F
[
f
(
t
)
]
=
F
(
ω
)
=
∫
−
∞
∞
f
(
t
)
.
e
−
j
ω
t
d
t
F
−
1
[
F
(
ω
)
]
=
f
(
t
)
=
1
2
π
∫
−
∞
∞
F
(
ω
)
.
e
+
j
ω
t
d
ω
{\displaystyle {\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
∏
(
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
π
)
{\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\pi }}\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)}
Demostraciones:
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
∏
(
t
T
)
=
{
1
,
−
T
╱
2
≤
t
≤
T
╱
2
0
,
|
t
|
>
T
╱
2
∏
(
t
−
t
0
T
)
=
{
1
,
−
T
╱
2
≤
t
−
t
0
≤
T
╱
2
0
,
|
t
−
t
0
|
>
T
╱
2
{\displaystyle {\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:
F
[
∏
(
t
T
)
]
=
∫
−
∞
∞
∏
(
t
T
)
⋅
e
−
j
ω
t
∂
t
=
∫
−
T
╱
2
T
╱
2
1
⋅
e
−
j
ω
t
∂
t
=
e
−
j
ω
t
−
j
ω
|
−
T
╱
2
T
╱
2
=
1
−
j
ω
(
e
−
j
ω
T
╱
2
−
e
+
j
ω
T
╱
2
)
=
2
2
1
j
ω
(
e
+
j
ω
T
╱
2
−
e
−
j
ω
T
╱
2
)
→
sin
x
=
e
j
x
−
e
−
j
x
2
j
=
2
sin
(
ω
T
╱
2
)
ω
{\displaystyle {\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().
sinc
(
t
)
=
sin
(
π
t
)
π
t
;
sinc
(
T
t
)
=
sin
(
T
π
t
)
T
π
t
Nulos:
sin
(
)
=
n
π
→
sin
(
T
π
t
)
=
0
→
T
π
t
=
n
π
→
t
=
n
╱
T
sinc
(
T
t
)
=
{
0
,
t
=
n
╱
T
,
∀
n
∈
Z
,
n
≠
0
1
,
t
=
0
,
{\displaystyle {\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:
2
sin
(
ω
T
╱
2
)
ω
|
ω
=
0
=
0
0
→
lim
ω
→
0
2
sin
(
ω
T
╱
2
)
ω
→
sin
x
≈
x
2
ω
T
╱
2
ω
=
T
{\displaystyle \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}
2
sin
(
ω
T
╱
2
)
ω
=
f
(
sinc()
)
→
ω
=
2
π
f
2
sin
(
(
2
π
f
)
T
╱
2
)
2
π
f
→
T
╱
T
T
sin
(
π
f
T
)
π
f
T
⏟
sinc
(
T
f
)
=
T
sinc
(
T
f
)
→
f
=
0
→
T
sinc
(
0
)
=
T
ω
=
2
π
f
↔
f
=
ω
╱
2
π
→
T
sinc
(
T
f
)
=
T
sinc
(
T
ω
2
π
)
{\displaystyle {\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}}}
Función triangular normalizada a T = 1
Λ
(
t
T
)
=
{
1
−
|
t
|
T
,
|
t
|
≤
T
0
,
|
t
|
>
T
F
[
Λ
(
t
T
)
]
=
∫
−
∞
∞
Λ
(
t
T
)
⏟
par
⋅
e
−
j
ω
t
⏟
cos
(
ω
t
)
⏟
par
−
j
sin
(
ω
t
)
⏟
impar
∂
t
=
2
∫
0
T
(
1
−
t
T
)
cos
(
ω
t
)
∂
t
=
→
∫
u
.
∂
v
{
u
=
1
−
t
T
∂
u
=
−
1
T
∂
t
∂
v
=
cos
(
ω
t
)
∂
t
v
=
sin(
ω
t)
ω
}
→
u
.
v
−
∫
v
∂
u
2
(
1
−
t
T
)
sin
(
ω
t
)
ω
|
0
T
+
2
∫
0
T
sin
(
ω
t
)
ω
T
∂
t
2
(
1
−
t
T
)
sin
(
ω
t
)
ω
|
0
T
+
−
2
cos
(
ω
t
)
ω
2
T
|
0
T
=
2
(
1
−
T
T
⏟
^
0
)
sin
(
ω
t
)
ω
+
−
2
cos
(
ω
t
)
ω
2
T
|
0
T
=
−
2
cos
(
ω
t
)
ω
2
T
|
0
T
=
−
2
ω
2
T
(
cos
(
ω
T
)
−
1
)
=
2
(
1
−
cos
(
ω
T
)
)
ω
2
T
{\displaystyle {\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
→
sin
2
x
=
1
−
cos
2
x
2
2
(
2
sin
2
(
ω
T
2
)
)
ω
2
T
=
4
sin
2
(
ω
T
2
)
ω
2
T
=
sin
2
(
ω
T
2
)
ω
2
T
4
=
T
sin
2
(
ω
T
2
)
ω
2
T
2
4
=
T
sin
2
(
ω
T
2
)
(
ω
T
2
)
2
=
→
sinc
(
T
ω
)
=
sin
(
π
T
ω
)
π
T
ω
T
⋅
sinc
2
(
T
ω
2
π
)
⟷
o tambien
ω
=
2
π
f
T
sinc
2
(
T
f
)
=
F
Λ
(
f
)
Sabiendo que
F
[
∏
(
t
T
)
]
=
T
sinc
(
T
f
)
=
F
(
f
)
→
F
2
(
f
)
=
T
⋅
F
Λ
(
f
)
↔
f
(
t
)
∗
f
(
t
)
=
T
⋅
f
Λ
(
t
)
→
∏
(
t
T
)
∗
∏
(
t
T
)
=
T
⋅
Λ
(
t
T
)
⇔
F
[
∏
(
t
T
)
∗
∏
(
t
T
)
]
=
T
2
sinc
2
(
T
f
)
{\displaystyle {\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
∏
(
t
T
)
{\displaystyle \prod {\left({\frac {t}{T}}\right)}}
Λ
(
t
T
)
{\displaystyle \Lambda \left({\frac {t}{T}}\right)}
2 T (desde –T a T). Es conveniente recordarlo, pues suele ser un error habitual.
La función sign(t) es una función auxiliar bastante utilizada en áreas de telecomunicación que, además, es fácilmente representable:
Sign(t)
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.}
El valor de sign(t) cuando t=0 es:
ℓ
=
lim
t
→
0
+
sign
(
t
)
+
lim
t
→
0
−
sign
(
t
)
2
=
1
+
(
−
1
)
2
=
0
{\displaystyle \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:
sign
(
t
)
=
{
1
,
t
>
0
−
1
,
t
<
0
F
[
sign
(
t
)
]
=
∫
−
∞
∞
sign
(
t
)
⋅
e
−
j
ω
t
∂
t
{\displaystyle {\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.
f
(
t
)
=
{
1
,
t
>
0
0
,
t
<
0
g
(
t
)
=
{
0
,
t
>
0
−
1
,
t
<
0
→
g
(
t
)
=
−
f
(
−
t
)
sign
(
t
)
=
f
(
t
)
−
f
(
−
t
)
F
[
f
(
t
)
]
=
∫
−
∞
∞
f
(
t
)
⋅
e
−
j
ω
t
∂
t
=
∫
0
∞
1
⋅
e
−
j
ω
t
∂
t
=
e
−
j
ω
t
−
j
ω
|
0
∞
=
1
j
ω
F
[
f
(
t
)
]
=
F
(
ω
)
→
F
[
f
(
−
t
)
]
=
F
(
−
ω
)
F
[
sign
(
t
)
]
=
F
[
f
(
t
)
−
f
(
−
t
)
]
=
1
j
ω
−
1
j
(
−
ω
)
=
2
j
ω
→
ω
=
2
π
f
1
j
π
f
{\displaystyle {\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}}}
La siguiente función, llamada Heaviside step function, o la función escalon unidad, se define:
función escalón considerando u (0) = 1/2
u
(
t
)
=
{
1
,
t
>
0
0
,
t
<
0
{\displaystyle 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:
ℓ
=
lim
t
→
t
+
u
(
t
)
+
lim
t
→
t
−
u
(
t
)
2
=
1
+
0
2
=
1
2
{\displaystyle \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:
F
[
u
(
t
)
]
=
∫
−
∞
∞
u
(
t
)
⋅
e
−
j
ω
t
∂
t
=
∫
0
∞
1
⋅
e
−
j
ω
t
∂
t
=
e
−
j
ω
t
−
j
ω
|
0
∞
=
1
j
ω
+
π
δ
(
ω
)
⏟
F
[
1
╱
2
]
{\displaystyle \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).
F
[
u
(
t
)
]
=
?
F
[
sign
(
t
)
]
=
2
j
ω
u
(
t
)
=
1
2
⋅
(
1
+
sign
(
t
)
)
→
F
[
u
(
t
)
]
=
F
[
1
╱
2
]
+
F
[
1
╱
2
⋅
sign
(
t
)
]
=
π
δ
(
ω
)
+
1
j
ω
{\displaystyle {\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:
E
j
:
f
(
t
)
=
e
−
a
t
u
(
t
)
=
{
e
−
a
t
,
t
>
0
0
,
t
<
0
;
a
>
0
F
[
f
(
t
)
]
=
F
(
ω
)
=
∫
−
∞
∞
f
(
t
)
e
−
j
ω
t
∂
t
=
∫
0
∞
e
−
a
t
e
−
j
ω
t
∂
t
=
e
−
a
t
e
−
j
ω
t
−
a
−
j
ω
|
0
∞
=
1
−
a
−
j
ω
(
e
−
∞
⏟
0
e
−
j
ω
∞
−
1
)
=
1
a
+
j
ω
=
F
(
ω
)
{\displaystyle {\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}}}
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:
δ
(
ω
)
=
{
∞
,
ω
=
0
0
,
ω
≠
0
{\displaystyle \delta (\omega )=\left\{{\begin{aligned}&\infty ,{\text{ }}\omega =0\\&0,{\text{ }}\omega \neq 0\\\end{aligned}}\right.}
Entre sus propiedades:
x
(
t
)
∗
δ
(
t
)
=
x
(
t
)
x
(
t
)
∗
δ
(
t
−
t
0
)
=
x
(
t
−
t
0
)
x
(
t
)
⋅
δ
(
t
0
)
=
x
(
t
0
)
→
x
(
t
)
⋅
δ
(
0
)
=
x
(
0
)
δ
(
t
)
=
δ
(
−
t
)
{\displaystyle {\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:
F
[
δ
(
t
)
]
=
1
↔
F
[
1
]
=
2
π
δ
(
−
ω
)
=
2
π
δ
(
ω
)
{\displaystyle \mathbb {F} [\delta (t)]=1\leftrightarrow F[1]=2\pi \delta (-\omega )=2\pi \delta (\omega )}
También tenemos que:
δ
(
a
t
)
=
1
|
a
|
δ
(
t
)
Debido a:
F
[
δ
(
a
t
)
]
→
F
[
f
(
a
t
)
=
1
|
a
|
F
(
ω
a
)
]
F
[
δ
(
t
)
]
=
1
≠
f
(
ω
)
1
|
a
|
⋅
1
→
F
−
1
[
1
a
]
=
1
a
⋅
δ
(
t
)
{\displaystyle {\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:
∂
u
(
t
)
∂
t
=
δ
(
t
)
↔
u
(
t
)
=
∫
−
∞
t
δ
(
τ
)
∂
τ
debido a:
F
[
∫
−
∞
t
f
(
τ
)
∂
τ
]
=
F
(
ω
)
j
ω
⇒
f
(
t
)
=
δ
(
t
)
→
F
(
ω
)
=
1
F
[
∫
−
∞
t
δ
(
τ
)
∂
τ
]
=
1
j
ω
⏟
F
[
u
(
t
)
]
{\displaystyle {\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:
∫
−
∞
∞
δ
(
t
)
∂
t
=
1
debido a:
F
[
δ
(t)
]
=1
→
1
|
ω
=
0
=
∫
−
∞
∞
f
(
t
)
∂
t
{\displaystyle \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().
F
[
∏
(
t
T
)
]
=
T
sinc
(
T
f
)
=
T
sinc
(
T
ω
╱
2
π
)
lim
T
→
∞
∏
(
t
T
)
=
1
F
[
lim
T
→
∞
∏
(
t
T
)
]
=
lim
T
→
∞
T
sinc
(
T
ω
2
π
)
⏟
δ
(
ω
2
π
)
=
lim
T
→
∞
T
sinc
(
T
f
)
⏟
δ
(
f
)
=
{
∞
,
f
=
0
0
,
f
≠
0
δ
(
f
)
=
lim
T
→
∞
T
sinc
(
T
f
)
=
δ
(
ω
╱
2
π
)
→
δ
(
a
ω
)
=
1
|
a
|
δ
(
ω
)
2
π
δ
(
ω
)
{\displaystyle {\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}}}
F
[
cos
(
ω
0
t
)
]
=
π
δ
(
ω
−
ω
0
)
+
π
δ
(
ω
+
ω
0
)
F
[
sin
(
ω
0
t
)
]
=
π
j
δ
(
ω
−
ω
0
)
−
π
j
δ
(
ω
+
ω
0
)
{\displaystyle {\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:
F
[
1
]
=
2
π
δ
(
ω
)
y
F
[
f
(
t
)
⋅
e
+
j
ω
0
t
]
=
F
(
ω
−
ω
0
)
F
[
1
⋅
e
+
j
ω
0
t
]
=
2
π
δ
(
ω
−
ω
0
)
F
[
cos
(
ω
0
t
)
]
=
F
[
e
j
ω
0
t
+
e
−
j
ω
0
t
2
]
=
1
2
[
2
π
δ
(
ω
−
ω
0
)
+
2
π
δ
(
ω
+
ω
0
)
]
{\displaystyle {\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:
F
[
e
+
j
ω
0
t
]
=
2
π
δ
(
ω
−
ω
0
)
F
[
sin
(
ω
0
t
)
]
=
F
[
e
j
ω
0
t
−
e
−
j
ω
0
t
2
j
]
=
1
2
j
[
2
π
δ
(
ω
−
ω
0
)
−
2
π
δ
(
ω
+
ω
0
)
]
{\displaystyle {\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}}}
∑
k
=
−
∞
∞
δ
(
t
−
k
T
s
)
{\displaystyle \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.
f
(
t
)
=
∑
k
=
−
∞
∞
δ
(
t
−
k
T
s
)
T
=
T
s
→
f
s
=
1
T
s
→
ω
s
=
2
π
f
s
=
2
π
T
s
f
(
t
)
=
∑
k
=
−
∞
∞
C
k
⋅
e
j
2
π
T
k
t
=
∑
k
=
−
∞
∞
C
k
⋅
e
j
2
π
T
s
k
t
C
k
=
1
T
∫
−
T
╱
2
T
╱
2
f
(
t
)
⋅
e
−
j
2
π
T
s
k
t
∂
t
=
1
T
s
∫
−
T
s
╱
2
T
s
╱
2
δ
(
t
)
⋅
e
−
j
2
π
T
s
k
t
∂
t
=
1
T
s
∫
−
T
s
╱
2
T
s
╱
2
δ
(
t
)
⋅
e
−
j
2
π
T
s
k
t
|
t
=
0
∂
t
=
1
T
s
∫
−
T
s
╱
2
T
s
╱
2
δ
(
t
)
∂
t
→
∫
−
∞
∞
δ
(
t
)
∂
t
=
1
→
C
k
=
1
T
s
∀
k
→
f
(
t
)
=
∑
k
=
−
∞
∞
1
T
s
e
j
2
π
T
s
k
t
=
1
T
s
∑
k
=
−
∞
∞
e
j
2
π
T
s
k
t
∑
k
=
−
∞
∞
δ
(
t
−
k
T
s
)
=
1
T
s
∑
k
=
−
∞
∞
e
j
2
π
T
s
k
t
{\displaystyle {\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}}}
Recordemos las propiedades:
F
[
e
+
j
ω
0
t
f
(
t
)
]
=
F
(
ω
−
ω
0
)
ω
0
=
2
π
T
s
,
F
[
1
]
=
2
π
δ
(
ω
)
→
F
[
e
+
j
ω
0
t
]
=
2
π
δ
(
ω
−
ω
0
)
F
[
1
T
s
∑
k
=
−
∞
∞
e
j
2
π
T
s
k
t
]
=
1
T
s
⋅
F
[
1
+
e
j
2
π
T
s
t
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{\displaystyle {\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}}}
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2742d7ac7d42df7e5bc18569e8d72763b306886c)
![{\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\pi }}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e095a2f07f3d4abd5002066e40ceecb66341c49c)
![{\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}&\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3301a15ceea7e972f6d6aa086848b3e25544d71)
.gif)
![{\displaystyle \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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a846290165d6c53ac81c3ca07ca191f5444e3b52)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a218a4a36713f13016e9d6be1fc806b0743b7b2e)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/30cd1826bb728b6b7a7d1ea6c73b4ca0c51ad145)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ae0b6dd9f027d4b00da3910b3305e5b437c8b1d)
.gif)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e5e749e1f103704cc6d6ef09e075396b6db3f6ef)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1969274234e1f3be61d97e31f914f501202574e8)
.gif)
![{\displaystyle \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}\;]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56914b8711dbdd1bdcd2aa3498c250d356af7756)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/caaeece2cf54530459c261a801cc2e6ec1d2949e)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7225451e68835aa5f89fb54d822b462d5075467d)
![{\displaystyle \mathbb {F} [\delta (t)]=1\leftrightarrow F[1]=2\pi \delta (-\omega )=2\pi \delta (\omega )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6605cb32e9ea14af93b9d6982b62a7172b1c4748)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b786107b2eb4768c6cbbb3a6ec5944c222eca036)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6a4bd0e167e1ecdc72a20a8b75226194aaae3bf)
![{\displaystyle \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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/62cf36af50c9e96f56718f3123e5a45aa1617def)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe34696c1d821f4a31e507d79783849caadc5ea6)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/157833d852150781325ff9613a26c3cec60aaa62)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9982b6efd2d01ee03d1ddda095adf1e46c9762bb)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/891127f4dbfa9e8e118652215659545d4823d41d)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab5c0fde6523f79b91bb9519d7daaf7114da8c91)