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Transients Analysis
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Solution to First Order Differential Equation
Consider the general Equation
dx(t )

 x(t )  K s f (t )
dt
Let the initial condition be x(t = 0) = x( 0 ), then we solve the
differential equation:
dx(t )

 x(t )  K s f (t )
dt
The complete solution consists of two parts:
• the homogeneous solution (natural solution)
• the particular solution (forced solution)
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The Natural Response
Consider the general Equation
dx(t )

 x(t )  K s f (t )
dt
Setting the excitation f (t) equal to zero,
dxN (t )
dxN (t )
x N (t ) dxN (t )
dt

 x N (t )  0 or

,

dt
dt

x N (t )

dxN (t )
dt
  ,

x N (t )

x N (t )   e t / 
It is called the natural response.
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The Forced Response
Consider the general Equation
dx(t )

 x(t )  K s f (t )
dt
Setting the excitation f (t) equal to F, a constant for t 0
dxF (t )

 xF (t )  K S F
dt
xF (t )  K S F for t  0
It is called the forced response.
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The Complete Response
Consider the general Equation
dx(t )

 x(t )  K s f (t )
dt
Solve for ,
The complete response is:
• the natural response +
• the forced response
  x(0)  x()
x  x N (t )  xF (t )
  e t /   K S F
  e  t /   x ( )
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for t  0
x(t  0)  x(0)    x()
The Complete solution:
x(t )  [ x(0)  x()] et /   x()
[ x(0)  x()] et /  called transient response
x ( ) called steady state response
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WHAT IS TRANSIENT RESPONSE
Figure 5.1
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Circuit with switched DC
excitation
A general model of the
transient analysis problem
Figu
re
5.2,
5.3
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In general, any circuit containing energy storage element
Figure 5.5, 5.6
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(a) Circuit at t = 0
(b) Same circuit a long time after the switch is closed
Figure 5.9,
5.10
The capacitor acts as open circuit for the steady state condition
(a long time after the switch is closed).
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(a) Circuit for t = 0
(b) Same circuit a long time before the switch is opened
The inductor acts as short circuit for the steady state condition
(a long time after the switch is closed).
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Why there is a transient response?
• The voltage across a capacitor cannot be
changed instantaneously.


VC (0 )  VC (0 )
• The current across an inductor cannot be
changed instantaneously.


I L (0 )  I L (0 )
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Example
Figure 5.12,
5.13
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5-6
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Transients Analysis
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 firstorder
circuits to the time constant.
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Transients
The solution of the differential equation
represents are response of the circuit. It
is called natural response.
The response must eventually die out,
and therefore referred to as transient
response.
(source free response)
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Discharge of a Capacitance through a Resistance
ic
iR
i  0,
iC  iR  0
dvC t  vC t 
C

0
dt
R
Solving the above equation
with the initial condition
Vc(0) = Vi
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Discharge of a Capacitance through a Resistance
dvC t  vC t 
C

0
dt
R
dvC t 
RC
 vC t   0
dt
vC t   Ke
vC t   Ke
st
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t RC

vC (0 )  Vi
st
RCKse  Ke  0
st
1
s
RC
 Ke
K
vC t   Vi e
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0 / RC
t RC
vC t   Vi e t
RC
Exponential decay waveform
RC is called the time constant.
At time constant, the voltage is 36.8%
of the initial voltage.
vC t   Vi (1  e t RC )
Exponential rising waveform
RC is called the time constant.
At time constant, the voltage is
63.2% of the initial voltage.
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RC CIRCUIT
t=0
V
i(t)
R
+
_
C
t=0
+
VC
Vu(t)
+
_
-
0-,
for t =
i(t) = 0
u(t) is voltage-step function
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i(t)
R
C
+
VC
-
Vu(t)
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RC CIRCUIT
iR  iC
vu(t )  vC
dvC
iR 
, iC  C
R
dt
dvC
RC
 vC  V , v u (t )  V for t  0
dt
Vu(t)
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Solving the differential equation
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Complete Response
Complete response
= natural response + forced response
• Natural response (source free response) is
due to the initial condition
• Forced response is the due to the external
excitation.
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Figure
5.17,
5.18
a). Complete, transient and steady
state response
b). Complete, natural, and forced
responses of the circuit
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5-8
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Circuit Analysis for RC Circuit
iR
iC
R
Vs
Apply KCL
+ VR -
C
+
Vc
-
iR  iC
vs  v R
dvC
iR 
, iC  C
R
dt
dvC
1
1

vR 
vs
dt
RC
RC
vs is the source applied.
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Solution to First Order Differential Equation
Consider the general Equation
dx(t )

 x(t )  K s f (t )
dt
Let the initial condition be x(t = 0) = x( 0 ), then we solve the
differential equation:
dx(t )

 x(t )  K s f (t )
dt
The complete solution consits of two parts:
• the homogeneous solution (natural solution)
• the particular solution (forced solution)
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The Natural Response
Consider the general Equation
dx(t )

 x(t )  K s f (t )
dt
Setting the excitation f (t) equal to zero,

dx N (t )
dx N (t )
x (t )
 x N (t )  0 or
 N
dt
dt

x N (t )   e t / 
It is called the natural response.
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The Forced Response
Consider the general Equation
dx(t )

 x(t )  K s f (t )
dt
Setting the excitation f (t) equal to F, a constant for t 0
dxF (t )

 xF (t )  K S F
dt
xF (t )  K S F for t  0
It is called the forced response.
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The Complete Response
Consider the general Equation
dx(t )

 x(t )  K s f (t )
dt
The complete response is:
• the natural response +
• the forced response
x  x N (t )  xF (t )
  e t /   K S F
  e  t /   x ( )
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Solve for ,
for t  0
x(t  0)  x(0)    x()
  x(0)  x()
The Complete solution:
x(t )  [ x(0)  x()] et /   x()
[ x(0)  x()] et /  called transient response
x ( ) called steady state response
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Example
iR
+ VR 100 k
ohms
100V
0.01
microF
iC
+
Vc
-
Initial condition Vc(0) = 0V
iR  iC
vs  vC
dvC
iR 
, iC  C
R
dt
dvC
RC
 vC  vs
dt
5
 6 dvC
10  0.0110
 vC  100
dt
3 dvC
10
 vC  100
dt
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iR
Example
+ VR iC
100 k
ohms
100V
0.01
microF
+
Vc
-
dx(t )

 x(t )  K s f (t )
dt
x  x N (t )  xF (t )
 e
10
3
dvC
 vC  100
dt
t

3
10
vc  100  Ae
As vc (0)  0, 0  100  A
and
t / 
Initial condition Vc(0) = 0V
 KS F
  e  t /   x ( )
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A  100

t
3
10
vc  100  100e
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Energy stored in capacitor
dv
p  vi  Cv
dt
dv
t
t
t
pdt

Cv
dt

C
to
to
to vdv
dt

1
 C v(t )2  v(to )2
2

If the zero-energy reference is selected at to, implying that the
capacitor voltage is also zero at that instant, then
1 2
wc (t )  Cv
2
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RC CIRCUIT
Power dissipation in the resistor is:
R
C
pR = V2/R = (Vo2 /R) e -2 t /RC
Total energy turned into heat in the resistor
WR  0 p R dt 
 Vo2 R(
2  2t / RC
Vo 0 e
dt
1
)e  2t / RC |
0
2 RC
1
2
 CVo
2
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R
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RL CIRCUITS
Initial condition
i(t = 0) = Io
i(t)
VR
+
R
L
di
vR  vL  0  Ri  L
dt
+
L di
VL
i  0
R dt
Solving the differential equation
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RL CIRCUITS
i(t)
VR
R
L
+
+
VL
-
Initial condition
i(t = 0) = Io
di R
 i0
dt L
i ( t ) di
di
R
  dt,

Io
i
L
i
R t
i
ln i |I o   t |o
L
R
ln i  ln I o   t
L
 Rt / L
i (t )  I o e
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
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
t
R
 dt
o
L
RL CIRCUIT
Power dissipation in the resistor is:
pR = i2R = Io2e-2Rt/LR
i(t)
Total energy turned into heat in the resistor
VR
+
R
L
+
VL
WR 
-



0
pR dt 
I o2 R( 
I o2 R


e 2 Rt / L dt
0
L 2 Rt / L 
)e
|0
2R
1 2
 LI o
2
It is expected as the energy stored in the inductor is
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1 2
LI o
2
i(t)
Vu(t)
+
_
Vu(t)
R
L
+
VL
-
RL CIRCUIT
di
Ri  L  V
dt
Ldi
 dt
V  Ri
Integrating both sides,
L
 ln(V  Ri )  t  k
R
L
i (0 )  0, thus k   ln V
R
L
 [ln(V  Ri )  ln V ]  t
R
V  Ri
 e  Rt / L
or
V
V V  Rt / L
i  e
, for t  0
R R

where L/R is the time constant
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DC STEADY STATE
The steps in determining the forced response for
RL or RC circuits with dc sources are:
1. Replace capacitances with open circuits.
2. Replace inductances with short circuits.
3. Solve the remaining circuit.
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