Chapter 5
Transient Analysis
Jaesung Jang
Complete response = Transient response + Steady-state response
Time Constant
First order and Second order Differential Equation
.
1
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
– Constant signals
– Sinusoidal signals
dt
0. 9
0. 8
0. 7
0. 6
V/Vs
• Steady-state region: the region where the
signals are not time dependent (time rate of
change of signals is equal to zero) or
d( )
periodic.
=0
1
0. 5
0. 4
0. 3
0. 2
0. 1
0
0
1
2
3
4
5
t (s e c )
6
7
8
9
10
2
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.
dx
+ x = Vs
dt
dx N
+ xN = 0
dt
dx F
+ x F = Vs
dt
• 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.
1
0.9
0.8
– RL or RC circuit.
– RLC or LC circuit.
0.6
V/Vs
• Second order: The largest order of the differential
equation is the second order.
0.7
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
t (s ec )
6
7
8
9
10
3
Writing Differential Equations
• Key laws: KVL & KCL for capacitor voltages or inductor currents
vR
= iC
R
KVL : − v S + v R + vC = 0 → −v S + iR R + vC = 0
KCL : iR = iC →
iC R + vC (t = 0) +
t
∫
0
iC (t ′)
dt ′ = v S
C
diC
i
dv
di
i
dv
R + C = S → C + C = S : Differential equation for iC
dt
C
dt
dt RC Rdt
dv
v − vC
dv
v
v
vR
= iC = C C = S
→ C + C = S : Differential equation for vC
R
dt
R
dt
RC RC
dx(t )
+ a0 x(t ) = b0 f (t )
dt
where x(t ) represents the capacitor voltage or the inductor current and
the constants a1 , a0 , and b0 represents combinations of circuit element parameters.
a1
→ First - order linear ordinary differential equation
4
Writing Differential Equations (cont.)
• Key laws: KVL & KCL
KCL : iR = iC = iL = i
KVL : − v S + v R + vC + v L = 0 → −vS + iR R + vC + v L = 0
iR + vC (t = 0 ) +
t
∫
0
2
i(t ′)
di
dt ′ + L = v S
C
dt
dv
di
i
d i dvS
d 2i Rdi
i
R+ +L 2 =
→ 2+
+
= S : Differential equation for i
dt
C
dt
Ldt LC Ldt
dt
dt
dv
v − vC − v L
dv
v
v
vR
1  d dv 
= iC = C C = S
→C C = S − C − L C C 
R
dt
R
dt
R
R R  dt
dt 
 d 2 vC
dvC
RC
= v S − vC − LC  2
 dt
dt





 d 2 vC 
 + RC dvC + vC = vS : Differential equation for vC
LC 
 dt 2 
dt


2
d x(t )
dx(t )
a2
+
a
+ a0 x(t ) = b0 f (t ) → Second - order linear ordinary differential equation
1
dt
dt 2
where x(t ) represents the capacitor voltage or the current and
the constants a 2 , a1 , a0 , and b0 represents combinations of circuit element parameters.
a2 d 2 x(t ) a1 dx(t )
b0
1 d 2 x(t ) 2ζ dx(t )
+
+ x(t ) =
+
+ x(t ) = K S f (t )
f (t ) → 2
a0 dt 2
a0 dt
a0
ωn dt
ω n dt 2
where the constants ω n = a0 a 2 , ζ = (a1 2) 1 a0 a2 and K S = b0 a0 termed the natural frequency, the damping ratio,
and the DC gain, respectively.
5
Examples of Writing Differential Equations
vR
= iL + i R2
R
KVL : − v S + v R + v L = 0 → v R = vS − v L
KCL : i R1 = iL + i R2 →
vR
v − vL
v (t ′)
v
dt ′ + L
= i L + i R2 → S
= iL (t = 0) + L
R
R
L
R
t
∫
0
Rv L (t ′)
Rv L (t ′)
v S − v L = RiL (t = 0 ) +
dt ′ + v L → v S = RiL (t = 0 ) +
dt ′ + 2v L
L
L
t
t
∫
∫
0
0
dvS R
dv
2dvL
dv
R
= vL +
→ 2 L + v L = S : Differential equation for v L
dt
L
dt
dt
L
dt
KCL : iR1 = iC + i L
KVL : − vS + v R1 + vC = 0 → vS = v R1 + vC
− vC + v R2 + v L = 0 → vC = v R2 + v L = L
diL
+ iL R2
dt


 d  diL

d 2 iL
di
 dvC



+ iL  R1 =  C  L
+ iL R2  + i L  R1 = LC 2 + R2C L + i L  R1
v R1 = i R1 R1 = (iC + iL )R1 =  C


dt
dt

 dt

 dt  dt



2


d
i
di
di
v S = v R1 + vC =  LC 2L + R2 C L + iL  R1 + L L + i L R2 →


dt
dt
dt


v S = R1 LC
R1 LC
d 2 iL
dt 2
d 2i L
dt 2
+ R1 R2C
+ (R1 R2 C + L )
diL
di
+ R1i L + L L + iL R2 →
dt
dt
diL
+ (R1 + R2 )iL = vS : Differential equation for iL
dt
6
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.
+
+
+
+
dvC vC
v
+
= S
dt
RC RC
vC = v S at the steady state
R1 LC
d 2 iL
2
dvC (t )
→ 0 as t → ∞
dt
di (t )
v L (t ) = L L → 0 as t → ∞
dt
At DC steady state,
all capacitors behave as open circuits
and all inductors hehave as short circuits.
iC (t ) = C
+ (R1 R2C + L )
di L
+ (R1 + R2 )iL = vS
dt
dt
vS
iL =
at the steady state
(R1 + R2 )
7
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
1
1
storage elements
WL (t ) = Li L2 (t ) WC (t ) = CvC2 (t )
2
2
– 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).
( ) ( )
(t = 0 ) = i (t = 0 )
vC t = 0 − = vC t = 0 +
iL
−
+
L
where the notation 0 − signifies " just before t = 0" and
0 + signifies " just after t = 0"
Discontinuous of capacitor voltage
-> infinite power at t=0.
8
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.
x(t ) = x(t = ∞ ) + [x(t = 0 ) − x(t = ∞ )]exp(− t τ )
• 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. [x(t = τ ) − x(t = 0)]
[x(t = ∞ ) − x(t = 0)]
= 1 − e −1 = 0.632
9
First Order Response (cont.)
• First-order circuit: one energy storage element + one
energy loss element (e.g. RC circuit, RL circuit)
a1
a dx (t )
b
dx (t )
dx(t )
+ a0 x (t ) = b0 f (t ) → 1
+ x(t ) = 0 f (t ) → τ
+ x (t ) = K S f (t )
dt
a0 dt
a0
dt
where τ = a1 a0 and K S = b0 a0 termed the time constant and DC gain, respectively.
Natural Response
dx (t )
dx (t ) − x N (t )
→ x N (t ) = x0 e −t τ where x0 is a constant.
τ N + x N (t ) = 0 → N =
τ
dt
dt
dx (t )
Forced Response due to DC (where f (t ) = F ) : F → 0
dt
dx (t )
τ F + x F (t ) = K S F t ≥ 0 → x F (t ) = K S F t ≥ 0
dt
Complete Response
x(t ) = x N (t ) + x F (t ) = x0 e −t τ + x(t = ∞ ) = x0 e −t τ + K S F (for DC)
x(t = 0 ) = x0 + x(t = ∞ ) → x0 = x(t = 0 ) − x(t = ∞ ) for t ≥ 0
10
Example: First Order Response 1
+
vR
= iC
R
KVL : − v S + v R + vC = 0 → −v S + i R R + vC = 0
Step1 : KCL : iR = iC →
dv
v − vC
dv
vR
dx(t )
= iC = C C = S
→ RC C + vC = v S t > 0 → τ
+ x(t ) = K S F
R
dt
R
dt
dt
(
)
(
)
Step2 : vC t = 0 − = 5 V = vC t = 0 + , vC (t = ∞ ) = 12V(= v S )
Step3 : x = vC , τ = RC = 1kΩ × 470µF = 0.47, K S = 1, F = vS
Step4 : vC (t ) = (vC (t = 0) − vC (t = ∞ ))e −t τ + vC (t = ∞ ) = 12 + (− 7 )e −t
0.47
11
Example: First Order Response 2
Step1 : KCL : iR = i L
+
KVL : − v B + v R + v L = 0 → −v B + i L R + L
di L
=0
dt
L di L
v
dx(t )
+ iL = B t > 0 → τ
+ x(t ) = K S F
R dt
R
dt
Step2 : iL t = 0 − = 0 A = iL t = 0 + , iL (t = ∞ ) = v B R = 12.5A
→
(
)
(
)
Step3 : x = i L , τ = L R = 0.1H 4Ω = 0.025, K S = 1 R , F = v B
Step4 : iL (t ) = (i L (t = 0) − i L (t = ∞ ))e −t τ + i L (t = ∞ ) = 12.5 + (− 12.5)e −t 0.025
12
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.
vC (t ) = V2 t ≤ 0
(
)
(
vC t = 0 − = V2 = vC t = 0 +
)
13
First Order Transient Response Using
Thevenin/Norton Theorem (cont.)
Page 11
dv
dx(t )
Step1 : RT C C + vC = VT t > 0 → τ
+ x(t ) = K S F
dt
dt
Step2 : vC t = 0 − = V2 = vC t = 0 + , vC (t = ∞ ) = VT
(
)
(
)
Step3 : x = vC , τ = RT C , K S = 1, F = VT
Step4 : vC (t ) = (vC (t = 0 ) − vC (t = ∞ ))e −t τ + vC (t = ∞ ) = (V2 − VT )e −t τ + VT
RT = R1 || R2 || R3
dvC
dx(t )
+ vC = v S t > 0 → τ
+ x(t ) = K S F
dt
dt
Step2 : vC t = 0 − = vC t = 0 + , vC (t = ∞ ) = v S
Step1 : RC
(
)
(
)
Step3 : x = vC , τ = RC , K S = 1, F = v S
Step4 : vC (t ) = (vC (t = 0 ) − vC (t = ∞ ))e −t τ + vC (t = ∞ )
V V 
VT = RT  1 + 2 
 R1 R2 
14
First Order Transient Response Using
Thevenin/Norton Theorem (cont.)
Example 5.10
0 (closing) < t < 50 ms
dvC
dx(t )
+ vC = VT t > 0 → τ
+ x(t ) = K S F
dt
dt
Step2 : vC t = 0 − = 0 = vC t = 0 + , vC (t = ∞ ) = VT
Step1 : RT C
(
)
(
)
Step3 : x = vC , τ = RT C , K S = 1, F = VT
Step4 : vC (t ) = (vC (t = 0) − vC (t = ∞ ))e −t τ + vC (t = ∞ ) = (− VT )e −t τ + VT
RT = (R1 || R2 ) + R3
VT =
R2
VB : voltage divider
R1 + R2
50 ms (open the switch again) < t
dvC
dx(t )
+ vC = 0 t > 0 → τ
+ x (t ) = K S F
dt
dt
Step2 : vC t = 0 − = vC (t = 50ms )(from the solution above ) = vC* = vC t = 0 + , vC (t = ∞ ) = 0
Step1 : RT C
(
)
Step3 : x = vC , τ = RT C , K S = 1, F = 0 where RT = R2 + R3
( )
(
)
( )
Step4 : vC (t ) = (vC (t = 0 ) − vC (t = ∞ ))e −t τ + vC (t = ∞ ) = vC* e −t τ → vC (t ) = vC* e − (t −0.05 ) τ
15
RC Charging & Discharging
Discharging
Charging: S1 closed & S2 opened
Discharging: S2 closed & S1 opened
Time constant (τ = RC)=0.1 sec
Charging
Note: Capacitor voltage is continuous,
but capacitor current is not (many jumps).
[x(t = τ ) − x(t = 0)] = 1 − e −1 = 0.632
[x(t = ∞ ) − x(t = 0)]
16
Second Order Transient Response
• Second-order circuit: two energy storage element w/wo
one energy loss element (e.g. RLC circuit, LC circuit)
KCL : iS = iC + iL →
vR
= i L + iC
RT
KVL : − vT + v R + v L = 0 → v R = vT − v L and − vT + v R + vC = 0 → v R = vT − vC
dv
vR
di 
d  di 
1 
= iL + iC →
 vT − L L  = iL + C C = i L + C  L L 
RT
RT 
dt
dt  dt 
dt 
di L 
d 2i L
vT
d 2iL
L di L
1 
= LC 2 +
+ iL
 vT − L
 = i L + LC 2 →
RT 
dt 
RT
RT dt
dt
dt
17
Second Order Transient Response (cont.)
a2
d 2 x(t )
dt 2
dx(t )
1 d 2 x(t ) 2ζ dx(t )
+ a1
+ a0 x(t ) = b0 f (t ) → 2
+
+ x(t ) = K S f (t )
2
ω n dt
dt
ω n dt
where the constants ω n = a0 a 2 , ζ = (a1 2 ) 1 a0 a 2 and K S = b0 a0 termed the natural frequency, the damping ratio,
and the DC gain, respectively.
• 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.
ω n = 1, ζ = 0.1 and K S = 1
• 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
18
Second Order Response
1 d 2 x(t )
ω n2 dt 2
+
2ζ dx(t )
+ x(t ) = K S f (t )
ω n dt
Natural Response
1 d 2 x N (t )
ω n2
dt 2
+
2ζ dx N (t )
+ x N (t ) = 0
ω n dt
x N (t ) = α1e s1t + α 2 e s2t
where s1, 2 = −ζω n ± ω n ζ 2 − 1
Case 1 : Real and distinct roots.(ζ > 1) → Overdamped response
→ Look like the first order system
s1, 2 = −ζω n ± ω n ζ 2 − 1
Case 2 : Real and repeated roots.(ζ = 1)
→ Critically overdamped response → Oscillation
s1, 2 = −ω n
Case 3 : Complex roots.(ζ < 1) → Underdamped response → Oscillation
s1, 2 = −ζω n ± jω n 1 − ζ 2
Forced Response due to DC (where f (t ) = F ) :
1 d 2 x F (t )
ω n2
dt 2
+
dx F (t )
→0
dt
2ζ dx F (t )
+ x F (t ) = K S f (t ) t ≥ 0 → x F (t ) = K S F t ≥ 0
ω n dt
Complete Response
x(t ) = x N (t ) + x F (t )
α1 and α 2 is constants that will be determined by the initial conditions.
19
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.
 −ζω +ω ζ 2 −1 t
 −ζω −ω ζ 2 −1 t




n
n
n
n
 + α e

= α1e 
2
t = α1e (−ωn )t + α 2te (−ωn )t + x F t
Case 1 : Real and distinct roots.(ζ > 1) : x(t )
Case 2 : Real and repeated roots.(ζ = 1) : x(
Case 3 : Complex roots.(ζ < 1) : x(t )
)
 −ζω + jω 1−ζ 2 t


n
n


= α1e
()
+α2
 −ζω − jω 1−ζ 2 t


n
n


e
+ x F (t )
+ x F (t )
– Apply the initial conditions to solve for the constants α1 and α 2 .
20
Example: Second Order Response
Step1 : KCL : iS = iC = iL →
vR
= iL + iC
RT
KVL : − v S + v R + v L + vC = 0 → v R + v L + vC = v S
dv
diL
i L (t ′)
d 2i L
di
i
iL R + L
dt ′ = v S → L 2 + R L + L = S = 0
+ vC (t = 0 ) +
dt
C
dt C
dt
dt
t
∫
(
)
0
(
) (
)
(
Step2 : vC t = 0 − = 5 V = vC t = 0 + , iL t = 0 − = 0 A = iL t = 0 +
(
)
iL t = 0 + R + L
Step3 : L
1
ω n2
d 2i L
dt 2
(
)
(
)
)
(
)
diL
di
di
t = 0 + + vC (t = 0 ) = v S → 1 L t = 0 + + 5 V = 25V → L t = 0 + = 20A/s
dt
dt
dt
diL i L
d 2iL
diL
1 d 2 x(t ) 2ζ dx(t )
+R
+ = 0 → LC 2 + RC
+ iL = 0 : 2
+
+ x (t ) = K S f (t )
dt C
dt
ω n dt
dt
ω n dt 2
= LC → ω n =
RCω n R
1
1
2ζ
RC
=
=
(
)
=
→
=
=
1000
rad/s
,
ζ
LC
2
2
ωn
10 −6
C 5000 10 − 6
=
= 2.5
L
2
1
→ Overdamped response
iL (t ) = α1e s1t + α 2 e s2 t
where s1, 2 = −ζω n ± ω n ζ 2 − 1
Complete Response (forced response = 0)
iL (t )
 −ζω +ω ζ 2 −1 t


n
n

= α1e 
(
+ α2
Step4 : Using 0 A = iL t = 0 +
(
)
)
 −ζω −ω ζ 2 −1  t


n
n

e
and
(
)
diL
t = 0 + = 20A/s, determine the constants α1 and α 2
dt
iL t = 0 + = 0 = α1 + α 2



 −ζω n + ωn ζ −1 t
diL
2
e  −ζω n −ω n
 + α  − ζω − ω
= α1  − ζω n + ω n ζ 2 − 1 e 
−
1
ζ

n
n
2




dt
diL
t = 0 + = 20 = α1  − ζω n + ω n ζ 2 − 1  + α 2  − ζω n − ω n ζ 2 − 1 




dt
(
)
2
ζ 2 −1 t

21
Overdamped and Underdamped Circuit
22