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Lecture 21&22
1
Welcome to ENEE 204
Lecture 21
Chapter 7:
TRANSIENT ANALYSIS
understand time varying signals
 differential equations
(transient and steady state solutions)
1
st
and 2
nd
order circuits
(Differential Equations)
Lecture 21&22
2
Transient Analysis – Simple Picture
(Series Connection)
1
st
order:
1 resistor and 1 cap (or 1 inductor) in series
– Solution without source
– Solution with source, dc or ac
2
nd
order
 Undamped: 1 cap and 1 inductor in series (LC)
 Damped: 1 cap, 1 inductor, 1 resistor in series
 underdamped
overdamped
 critically dampled
…with and without sources
t = 0
v
s
(t)
t = 0
Lecture 21&22
3
Transient Analysis  Big Picture II
(Parallel)
1
st
order: 1 resistor and 1 cap (or 1 inductor) in parallel
– Solution without source
– Solution with source, dc or ac
2
nd
order:
 Undamped: 1 cap and 1 inductor in parallel (LC)
 Damped: 1 cap, 1 inductor, 1 resistor in parallel
 underdamped
 critically dampled
 with and without sources
Lecture 21&22
4
Main steps in transient analysis:
1. Given a circuit, use KVL or KCL equations.
2. Use terminal relations for the elements.
3. Use 1 and 2 to derive a differential equation.
4. Solve the differential equation.
5. Use initial conditions*
to determine the constants.
Main Goal: To derive and solve the differential
equations that describe the time dependence
of the voltage,
v(t)
and current,
i(t)
from the
initial state
(t=0)
to
steady state
t
∞
→
*initial conditions are defined before hand, do not
worry on how the circuit got to that state.
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Lecture 21&22
5
1
st
Order Circuit
0
=
dt
(t)
dv
RC
+
(t)
v
C
C
RC
t
c
Ae
t
v
/
)
(
−
=
0
)
(
)
(
=
+
t
v
t
v
C
R
t = 0

+
v
R
(t)
v
C
(t)
i
(t)
R
C
Terminal relations:
+
_
KVL:
dt
(t)
dv
C
=
(t)
C
i
R
t
i
t
v
R
)
(
)
(
=
Differential equation of the circuit:
Solution of the circuit:
Initial condition:
o
V
t
v
=
=
)
0
(
0
/
0
)
0
(
V
Ae
v
RC
c
=
=
−
1
RC
t
o
c
e
V
t
v
/
)
(
−
=
Solution of the form:
Lecture 21&22
6
Graph of
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
RC
RC
t
o
C
e
V
=
(t)
v
−
t
1.0
V
0
(t)
v
C
e
1
2RC
3RC
4RC
e
2
e
3
RC
t
o
c
e
V
t
v
/
)
(
−
=
This gives the familiar result that the voltage on a
capacitor, initially charged to voltage, Vo, will
discharge exponentially with time. The rate of decay is
given by the RC time constant.
Lecture 21&22
7
Another example: an inductor charged with
current is connected in series with a resistor.
Initial condition
(switch closed)
L
(0
−
) = I
o
What is
i(t), t>0?
_
+
v
R
(t)
_
+
v
l
(t)
i
(t)
L
R
0
i(t)=
L
R
dt
di(t)
+
0
=
dt
di(t)
i(t)R + L
0
)
(
)
(
=
+
t
v
t
v
L
R
KVL:
dt
di(t)
(t) = L
v
L
R
t
i
t
v
R
)
(
)
(
=
T.R’s:
(
)
L
R
t
Ae
t
i
/
)
(
−
=
Solution of the form:
Initial condition:
o
L
R
o
I
A
Ae
)=I
i(
=
→
=
⎟
⎠
⎞
⎜
⎝
⎛
−
0
0
(
)
L
R
t
o
e
I
t
i
/
)
(
−
=
Solution:
Lecture 21&22
8
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
L/R
2L/R
3L/R
4L/R
t
1.0
I
0
(t)
L
e
1
e
2
e
3
Graph of the answer
L
Rt
o
e
I
=
(t)
−
i
3
Lecture 21&22
9
1
st
Order
circuits with no driving sources
and excited only by initial conditions
produce exponential decay in time
τ
=
RC
for capacitor
τ
=
L/R
for inductor
L
Rt
o
L
e
I
=
(t)
−
i
RC
t
o
C
e
V
=
(t)
v
−
The rate of decay is determined by the time constant
τ
:
Lecture 21&22
10
1
st
Order excited by a voltage source.
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This note was uploaded on 04/07/2008 for the course ENEE 204 taught by Professor Gomez during the Fall '04 term at Maryland.
 Fall '04
 Gomez
 Steady State

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