2012/10/24
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.
1
2012/10/24
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.
i
_
+ v
–Continuity properties:
•Capacitor voltage: vC (0 ) vC (0 ) (VS-like)
•Inductor current: iL (0 ) iL (0 ) (IS-like)
Example
Q : Find
(a) i(0 ), v(0 ),
(b) di(0 ) dt , dv(0 ) dt ,
(c) i(), v().
Sol : (a) Apply dc analysis for t 0.
12
i (0 ) 
2 A
4 2
v(0 ) 2i(0 ) 4 V

i(0 ) i(0 ) 2 A

 

v(0 ) v(0 ) 4 V

2
2012/10/24
Cont
’
d
Sol : (c)
Apply dc analysis
for t 0.
 i () 0 A
v() 12 V
Cont
’
d
dv(0 )
:
dt
dv
dv iC
C
iC 

dt
dt C
Since the inductor current cannot
Sol : (b) To find
change abruptly.
 The inductor can be treated as
a current source in this case.
We can easily find
iC (0 ) i(0 ) 2 A
dv(0 ) iC (0 )


20 V/s
dt
C
2A
t=
0+
3
2012/10/24
Cont
’
d
di(0 )
:
dt
di
di v
L vL   L
dt
dt L
Since the capacitor voltage
Sol : (b) To find
cannot change abruptly.
 The capacitor can be treated as
a voltage source in this case.
To obtain vL (0 ), applying KVL gives
12 4i(0 ) vL (0 ) vC (0 ) 0
 vL (0 ) 12 8 4 0
Thus we have
di(0 ) vL (0 )
0


0 A/s
dt
L
0.25
t = 0+
The Source-Free Series RLC Circuit
Assumed initial conditions :
i 
0 I 0


1 0

v

0

idt V0
C


C

Applying KVL gives
(1a)
(1b)
di 1 t

idt 0
(2)

dt C 
d 2i R di
i
 2
 0
(3)
dt
L dt LC
To solve (3), di(0) dt is required.
(1) and (2) gives
Ri L
di (0)
V0 0
dt
di(0)
1

 
RI 0 V0  (4)
dt
L
Ri (0) L
d 2i R di
i

 0
2
dt
L dt LC
Initial conditions :
0 I 0
i 

di (0) 1 
RI 0 V0 

L
 dt
4
2012/10/24
Cont
’
d
d 2i R di
i

 0
2
dt
L dt LC
Initial conditions :
2
R
R  1
s1    
2L
2 L  LC
2
R
R  1
s2    
2L
2 L  LC
0 I 0
i 

di
 (0) 1 
RI 0 V0 

L
 dt
Let i Ae st : A and s are constants.
AR
A
 As 2 e st  se st  e st 0
L
LC
1 
 R
 Ae st s 2  s  0
LC 
 L
R
1
Characteristic
 s 2  s  0
equation
L
LC
2
2

s1   0 Natural

2
2 frequencies

s2   0
R
Damping



factor

where  2 L
1 Resonant

0 

LC frequency

(or undamped natural
frequency)
Summary
Characteristic equation :
s 2 2s 02 0

 2 02
s1 

 2 02

s2 
R




where  2 L
1

0 

LC

Two solutions (if s1 s2 ) :
i1 A1e s1t , i2 A2e s2t
A general solution :
i (t ) A1e s1t A2e s2t
where A1 and A2 are determined from
the initial conditions.
•Three cases discussed
–Overdamped case (distinct real roots)
: > 0
–Critically damped case (repeated real root) : = 0
–Underdamped case (complex-conjugate roots): < 0
5
2012/10/24
Overdamped Case (> 0)
R
1
4L

C  2
2L
R
LC
Both s1 and s2 are negative and real.
 i (t ) A1e s1t A2 e s2t
i(t)
e s1t
e s 2t
t
Critically damped Case (= 0)
Let C 4 L R 2
 s1 s2 L 2 R
i (t ) A1e t A2 e t A3e t
Single constant can' t satisfy two initial
conditions!
Back to the original differential equation.
d 2i
di
2 2i 0
2
dt
dt
d di
 di

  i  i 0
dt dt
 dt

di
Let f  i
dt
df

f 0  f A1e t
dt
di
i A1e t
dt
di
et eti A1
dt
d t

e i A1
dt
 et i A1t A2
 
 i(t ) 
A1t A2 
e t
6
2012/10/24
Critically damped Case (Cont
’
d)
i (t ) 
A1t A2 
e t
i(t)
e t
te t
1

t
Underdamped Case (< 0)
Let
C 4 L R 2





 02 2 jd
s1 

 02 2 jd

s2 
where d  02 2
i (t ) B1e (jd ) t B2 e (jd )t
e t ( B1e jd t B2 e jd t )

e t 
B1 
cos d t j sin d t 
B2 
cos d t j sin d t 
B1 B2 cos d t j B1 B2 sin d t 
e t 
A B1 B2
 i(t ) e t 
A1 cos d t A2 sin d t  where  1
B1 B2 
A2 j 
7
2012/10/24
Underdamped Case (Cont
’
d)
R
i (t ) 
A1 cos d t A2 sin d t 
e t , α
2L
i(t)
e t
2
d
t
Finding The Constants A1,2
To determine A1 and A2 ,
we need i (0) and di (0) /dt.
1. i (0) I 0
2. KVL at t 0 gives
di (0)
L
RI 0 V0 0
dt
di (0)
1
or
 ( RI 0 V0 )
dt
L
8
2012/10/24
Conclusions
•The concept of damping
–The gradual loss of the initial stored energy
–Due to the resistance R
•Oscillatory response is possible.
–The energy is transferred between L and C.
–Ringing denotes the damped oscillation in the
underdamped case.
•With the same initial conditions, the
overdamped case has the longest settling time.
The underdamped case has the fastest decay.
(If a constant 0 is assumed.)
Example
Find i(t).
(6+3)
t<0
t>0
9
2012/10/24
Example (Cont
’
d)
t<0
t>0
10
(a) i (0) 
1 A, v (0) 6i (0) 6 V
4 6
R
1
1
(b) α 9 , ω0 

10
2L
LC
0.01
Initial conditions :
i( 0 ) 1


di (0) 1 
Ri (0) v(0) 
 dt
L


2
9(1) 6 6

-
 s1,2 α α2 ω02 9  81 100
9 j 4.359
 i (t ) e
9 t
A1 cos 4.359t A2 sin 4.359t 
A 1
 1
A2 0.6882
The Source-Free Parallel RLC Circuit
Assumed initial conditions :
1 0


i 
0 I 0  v(t )dt

L 


v 0 V0

(1a)
(1b)
Applying KCL gives
v 1 t
dv
 vdt C
0 (2)


R L
dt
d 2v
1 dv
v
 2 
 0 (3)
dt
RC dt LC
Let v (t ) Ae st , the characteristic
equation becomes
1
1
s 2  s  0
RC
LC
s1, 2  2 02
1




where  2 RC
1
0 

LC

10
2012/10/24
Summary
•Overdamped case
: > 0
v(t ) A1e s1t A2 e s2t
•Critically damped case
v(t ) 
A1 A2t 
e t
•Underdamped case
s1, 2 
jd
: = 0
: < 0
where d  02 2
v(t ) e t 
A1 cos d t A2 sin d t 
Finding The Constants A1,2
To determine A1 and A2 ,
we need v(0) and dv(0) /dt.
1. v(0) V0
2. KCL at t 0 gives
V0
dv(0)
I 0 C
0
R
dt
dv(0)
(V RI 0 )
or
 0
dt
RC
11
2012/10/24
Comparisons
•Series RLC Circuit
•Parallel RLC Circuit
s1, 2  2 02
s1, 2 
 2 02
R




where  2 L
1

0 

LC

Initial conditions :
1




where  2 RC
1
0 

LC

Initial conditions :
i (0) I 0


di(0)

V RI 0 
 0

L
 dt
v(0) V0


dv(0)

V RI 0 
 0

RC
 dt
Example 1
Find v(t) for t > 0.
v(0) = 5 V, i(0) = 0
Consider three cases:
R = 1.923 
R=5
R =6.25 
Case 1 : R 1.923 
1
1
α
26 , ω0 
10
2 RC
LC
 s1,2 α α2 ω02 2, 50
v(t ) A1e 2t A2 e 50t
Initial conditions :
v (0) 5


dv(0) v(0) Ri (0) 260

RC
 dt
A 0.2083
 1
A2 5.208
12
2012/10/24
Example 1 (Cont
’
d)
Case 2 : R 5 
Initial conditions :
1
1
α
10, ω0 
10
2 RC
LC
v (0) 5


dv(0) v(0) Ri (0) 100

RC
 dt
A 5
 1
A2 50
 s1,2 α α2 ω02 10
v(t ) 
A1 A2t 
e 10t
Case 3 : R 6.25 
Initial conditions :
1
1
α
8, ω0 
10
2 RC
LC
v (0) 5


dv(0) v(0) Ri (0) 80

RC
 dt
A 5
 1
A2 6.667
 s1,2 α α2 ω02 8 j 6
v(t ) 
A1 cos 6t A2 sin 6t 
e 8t
Example 1 (Cont
’
d)
13
2012/10/24
Example 2
Find v(t).
Get x(0).
Get x(), dx(0)/dt, s1,2, A1,2.
t<0
t>0
Example 2 (Cont
’
d)
t>0
1

α
500

 2 RC

1

ω0 
354

LC

 s1,2 α α2 ω02
854, 146
v(t ) A1e 854t A2 e 146t
t<0
From the initial conditions :
50

v(0) 
(40) 25 V

30 50

i (0)  40 0.5 A

30 50

dv
(
0
)
v(0) Ri(0) 25 50 0.5



0

RC
50 20 10 6
 dt
A1 5.156

A2 30.16
14
2012/10/24
Step Response of A Series RLC Circuit
Applying KVL for t 0,
di
v VS (1)
dt
dv
But i C
dt
2
d v R dv
v
V
 2 
  S (2)
dt
L dt LC LC
Ri L
(2) has the same form as
in the source - free case.
v(t ) vt (t ) vss (t )
where
vt : the transient response


vss : the steady - state response

Characteristic Equation
d 2 v R dv v VS


0
dt 2 L dt
LC
Let v ' v VS ,
d 2 v ' R dv ' v '
 2 
 0
dt
L dt LC
The characteristic equation becomes
R
1
 s 2  s  0
L
LC
Same as in the source - free case.
15
2012/10/24
Summary
 v(t ) vt (t ) vss (t )
vt () 0

where 
vss () v() VS

A1e s1t A2 e s2t
(Overdamped)

A1 A2t 
vt (t ) 
e t
(Critically damped)

A1 cos d t A2 sin d t 
e t (Underdamped)

where A1, 2 are obtained from v(0) and dv(0) /dt.
Example
Find v(t), i(t) for t > 0.
Consider three cases:
R=5
R=4
R =1 
Get x(0).
t<0
Get x(), dx(0)/dt, s1,2, A1,2.
t>0
16
2012/10/24
Case 1: R = 5 
t>0
t<0
5
 R
α 
2.5

 2 L 2(1)

1

ω0 
2

LC

vss v() 24 V
Initial conditions:
 s1,2 α α2 ω02
1, 4

v(t ) vss  A1e t A2 e 4t
i (t ) C

dv
dt
24

i (0) 
4 A , v(0) 1i (0) 4 V


5 1

i (0) C dv(0)  dv(0) 4 16

dt
dt
C

A 64 3
 1
A2 4 3
Case 2: R = 4 
4
 R
α 
2

 2 L 2(1)

1

ω0 
2

LC

 s1,2 α2
v(t ) vss 
A1 A2t 
e 2t
i(t ) C
dv
dt
vss v() 24 V
Initial conditions :
24

i ( 0) 
4.8 A , v(0) 1i (0) 4.8 V

4 1

dv(0)
dv(0) 4.8
i (0) C

 19.2
dt
dt
C

A 19.2
 1
A2 19.2
17
2012/10/24
Case 3: R = 1 
1
 R
α 
0.5

 2 L 2(1)

1

ω0 
2

LC

 s1,2 0.5 j1.936
A1 cos 1.936t 0.5t
v(t ) vss 
e
A sin 1.936t 

 2

i (t ) C
vss v() 24
Initial conditions :
24

i ( 0) 
12 A , v(0) 1i (0) 12 V

1 1

dv(0)
dv(0) 12

i (0) C

 48
dt
dt
C

A 12
 1
A2 21.694
dv
dt
Example (Cont
’
d)
18
2012/10/24
Step Response of A Parallel RLC Circuit
Applying KCL for t 0,
v
dv
i C
I S
(1)
R
dt
di
But v L
dt
2
d i
1 di
i
I
 2
  S (2)
dt
RC dt LC LC
(2) has the same form as
in the source - free case.
i(t ) it (t ) iss (t )
where
it : the transient response


iss : the steady - state response

Characteristic Equation
d 2i
1 di i I S


0
dt 2 RC dt
LC
Let i ' i I S ,
d 2i '
1 di '
i'
 2 
 0
dt
RC dt LC
The characteristic equation becomes
1
1
 s 2  s  0
RC
LC
Same as in the source - free case.
19
2012/10/24
Summary
 i (t ) it (t ) iss (t )
it () 0

where 
iss (t ) i () I S

(Overdamped)
A1e s1t A2e s2t

(Critically damped)
it (t ) 
A1 A2t 
e t

A1 cos d t A2 sin d t 
e t (Underdamped)

where A1, 2 are obtained from i (0) and di (0) /dt.
General Second-Order Circuits
•Steps required to determine the step response.
–Determine x(0), dx(0)/dt, and x().
–Find the transient response xt(t).
•Apply KCL and KVL to obtain the differential equation.
•Determine the characteristic roots (s1,2).
•Obtain xt(t) with two unknown constants (A1,2).
–Obtain the steady-state response xss(t) = x().
–Use x(t) = xt(t) + xss(t) to determine A1,2 from the
two initial conditions x(0) and dx(0)/dt.
20
2012/10/24
Example
Find v, i
for t > 0.
Get x(0).
Get x(), dx(0)/dt, s1,2, A1,2.
t<0
t>0
Example (Cont
’
d)
t<0
t>0
Initial conditions :

v(0 ) v(0 ) 12 V (1a)

 

i (0 ) i (0 ) 0
(1b)

Applying KCL at node a (t 0),

v (0 )
i (0 ) iC (0 ) 
2

 iC (0 ) 6 A
dv (0 ) iC (0 )

12 V/s (1c)
dt
C
Final values for t  :

12

i() 
2 A

4 2


v() 2i () 4 V

21
2012/10/24
Example (Cont
’
d)
Applying KCL at node a gives
v 1 dv
i 
(2)
2 2 dt
Applying KVL to the left mesh gives
di
4i 1 v 12
(3)
dt
Substituti ng (2) into (3) gives
dv 1 dv 1 d 2 v
2v 2 

v 12
dt 2 dt 2 dt 2
d 2v
dv
 2 5 6v 24 (4)
dt
dt
Characteristic equation :
s 5s 6 0
2
t>0
 s 2, 3
v (t ) vss vt (t )
vss v() 4

where 
vt (t ) A1e 2t A2e 3t

From (1a) and (1c) we obtain
 A1 12, A2 8
i (t ) can be obtain by using (2)
Duality
•Duality means the same characterizing equations with
dual quantities interchanged.
Table for dual pairs
Resistance R
Inductance L
Voltage v
Voltage source
Node
Series path
Open circuit
KVL
Thevenin
Conductance G
Capacitance C
Current i
Current source
Mesh
Parallel path
Short circuit
KCL
Norton
22
2012/10/24
A Case Study
i
i1
….
+ v1 -
+ v2 -
+ vn -
i2
in
….
v1 f1 (i )
i1 f1 (v)
v2 f 2 (i )
i2 f 2 (v)


vn f n (i )
in f n (v)
KVL : v1 v2  vn 0
+
v
_
KCL : i1 i2  in 0
Element Transformations
v Ri  i Rv
(Conductance R)
i C
dv
di
 v C
dt
dt
(Inductance C )
v L
di
dv
 i L
dt
dt
(Capacitance L)
v VS
 i VS
(Current VS )
i I S
 v I S
(Voltage I S )
23
2012/10/24
Example 1
•Series RLC Circuit
•Parallel RLC Circuit
R
1
s 2  s  0
L
LC
R
1
s 2  s  0
L
LC
Example 2
24
2012/10/24
Application: Smoothing Circuits
Output
from a
D/A
converter
vs v0
25