1 Chapter 5 Transient Analysis Jaesung Jang Complete response = Transient response + Steady-state response Time Constant First order and Second order Differential Equation .
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1
Chapter 5Transient Analysis
Jaesung Jang
Complete response = Transient response + Steady-state responseTime Constant
First order and Second order Differential Equation
.
2
Transient Analysis• The difference of analysis of circuits with
energy storage elements (inductors or capacitors) & time-varying signals with resistive circuits is that the equations resulting from KVL and KCL are now differential equations rather than algebraic linear equations resulting from the resistive circuits.
• Transient region: the region where the signals are highly dependent on time. (temporary)
– No voltage or current sources– Transient Analysis
• Steady-state region: the region where the signals are not time dependent (time rate of change of signals is equal to zero) or periodic.
– Constant signals– Sinusoidal signals 0 1 2 3 4 5 6 7 8 9 10
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
t (s ec )
V/V
s
( )0=
dt
d
3
Solution of Ordinary Differential Equation• Transient solution (xN) is a solution of the
homogeneous equation: transient (natural) response. -> temporary behavior without the source.
• Steady-state (particular) solution (xF) is a solution due to the source: steady-state (forced ) response.
• Complete response = transient (natural) response + steady-state (forced ) response -> x = xN + xF
• First order: The largest order of the differential equation is the first order.– RL or RC circuit.
• Second order: The largest order of the differential equation is the second order.– RLC or LC circuit.
sVxdt
dx =+
0=+ NN x
dt
dx
sFF Vx
dt
dx =+
0 1 2 3 4 5 6 7 8 9 100
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
t (sec)
V/V
s
4
Writing Differential Equations• Key laws: KVL & KCL for capacitor voltages or inductor currents
( ) ( )
CSCCCSC
CR
CSCCSCC
S
tC
CC
CRSCRS
CR
CR
vRC
v
RC
v
dt
dv
R
vv
dt
dvCi
R
v
iRdt
dv
RC
i
dt
di
dt
dv
C
iR
dt
di
vtdC
titvRi
vRivvvv
iR
vii
for equation alDifferenti:
for equation alDifferenti:
0
00 :KVL
:KCL
0
=+→−
===
=+→=+
=′′
+=+
=++−→=++−
=→=
∫
( ) ( ) ( )( )
equation aldifferentiordinary linear order -First
.parameterselement circuit of nscombinatio represents and ,, constants the
andcurrent inductor or the voltagecapacitor therepresents where
001
001
→
=+
baa
tx
tfbtxadt
tdxa
5
Writing Differential Equations (cont.)• Key laws: KVL & KCL
( ) ( )
CSCCC
CCS
C
CCSCLCSCC
R
SS
S
t
C
LCRSLCRS
LCR
vvvdt
dvRC
dt
vdLC
dt
vdLCvv
dt
dvRC
dt
dvC
dt
dL
RR
v
R
v
dt
dvC
R
vvv
dt
dvCi
R
v
iLdt
dv
LC
i
Ldt
Rdi
dt
id
dt
dv
dt
idL
C
iR
dt
di
vdt
diLtd
C
titviR
vvRivvvvv
iiii
for equation alDifferenti:
1
for equation alDifferenti:
0
00 :KVL
:KCL
2
2
2
2
2
2
2
2
0
=++
−−=
−−=→−−
===
=++→=++
=+′′+=+
=+++−→=+++−===
∫
( ) ( ) ( ) ( )( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( )ly.respective gain, DC theand
ratio, damping thefrequency, natural thetermed and 12, constants thewhere
21
.parameterselement circuit of nscombinatio represents and ,,, constants the
andcurrent or the voltagecapacitor therepresents where
equation aldifferentiordinary linear order -Second
0020120
2
2
20
0
0
12
2
0
2
0012
0012
2
2
abKaaaaa
tfKtxdt
tdx
dt
txdtf
a
btx
dt
tdx
a
a
dt
txd
a
a
baaa
tx
tfbtxadt
tdxa
dt
txda
Sn
Snn
===
=++→=++
→=++
ζω
ωζ
ω
6
Examples of Writing Differential Equations
( ) ( )
( ) ( ) ( ) ( )
LS
LLL
LS
L
tL
LSL
tL
LLS
Lt
LL
LSRL
R
LSRLRS
RLR
RLR
vdt
dvv
L
R
dt
dv
dt
dvv
L
R
dt
dv
vtdL
tRvtRivvtd
L
tRvtRivv
R
vtd
L
tvti
R
vvii
R
v
vvvvvv
iiR
viii
for equation alDifferenti:22
200
0
0 :KVL
:KCL
00
02
221
=+→+=
+′′
+==→+′′
+==−
+′′
+==−
→+=
−=→=++−
+=→+=
∫∫
∫
( )
( ) ( ) LSLLL
LL
LLL
S
LL
LLL
CRS
LLL
LLL
LC
LCRR
LL
LRCLRC
CRSCRS
LCR
iviRRdt
diLCRR
dt
idLCR
Ridt
diLiR
dt
diCRR
dt
idLCRv
Ridt
diLRi
dt
diCR
dt
idLCvvv
Ridt
diCR
dt
idLCRiRi
dt
diL
dt
dCRi
dt
dvCRiiRiv
Ridt
diLvvvvvv
vvvvvv
iii
for equation alDifferenti:
0
0 :KVL
:KCL
21212
2
1
21212
2
1
2122
2
122
2
12111
2
1
11
22
11
1
=++++
→++++=
→++
++=+=
++=
+
+=
+=+==
+=+=→=++−
+=→=++−
+=
7
DC steady state solution: Final Condition• Steady state solution due to AC (sinusoidal waveforms) is in Chap. 6 (frequency
response).• DC steady state solution: response of a circuit that have been connected to a DC
source for a long time or response of a circuit long after a switch has been activated.– All the time derivatives are equal to zero at the steady state.
• Capacitors: insulators (very large resistances) are inside the capacitors.• Inductors: Induction works only when the change in electric fields happens.
statesteady at the SC
SCC
vvRC
v
RC
v
dt
dv
=
=+ ( ) ( )
( ) statesteady at the 21
21212
2
1
RR
vi
viRRdt
diLCRR
dt
idLCR
SL
SLLL
+=
=++++
( ) ( )
( ) ( )
circuits.short as hehave inductors all and
circuitsopen as behave capacitors all
state,steady DCAt
as 0
as 0
∞→→=
∞→→=
tdt
tdiLtv
tdt
tdvCti
LL
CC
+
+ +
+
8
DC steady state solution: Initial Condition• Initial condition: response of a circuit before a switch is first activated.
– Since power equals energy per unit time, finite power requires continuous change in energy.
• Primary variables: capacitor voltages and inductor currents-> energy storage elements– Capacitor voltages and inductor currents cannot change instantaneously but
should be continuous. -> continuity of capacitor voltages and inductor currents
– The value of an inductor current or a capacitor voltage just prior to the closing (or opening) of a switch is equal to the value just after the switch has been closed (or opened).
( ) ( )( ) ( )
0"after just " signifies 0
and 0" beforejust " signifies 0notation thewhere
00
00
=
=
===
===
+
−
+−
+−
t
t
titi
tvtv
LL
CC
Discontinuous of capacitor voltage-> infinite power at t=0.
( ) ( ) ( ) ( )tCvtWtLitW CCLL22
2
1
2
1 ==
9
First Order Response• First-order circuit: one energy storage element + one energy loss
element (e.g. RC circuit, RL circuit)
• Procedures– Write the differential equation of the circuit for t=0+, that is, immediately
after the switch has changed. The variable x(t) in the differential equation will be either a capacitor voltage or an inductor current. You can reduce the circuit to Thevenin or Norton equivalent form.
– Identify the initial conditions x(t=0+) [= x(t=0-)] and final conditions x(t=∞).– Solve the differential equation.– Write the complete solution for the circuit in the form.
• The time constant (τ) is a measure of how fast capacitor voltages or inductor currents react to the input (voltage or current source). It is a period of time during which capacitor voltages or inductor currents change by 63.2% to get to the steady state.
( ) ( ) ( ) ( )[ ] ( )τttxtxtxtx −∞=−=+∞== exp0
( ) ( )[ ]( ) ( )[ ] 632.01
0
0 1 =−==−∞==−= −e
txtx
txtx τ
10
First Order Response (cont.)• First-order circuit: one energy storage element + one
energy loss element (e.g. RC circuit, RL circuit)( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( )( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 0for 00
DCfor
Response Complete
0 0
0: where DC todue Response Forced
constant. a is where0
Response Natural
ly.respective gain, DC andconstant time the termed and where
00
00
00
0001
0
0
0
1001
≥∞=−==→∞=+==+=∞=+=+=
≥=→≥=+
→=
=→−
=→=+
==
=+→=+→=+
−−
−
ttxtxxtxxtx
FKextxextxtxtx
tFKtxtFKtxdt
tdxdt
tdxFtf
xextxtx
dt
tdxtx
dt
tdx
abKaa
tfKtxdt
tdxtf
a
btx
dt
tdx
a
atfbtxa
dt
tdxa
Stt
FN
SFSFF
F
tN
NNN
N
S
S
ττ
τ
τ
ττ
τ
τ
11
Example: First Order Response 1
( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( )( ) ( ) ( ) 47.07120 :Step4
,1 ,47.0F470k1 , :Step3
V12,0 V 50 :Step2
0
00 :KVL
:KCL :Step1
tC
tCCC
SSC
SCCC
SSCCCSC
CR
CRSCRS
CR
CR
etvetvtvtv
vFKRCvx
vtvtvtv
FKtxdt
tdxtvv
dt
dvRC
R
vv
dt
dvCi
R
v
vRivvvv
iR
vii
−−
+−
−+=∞=+∞=−==
===×Ω=====∞=====
=+→>=+→−
===
=++−→=++−
=→=
τ
µτ
τ
+
12
Example: First Order Response 2
( ) ( )
( ) ( ) ( )
( ) ( ) ( )( ) ( ) ( ) 025.05.125.120 :Step4
,1 ,025.04H1.0 , :Step3
A5.12,0A 00 :Step2
0
00 :KVL
:KCL :Step1
tL
tLLL
BSL
BLLL
SB
LL
LLBLRB
LR
etietititi
vFRKRLix
Rvtititi
FKtxdt
tdxt
R
vi
dt
di
R
Ldt
diLRivvvv
ii
−−
+−
−+=∞=+∞=−==
===Ω=====∞=====
=+→>=+→
=++−→=++−
=
τ
τ
τ
+
13
First Order Transient Response Using Thevenin/Norton Theorem
• One must be careful to determine the equivalent circuits before and after the switch changes position.
– it is possible that equivalent circuit seen by the load before activating the switch is different from the circuit seen after closing the switch.
( ) ( ) ( )+− ====≤= 00 0 22 tvVtvtVtv CCC
14
First Order Transient Response Using Thevenin/Norton Theorem (cont.)
( ) ( )
( ) ( ) ( )
( ) ( ) ( )( ) ( )∞=+∞=−==
=====∞====
=+→>=+
−
+−
tvetvtvtv
vFKRCvx
vtvtvtv
FKtxdt
tdxtvv
dt
dvRC
Ct
CCC
SSC
SCCC
SSCC
τ
τ
τ
0 :Step4
,1 , , :Step3
,00 :Step2
0 :Step1( ) ( )
( ) ( ) ( )
( ) ( ) ( )( ) ( ) ( )
+==
+−=∞=+∞=−==
=====∞=====
=+→>=+
−−
+−
2
2
1
1321
2
2
||||
0 :Step4
,1 , , :Step3
,0 0 :Step2
0 :Step1
R
V
R
VRVRRRR
VeVVtvetvtvtv
VFKCRvx
VtvtvVtv
FKtxdt
tdxtVv
dt
dvCR
TTT
Tt
TCt
CCC
TSTC
TCCC
STCC
T
ττ
τ
τ
Page 11
15
First Order Transient Response Using Thevenin/Norton Theorem (cont.)
( ) ( )
( ) ( ) ( )
( ) ( ) ( )( ) ( ) ( )( ) divider voltage: ||
0 :Step4
,1 , , :Step3
,0 00 :Step2
0 :Step1
21
2321 BTT
Tt
TCt
CCC
TSTC
TCCC
STCC
T
VRR
RVRRRR
VeVtvetvtvtv
VFKCRvx
Vtvtvtv
FKtxdt
tdxtVv
dt
dvCR
+=+=
+−=∞=+∞=−==
=====∞=====
=+→>=+
−−
+−
ττ
τ
τ
Example 5.100 (closing) < t < 50 ms
( ) ( )
( ) ( )( ) ( ) ( )
( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 0 :Step4
where0 ,1 , , :Step3
0,0abovesolution thefrom500 :Step2
0 0 :Step1
05.0**
32
*
τττ
τ
τ
−−−−
+−
=→=∞=+∞=−==
+======∞=======
=+→>=+
tCC
tCC
tCCC
TSTC
CCCCC
SCC
T
evtvevtvetvtvtv
RRRFKCRvx
tvtvvmstvtv
FKtxdt
tdxtv
dt
dvCR
50 ms (open the switch again) < t
16
RC Charging & Discharging
Charging: S1 closed & S2 openedDischarging: S2 closed & S1 openedTime constant (τ = RC)=0.1 sec Charging
Discharging
Note: Capacitor voltage is continuous,but capacitor current is not (many jumps).
( ) ( )[ ]( ) ( )[ ] 632.01
0
0 1 =−==−∞==−= −e
txtx
txtx τ
17
Second Order Transient Response• Second-order circuit: two energy storage element w/wo
one energy loss element (e.g. RLC circuit, LC circuit)
LL
T
L
T
TLL
LT
T
LL
CL
LT
TCL
T
R
CTRCRTLTRLRT
CLT
RLCS
idt
di
R
L
dt
idLC
R
v
dt
idLCi
dt
diLv
R
dt
diL
dt
dCi
dt
dvCi
dt
diLv
Rii
R
v
vvvvvvvvvvvv
iiR
viii
++=→+=
−
+=+=
−→+=
−=→=++−−=→=++−
+=→+=
2
2
2
21
1
0 and 0 :KVL
:KCL
18
Second Order Transient Response (cont.)( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( )ly.respective gain, DC theand
ratio, damping thefrequency, natural thetermed and 12, constants thewhere
21
0020120
2
2
20012
2
2
abKaaaaa
tfKtxdt
tdx
dt
txdtfbtxa
dt
tdxa
dt
txda
Sn
Snn
===
=++→=++
ζω
ωζ
ω
1 and 1.0,1 === Sn Kζω
• The final value of 1 is predicted by the DC gain KS=1, which tells us about the steady state.
• The period of oscillation of the response is related to the natural frequency wn=1 leads to T=2 pi/wn = 6.28 sec.
• The reduction in amplitude of the oscillation is governed by the damping ratio. With large damping ratio, the response not overshoots (oscillates) but looks like the first order response.
• Damping -> friction effect
19
Second Order Response( ) ( ) ( ) ( )
( ) ( ) ( )
( )( )
( )
( )
( )( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )txtxtx
tFKtxttfKtxdt
tdx
dt
txd
dt
tdxFtf
js
s
s
seetx
txdt
tdx
dt
txd
tfKtxdt
tdx
dt
txd
FN
SFSFF
n
F
n
F
nn
n
nn
nntsts
N
NN
n
N
n
Snn
+=
≥=→≥=++
→=
−±−=
→→<
−=→→
=
−±−=
→→>
−±−=+=
=++
=++
Response Complete
0 0 21
0: where DC todue Response Forced
1
nOscillatioresponse dUnderdampe1roots.Complex :3 Case
nOscillatioresponse overdamped Critically
1roots. repeated and Real:2 Case
1
systemorder first thelikeLook
response Overdamped1roots.distinct and Real :1 Case
1 where
021
Response Natural
21
2
2
2
22,1
2,1
22,1
22,121
2
2
2
2
2
2
21
ωζ
ω
ζωζω
ζω
ζζωζω
ζζωζωαα
ωζ
ω
ωζ
ω
.conditions initial by the determined be that willconstants is and 21 αα
20
Second Order Response (cont.)• Procedures
– Write the differential equation of the circuit for t=0+, that is, immediately after the switch has changed. The variable x(t) in the differential equation will be either a capacitor voltage or an inductor current. You can reduce the circuit to Thevenin or Norton equivalent form. Rewrite the equation as the standard form.
– Identify the initial conditions x(t=0+) and dx/dt(t=0+) using the continuity of capacitor voltages and inductor currents.
– Write the complete solution for the circuit in the form.
– Apply the initial conditions to solve for the constants . and 21 αα
( ) ( ) ( )( ) ( ) ( ) ( ) ( )
( ) ( ) ( )txeetx
txteetx
txeetx
F
tjtj
Ftt
F
tt
nnnn
nn
nnnn
++=<
++==
++=>
−−−
−+−
−−
−−−
−+−
:1roots.Complex :3 Case
:1roots. repeated and Real:2 Case
:1roots.distinct and Real :1 Case
22
22
1
2
1
1
21
1
2
1
1
ζωζωζωζω
ωω
ζωζωζωζω
ααζ
ααζ
ααζ
21
Example: Second Order Response
( ) ( )
( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( )
( )
( )( ) ( )
( )
( )
−−−+
−+−===
−−−+
−+−=
+===
====
+=
=−±−=+=
→
====→====→=
=++=++→=++
==→=+=→==+=+=
========
==++→=′′
+=++
=++→=+++−
+=→==
+
−−−
−+−
+
++
−−−
−+−
−
−
++++
+−+−
∫
11200
11
00
and constants thedetermine A/s,200 and 0A 0 Using:Step4
0) response (forced Response Complete
1 where
response Overdamped
5.21
10
2
5000
22
2,rad/s 1000
10
111
21:00 :Step3
A/s200V25V 501000
0A 00,0V 50 :Step2
00
0 :KVL
:KCL :Step1
22
21
12
2
12
1
21
21
1
2
1
1
22,121
6
62
2
2
22
2
2
2
2
2
0
22
22
21
ζωζωαζωζωα
ζωζωαζωζωα
αα
αα
αα
ζωζωαα
ωζω
ζωω
ωζ
ω
ζωζωζωζω
ζωζωζωζω
nnnnL
t
nn
t
nnL
L
LL
tt
L
nntsts
L
n
nn
n
Snn
LLLLLL
LLSC
LL
LLCC
SLLLS
tL
CL
L
SCLRCLRS
CLT
RLCS
tdt
di
eedt
di
ti
tdt
diti
eeti
seeti
L
CRRCRC
LCLC
tfKtxdt
tdx
dt
txdi
dt
diRC
dt
idLC
C
i
dt
diR
dt
idL
tdt
dit
dt
divtvt
dt
diLRti
tititvtv
dt
dv
C
i
dt
diR
dt
idLvtd
C
titv
dt
diLRi
vvvvvvvv
iiR
viii
nnnn
nnnn
22
Overdamped and Underdamped Circuit
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