Transients Analysis
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
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)
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
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.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
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.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
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 ( )
BASIC ELECTRONIC ENGINEERING
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
Department of Electronic Engineering
WHAT IS TRANSIENT RESPONSE
Figure 5.1
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Circuit with switched DC
excitation
A general model of the
transient analysis problem
Figu
re
5.2,
5.3
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
In general, any circuit containing energy storage element
Figure 5.5, 5.6
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
(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).
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
(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).
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
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 )
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Example
Figure 5.12,
5.13
BASIC ELECTRONIC ENGINEERING
5-6
Department of Electronic Engineering
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.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
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)
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
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
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
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
BASIC ELECTRONIC ENGINEERING
t RC

vC (0 )  Vi
st
RCKse  Ke  0
st
1
s
RC
 Ke
K
vC t   Vi e
Department of Electronic Engineering
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.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
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
BASIC ELECTRONIC ENGINEERING
i(t)
R
C
+
VC
-
Vu(t)
Department of Electronic Engineering
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)
BASIC ELECTRONIC ENGINEERING
Solving the differential equation
Department of Electronic Engineering
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.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Figure
5.17,
5.18
a). Complete, transient and steady
state response
b). Complete, natural, and forced
responses of the circuit
BASIC ELECTRONIC ENGINEERING
5-8
Department of Electronic Engineering
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.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
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)
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
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.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
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.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
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 ( )
BASIC ELECTRONIC ENGINEERING
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
Department of Electronic Engineering
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
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
iR
Example
+ VR iC
100 k
ohms
100V
0.01
microF
+
Vc
-
Initial condition Vc(0) = 0V
 3 dvC
10
 vC  100
dt
dx(t )

 x(t )  K s f (t )
dt
x  x N (t )  xF (t )
 e
 e
t / 
t
10 3
As vc (0)  0, 0  100  A
and
t / 
vc  100  Ae

 KS F
 x ( )
BASIC ELECTRONIC ENGINEERING
A  100
vc  100  100e
Department of Electronic Engineering

t
10 3
Energy stored in capacitor
dv
p  vi  Cv
dt
t
to
pdt

dv
t
 t Cv dt
o
dt
 C tt vdv
1
 C v(t )2  v(to )2
2

o
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
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
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
BASIC ELECTRONIC ENGINEERING
R
Department of Electronic Engineering
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
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
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
BASIC ELECTRONIC ENGINEERING

Department of Electronic Engineering

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
p R 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
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
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
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
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.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering