)\right]={\frac {2}{j\omega }}{\xrightarrow[{\omega =2\pi f}]{}}{\frac {1}{j\pi f}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5679a89917c7cdfc4cdd817c2ed0122f243c579a)
![{\displaystyle \mathbb {F} [u(t)]={\frac {1}{j\omega }}+\underbrace {\pi \delta (\omega )} _{\text{a veces se omite}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a361679c8c7869f3fa92396b8141b110988dd8c8)
![{\displaystyle \mathbb {F} [\delta (t)]=1\leftrightarrow F[1]=2\pi \delta (-\omega )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/964e0a3abe0d88abcaa217061802c329595d179d)
![{\displaystyle \mathbb {F} \left[\cos(\omega _{0}t)\right]=\pi \delta (\omega -\omega _{0})+\pi \delta (\omega +\omega _{0})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/835f5d5df443cffd86fb21ab66578ca1520f5079)
![{\displaystyle \mathbb {F} \left[\sin(\omega _{0}t)\right]={\frac {\pi }{j}}\delta (\omega -\omega _{0})-{\frac {\pi }{j}}\delta (\omega +\omega _{0})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a22c74976a1c158168f11a273a342c5fcfe0237)
![{\displaystyle \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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0fc52fca5e138f8d4980e63733052b49356e7bc5)
![{\displaystyle {\begin{aligned}&x(t)=x(t+T_{s})\\&x_{0}(t):{\text{ función fundamental}}\\&x(t)=\sum \limits _{k=-\infty }^{\infty }{x_{0}\left(t-kT_{s}\right)}=x_{0}(t)*\sum \limits _{k=-\infty }^{\infty }{\delta \left(t-kT_{s}\right)}\\&\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}}\to C_{k}={\frac {1}{T_{s}}}{\text{ }}\forall k{\text{ (Propiedades de Trans}}{\text{. Fourier)}}\\&{\text{Recordemos las propiedades: }}\\&\left\{{\begin{aligned}&\mathbb {F} [e^{+j\omega _{0}t}f(t)]=F(\omega -\omega _{0})\\&\mathbb {F} [1]=2\pi \delta (\omega )\to \mathbb {F} [e^{+j\omega _{0}t}]=2\pi \delta (\omega -\omega _{0})\\&\mathbb {F} \left[f(t)*g(t)\right]=F(\omega )\cdot G(\omega )\\\end{aligned}}\right\}\\&\mathbb {F} \left[x(t)\right]=\mathbb {F} \left[x_{0}(t)*\sum \limits _{k=-\infty }^{\infty }{\delta \left(t-kT_{s}\right)}\right]=\mathbb {F} \left[x_{0}(t)*{\frac {1}{T_{s}}}\sum \limits _{k=-\infty }^{\infty }{e^{j{\frac {2\pi }{T_{s}}}kt}}\right]\\&X(\omega )=X_{0}(\omega ).{\frac {1}{T_{s}}}\sum \limits _{k=-\infty }^{\infty }{2\pi \delta \left(\omega -k{\frac {2\pi }{T_{s}}}\right)}={\frac {2\pi }{T_{s}}}\sum \limits _{k=-\infty }^{\infty }{X_{0}\left({\frac {k2\pi }{T_{s}}}\right)\cdot \delta \left(\omega -k{\frac {2\pi }{T_{s}}}\right)}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/695a001b3cbcd61194c053eca52f7bedc6b095db)
