2. EL2093
MTHTransient
• Terjadi bila sistem dengan energi potersial mapan
mengalami gangguan dan mencari kesetimbangan baru
• Contoh, dua tanki dihubungkan dengan keran yang
dibuka pada t=0
• Apa yang menentukan kecepatan aliran?
2
h1(t)
h2(t)
3. EL2093
MTHTransient
• Kapasitor yang mempunyai beda potensial V, pada t=0
dihubungkan ke resistor
3
• t=0
• muatan Q = Cv
• arus i = v/R
• arus i mengurangi Q
• kapasitansi tetap C
• arus i menurunkan v
• turunnya v menurunkan
arus i
• Arus i dan tegangan v turun
• Perubahan arus i dan v mengecil sejalan dengan waktu
4. EL2093
MTHTransient
• Perhatikan rangkaian berikut
• Dari rangkaian diketahui juga
• Dengan demikian
atau sehingga
4
v(t)
Q(t)=Cv(t)
iR(t)=v(t)/R
R
C1
iR(t)
+
-
iC(t)
iC(t)=-iR(t)
iC(t)=-v(t)/R
iC(t)=dQ(t)/dt v(t)=Q(t)/C
dQ/dt=-Q(t)/CR
v(t) = V0exp(-t/CR)
Cdv/dt=-Cv/CR
dv/dt=-v/CR
9. EL2093
MTHTransient
• Bagaimana dengan rangkaian berikut
• Arus mengalir bila tegangan v1≠v2
• Muatan DQ keluar C1 sama dengan DQ diterima C2
9
v1(t) v2(t)
Q1(t) Q2(t)
R
C1 C2
iR(t)
10. EL2093
MTHTransient
• Susun ulang ekivalennya
• Bentuk solusi dapat dipastikan
• Pertanyaannya Vawal dan Vakhir
10
v1(t)
v2(t)
Q1(t)
Q2(t)
R
C1
C2
iR(t)
+
-
+
-
v(t)
Q(t)
R
C
iR(t)
+
-
v(t)=(Vawal -Vakhir)exp(-t/CR)+Vakhir
11. EL2093
MTHAnalisis Transient (Orde 1)
• Persamaan diferensial orde saru
– dari persamaan muatan
– dari KCL
• Solusi langsung persamaan tiga
– Tegangan awal
– Tegangan akhir
– Konstanta waktu
• Transformasi Laplace
11
12. EL2093
MTHTransient Tanpa Sumber
Untuk rangkaian RC tanpa sumber solusi berbentuk:
12
1. Tegangan awal kapasitor v(0) = V0.
2. Konstanta waktu = RC.
/
0)( t
eVtv
CRdimana
13. EL2093
MTHContoh
Example 1
Refer to the circuit below, determine vC, vx, and io for t ≥
0.
Assume that vC(0) = 30 V.
13
• Please refer to lecture or textbook for more detail elaboration.
Answer: vC = 30e–0.25t V ; vx = 10e–0.25t ; io = –2.5e–0.25t A
14. EL2093
MTHContoh
Example 2
The switch in circuit below is opened at t = 0, find v(t)
for t ≥ 0.
14
• Please refer to lecture or textbook for more detail elaboration.
Answer: V(t) = 8e–2t V
15. EL2093
MTHRangkaian RL Tanpa Sumber
• A first-order RL circuit consists of a inductor L (or its
equivalent) and a resistor (or its equivalent)
15
0 RL vvBy KVL
0 iR
dt
di
L
I-V induktor Hukum Ohm
dt
L
R
i
di
LtR
eIti /
0)(
17. EL2093
MTHRangkaian RL Tanpa Sumber
17
/
0)( t
eIti
R
L
RL tanpa sumber
dimana /
0)( t
eVtv
RC
RC tanpa sumber
dimana
Perbandingan rangkaian RL dan RC tanpa sumber
18. EL2093
MTHRangkaian RL Tanpa Sumber
Bentuk solusi rangkaian RL tanpa sumber:
18
1. Arus awal induktor i(0) = I0.
2. Konstanta waktu = L/R.
/
0)( t
eIti
R
L
dimana
20. EL2093
MTHKonstanta Waktu (1)
• Kapasitor: energi W=½
CV2
– di simpan dalam medan
listrik – polaritas muatan
– sebanding dengan
kapasitasi: semakin besar
kapasitansi semakin besar
energi tersimpan
– lama perpindahan
sebanding dengan
kapasitansi
• Induktor: energi
W=½ LI2
– di simpan dalam medan
magnit – fluks magnit
– sebanding dengan
induktansi: semakin besar
kapasitansi semakin besar
energi tersimpan
– lama perpindahan
sebanding dengan
induktansi
20
21. EL2093
MTHKonstanta Waktu (2)
• Kapasitor:
– besaran energi berubah bila bila muatan listrik
berubah
– arus sebanding dengan perubahan muatan (i=dq/dt):
semakin besar arus semakin cepat besaran energi
berubah
– arus berbanding terbalik dengan resistansi (i=V/R):
perubahan muatan cepat untuk arus besar atau
resistansi kecil
– konstanta waktu kecil (cepat) bila resistansi kecil atau
konstanta waktu sebanding dengan resistansi
21
22. EL2093
MTHKonstanta Waktu (3)
• Induktor:
– besaran energi berubah bila bila fluks magnet
berubah
– tegangan sebanding dengan perubahan fluks
(v=df/dt): semakin besar tegangan semakin cepat
besaran energi berubah
– Tegangan sebanding dengan resistansi (v=iR):
perubahan fluks cepat untuk tegangan besar atau
resistansi besar
– konstanta waktu kecil (cepat) bila resistansi besar
atau konstanta waktu berbanding terbalik dengan
resistansi
22
23. EL2093
MTHKonstanta Waktu (4)
RC
• Jumlah energi
– W sebanding C
– sebanding C
• Laju perubahan
– berbanding terbalik DQ
– DQ sebanding i
– i berbanding terbalik R
– sebanding R
• Konstanta Waktu: CR
RL
• Jumlah energi
– W sebanding L
– sebanding L
• Laju perubahan
– berbanding terbalik Df
– Df sebanding v
– v sebanding R
– berbanding terbalik R
• Kontanta waktu: L/R
23
24. EL2093
MTHContoh
Example 3
Find i and vx in the circuit.
Assume that i(0) = 5 A.
24
• Please refer to lecture or textbook for more detail elaboration.
Answer: i(t) = 5e–53t A
25. EL2093
MTHContoh
Example 4
For the circuit, find i(t) for t > 0.
25
• Please refer to lecture or textbook for more detail elaboration.
Answer: i(t) = 2e–2t A
26. EL2093
MTH7.3 Unit-Step Function (1)
• The unit step function u(t) is 0 for negative values of
t and 1 for positive values of t.
26
0,1
0,0
)(
t
t
tu
o
o
o
tt
tt
ttu
,1
,0
)(
o
o
o
tt
tt
ttu
,1
,0
)(
28. EL2093
MTH
7.4 The Step-Response
of a RC Circuit (1)
• The step response of a circuit is its behavior when the excitation
is the step function, which may be a voltage or a current source.
28
• Initial condition:
v(0-) = v(0+) = V0
• Applying KCL,
or
• Where u(t) is the unit-step function
0
)(
R
tuVv
dt
dv
c s
)(tu
RC
Vv
dt
dv s
29. EL2093
MTH
7.4 The Step-Response
of a RC Circuit (2)
• Integrating both sides and considering the initial
conditions, the solution of the equation is:
29
0)(
0
)( /
0
0
teVVV
tV
tv t
ss
Final value
at t -> ∞
Initial value
at t = 0
Source-free
Response
Complete Response = Natural response + Forced Response
(stored energy) (independent source)
= V0e–t/τ + Vs(1–e–t/τ)
30. EL2093
MTH
7.4 The Step-Response
of a RC Circuit (3)
Three steps to find out the step response of an
RC circuit:
30
1. The initial capacitor voltage v(0).
2. The final capacitor voltage v() — DC
voltage across C.
3. The time constant .
/
)]()0([)()( t
evvvtv
Note: The above method is a short-cut method. You may also
determine the solution by setting up the circuit formula directly
using KCL, KVL , ohms law, capacitor and inductor VI laws.
31. EL2093
MTH
7.4 The Step-Response
of a RC Circuit (4)
Example 5
Find v(t) for t > 0 in the circuit in below. Assume the
switch has been open for a long time and is closed at t
= 0.
Calculate v(t) at t = 0.5.
31
• Please refer to lecture or textbook for more detail elaboration.
Answer: and v(0.5) = 0.5182V515)( 2
t
etv
32. EL2093
MTH
7.5 The Step-response
of a RL Circuit (1)
• The step response of a circuit is its behavior when the excitation
is the step function, which may be a voltage or a current source.
32
• Initial current
i(0-) = i(0+) = Io
• Final inductor current
i(∞) = Vs/R
• Time constant = L/R
)()()( tue
R
V
I
R
V
ti
t
s
o
s
33. EL2093
MTH
7.5 The Step-Response
of a RL Circuit (2)
Three steps to find out the step response of an
RL circuit:
33
1. The initial inductor current i(0) at t = 0+.
2. The final inductor current i().
3. The time constant .
Note: The above method is a short-cut method. You may also
determine the solution by setting up the circuit formula directly
using KCL, KVL , ohms law, capacitor and inductor VI laws.
/
)]()0([)()( t
eiiiti
34. EL2093
MTH
7.5 The Step-Response
of a RL Circuit (4)
Example 6
The switch in the circuit shown below has been closed
for a long time. It opens at t = 0.
Find i(t) for t > 0.
34
• Please refer to lecture or textbook for more detail elaboration.
Answer: t
eti 10
2)(