Chapter 4Transients
2008.9
Electrical Engineering and Electronics II Electrical Engineering and Electronics II
Scott
•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.
Introduction
Initial state and DC Steady State
First-order RC Circuits
First-order RL Circuits
Summary
•Main Contents
t
E
Cu New steady statetransient
C
Old steady statesteady state
K R
E+
_ CuSwitch K is closed
4.1 Introduction
New steady steady statestate
R
Us+
_ Cu
Conception of steady state and transient state
When t=0 , uc(0)=0
When t=∞, uc(∞)=Us
Old steady state
Why the transient response happens?
No transient
I
Resistance circuit
t = 0
E R
+
_
I
K
•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.
CW Charging or discharging Cu Change
gradually
Electric field energy )( 2
2
1CCuWc
E
K R
+
_ CuC
E
t
Cu
Magnetic field energy )( 2
2
1LL LiW
LWLi Change
gradually
K R
E+
_
t=0iL
t
LiE/R
Energy can not change instantly because of accumulating or decaying period.
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
change gradually;
2.Changing circuit, such as switching source.
LC iu ,
4.2 Initial state and steady state
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.
)0()0(
)0()0(
CC
LL
WW
WW
)0()0(
)0()0(
CC
LL
uu
iiFrom t=0–to t=0+,iL 、 uC
change continuously.
t=0tt=0
-
t=0+
The law of changing circuit
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 valueExercise 1: Assuming old circuit is in DC steady state
before switch K is closed. how to get uC(0+),iR(0+)?
iR
R14k
12V
K
t=08kR2
2FuC
Solution:
When t=0-, capacitor is considered as open circuit, we get equivalent circuit. R1
4k
12V uC(0–)8k
t=0-
8(0 ) 12 8V
4 8Cu
R14k
12V uC(0–)8k
iR
R14k
12V
K
t=08kR2 2F
uC
8(0 ) 12 8V
4 8Cu
Vuu CC 8)0()0(
substituting voltage source for uC(0+)
iR(0+)
8kR2
+
– u
C(0+)
t=0 +2
(0 ) 8(0 ) 1m A
8C
R
ui
R
How to get initial value
•Exercise 2: Given by Exercise 2: Given by RR11=4Ω, =4Ω, RR22=6Ω, =6Ω, RR33=3Ω, =3Ω, CC=0.1µF, =0.1µF,
LL=1mH, =1mH, UUSS=36V, switch S is closed for a long time. =36V, switch S is closed for a long time.
Open the switch S wOpen the switch S whenhen t=0, how to get the initial values t=0, how to get the initial values
of all elements?of all elements?
How to get initial value
Equivalent circuit of First-order circuit
Two parts: one (equivalent) capacitor or inductor; a two terminal network with resistance and sources.
N L N Cor
First-order circuit
Only one (equivalent) capacitor or inductor is included in a linear circuit.
4.3 First-order RC Circuits
According to Thevenin Law
N L N Cor
RU LuL
iL
+
-
RU CuC
iC
+
-
4.3 First-order RC Circuits
Differential equation of first-order RC circuit
R
U LuL
iL
+
-
R
U CuC
iC
+
-
Uuu CR
Uudt
duRC C
C
Uuu LR
Udt
tdiLtRi L
L )(
)(
R
Uti
dt
tdi
R
LL
L )()(
)0( tS
R
C
Ci
Cu
SU
2
1Ru
0)0( Cu
Solution:
0
Cu Ru i
0
t)(tf
0 0 0
0 SU
SU
R
U S
0 0
First-order RC Circuits•Example: to find the transient response after changing circuit when t=0.
)0( tS
R
C
Ci
Cu
SU
2
1Ru
SCR Uuu
dt
duCiRiu C
R
SCC Uudt
duRC
0)0( Cu 0)0( Cu 0)0()0( CC uu
First-order RC Circuits
)0( tS
R
C
Ci
Cu
SU
2
1Ru
"'CCC uuu
stC Aeu '
"Cu
——homogeneous solution
——particular solution
SCC Uudt
duRC
First-order RC Circuits
homogeneous solution
)0( tS
R
C
Ci
Cu
SU
2
1Ru
SCC Uudt
duRC
01 RCs
RCs
1
tRC
C Aeu1
First-order RC Circuits
)0( tS
R
C
Ci
Cu
SU
2
1Ru
SCC Uudt
duRC
Therefore
SCC Uuu )("
Then, the final solution is
Sst
CCC UAeuuu "'
Particular solution
First-order RC Circuits
The solution of differential equation
Sst
CCC UAeuuu "'
Substituting the initial condition:
0)0( 0"' Ss
CCC UAeuuu
SCC UuuA )()0(
tRC
SS
tRC
CCCC
eUU
euuutu1
1
)]()0([)()(
First-order RC Circuits
The solution of differential equation
RC ——Time constant
t
CCCC euuutu
)]()0([)()(
)(Cu
)0( Cu
——Steady state value
——Initial value
First-order RC Circuits
Solution of other parameters
( ) ( ) (0 )
( ) [ (0 ) ( )]
t t
R S C S R
t
R R R
u t U u t U e u e
u u u e
( )
( ) (0 )
( ) [ (0 ) ( )]
t tSR
t
Uu ti t e i e
R R
i i i e
Three elements method
Three elements: 1.steady state value f(∞); 2.time constant τ; 3. initial value f(0+).
Formula of Three element method :
t
effftf
)]()0([)()(
f(∞)——steady state value
τ——time constant
f(0+)——initial value
τ=RC ——time constant of RC circuitτ= ?? —— time constant of RL circuit
4.3 First-order RL Circuits
4.3 First-order RL Circuits
Time constant
τ=RC
τ=L/R
Uudt
duRC C
C
R
Uti
dt
tdi
R
LL
L )()(
RU LuL
iL
+
-
RU CuC
iC
+
-
4.3 First-order RL Circuits
• Time constant reflects the length of transient period.
t 2 3 4 5 6 7
e-t/ 36.8% 13.5% 5% 1.8% 0.3% 0.25% 0.09%
•After about five time constants, the transient response is over.
•After one time constants, the transient response is equal to 36.8 percent of its initial value.
The curves versus time
SU
)(tuC
)(tuR
R
U S
0t
)()( titu
)(ti
SU632.0
)0(368.0 RU
2
The initial slop intersects the final value at one time constant.
Mounting curve
Decaying curve
• Time constant reflects the length of transient period.
•Three element method
Initial value: t=0-→t=0+ f(0+) Steady state value: t =∞ f(∞) Time constant : τ=RC τ=L/R Substituting three elements
Draw the curve versus time
t
effftf
)]()0([)()(
Steps
Limited Condition:
1) first-order circuit2) DC source
•Example 4.2 Find voltage of v(t) and current i(t) in this circuit for t>0.
Answer:( ) 2 2 (A), ( ) 100 (V)
0.12(ms)
50
t t
i t e v t e
L
R
( ) 2 2 (A), ( ) 100 (V)
0.12(ms)
50
t t
i t e v t e
L
R
•Example 4.3 Find voltage of v(t) and current i(t) in this circuit for t>0.
Answer: 1 1
2
( ) , ( )t t
s SV LVi t e v t e
R R
L
R
1 1
2
( ) , ( )t t
s SV LVi t e v t e
R R
L
R
•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:
)(2.0
)(20)(),(22)(),(2)(
s
VetvAetiAetitt
L
t
R
•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, 0 1, 0( ) ( ) 5( )
0.5 0.5 , 0 100 , 0t t
t ti t v t ms
e t e t
P4.8 P4.18 P4.26 P4.30
•Homework 4
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