unit6 RL-transient.ppt

Department of Electronic Engineering
BASIC ELECTRONIC ENGINEERING
Transients Analysis
Presented by
ReddyPrasad Reddivari
Assistant Professor
EEE Dept.

Solution to First Order Differential Equation
)
(
)
(
)
(
t
f
K
t
x
dt
t
dx
s



Consider the general Equation
Let the initial condition be x(t = 0) = x( 0 ), then we solve the
differential equation:
)
(
)
(
)
(
t
f
K
t
x
dt
t
dx
s



The complete solution consists of two parts:
• the homogeneous solution (natural solution)
• the particular solution (forced solution)

The Natural Response






/
)
(
,
)
(
)
(
)
(
)
(
,
)
(
)
(
0
)
(
)
(
t
N
N
N
N
N
N
N
N
N
e
t
x
dt
t
x
t
dx
dt
t
x
t
dx
t
x
dt
t
dx
or
t
x
dt
t
dx












Setting the excitation f (t) equal to zero,
)
(
)
(
)
(
t
f
K
t
x
dt
t
dx
s



It is called the natural response.

The Forced Response
0
)
(
)
(
)
(




t
for
F
K
t
x
F
K
t
x
dt
t
dx
S
F
S
F
F

Setting the excitation f (t) equal to F, a constant for t 0
)
(
)
(
)
(
t
f
K
t
x
dt
t
dx
s



It is called the forced response.

The Complete Response
)
(
)
(
)
(
/
/









x
e
F
K
e
t
x
t
x
x
t
S
t
F
N




The complete response is:
• the natural response +
• the forced response
)
(
)
(
)
(
t
f
K
t
x
dt
t
dx
s



Solve for ,
)
(
)
0
(
)
(
)
0
(
)
0
(
0









x
x
x
x
t
x
t
for


The Complete solution:
)
(
)]
(
)
0
(
[
)
( /




 
x
e
x
x
t
x t 

/
)]
(
)
0
(
[ t
e
x
x 

 called transient response
)
(
x called steady state response

WHAT IS TRANSIENT RESPONSE
Figure 5.1

Figu
re
5.2,
5.3
Circuit with switched DC
excitation
A general model of the
transient analysis problem

In general, any circuit containing energy storage element
Figure 5.5, 5.6

Figure 5.9,
5.10
(a) Circuit at t = 0
(b) Same circuit a long time after the switch is closed
The capacitor acts as open circuit for the steady state condition
(a long time after the switch is closed).

(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).

Why there is a transient response?
• The voltage across a capacitor cannot be
changed instantaneously.
)
0
(
)
0
( 

 C
C V
V
• The current across an inductor cannot be
changed instantaneously.
)
0
(
)
0
( 

 L
L I
I

Figure 5.12,
5.13
5-6
Example

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 first-order
circuits to the time constant.

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)

Discharge of a Capacitance through a Resistance
ic iR 0
,
0 


 R
C i
i
i
    0


R
t
v
dt
t
dv
C C
C
Solving the above equation
with the initial condition
Vc(0) = Vi

Discharge of a Capacitance through a Resistance
    0


R
t
v
dt
t
dv
C C
C
    0

 t
v
dt
t
dv
RC C
C
  st
C Ke
t
v 
0

 st
st
Ke
RCKse
RC
s
1


  RC
t
C Ke
t
v 

K
Ke
V
v
RC
i
C




/
0
)
0
(
  RC
t
i
C e
V
t
v 


  RC
t
i
C e
V
t
v 
 Exponential decay waveform
RC is called the time constant.
At time constant, the voltage is 36.8%
of the initial voltage.
  )
1
( RC
t
i
C e
V
t
v 


Exponential rising waveform
RC is called the time constant.
At time constant, the voltage is
63.2% of the initial voltage.

RC CIRCUIT
for t = 0-, i(t) = 0
u(t) is voltage-step function
Vu(t)
R
C
+
VC
-
i(t)
t = 0
+
_
V
R
C
+
VC
-
i(t)
t = 0
+
_
Vu(t)

RC CIRCUIT
Vu(t)
0
)
(
,
,
)
(








t
for
V
t
u
v
V
v
dt
dv
RC
dt
dv
C
i
R
v
t
vu
i
i
i
C
C
C
C
C
R
C
R
Solving the differential equation

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.

Figure
5.17,
5.18
5-8
a). Complete, transient and steady
state response
b). Complete, natural, and forced
responses of the circuit

Circuit Analysis for RC Circuit
Vs
+
Vc
-
+ VR -
R
C
iR
iC
s
R
C
C
C
R
s
R
C
R
v
RC
v
RC
dt
dv
dt
dv
C
i
R
v
v
i
i
i
1
1
,






Apply KCL
vs is the source applied.

Solution to First Order Differential Equation
)
(
)
(
)
(
t
f
K
t
x
dt
t
dx
s



Let the initial condition be x(t = 0) = x( 0 ), then we solve the
differential equation:
)
(
)
(
)
(
t
f
K
t
x
dt
t
dx
s



The complete solution consits of two parts:
• the homogeneous solution (natural solution)
• the particular solution (forced solution)

The Natural Response




/
)
(
)
(
)
(
0
)
(
)
(
t
N
N
N
N
N
e
t
x
t
x
dt
t
dx
or
t
x
dt
t
dx






Setting the excitation f (t) equal to zero,
)
(
)
(
)
(
t
f
K
t
x
dt
t
dx
s



It is called the natural response.

Example
+
Vc
-
+ VR -
100 k
ohms
0.01
microF
iR
iC
100V
Initial condition Vc(0) = 0V
s
C
C
C
C
C
s
R
C
R
v
v
dt
dv
RC
dt
dv
C
i
R
v
v
i
i
i






,
100
10
100
10
01
.
0
10
3
6
5








C
C
C
C
v
dt
dv
v
dt
dv

Example
+
Vc
-
+ VR -
100 k
ohms
0.01
microF
iR
iC
100V
Initial condition Vc(0) = 0V
and
3
3
10
10
100
100
100
100
0
,
0
)
0
(
100













t
c
c
t
c
e
v
A
A
v
As
Ae
v
)
(
)
(
)
(
t
f
K
t
x
dt
t
dx
s



)
(
)
(
)
(
/
/









x
e
F
K
e
t
x
t
x
x
t
S
t
F
N




100
10 3



C
C v
dt
dv

Energy stored in capacitor
   
 
2
2
)
(
)
(
2
1
o
t
t
t
t
t
t
t
v
t
v
C
vdv
C
dt
dt
dv
Cv
pdt
dt
dv
Cv
vi
p
o
o
o



 
 


If the zero-energy reference is selected at to, implying that the
capacitor voltage is also zero at that instant, then
2
2
1
)
( Cv
t
wc 

R C
Power dissipation in the resistor is:
pR = V2/R = (Vo
2 /R) e -2 t /RC
2
0
/
2
2
0
/
2
2
0
2
1
|
)
2
1
(
o
RC
t
o
RC
t
o
R
R
CV
e
RC
R
V
R
dt
e
V
dt
p
W









 

RC CIRCUIT
Total energy turned into heat in the resistor

RL CIRCUITS
L
R
-
VR
+
+
VL
-
i(t)
Initial condition
i(t = 0) = Io
equation
al
differenti
the
Solving
i
dt
di
R
L
dt
di
L
Ri
v
v L
R
0
0







RL CIRCUITS
L
R
-
VR
+
+
VL
-
i(t)
Initial condition
i(t = 0) = Io L
Rt
o
o
t
o
i
I
t
o
t
i
I
e
I
t
i
t
L
R
I
i
t
L
R
i
dt
L
R
i
di
dt
L
R
i
di
i
L
R
dt
di
o
o
/
)
(
)
(
ln
ln
|
|
ln
,
0
















RL CIRCUIT
L
R
-
VR
+
+
VL
-
i(t)
Power dissipation in the resistor is:
pR = i2R = Io
2e-2Rt/LR
Total energy turned into heat in the resistor
2
0
/
2
2
0
/
2
2
0
2
1
|
)
2
(
o
L
Rt
o
L
Rt
o
R
R
LI
e
R
L
R
I
dt
e
R
I
dt
p
W












It is expected as the energy stored in the inductor is 2
2
1
o
LI

RL CIRCUIT
Vu(t)
R
L
+
VL
-
i(t)
+
_
Vu(t)
k
t
Ri
V
R
L
sides
both
g
Integratin
dt
Ri
V
Ldi
V
dt
di
L
Ri








)
ln(
,
0
,
]
ln
)
[ln(
ln
,
0
)
0
(
/
/















t
for
e
R
V
R
V
i
or
e
V
Ri
V
t
V
Ri
V
R
L
V
R
L
k
thus
i
L
Rt
L
Rt
where L/R is the time constant

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.

THANK YOU

unit6 RL-transient.ppt

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