Second order circuits (Step response & natural response) | Network Theory (Electric Circuits) - Electrical Engineering (EE) PDF Download

 Page 1


1
Second-Order Circuits
• Introduction
• Finding Initial and Final Values
• The Source-Free Series RLC Circuit
• The Source-Free Parallel RLC Circuit
• Step Response of a Series RLC Circuit
• Step Response of a Parallel RLC Circuit
• General Second-Order Circuits
• Duality
• Applications
Introduction
• A second-order circuit is characterized by a
second-order differential equation.
• It consists of resistors and the equivalent of
two energy storage elements.
Page 2


1
Second-Order Circuits
• Introduction
• Finding Initial and Final Values
• The Source-Free Series RLC Circuit
• The Source-Free Parallel RLC Circuit
• Step Response of a Series RLC Circuit
• Step Response of a Parallel RLC Circuit
• General Second-Order Circuits
• Duality
• Applications
Introduction
• A second-order circuit is characterized by a
second-order differential equation.
• It consists of resistors and the equivalent of
two energy storage elements.
2
Finding Initial and Final Values
• v(0), i(0), dv(0)/dt, di(0)/dt, v( ? ), and i( ? )
• Two key points:
– v and i are defined according to the passive sign
convention.
– Continuity properties:
• Capacitor voltage: (V
S
-like)
• Inductor current: (I
S
-like) ) 0 ( ) 0 (
) 0 ( ) 0 (
? ?
? ?
?
?
L L
C C
i i
v v
v
i
+
_
Example
). ( , ) ( (c)
, ) 0 ( , ) 0 ( (b)
), 0 ( , ) 0 ( (a)
Find : Q
? ?
? ?
? ?
v i
dt dv dt di
v i
?
?
?
?
?
? ?
? ?
?
? ?
?
?
?
?
? ?
? ?
? ?
?
V 4 ) 0 ( ) 0 (
A 2 ) 0 ( ) 0 (
V 4 ) 0 ( 2 ) 0 (
A 2
2 4
12
) 0 (
0. for t analysis dc Apply (a) : Sol
v v
i i
i v
i
Page 3


1
Second-Order Circuits
• Introduction
• Finding Initial and Final Values
• The Source-Free Series RLC Circuit
• The Source-Free Parallel RLC Circuit
• Step Response of a Series RLC Circuit
• Step Response of a Parallel RLC Circuit
• General Second-Order Circuits
• Duality
• Applications
Introduction
• A second-order circuit is characterized by a
second-order differential equation.
• It consists of resistors and the equivalent of
two energy storage elements.
2
Finding Initial and Final Values
• v(0), i(0), dv(0)/dt, di(0)/dt, v( ? ), and i( ? )
• Two key points:
– v and i are defined according to the passive sign
convention.
– Continuity properties:
• Capacitor voltage: (V
S
-like)
• Inductor current: (I
S
-like) ) 0 ( ) 0 (
) 0 ( ) 0 (
? ?
? ?
?
?
L L
C C
i i
v v
v
i
+
_
Example
). ( , ) ( (c)
, ) 0 ( , ) 0 ( (b)
), 0 ( , ) 0 ( (a)
Find : Q
? ?
? ?
? ?
v i
dt dv dt di
v i
?
?
?
?
?
? ?
? ?
?
? ?
?
?
?
?
? ?
? ?
? ?
?
V 4 ) 0 ( ) 0 (
A 2 ) 0 ( ) 0 (
V 4 ) 0 ( 2 ) 0 (
A 2
2 4
12
) 0 (
0. for t analysis dc Apply (a) : Sol
v v
i i
i v
i
3
C o n t ’ d
V 12 ) (
A 0 ) (
. 0 for
analysis dc Apply
(c) : Sol
? ?
? ? ?
?
v
i
t
C o n t ’ d
V/s 20
0 0
A 2 0 0
find easily can We
case. in this source current a
as treated be can inductor The
abruptly. change
cannot current inductor the Since
:
0
find To (b) : Sol
? ? ?
? ?
?
?
?
? ? ?
? ?
? ?
?
C
i
dt
dv
i i
C
i
dt
dv
i
dt
dv
C
dt
dv
C
C
C
C
) ( ) (
) ( ) (
) (
?
t = 0
+
2 A
Page 4


1
Second-Order Circuits
• Introduction
• Finding Initial and Final Values
• The Source-Free Series RLC Circuit
• The Source-Free Parallel RLC Circuit
• Step Response of a Series RLC Circuit
• Step Response of a Parallel RLC Circuit
• General Second-Order Circuits
• Duality
• Applications
Introduction
• A second-order circuit is characterized by a
second-order differential equation.
• It consists of resistors and the equivalent of
two energy storage elements.
2
Finding Initial and Final Values
• v(0), i(0), dv(0)/dt, di(0)/dt, v( ? ), and i( ? )
• Two key points:
– v and i are defined according to the passive sign
convention.
– Continuity properties:
• Capacitor voltage: (V
S
-like)
• Inductor current: (I
S
-like) ) 0 ( ) 0 (
) 0 ( ) 0 (
? ?
? ?
?
?
L L
C C
i i
v v
v
i
+
_
Example
). ( , ) ( (c)
, ) 0 ( , ) 0 ( (b)
), 0 ( , ) 0 ( (a)
Find : Q
? ?
? ?
? ?
v i
dt dv dt di
v i
?
?
?
?
?
? ?
? ?
?
? ?
?
?
?
?
? ?
? ?
? ?
?
V 4 ) 0 ( ) 0 (
A 2 ) 0 ( ) 0 (
V 4 ) 0 ( 2 ) 0 (
A 2
2 4
12
) 0 (
0. for t analysis dc Apply (a) : Sol
v v
i i
i v
i
3
C o n t ’ d
V 12 ) (
A 0 ) (
. 0 for
analysis dc Apply
(c) : Sol
? ?
? ? ?
?
v
i
t
C o n t ’ d
V/s 20
0 0
A 2 0 0
find easily can We
case. in this source current a
as treated be can inductor The
abruptly. change
cannot current inductor the Since
:
0
find To (b) : Sol
? ? ?
? ?
?
?
?
? ? ?
? ?
? ?
?
C
i
dt
dv
i i
C
i
dt
dv
i
dt
dv
C
dt
dv
C
C
C
C
) ( ) (
) ( ) (
) (
?
t = 0
+
2 A
4
C o n t ’ d
t = 0
+
A/s 0
25 0
0 0 0
have we Thus
0 4 8 12 0
0 0 0 0 4 12
gives KVL applying , 0 obtain To
case. in this source voltage a
as treated be can capacitor The
abruptly. change cannot
voltage capacitor the Since
0
find To (b) : Sol
? ? ?
?
? ? ? ? ?
? ? ? ? ?
?
?
?
? ? ?
? ?
?
? ? ?
?
?
.
) ( ) (
) (
) ( ) ( ) (
) (
:
) (
L
v
dt
di
v
v v i
v
L
v
dt
di
v
dt
di
L
dt
di
L
L
C L
L
L
L
?
The Source-Free Series RLC Circuit
? ?
? ?
? ? ) 4 (
1 ) 0 (
0
) 0 (
) 0 (
gives (2) and (1)
required. is ) 0 ( (3), solve To
(3) 0
(2) 0
1
gives KVL Applying
(1b)
1
0
(1a) 0
: conditions initial Assumed
0 0
0
2
2
0
0
0
V RI
L dt
di
V
dt
di
L Ri
dt di
LC
i
dt
di
L
R
dt
i d
idt
C dt
di
L Ri
V idt
C
v
I i
t
C
? ? ? ?
? ? ?
? ? ? ?
? ? ?
?
?
?
?
?
? ?
?
?
?
? ?
? ?
? ?
? ?
?
?
?
?
?
? ? ?
?
? ? ?
0 0
0
2
2
1 ) 0 (
0
: conditions Initial
0
V RI
L dt
di
I i
LC
i
dt
di
L
R
dt
i d
Page 5


1
Second-Order Circuits
• Introduction
• Finding Initial and Final Values
• The Source-Free Series RLC Circuit
• The Source-Free Parallel RLC Circuit
• Step Response of a Series RLC Circuit
• Step Response of a Parallel RLC Circuit
• General Second-Order Circuits
• Duality
• Applications
Introduction
• A second-order circuit is characterized by a
second-order differential equation.
• It consists of resistors and the equivalent of
two energy storage elements.
2
Finding Initial and Final Values
• v(0), i(0), dv(0)/dt, di(0)/dt, v( ? ), and i( ? )
• Two key points:
– v and i are defined according to the passive sign
convention.
– Continuity properties:
• Capacitor voltage: (V
S
-like)
• Inductor current: (I
S
-like) ) 0 ( ) 0 (
) 0 ( ) 0 (
? ?
? ?
?
?
L L
C C
i i
v v
v
i
+
_
Example
). ( , ) ( (c)
, ) 0 ( , ) 0 ( (b)
), 0 ( , ) 0 ( (a)
Find : Q
? ?
? ?
? ?
v i
dt dv dt di
v i
?
?
?
?
?
? ?
? ?
?
? ?
?
?
?
?
? ?
? ?
? ?
?
V 4 ) 0 ( ) 0 (
A 2 ) 0 ( ) 0 (
V 4 ) 0 ( 2 ) 0 (
A 2
2 4
12
) 0 (
0. for t analysis dc Apply (a) : Sol
v v
i i
i v
i
3
C o n t ’ d
V 12 ) (
A 0 ) (
. 0 for
analysis dc Apply
(c) : Sol
? ?
? ? ?
?
v
i
t
C o n t ’ d
V/s 20
0 0
A 2 0 0
find easily can We
case. in this source current a
as treated be can inductor The
abruptly. change
cannot current inductor the Since
:
0
find To (b) : Sol
? ? ?
? ?
?
?
?
? ? ?
? ?
? ?
?
C
i
dt
dv
i i
C
i
dt
dv
i
dt
dv
C
dt
dv
C
C
C
C
) ( ) (
) ( ) (
) (
?
t = 0
+
2 A
4
C o n t ’ d
t = 0
+
A/s 0
25 0
0 0 0
have we Thus
0 4 8 12 0
0 0 0 0 4 12
gives KVL applying , 0 obtain To
case. in this source voltage a
as treated be can capacitor The
abruptly. change cannot
voltage capacitor the Since
0
find To (b) : Sol
? ? ?
?
? ? ? ? ?
? ? ? ? ?
?
?
?
? ? ?
? ?
?
? ? ?
?
?
.
) ( ) (
) (
) ( ) ( ) (
) (
:
) (
L
v
dt
di
v
v v i
v
L
v
dt
di
v
dt
di
L
dt
di
L
L
C L
L
L
L
?
The Source-Free Series RLC Circuit
? ?
? ?
? ? ) 4 (
1 ) 0 (
0
) 0 (
) 0 (
gives (2) and (1)
required. is ) 0 ( (3), solve To
(3) 0
(2) 0
1
gives KVL Applying
(1b)
1
0
(1a) 0
: conditions initial Assumed
0 0
0
2
2
0
0
0
V RI
L dt
di
V
dt
di
L Ri
dt di
LC
i
dt
di
L
R
dt
i d
idt
C dt
di
L Ri
V idt
C
v
I i
t
C
? ? ? ?
? ? ?
? ? ? ?
? ? ?
?
?
?
?
?
? ?
?
?
?
? ?
? ?
? ?
? ?
?
?
?
?
?
? ? ?
?
? ? ?
0 0
0
2
2
1 ) 0 (
0
: conditions Initial
0
V RI
L dt
di
I i
LC
i
dt
di
L
R
dt
i d
5
C o n t ’ d
? ?
? ?
0
1
0
1
0
constants. are and : Let
1 ) 0 (
0
: conditions Initial
0
2
2
2
0 0
0
2
2
? ? ? ?
? ?
?
?
?
?
?
? ? ?
? ? ? ?
?
?
?
?
?
?
? ? ?
?
? ? ?
LC
s
L
R
s
LC
s
L
R
s Ae
e
LC
A
se
L
AR
e As
s A Ae i
V RI
L dt
di
I i
LC
i
dt
di
L
R
dt
i d
st
st st st
st
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ? ?
? ? ? ?
? ?
?
?
?
?
?
? ? ?
? ?
?
?
?
?
?
? ? ?
LC
L
R
s
s
LC L
R
L
R
s
LC L
R
L
R
s
1
2
where
1
2 2
1
2 2
0
2
0
2
2
2
0
2
1
2
2
2
1
?
?
? ? ?
? ? ?
Characteristic
equation
Natural
frequencies
Damping
factor
Resonant
frequency
(or undamped natural
frequency)
Summary
. conditions initial the
from determined are and where
) (
: solution general A
,
: ) (if solutions Two
2 1
2 1
2 2 1 1
2 1
2 1
2 1
A A
e A e A t i
e A i e A i
s s
t s t s
t s t s
? ?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ? ?
? ? ? ?
? ? ?
LC
L
R
s
s
s s
1
2
where
0 2
: equation stic Characteri
0
2
0
2
2
2
0
2
1
2
0
2
?
?
? ? ?
? ? ?
? ?
• Three cases discussed
– Overdamped case (distinct real roots) : ?> ?
0
– Critically damped case (repeated real root) : ?= ?
0
– Underdamped case (complex-conjugate roots): ?< ?
0