problema 5 (a)y(b)
Transcript of problema 5 (a)y(b)
-
8/17/2019 problema 5 (a)y(b)
1/3
(a)
Planteamos la ecuación para el potencial E(t) para t>0
E ( t )=10e−3 t u ( t )+5 e−3 (t −2) u (t −2)
Hallamos la Transformada de Laplace:
L [ E (t ) ] ( s )= L [10e−3 t u ( t )+5e−3 (t −2 )u ( t −2 ) ](s)
Por la propiedad de linealidad
L [ E (t ) ] ( s)= L [10e−3 t u ( t ) ] (s )+ L [5e−3 (t −2) u ( t −2 ) ](s)
Aplicamos la formula: L [ f ( t −a ) u (t −a ) ]=e−as
F (s )
f ( t )=10e−3 t
g ( t )=5 e−3 (t −2)→ g ( t +2 )=5e−3 t → →h ( t )=g (t +2 )=5e−3 t
-
8/17/2019 problema 5 (a)y(b)
2/3
∴ L [ E ( t ) ] (s )= L [10e−3 t u ( t ) ] ( s)+ L [5e−3 ( t −2) u ( t −2 ) ] ( s)
L [ E (t ) ] ( s )= L [ f ( t ) u ( t ) ] (s )+ L [h (t −2 ) u (t −2 ) ] ( s )
L [ E (t ) ] ( s)=10 L [e−3 t ] (s )+5e−2 s L [e−3 t ] ( s)
L [ E (t ) ] ( s)= 10s+3
+5e
−2 s
s+3=
10+5e−2 s
s+3
(b)
Aplicamos la 1ra Ley de irc!o" a la primera malla
E ( t )=2 z ( t )+1 dx ( t )
dt =2 z ( t )+
dx (t )dt
Aplicamos la 1ra Ley de irc!o" a la se#unda malla
3 y ( t )+1dy (t )dt
−1 dx ( t )dt
=0
dx (t )dt
=3 y ( t )+dy (t )
dt
Aplicamos la $da Ley de irc!o" al nodo
z ( t )= x (t )+ y (t )
%inalmente planteamos el sistema de ecuaciones donde no !ay corriente en
t&0
-
8/17/2019 problema 5 (a)y(b)
3/3
{ E (t )=2 z ( t )+ x ' (t ) x
' (t )=3 y (t )+ y ' (t ) z ( t )= x (t )+ y( t )
z (0 )= y (0 )= x (0 )=0
z' (0)= y ' (0)= x ' (0 )=0