Por: Daniel Sanchez 06-40306
La parte par es:
z
1
p
a
r
(
t
)
=
0
,
5
[
u
(
t
+
10
)
u
(
10
−
t
)
+
r
a
(
t
−
4
)
−
2
r
a
(
t
−
5
)
+
r
a
(
t
−
6
)
+
r
a
(
t
+
6
)
−
2
r
a
(
t
+
5
)
+
r
a
(
t
+
4
)
]
{\displaystyle z1_{par}(t)=0,5[u(t+10)u(10-t)+r_{a}(t-4)-2r_{a}(t-5)+r_{a}(t-6)+r_{a}(t+6)-2r_{a}(t+5)+r_{a}(t+4)]\,}
La parte impar es:
z
1
i
m
p
a
r
(
t
)
=
0
,
5
[
−
u
(
t
+
10
)
+
2
u
(
t
)
−
u
(
t
−
10
)
+
r
a
(
t
−
4
)
−
2
r
a
(
t
−
5
)
+
r
a
(
t
−
6
)
−
r
a
(
t
+
6
)
+
2
r
a
(
t
+
5
)
−
r
a
(
t
+
4
)
]
{\displaystyle z1_{impar}(t)=0,5[-u(t+10)+2u(t)-u(t-10)+r_{a}(t-4)-2r_{a}(t-5)+r_{a}(t-6)-r_{a}(t+6)+2r_{a}(t+5)-r_{a}(t+4)]\,}
Las graficas son:
Parte Par
Problema 4 02 08
editar
Para las señales que se listan a continuación, indique cuáles son de energía y cuáles de potencia (en el caso de que lo fueran).
q
1
(
t
)
=
1
1
+
t
2
{\displaystyle q_{1}(t)={\frac {1}{1+t^{2}}}\,}
q
2
(
t
)
=
1
+
cos
(
π
t
)
u
(
t
)
{\displaystyle q_{2}(t)=1+\cos(\pi t)u(t)\,}
Problema 5 02 08
editar
Evalúe las siguientes operaciones:
∫
−
∞
∞
(
4
−
t
2
)
δ
(
t
+
3
)
d
t
{\displaystyle \int _{-\infty }^{\infty }(4-t^{2})\delta (t+3)dt\,}
y
4
(
t
)
=
∫
−
∞
t
[
4
−
τ
2
]
δ
(
τ
+
3
)
d
τ
{\displaystyle y_{4}(t)=\int _{-\infty }^{t}[4-\tau ^{2}]\delta (\tau +3)d\tau \,}
∫
−
3
6
(
6
−
t
2
)
[
δ
(
t
+
4
)
+
2
δ
(
2
t
+
4
)
]
d
t
{\displaystyle \int _{-3}^{6}(6-t^{2})[\delta (t+4)+2\delta (2t+4)]dt\,}
∫
−
∞
∞
[
e
−
5
t
+
cos
(
10
π
t
)
]
δ
˙
(
t
)
d
t
{\displaystyle \int _{-\infty }^{\infty }[e^{-5t}+\cos(10\pi t)]{\dot {\delta }}(t)dt\,}
y
5
(
t
)
=
∫
−
∞
t
[
e
−
5
τ
+
cos
(
10
π
τ
)
]
δ
˙
(
τ
)
d
τ
{\displaystyle y_{5}(t)=\int _{-\infty }^{t}[e^{-5\tau }+\cos(10\pi \tau )]{\dot {\delta }}(\tau )d\tau \,}
Solucion del Problema # 5:
Marianela Mendoza 06-39906 .
Estas operaciones se resolvieron aplicando propiedades del impulso y el doblete.
Impulso: Las integrales que llevan impulsos multiplicando funciones son sencillas de resolver, esto se debe a sus propiedades.
δ
(
t
−
t
o
)
.
f
(
t
)
=
δ
(
t
−
t
o
)
.
f
(
t
o
)
{\displaystyle \delta (t-t_{o}).f(t)=\delta (t-t_{o}).f(t_{o})\,}
∫
−
∞
∞
[
δ
(
t
−
t
o
)
.
f
(
t
)
]
d
t
=
f
(
t
o
)
.
∫
−
∞
∞
δ
(
t
−
t
o
)
d
t
=
f
(
t
o
)
{\displaystyle \int _{-\infty }^{\infty }[\delta (t-t_{o}).f(t)]dt=f(t_{o}).\int _{-\infty }^{\infty }\delta (t-t_{o})dt=f(t_{o})\,}
δ
(
b
(
t
−
t
o
)
)
=
δ
(
t
−
t
o
)
|
b
|
{\displaystyle \delta (b(t-t_{o}))={\frac {\delta (t-t_{o})}{\left|b\right|}}\,}
De la manera siguiente se obtuvieron los resultados: Procedimeinto: Se multiplica la delta por la funcion (es decir, evalua la funcion en donde existe la delta), y luego, lo que queda es una constante multiplicando la delta, finalmente la integral de lo que queda es esa constante, tal como se muestra en los ejercicios.
∫
−
∞
∞
(
4
−
t
2
)
δ
(
t
+
3
)
d
t
=
−
5
∫
−
∞
∞
δ
(
t
+
3
)
d
t
=
−
5
{\displaystyle \int _{-\infty }^{\infty }(4-t^{2})\delta (t+3)dt=-5\int _{-\infty }^{\infty }\delta (t+3)dt=-5\,}
y
4
(
t
)
=
∫
−
∞
t
[
4
−
τ
2
]
δ
(
τ
+
3
)
d
τ
=
5
∫
−
∞
t
δ
(
τ
+
3
)
d
τ
;
y
4
(
t
)
=
5
,
(
t
>
3
)
{\displaystyle y_{4}(t)=\int _{-\infty }^{t}[4-\tau ^{2}]\delta (\tau +3)d\tau =5\int _{-\infty }^{t}\delta (\tau +3)d\tau ;y_{4}(t)=5,(t>3)\,}
∫
−
3
6
(
6
−
t
2
)
[
δ
(
t
+
4
)
+
2
δ
(
2
t
+
4
)
]
d
t
=
∫
−
3
6
−
10
δ
(
t
+
4
)
+
2
δ
(
t
+
2
)
d
t
=
2
{\displaystyle \int _{-3}^{6}(6-t^{2})[\delta (t+4)+2\delta (2t+4)]dt=\int _{-3}^{6}-10\delta (t+4)+2\delta (t+2)dt=2\,}
Doblete: Propiedades del doblete, mostradas a continuacion son las herramientas que se utilizaron para la resolucion de las siguientes integrales.
∫
−
∞
∞
[
δ
˙
(
t
)
]
d
t
=
0
{\displaystyle \int _{-\infty }^{\infty }[{\dot {\delta }}(t)]dt=0\,}
δ
˙
(
t
)
.
f
(
t
)
=
δ
˙
(
t
)
.
f
(
o
)
−
f
˙
(
t
)
.
δ
(
t
)
{\displaystyle {\dot {\delta }}(t).f(t)={\dot {\delta }}(t).f(o)-{\dot {f}}(t).\delta (t)\,}
∫
−
∞
∞
[
δ
˙
(
t
)
.
f
(
t
)
]
d
t
=
∫
−
∞
∞
[
δ
˙
(
t
)
.
f
(
o
)
]
d
t
−
∫
−
∞
∞
[
f
˙
(
t
)
.
δ
(
t
)
]
d
t
=
−
f
˙
(
o
)
{\displaystyle \int _{-\infty }^{\infty }[{\dot {\delta }}(t).f(t)]dt=\int _{-\infty }^{\infty }[{\dot {\delta }}(t).f(o)]dt-\int _{-\infty }^{\infty }[{\dot {f}}(t).\delta (t)]dt=-{\dot {f}}(o)\,}
δ
˙
(
b
t
)
=
δ
(
t
)
b
|
b
|
{\displaystyle {\dot {\delta }}(bt)={\frac {\delta (t)}{b\left|b\right|}}\,}
Igual que en la parte anterior, aplicando estas propiedades se resuelven las integrales.
∫
−
∞
∞
[
e
−
5
t
+
cos
(
10
π
t
)
]
δ
˙
(
t
)
d
t
=
5
{\displaystyle \int _{-\infty }^{\infty }[e^{-5t}+\cos(10\pi t)]{\dot {\delta }}(t)dt=5\,}
y
5
(
t
)
=
∫
−
∞
t
[
e
−
5
τ
+
cos
(
10
π
τ
)
]
δ
˙
(
τ
)
d
τ
=
∫
−
∞
t
2
δ
˙
(
τ
)
+
5
δ
(
t
)
d
τ
=
5
(
t
=
0
)
;
0
(
t
≠
0
)
{\displaystyle y_{5}(t)=\int _{-\infty }^{t}[e^{-5\tau }+\cos(10\pi \tau )]{\dot {\delta }}(\tau )d\tau =\int _{-\infty }^{t}2{\dot {\delta }}(\tau )+5\delta (t)d\tau =5(t=0);0(t\neq 0)\,}
En un sistema lineal e invariante en el tiempo, la relación entre la entrada (
x
(
t
)
{\displaystyle x(t)}
y
(
t
)
{\displaystyle y(t)}
y
(
t
)
=
∫
−
∞
t
e
−
(
t
−
τ
)
x
(
−
0.5
τ
+
1
)
d
τ
{\displaystyle y(t)=\int _{-\infty }^{t}e^{-(t-\tau )}x(-0.5\tau +1)d\tau }
Determine:
La respuesta al impulso (
δ
(
t
)
{\displaystyle \delta (t)\,}
La respuesta al pulso (
Π
1
(
t
)
{\displaystyle \Pi _{1}(t)\,}
Solución Problema 1
editar
Propuesta de solucion:
Damian Vigouroux 17/05/2007
05-39044
Parte 1.
h
(
t
)
=
∫
−
∞
t
e
−
(
t
−
τ
)
δ
(
−
0.5
τ
+
1
)
d
τ
{\displaystyle h(t)=\int _{-\infty }^{t}e^{-(t-\tau )}\delta (\,-0.5\tau +1)d\tau }
δ
(
−
0.5
τ
+
1
)
=
δ
(
−
0.5
(
τ
−
2
)
)
=
2
δ
(
τ
−
2
)
{\displaystyle \delta (\,-0.5\tau +1)=\delta (\,-0.5(\tau -2))=2\delta (\,\tau -2)}
Por lo tanto :
e
−
(
t
−
τ
)
δ
(
−
0.5
τ
+
1
)
=
2
e
−
(
t
−
τ
)
δ
(
τ
−
2
)
=
2
e
−
(
t
−
2
)
δ
(
τ
−
2
)
{\displaystyle e^{-(t-\tau )}\delta (\,-0.5\tau +1)=2e^{-(t-\tau )}\delta (\,\tau -2)=2e^{-(t-2)}\delta (\,\tau -2)}
Entonces la respuesta al impulso será:
h
(
t
)
=
∫
−
∞
t
2
e
−
(
t
−
2
)
δ
(
τ
−
2
)
d
τ
=
2
e
−
(
t
−
2
)
∫
−
∞
t
δ
(
τ
−
2
)
d
τ
=
2
e
−
(
t
−
2
)
u
(
t
−
2
)
{\displaystyle h(t)=\int _{-\infty }^{t}2e^{-(t-2)}\delta (\,\tau -2)d\tau =2e^{-(t-2)}\int _{-\infty }^{t}\delta (\,\tau -2)d\tau =2e^{-(t-2)}u(t-2)}
Parte 2. Conociendo la respuesta al impulso del sistema (LTI) podemos carecterizar su comportamiento ante cualquier señar de entrada específicamente a:
Π
1
(
t
)
{\displaystyle \Pi _{1}(t)\,}
y
(
t
)
=
x
(
t
)
∗
h
(
t
)
=
∫
−
∞
+
∞
x
(
τ
)
h
(
t
−
τ
)
d
τ
{\displaystyle y(t)=x(t)*h(t)=\int _{-\infty }^{+\infty }x(\tau )h(t-\tau )d\tau }
x
(
t
)
=
Π
1
(
t
)
=
1
{\displaystyle x(t)=\Pi _{1}(t)\,=1}
Sea t < 1.5
Un análisis gráfico del producto
y
(
t
)
=
x
(
τ
)
.
h
(
t
−
τ
)
{\displaystyle y(t)=x(\tau ).h(t-\tau )}
Sea 1.5 < t < 2.5
En este caso
y
(
t
)
=
∫
−
∞
+
∞
x
(
τ
)
h
(
t
−
τ
)
d
τ
=
∫
−
0.5
t
−
2
2
e
−
(
t
−
τ
−
2
)
d
τ
=
2
e
−
(
t
−
2
)
∫
−
0.5
t
−
2
e
τ
d
τ
=
2
(
1
−
e
−
(
t
−
1.5
)
)
{\displaystyle y(t)=\int _{-\infty }^{+\infty }x(\tau )h(t-\tau )d\tau =\int _{-0.5}^{t-2}2e^{-(t-\tau -2)}d\tau =2e^{-(t-2)}\int _{-0.5}^{t-2}e^{\tau }d\tau =2(1-e^{-(t-1.5)})}
Sea 2.5 > t
Definiendo la region común, distinta de cero de las gráficas:
y
(
t
)
=
∫
−
0.5
0.5
2
e
−
(
t
−
τ
−
2
)
d
τ
=
2
e
−
(
t
−
2
)
∫
−
0.5
0.5
e
τ
d
τ
=
2
(
e
0.5
−
e
−
0.5
)
e
−
(
t
−
2
)
{\displaystyle y(t)=\int _{-0.5}^{0.5}2e^{-(t-\tau -2)}d\tau =2e^{-(t-2)}\int _{-0.5}^{0.5}e^{\tau }d\tau =2(e^{0.5}-e^{-0.5})e^{-(t-2)}}
Luego la respuesta a
x
(
t
)
=
Π
1
(
t
)
{\displaystyle x(t)=\Pi _{1}(t)}
Repita el problema anterior si la relación entrada salida cambia a:
y
(
t
)
=
∫
−
∞
∞
e
−
(
t
−
τ
)
x
(
−
0.5
τ
+
1
)
d
τ
{\displaystyle y(t)=\int _{-\infty }^{\infty }e^{-(t-\tau )}x(-0.5\tau +1)d\tau }
Solución Problema 2
editar
Realizado por : José Velásquez #05-39032
1) Tenemos que la respuesta al impulso (
δ
(
t
)
{\displaystyle \delta (t)\,}
h
(
t
)
=
∫
−
∞
∞
e
−
(
t
−
τ
)
δ
(
−
0.5
τ
+
1
)
d
τ
{\displaystyle h(t)=\int _{-\infty }^{\infty }e^{-(t-\tau )}\delta (\,-0.5\tau +1)d\tau }
h
(
t
)
{\displaystyle h(t)}
δ
[
α
(
t
−
β
)
]
=
|
1
α
|
δ
(
t
−
β
)
{\displaystyle \delta [\alpha (t-\beta )]=\left|{\frac {1}{\alpha }}\right\vert \delta (t-\beta )\,}
h
(
t
)
=
∫
−
∞
∞
2
e
−
(
t
−
τ
)
δ
(
τ
−
2
)
d
τ
{\displaystyle h(t)=\int _{-\infty }^{\infty }2e^{-(t-\tau )}\delta (\tau -2)d\tau }
∫
−
∞
∞
x
(
t
)
δ
(
t
−
α
)
d
t
=
x
(
α
)
{\displaystyle \int _{-\infty }^{\infty }x(t)\delta (t-\alpha )dt=x(\alpha )}
h
(
t
)
=
∫
−
∞
∞
2
e
−
(
t
−
τ
)
δ
(
τ
−
2
)
d
τ
=
2
e
−
(
t
−
2
)
{\displaystyle h(t)=\int _{-\infty }^{\infty }2e^{-(t-\tau )}\delta (\tau -2)d\tau =2e^{-(t-2)}}
2) Tenemos que la respuesta al pulso
Π
1
(
t
)
{\displaystyle \Pi _{1}(t)\,}
y
(
t
)
=
x
(
t
)
∗
h
(
t
)
=
h
(
t
)
∗
x
(
t
)
=
∫
−
∞
+
∞
h
(
τ
)
x
(
t
−
τ
)
d
τ
{\displaystyle y(t)=x(t)*h(t)=h(t)*x(t)=\int _{-\infty }^{+\infty }h(\tau )x(t-\tau )d\tau }
y
(
t
)
=
∫
t
−
0.5
t
+
0.5
2
e
−
(
τ
−
2
)
d
τ
=
2
(
e
2.5
−
e
1.5
)
e
−
t
{\displaystyle y(t)=\int _{t-0.5}^{t+0.5}2e^{-(\tau -2)}d\tau =2(e^{2.5}-e^{1.5})e^{-t}}
y
(
t
)
=
2
(
e
2.5
−
e
1.5
)
e
−
t
∀
t
∈
(
−
∞
,
+
∞
)
{\displaystyle y(t)=2(e^{2.5}-e^{1.5})e^{-t}\quad \forall \ t\ \in (-\infty ,+\infty )}
Π
1
(
t
)
{\displaystyle \Pi _{1}(t)\,}
Considere el sistema modulador que se muestra en la figura, en el que la moduladora es
g
(
t
)
=
sin
(
2
π
t
)
{\displaystyle g(t)=\sin(2\pi t)\,}
Determine la respuesta del sistema (
y
(
t
)
{\displaystyle y(t)}
x
(
t
)
=
∑
k
=
−
∞
∞
δ
(
t
−
4
k
)
{\displaystyle x(t)=\sum _{k=-\infty }^{\infty }\delta (t-4k)}
x
(
t
)
=
∑
k
=
−
∞
∞
Π
1
(
t
−
8
k
)
{\displaystyle x(t)=\sum _{k=-\infty }^{\infty }\Pi _{1}(t-8k)}
Solución Problema 3
editar
Realizado por : José Velásquez #05-39032
1) La respuesta del sistema (y(t)) cuando en la entrada se aplica:
x
(
t
)
=
∑
k
=
−
∞
∞
δ
(
t
−
4
k
)
{\displaystyle x(t)=\sum _{k=-\infty }^{\infty }\delta (t-4k)}
g
(
t
)
=
sin
(
2
π
t
)
{\displaystyle g(t)=\sin(2\pi t)\,}
X
I
(
t
)
=
X
(
t
)
∑
k
=
−
∞
∞
δ
(
t
−
k
t
s
)
=
∑
k
=
−
∞
∞
X
(
k
t
s
)
δ
(
t
−
k
t
s
)
{\displaystyle {}_{\!}X_{I}(t)=X(t)\sum _{k=-\infty }^{\infty }\delta (t-k\,{}_{\!}t_{s})=\sum _{k=-\infty }^{\infty }X(k\,{}_{\!}t_{s})\,\delta (t-k\,{}_{\!}t_{s})}
y
(
t
)
=
g
(
t
)
x
(
t
)
=
sin
(
2
π
t
)
∑
k
=
−
∞
∞
δ
(
t
−
4
k
)
=
∑
k
=
−
∞
∞
sin
(
2
π
4
k
)
δ
(
t
−
4
k
)
{\displaystyle y(t)=g(t)\,x(t)=\sin(2\pi t)\sum _{k=-\infty }^{\infty }\delta (t-4k)=\sum _{k=-\infty }^{\infty }\sin(2\pi \,4k)\,\delta (t-4k)}
y
(
t
)
=
∑
k
=
−
∞
∞
sin
(
8
π
k
)
δ
(
t
−
4
k
)
{\displaystyle y(t)=\sum _{k=-\infty }^{\infty }\sin(8\pi k)\,\delta (t-4k)}
g
(
t
)
=
sin
(
2
π
t
)
{\displaystyle g(t)=\sin(2\pi t)\,}
t
=
4
k
∀
k
(
e
n
t
e
r
o
)
∈
(
−
∞
,
+
∞
)
{\displaystyle t=4k\quad \forall \ k\,(entero)\ \in (-\infty ,+\infty )}
2) La respuesta del sistema (y(t)) cuando en la entrada se aplica:
x
(
t
)
=
∑
k
=
−
∞
∞
Π
1
(
t
−
8
k
)
=
∑
k
=
−
∞
∞
[
u
(
t
−
8
k
+
0.5
)
−
u
(
t
−
8
k
−
0.5
)
]
{\displaystyle x(t)=\sum _{k=-\infty }^{\infty }\Pi _{1}(t-8k)=\sum _{k=-\infty }^{\infty }[u(t-8k+0.5)-u(t-8k-0.5)]}
y
(
t
)
=
g
(
t
)
x
(
t
)
=
sin
(
2
π
t
)
∑
k
=
−
∞
∞
[
u
(
t
−
8
k
+
0.5
)
−
u
(
t
−
8
k
−
0.5
)
]
{\displaystyle y(t)=g(t)\,x(t)=\sin(2\pi t)\sum _{k=-\infty }^{\infty }[u(t-8k+0.5)-u(t-8k-0.5)]}
ω
o
=
2
π
/
T
{\displaystyle {}_{\!}\omega _{o}=2\pi /T}
T
=
2
π
ω
o
⇒
T
=
1
{\displaystyle T={\frac {2\pi }{\omega _{o}}}\Rightarrow T=1}
y
(
t
)
=
sin
(
2
π
t
)
s
i
8
k
−
0.5
≤
t
≤
8
k
+
0.5
∀
k
(
e
n
t
e
r
o
)
∈
(
−
∞
,
+
∞
)
{\displaystyle y(t)=\sin(2\pi t)\ si\ 8k-0.5\leq t\ \leq 8k+0.5\ \forall \ k\,(entero)\ \in (-\infty ,+\infty )}
Considere las señales
x
(
t
)
=
e
−
t
2
8
2
2
π
{\displaystyle x(t)={\frac {e^{-{\frac {t^{2}}{8}}}}{2{\sqrt {2\pi }}}}\,}
t
∈
(
−
∞
,
∞
)
{\displaystyle t\in (-\infty ,\infty )}
Π
(
t
)
=
{
0
t
<
−
0.1
1
−
0.1
≤
t
≤
0.1
0
t
>
0.1
{\displaystyle \Pi (t)=\left\{{\begin{array}{cc}0&t<-0.1\\1&-0.1\leq t\leq 0.1\\0&t>0.1\end{array}}\right.\,}
y sea
y
(
t
)
=
∑
k
=
−
6
6
x
(
k
T
)
Π
(
t
−
k
T
)
{\displaystyle y(t)=\sum _{k=-6}^{6}x(kT)\Pi (t-kT)\,}
Calcule y grafique
x
(
t
)
,
Π
(
t
)
,
y
(
t
)
{\displaystyle x(t),\Pi (t),y(t)\,}
T=1
T=0.2 Solución 1 Problema 4
editar
Realizado por: Jesús Querales #05-38758
T
=
1
{\displaystyle T=1\,}
La señal
x
(
t
)
{\displaystyle x(t)\,}
1
2
2
π
≈
0.2
{\displaystyle {\frac {1}{2{\sqrt {2\pi }}}}\approx 0.2}
x
(
t
)
{\displaystyle x(t)\,}
El pulso
Π
(
t
)
=
u
(
t
+
0.1
)
−
u
(
t
−
0.1
)
{\displaystyle \Pi (t)=u(t+0.1)-u(t-0.1)\,}
Archivo:Fig2 problema4.gif
La señal
y
(
t
)
=
∑
k
=
−
6
6
x
(
k
)
Π
(
t
−
k
)
{\displaystyle y(t)=\sum _{k=-6}^{6}x(k)\Pi (t-k)\,}
x
(
k
)
=
e
−
k
2
8
2
2
π
{\displaystyle x(k)={\frac {e^{-{\frac {k^{2}}{8}}}}{2{\sqrt {2\pi }}}}\,}
Π
(
t
−
k
)
=
u
(
t
−
k
+
0.1
)
−
u
(
t
−
k
−
0.1
)
{\displaystyle \Pi (t-k)=u(t-k+0.1)-u(t-k-0.1)\,}
y
(
t
)
=
∑
k
=
−
6
6
e
−
k
2
8
2
2
π
[
u
(
t
−
k
+
0.1
)
−
u
(
t
−
k
−
0.1
)
]
{\displaystyle y(t)=\sum _{k=-6}^{6}{\frac {e^{-{\frac {k^{2}}{8}}}}{2{\sqrt {2\pi }}}}\ [u(t-k+0.1)-u(t-k-0.1)]}
Considere la señal
x
(
t
)
=
cos
(
π
2
t
)
+
cos
(
π
t
)
+
cos
(
3
π
t
)
{\displaystyle x(t)=\cos({\frac {\pi }{2}}t)+\cos(\pi t)+\cos(3\pi t)\,}
La frecuencia y el período fundamentales
La energía, potencia, promedio, valor RMS y área absoluta en un período Solución Problema 5
editar
Realizado por : José Velásquez #05-39032
1. La frecuencia fundamental está dada por:
ω
o
=
G
C
D
(
π
2
,
π
,
3
π
)
=
π
2
{\displaystyle \omega _{o}=GCD({\frac {\pi }{2}},\pi ,3\pi )={\frac {\pi }{2}}}
T
=
2
π
ω
o
⇒
T
=
2
π
π
2
=
4
{\displaystyle T={\frac {2\pi }{\omega _{o}}}\Rightarrow T={\frac {2\pi }{\frac {\pi }{2}}}=4}
2. Para el análisis y la compresión de esta parte se presentan tres gráficas ilustrativas. En cada imagen se sombrea el área debajo la curva en un período
T
{\displaystyle {\mbox{T}}\,}
x
(
t
)
=
cos
(
π
2
t
)
+
cos
(
π
t
)
+
cos
(
3
π
t
)
{\displaystyle x(t)=\cos({\frac {\pi }{2}}t)+\cos(\pi t)+\cos(3\pi t)\,}
x
1
(
t
)
=
[
cos
(
π
2
t
)
+
cos
(
π
t
)
+
cos
(
3
π
t
)
]
2
{\displaystyle x_{1}(t)=[\cos({\frac {\pi }{2}}t)+\cos(\pi t)+\cos(3\pi t)]^{2}}
x
2
(
t
)
=
|
cos
(
π
2
t
)
+
cos
(
π
t
)
+
cos
(
3
π
t
)
|
{\displaystyle x_{2}(t)=\left|\cos({\frac {\pi }{2}}t)+\cos(\pi t)+\cos(3\pi t)\right\vert }
La Energía en un período viene dada por:
E
x
=
∫
T
|
x
(
t
)
|
2
d
t
=
∫
0
4
[
cos
(
π
2
t
)
+
cos
(
π
t
)
+
cos
(
3
π
t
)
]
2
d
t
=
6
{\displaystyle {}_{\!}E_{x}=\int _{T}\left|x(t)\right\vert ^{2}\,dt=\int _{0}^{4}[\cos({\frac {\pi }{2}}t)+\cos(\pi t)+\cos(3\pi t)]^{2}\,dt=6}
La Potencia en un período viene dada por:
P
x
=
1
T
∫
T
|
x
(
t
)
|
2
d
t
=
E
x
T
=
6
4
=
3
2
{\displaystyle {}_{\!}P_{x}={\frac {1}{T}}\int _{T}\left|x(t)\right\vert ^{2}\,dt={\frac {{}_{\!}E_{x}}{T}}={\frac {6}{4}}={\frac {3}{2}}}
El Promedio en un período viene dado por:
X
pro
=
1
T
∫
T
x
(
t
)
d
t
=
∫
0
4
[
cos
(
π
2
t
)
+
cos
(
π
t
)
+
cos
(
3
π
t
)
]
d
t
=
0
{\displaystyle {}_{\!}X_{\mbox{pro}}={\frac {1}{T}}\int _{T}x(t)\,dt=\int _{0}^{4}[\cos({\frac {\pi }{2}}t)+\cos(\pi t)+\cos(3\pi t)]\,dt=0}
El valor RMS viene dado por:
X
rms
=
P
x
=
3
2
{\displaystyle {}_{\!}X_{\mbox{rms}}={\sqrt[{}]{{}_{\!}P_{x}}}={\sqrt[{}]{\frac {3}{2}}}}
A
abs
=
1
T
∫
T
|
x
(
t
)
|
d
t
=
1
4
∫
0
4
|
cos
(
π
2
t
)
+
cos
(
π
t
)
+
cos
(
3
π
t
)
|
d
t
=
3
,
9847
4
=
0
,
996
{\displaystyle {}_{\!}A_{\mbox{abs}}={\frac {1}{T}}\int _{T}\left|x(t)\right\vert \,dt={\frac {1}{4}}\int _{0}^{4}\left|\cos({\frac {\pi }{2}}t)+\cos(\pi t)+\cos(3\pi t)\right\vert \,dt={\frac {3,9847}{4}}=0,996}
Evalúe las siguientes expresiones:
∫
−
∞
∞
(
5
−
8
t
2
)
δ
(
−
3
t
−
9
)
d
t
{\displaystyle \int _{-\infty }^{\infty }(5-8t^{2})\delta (-3t-9)dt\,}
x
(
t
)
=
∫
−
∞
t
(
5
−
8
τ
2
)
δ
(
−
3
τ
−
9
)
d
τ
{\displaystyle x(t)=\int _{-\infty }^{t}(5-8\tau ^{2})\delta (-3\tau -9)d\tau \,}
∫
−
∞
∞
[
e
−
5
t
−
sin
(
5
π
t
)
]
δ
˙
(
−
2
t
+
8
)
d
t
{\displaystyle \int _{-\infty }^{\infty }[e^{-5t}-\sin(5\pi t)]{\dot {\delta }}(-2t+8)dt\,}
x
(
t
)
=
∫
−
∞
t
[
e
−
5
τ
−
sin
(
5
π
τ
)
]
δ
˙
(
−
2
τ
+
8
)
d
τ
{\displaystyle x(t)=\int _{-\infty }^{t}[e^{-5\tau }-\sin(5\pi \tau )]{\dot {\delta }}(-2\tau +8)d\tau \,}
Solución Problema 6
editar
Realizado por: Orlando Díaz #05-38117
1)
Utilizando las propiedades de la
δ
(
t
)
{\displaystyle \delta (t)\,}
∫
−
∞
∞
(
5
−
8
t
2
)
δ
(
−
3
t
−
9
)
d
t
=
1
3
∫
−
∞
∞
(
5
−
8
t
2
)
δ
(
t
+
3
)
d
t
{\displaystyle \int _{-\infty }^{\infty }(5-8t^{2})\delta (-3t-9)dt={\frac {1}{3}}\int _{-\infty }^{\infty }(5-8t^{2})\delta (t+3)dt\,}
δ
(
t
)
{\displaystyle \delta (t)\,}
∫
−
∞
∞
x
(
t
)
δ
(
t
−
α
)
d
t
=
x
(
α
)
{\displaystyle \int _{-\infty }^{\infty }x(t)\delta (t-\alpha )dt=x(\alpha )}
1
3
∫
−
∞
∞
(
5
−
8
t
2
)
δ
(
t
+
3
)
d
t
=
1
3
(
5
−
8
(
−
3
)
2
)
=
−
67
3
{\displaystyle {\frac {1}{3}}\int _{-\infty }^{\infty }(5-8t^{2})\delta (t+3)dt={\frac {1}{3}}(5-8(-3)^{2})=-{\frac {67}{3}}}
2)
Trabajaremos primero con la parte interna de la integral, por propiedades obtenemos:
(
5
−
8
τ
2
)
δ
(
−
3
τ
−
9
)
=
1
3
(
5
−
8
τ
2
)
δ
(
τ
+
3
)
{\displaystyle (5-8\tau ^{2})\delta (-3\tau -9)={\frac {1}{3}}(5-8\tau ^{2})\delta (\tau +3)}
δ
(
t
)
{\displaystyle \delta (t)}
0
{\displaystyle 0}
t
≠
o
{\displaystyle t\neq o}
1
{\displaystyle 1}
t
=
0
{\displaystyle t=0}
δ
(
τ
+
3
)
{\displaystyle \delta (\tau +3)}
τ
=
−
3
{\displaystyle \tau =-3}
(
5
−
8
τ
2
)
{\displaystyle (5-8\tau ^{2})}
δ
(
τ
)
{\displaystyle \delta (\tau )}
x
(
τ
)
{\displaystyle x(\tau )}
τ
=
−
3
{\displaystyle \tau =-3}
1
3
(
5
−
8
(
−
3
)
2
)
δ
(
τ
+
3
)
=
−
67
3
δ
(
τ
+
3
)
{\displaystyle {\frac {1}{3}}(5-8(-3)^{2})\delta (\tau +3)={\frac {-67}{3}}\delta (\tau +3)}
∫
−
∞
t
−
67
3
δ
(
τ
+
3
)
d
τ
=
−
67
3
∫
−
∞
t
δ
(
τ
+
3
)
d
τ
{\displaystyle \int _{-\infty }^{t}{\frac {-67}{3}}\delta (\tau +3)d\tau ={\frac {-67}{3}}\int _{-\infty }^{t}\delta (\tau +3)d\tau }
El impulso es la derivada de la función escalón, así que la integral de un impulso es un escalón que comienza donde existe el impulso, por tanto la respuesta es:
∫
−
∞
t
1
3
(
5
−
8
τ
2
)
δ
(
τ
+
3
)
d
τ
=
−
67
3
U
(
t
+
3
)
{\displaystyle \int _{-\infty }^{t}{\frac {1}{3}}(5-8\tau ^{2})\delta (\tau +3)d\tau ={\frac {-67}{3}}U(t+3)}
3)
El doblete
δ
′
(
t
)
{\displaystyle \delta ^{\prime }(t)}
[
e
−
5
t
−
s
e
n
(
5
π
t
)
]
δ
′
(
−
2
t
+
8
)
=
1
−
4
[
e
−
5
t
−
s
e
n
(
5
π
t
)
]
δ
′
(
t
−
4
)
{\displaystyle [e^{-5t}-sen(5\pi t)]\delta ^{\prime }(-2t+8)={\frac {1}{-4}}[e^{-5t}-sen(5\pi t)]\delta ^{\prime }(t-4)}
∫
−
∞
∞
x
(
t
)
δ
′
(
t
−
α
)
d
t
=
−
x
′
(
α
)
{\displaystyle \int _{-\infty }^{\infty }x(t)\delta ^{\prime }(t-\alpha )dt=-x^{\prime }(\alpha )}
1
−
4
∫
−
∞
∞
[
e
−
5
t
−
sin
(
5
π
t
)
]
δ
˙
(
t
−
4
)
d
t
=
−
1
4
[
−
5
e
−
5
t
−
5
π
c
o
s
(
5
π
t
)
]
T
=
4
=
5
e
−
20
4
+
5
π
4
{\displaystyle {\frac {1}{-4}}\int _{-\infty }^{\infty }[e^{-5t}-\sin(5\pi t)]{\dot {\delta }}(t-4)dt=-{\frac {1}{4}}[-5e^{-5t}-5\pi cos(5\pi t)]_{T=4}={\frac {5e^{-20}}{4}}+{\frac {5\pi }{4}}}
4)
Realizado por Julio Aguilar Carnet: 06-39117
El doblete
δ
′
(
t
)
{\displaystyle \delta ^{\prime }(t)}
cero (0) para todo "t" distinto de cero y
∫
−
∞
∞
δ
′
(
t
)
d
t
=
0
{\displaystyle \int _{-\infty }^{\infty }\delta ^{\prime }(t)dt=0}
(1)
x
(
t
)
δ
′
(
t
−
α
)
=
x
(
α
)
δ
′
(
t
−
α
)
−
x
(
α
)
′
δ
(
t
−
α
)
{\displaystyle x(t)\delta ^{\prime }(t-\alpha )=x(\alpha )\delta ^{\prime }(t-\alpha )-x(\alpha )^{\prime }\delta (t-\alpha )}
∫
−
∞
∞
x
(
t
)
δ
′
(
t
−
α
)
d
t
=
−
x
′
(
α
)
{\displaystyle \int _{-\infty }^{\infty }x(t)\delta ^{\prime }(t-\alpha )dt=-x^{\prime }(\alpha )}
(3)
δ
′
(
a
t
+
b
)
=
1
a
|
a
|
δ
′
(
t
+
b
a
)
{\displaystyle \delta ^{\prime }(at+b)={\frac {1}{a|a|}}\delta ^{\prime }(t+{\frac {b}{a}})}
Aplicando la tercera propiedad se tiene que:
[
e
−
5
t
−
s
e
n
(
5
π
t
)
]
δ
′
(
−
2
t
+
8
)
=
1
−
4
[
e
−
5
t
−
s
e
n
(
5
π
t
)
]
δ
′
(
t
−
4
)
{\displaystyle [e^{-5t}-sen(5\pi t)]\delta ^{\prime }(-2t+8)={\frac {1}{-4}}[e^{-5t}-sen(5\pi t)]\delta ^{\prime }(t-4)}
y luego la primera propiedad tenemos que:
1
−
4
[
e
−
5
t
−
s
e
n
(
5
π
t
)
]
δ
′
(
t
−
4
)
=
e
−
20
−
4
δ
′
(
t
−
4
)
−
[
−
5
e
−
20
−
5
π
4
]
δ
(
t
−
4
)
{\displaystyle {\frac {1}{-4}}[e^{-5t}-sen(5\pi t)]\delta ^{\prime }(t-4)={\frac {e^{-20}}{-4}}\delta ^{\prime }(t-4)-[{\frac {-5e^{-20}-5\pi }{4}}]\delta (t-4)}
Integrando y aplicando la definicion se tiene que la primera parte es igual a cero:
∫
−
∞
t
δ
′
(
t
−
4
)
=
0
{\displaystyle \int _{-\infty }^{t}\delta ^{\prime }(t-4)=0}
La segunda parte queda:
∫
−
∞
t
[
5
e
−
20
+
5
π
4
]
δ
(
t
−
4
)
=
[
5
e
−
20
+
5
π
4
]
U
(
t
−
4
)
{\displaystyle \int _{-\infty }^{t}[{\frac {5e^{-20}+5\pi }{4}}]\delta (t-4)=[{\frac {5e^{-20}+5\pi }{4}}]U(t-4)}
![{\displaystyle Potencia={\frac {1}{2}}[\int _{-1}^{0}|t+1|^{2}\,dt+\int _{0}^{1}|1-t|^{2}\,dt]={\frac {1}{6}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a46db35cb754222e367fd594aac61fd5a7addd4)
![{\displaystyle z1_{par}(t)=0,5[u(t+10)u(10-t)+r_{a}(t-4)-2r_{a}(t-5)+r_{a}(t-6)+r_{a}(t+6)-2r_{a}(t+5)+r_{a}(t+4)]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7fa280efec17e0c034ea1134ad1ded85ad5f5830)
![{\displaystyle z1_{impar}(t)=0,5[-u(t+10)+2u(t)-u(t-10)+r_{a}(t-4)-2r_{a}(t-5)+r_{a}(t-6)-r_{a}(t+6)+2r_{a}(t+5)-r_{a}(t+4)]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3fb33858bc482638b31a049061407276dbc052c4)
![{\displaystyle y_{4}(t)=\int _{-\infty }^{t}[4-\tau ^{2}]\delta (\tau +3)d\tau \,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0894cca3e96c44e308edbbf253d8a646a03df5f3)
![{\displaystyle \int _{-3}^{6}(6-t^{2})[\delta (t+4)+2\delta (2t+4)]dt\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1979c8acc8eb11496e55d6187efc36b4e4079170)
![{\displaystyle \int _{-\infty }^{\infty }[e^{-5t}+\cos(10\pi t)]{\dot {\delta }}(t)dt\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/743ac12018335afe1d99996fe3f05d5103a314a3)
![{\displaystyle y_{5}(t)=\int _{-\infty }^{t}[e^{-5\tau }+\cos(10\pi \tau )]{\dot {\delta }}(\tau )d\tau \,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/130dc24a3dec6ef8523a525a0d4c3033df9464b8)
![{\displaystyle \int _{-\infty }^{\infty }[\delta (t-t_{o}).f(t)]dt=f(t_{o}).\int _{-\infty }^{\infty }\delta (t-t_{o})dt=f(t_{o})\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d01ed5e952d0b98fe45e59f1ed29bbea2ab4af5)
![{\displaystyle y_{4}(t)=\int _{-\infty }^{t}[4-\tau ^{2}]\delta (\tau +3)d\tau =5\int _{-\infty }^{t}\delta (\tau +3)d\tau ;y_{4}(t)=5,(t>3)\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a8a062aa2d0ec064a2bebc442081ed8d8c93a5b)
![{\displaystyle \int _{-3}^{6}(6-t^{2})[\delta (t+4)+2\delta (2t+4)]dt=\int _{-3}^{6}-10\delta (t+4)+2\delta (t+2)dt=2\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d2da069507e2997c1bd6f6ad80e781c5d0635e6)
![{\displaystyle \int _{-\infty }^{\infty }[{\dot {\delta }}(t)]dt=0\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f86cf9edcf9e4fd0381fadc69b9a46833c632be8)
![{\displaystyle \int _{-\infty }^{\infty }[{\dot {\delta }}(t).f(t)]dt=\int _{-\infty }^{\infty }[{\dot {\delta }}(t).f(o)]dt-\int _{-\infty }^{\infty }[{\dot {f}}(t).\delta (t)]dt=-{\dot {f}}(o)\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41eee6d81ef3f9da168beeee14c96860de52de07)
![{\displaystyle \int _{-\infty }^{\infty }[e^{-5t}+\cos(10\pi t)]{\dot {\delta }}(t)dt=5\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e547f8867aa4ab1d3963afbbb68a2e4bf25aa5c)
![{\displaystyle y_{5}(t)=\int _{-\infty }^{t}[e^{-5\tau }+\cos(10\pi \tau )]{\dot {\delta }}(\tau )d\tau =\int _{-\infty }^{t}2{\dot {\delta }}(\tau )+5\delta (t)d\tau =5(t=0);0(t\neq 0)\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c44de5cb71eb0c4f9f890bdcdc4e9249f255bf4c)
![{\displaystyle \delta [\alpha (t-\beta )]=\left|{\frac {1}{\alpha }}\right\vert \delta (t-\beta )\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18805ce27d72343233a633598ee7b5bb012b11c7)
![{\displaystyle x(t)=\sum _{k=-\infty }^{\infty }\Pi _{1}(t-8k)=\sum _{k=-\infty }^{\infty }[u(t-8k+0.5)-u(t-8k-0.5)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69e120a5603e58ae16575d4e2e0a6b671bfa351a)
![{\displaystyle y(t)=g(t)\,x(t)=\sin(2\pi t)\sum _{k=-\infty }^{\infty }[u(t-8k+0.5)-u(t-8k-0.5)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3ed1a3f485074c32a99d2e90e713c7766847ef4)
![{\displaystyle y(t)=\sum _{k=-6}^{6}{\frac {e^{-{\frac {k^{2}}{8}}}}{2{\sqrt {2\pi }}}}\ [u(t-k+0.1)-u(t-k-0.1)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/58812ec80243209c147b49ed7e6798be204a964a)
![{\displaystyle x_{1}(t)=[\cos({\frac {\pi }{2}}t)+\cos(\pi t)+\cos(3\pi t)]^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/affc6639e5ac313e3e08c89dad2c4d02c96bf361)
![{\displaystyle {}_{\!}E_{x}=\int _{T}\left|x(t)\right\vert ^{2}\,dt=\int _{0}^{4}[\cos({\frac {\pi }{2}}t)+\cos(\pi t)+\cos(3\pi t)]^{2}\,dt=6}](https://wikimedia.org/api/rest_v1/media/math/render/svg/28a3c2aad5fe072b2d99aaa9e2e536621aab76b8)
![{\displaystyle {}_{\!}X_{\mbox{pro}}={\frac {1}{T}}\int _{T}x(t)\,dt=\int _{0}^{4}[\cos({\frac {\pi }{2}}t)+\cos(\pi t)+\cos(3\pi t)]\,dt=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9516314313f6d4db4952e9d396401c7063a2d360)
![{\displaystyle {}_{\!}X_{\mbox{rms}}={\sqrt[{}]{{}_{\!}P_{x}}}={\sqrt[{}]{\frac {3}{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/856251c867d7d0eec5e9bc453507bcf02e5329eb)
![{\displaystyle \int _{-\infty }^{\infty }[e^{-5t}-\sin(5\pi t)]{\dot {\delta }}(-2t+8)dt\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/03ae20314f880a74c6fd95a0aebf93f876174da4)
![{\displaystyle x(t)=\int _{-\infty }^{t}[e^{-5\tau }-\sin(5\pi \tau )]{\dot {\delta }}(-2\tau +8)d\tau \,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b3db52898a0145bba04d5ccb1425ea8e4e64bd9)
![{\displaystyle [e^{-5t}-sen(5\pi t)]\delta ^{\prime }(-2t+8)={\frac {1}{-4}}[e^{-5t}-sen(5\pi t)]\delta ^{\prime }(t-4)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c5b0cfbdddd84c482cf82409016f6cec56defb95)
![{\displaystyle {\frac {1}{-4}}\int _{-\infty }^{\infty }[e^{-5t}-\sin(5\pi t)]{\dot {\delta }}(t-4)dt=-{\frac {1}{4}}[-5e^{-5t}-5\pi cos(5\pi t)]_{T=4}={\frac {5e^{-20}}{4}}+{\frac {5\pi }{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed450b499d4b63823eac78802a5b90e39508252e)
![{\displaystyle {\frac {1}{-4}}[e^{-5t}-sen(5\pi t)]\delta ^{\prime }(t-4)={\frac {e^{-20}}{-4}}\delta ^{\prime }(t-4)-[{\frac {-5e^{-20}-5\pi }{4}}]\delta (t-4)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a05b7e591383221e113970ed1170981d05ca828)
![{\displaystyle \int _{-\infty }^{t}[{\frac {5e^{-20}+5\pi }{4}}]\delta (t-4)=[{\frac {5e^{-20}+5\pi }{4}}]U(t-4)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3133c5b533aa20a1e916bce56100854b3f171fee)
