Analisis Rangkaian Listrik Jilid-2 - Ee-cafe.org

Analisis Transien Rangkaian Orde-1
Tanggapan lengkap : v = 4cos10t
+ 8sin10t
+ A e
Kondisi awal v(0
) = 10 →10
= 4 + A → A
Jadi : v = 4cos10t
+ 8sin10t
+ 6 e
+
Arus kapasitor : i
C
= C
dv
dt
=
1
30
0
−5
t
V
= 6
−5
t
( − 40sin10t
+ 80cos10t
− 30 e )
= −1,33sin10
t + 2,33cos10t
− e
0
−5t
0
−5
t
A
CONTOH-1.12: Carilah tegangan kapasitor pada contoh 1.10. jika v s =
50cos(10t + θ)u(t) V dan kondisi awalnya adalah v(0 + ) = 10 V.
Solusi :
dv
Persamaan rangkaian : + 5v
= 100cos(10t
+ θ)
dt
= 100cosθcos10t
− 100sin θsin10t
−5
t
Tanggapan alami : va
= A0e
(sama seperti contoh 4.10.)
Dugaan tanggapan paksa : v p = Ac
cos10t
+ As
sin10t
Substitusi tanggapan paksa dugaan ini ke persamaan rangkaian
memberikan:
− 10Ac
sin10t
+ 10As
cos10t
+ 5Ac
cos10t
+ 5As
sin10t
= 100cosθcos10t
− 100sin θsin10t
→ −10Ac
+ 5As
= −100sin
θ dan 10As
+ 5Ac
= 100cosθ
→ As
= −20sin
θ + 2Ac
dan − 200sin θ + 20Ac
+ 5Ac
= 100cosθ
⇒ Ac
= 4cosθ + 8sin θ dan As
= −4sin
θ + 8cosθ
−5
t
Tanggapan lengkap : v = 4cos(10t
+ θ)
+ 8sin(10t
+ θ)
+ A0e
+
Kondisi awal v(0
) = 10 → 10 = 4cosθ + 8sin θ + A0
→ A0
= 10 − (4cosθ + 8sin θ)
−5
t
Jadi : v = 4cos(10t
+ θ)
+ 8sin(10t
+ θ)
+ (10 − 4cosθ − 8sin θ)
e
24 Sudaryatno Sudirham, Analisis Rangkaian Listrik (2)