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Electrical Engineering and Electronics II
Chapter 4
Transients
Scott
2008.9
•Main Contents
1. Solve first-order RC or RL circuits.
2. Understand the concepts of transient
response and steady-state response.
3. Relate the transient response of first-order
circuits to the time constant.
4. Solve RLC circuits in dc steady-state
conditions.
•Main Contents
 Introduction
 Initial
state and DC Steady State
 First-order
RC Circuits
 First-order
RL Circuits
 Summary
4.1 Introduction
 Conception of steady state and transient state R
K
R
+
+
_
E
uC
uC
Us
_
C
New steady state
When t=∞, uc(∞)=Us
Old steady state
When t=0，uc(0)=0
utransient
C
New steady state
E
Old steady state
t
 Why the transient response happens?
Resistance circuit
K
t=0
I
+
E
_
I
R
No transient
•Resistor is a energy-consumption element, current is
proportional to voltage, no transient response will happen
even if changing source
Energy
can not change instantly because
of accumulating or decaying period.
1
2
Electric field energy（Wc  Cu ）
C
2
or
Change
u
WC Charging
discharging
C
gradually
K
+
_E
R
uC
E
uC
C
t
Energy
can not change instantly because
of accumulating or decaying period.
1 2
Magnetic field energy （W 
LiL ）
L
2
WL
i Change
L gradually
K
R
iL
E/R
+ t=0
E
_
iL
t
Transients
•The time-varying currents and voltages resulting from
the sudden application of sources, usually due to
switching.
•By writing circuit equations, we obtain
integrodifferential equations.
The
causes of transients:
1. Energy storage elements
-inductors and capacitors
uC , iL
change gradually；
2.Changing circuit, such as
switching source.
4.2 Initial state and steady state
t=0
t=0-
t=0+
t
Assume changing circuit when t=0, then t=0– is end point of old
steady state; t=0+ is the start point of transient state.
WL (0  )  WL (0  )

WC (0  )  WC (0  )
 i L ( 0 )  i L ( 0 )

uC (0 )  uC (0 )
The law of
changing circuit
From t=0–to t=0+,iL 、 uC
change continuously.
DC Steady State Response

•The steps in determining the forced response
or steady state response for RLC circuits with
dc sources are:
1. Replace capacitances with open circuits.
2. Replace inductances with short circuits.
3. Solve the remaining circuit.
Example 4.1 Find steady-state values of vx and ix
in this circuit for t>>0.
Answer: vx =5V, ix = 1A
t>>0
Exercise 4.3 Find steady-state values of labeled
currents and voltages for t>>0.
Answer: va =50V, ia = 2A
i1 = 2A, i2=1A, i3=1A
How to get initial value

Exercise 1: Assuming old circuit is in DC steady state
before switch K is closed. how to get uC(0+),iR(0+)?
Solution:
When t=0-, capacitor is
considered as open circuit,
we get equivalent circuit.
8
uC (0 ) 
12  8 V
48
R1
4k
K
t=0
12V
iR
8k
R2
2mF
uC
R1
4k
12V
8k
uC(0–)
t=0-
How to get initial value

R1
4k
K
t=0
12V
iR
8k
R2
2mF
R1
4k
12V
uC
8
uC (0 ) 
12  8 V
48
u C (0  )  u C (0  )  8V
substituting voltage source
for uC(0+)
8k
uC(0–)
uC (0) 8
iR (0) 
  1m A
R2
8
i (0+)
R
R 8k
2
+
u (0+)
–C
t=0＋
How to get initial value

•Exercise 2: Given by R1=4Ω, R2=6Ω, R3=3Ω, C=0.1µF,
L=1mH, US=36V, switch S is closed for a long time.
Open the switch S when t=0, how to get the initial values
of all elements?
4.3 First-order RC Circuits


First-order circuit
Only one (equivalent) capacitor or inductor is
included in a linear circuit.
Equivalent circuit of First-order circuit
Two parts: one (equivalent) capacitor or inductor;
a two terminal network with resistance and sources.
N
L
or
N
C
4.3 First-order RC Circuits

According to Thevenin Law
L
N
C
N
or
iC
iL
+
-
R
U
+
uL
L
-
R
U
uC
C

Differential equation of first-order RC circuit
iL
+
iC
R
U
uL
L
-
uR  uL  U
di L ( t )
RiL ( t )  L
U
dt
L diL (t )
U
 iL (t ) 
R dt
R
+
R
U
uC
C
-
uR  uC  U
duC
RC
 uC  U
dt

First-order RC Circuits
•Example: to find the transient response after
changing circuit when t=0.
S (t  0)
iC
2

Solution:
1
R
uR
uC
uR
i
0
0
0
0
0
US
US
R

US
0
0
t
0
US

f (t )
C
uC (0 )  0
uC

First-order RC Circuits
uR  uC  U S
S (t  0)
iC
2

1
R
uR
C
uC
US

uC (0 )  0
u R  Ri
duC
iC
dt
duC
RC
 uC  U S
dt
uC (0 )  uC (0 )  0

First-order RC Circuits
duC
RC
 uC  U S
dt
S (t  0)
iC
2

1
R
uR
C
uC
US

uC  u  u
'
st
uC  Ae
'
C
"
C
——homogeneous solution
u
"
C
——particular solution

First-order RC Circuits

homogeneous solution
duC
RC
 uC  U S
dt
S (t  0)
iC
2

1
R
uR
US

C
uC
RCs  1  0
1
s
RC
uC 
1
 t
Ae RC

First-order RC Circuits
Particular solution

S (t  0)
iC
2

1
Therefore
R
uR
C
uC
US

duC
RC
 uC  U S
dt
u  uC ( )  U S
"
C
Then, the final solution is
uC  u  u  Ae  U S
'
C
"
C
st

First-order RC Circuits

The solution of differential equation
uC  u  u  Ae  U S
'
C
"
C
st
Substituting the initial condition：
uC (0 )  u  u  Ae  U S  0
'
C
"
C
s0
A  uC (0)  uC ()  U S
uC (t )  uC ()  [uC (0)  uC ()]e
 U S U S
1

t
e RC

1
t
RC

First-order RC Circuits

The solution of differential equation
  RC
——Time constant
uC (t )  uC ()  [uC (0)  uC ()]e
uC ()
——Steady state value
uC (0)
——Initial value

t


Three elements method

Solution of other parameters
t
uR (t )  U S  uC (t )  U S e


 uR (0)e
 uR ()  [uR (0)  uR ()]e
t

uR (t ) U S t
i (t ) 

e  i (0)e 
R
R
 i ()  [i(0)  i()]e



t

t

t

Three elements: 1.steady state value f(∞);
2.time constant τ; 3. initial value f(0+).
4.3 First-order RL Circuits

Formula of Three element method：
f ( t )  f ( )  [ f (0 )  f ( )]e

t

f(∞)——steady state value
f(0+)——initial value
τ——time constant
τ=RC ——time constant of RC circuit
τ= ？？ ——time constant of RL circuit
4.3 First-order RL Circuits
4.3 First-order RL Circuits

Time constant
τ=RC
iC
+
R
uC
U
-
C
iL
+
-
R
U
uL
L
duC
RC
 uC  U
dt
L diL (t )
U
 iL (t ) 
R dt
R
τ=L/R
• Time constant reflects the length of transient period.
t
e-t/

2
36.8% 13.5%
3
4
5
5%
1.8%
0.3%
6
7
0.25% 0.09%
•After one time constants, the transient response is equal to
36.8 percent of its initial value.
•After about five time constants, the transient response is over.
• Time constant reflects the length of transient period.

The curves versus time
u(t )
US
i (t )
Mounting
curve
uC (t )
0.632U S
US
R
0.368U R (0)
uR (t )i (t )
0
 2
The initial slop
intersects the final
value at one time
constant.
Decaying
curve
t
•Three element method





Steps
Initial value: t=0-→t=0+ f(0+)
Steady state value: t =∞ f(∞)
Time constant : τ=RC τ=L/R
Substituting three elements
f ( t )  f ( )  [ f (0 )  f ( )]e

Draw the curve versus time

Limited Condition:
1) first-order circuit
2) DC source

t

•Example 4.2 Find voltage of v(t) and current i(t) in this
circuit for t>0.

Answer:
t

t
i (t )  2  2e  (A), v(t )  100e  (V)
L 0.1
 
 2(ms)
R 50

t

t
i (t )  2  2e  (A), v(t )  100e  (V)
L 0.1
 
 2(ms)
R 50
•Example 4.3 Find voltage of v(t) and current i(t) in this
circuit for t>0.
Answer:
Vs  t
LVS  t
i (t )  e , v(t )  
e
R1
R1
L

R2
Vs  t
LVS  t
i (t )  e , v(t )  
e
R1
R1
L

R2
•Exercise 4.5 Find voltage of v(t) and current iR(t) , iL(t) in
this circuit for t>0, assume that iL(0)=0.
Answer:

t

t

t
iR (t )  2e  ( A), iL (t )  2  2e  ( A), v(t )  20e  (V )
  0.2( s)
•Exercise 4.5 Find voltage of v(t) and current i(t), v(t) in
this circuit for t>0, assume that the switch has been closed
for a very long time prior to t=0.
Answer:
1, t  0

i (t )  
 t
0.5  0.5e  , t  0
 1, t  0
v(t )  
 t
100e  , t  0
  5(ms)
•Homework 4




P4.8
P4.18
P4.26
P4.30