Second-Order Circuits
Instructor: Chia-Ming Tsai
Electronics Engineering
National Chiao Tung University
Hsinchu, Taiwan, R.O.C.
Contents
•
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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
Finding Initial and Final Values
• v and i are defined according to the passive
sign convention
i
_
v
+
• Continuity properties
– Capacitor voltage
– Inductor current


vC (0 )  vC (0 )


iL (0 )  iL (0 )
Example
Find
(a) i (0  ), v(0  ),
(b) di(0  ) dt , dv(0  ) dt ,
(c) i (), v().
Sol : (a)
12
i (0 ) 
2A
42
v(0  )  2i (0  )  4 V

 i (0  )  i (0  )  2 A
 v (0  )  v (0  )  4 V
Example (Cont’d)
Sol : (b)
iC (0  )  i (0  )  2 A
dv
dv iC
C
 iC ,

dt
dt C
dv(0  ) iC (0  )
2


 20 V/s
dt
C
0 .1
di
di vL
 L  vL ,

dt
dt L
T o obt ain vL , applyingKVL gives
 12  4i (0  )  vL (0  )  v(0  )  0
 vL (0  )  12  8  4  0
di(0  ) vL (0  )

dt
L
0

 0 A/s
0.25
Example (Cont’d)
Sol : (c)
i ( )  0 A
v()  12 V
The Source-Free Series RLC Circuit
Assumed initial conditions:
i 0  I 0

v 0  1 0 idt  V
0
 C
C  
Applying KVL gives
di 1 t
Ri  L   idt  0
dt C 
d 2i R di
i
 2

0
dt
L dt LC
(1) and (2) gives
di(0)
Ri(0)  L
 V0  0
dt
di(0)
1

  RI0  V0 
dt
L
(1a)
(1b)
(2)
(3)
(4)
d 2i R di
i


0
2
dt
L dt LC
Init ial condit ions:
 i 0   I 0

 di(0)   1 RI  V 
0
0
 dt
L
Cont’d
d 2i R di
i


0
2
dt
L dt LC
Initial conditions:
i 0  I 0

 di(0)   1 RI  V 
0
0
 dt
L
Let i  Aest : A and s are constants
AR st
A st
2 st
 As e 
se 
e 0
L
LC
R
1 
st  2
 Ae  s  s 
0
L
LC 

R
1
Characteristic
 s2  s 
0
equation
L
LC
2
s1  
R
1
 R 
 


2L
 2 L  LC
2
s2  
R
1
 R 
 


2L
 2 L  LC
 s     2   2 Natural
1
0

2
2
s2      0 frequencies
R

Damping



factor
2L
where 
1 Resonant
0 

LC frequency
Summary
 s     2   2
1
0

2
2
s







 2
0
R

  2 L
where 
1
0 

LC
T wo solutions (if s1  s2 ) :
i1  A1e , i2  A2 e
s1t
s2t
A general solution:
i (t )  A1e s1t  A2e s2t
• Three cases discussed
– Overdamped case
– Critically damped case
– Underdamped case
:  > 0
:  = 0
:  < 0
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
s2t
t
Critically damped Case ( = 0)
C  4L R 2
 s1  s2     L 2 R
di
Let f   i
dt
df
t
t
t

 f  0  f  A1e t
i (t )  A1e  A2 e  A3e
dt
Single constantcan't satisfy two init ial
di
t


i

A
e
1
condit ions!
dt
t di
e
 eti  A1
dt
Back totheoriginaldifferential equation.
d t
d 2i
di

e i  A1
2

2



i

0
dt
dt 2
dt
t

e
i  A1t  A2
d  di

 di

   i      i   0
t



i
(
t
)

A
t

A
e
1
2
dt  dt

 dt

 
Critically damped Case (Cont’d)
i(t )   A1t  A2 e
t
i(t)
e t
te
1

t
t
Underdamped Case ( < 0)
C  4L R 2




 s      2   2    j
1
0
d

2
2
s2     0      jd
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 
 e t B1  B2  cosd t  j B1  B2 sin d t 
 i (t )  e
t
 A1 cosd t  A2 sin d t 
 A1  B1  B2
where 
 A2  j B1  B2 
Underdamped Case (Cont’d)
i(t )   A1 cosd t  A2 sin d t e
i(t)
e
t
t
2
d
t
Finding The Constants A1,2
T o determineA1 and A2 ,
we need i (0) and di(0) /dt.
 i ( 0)  I 0
 di(0)
 RI0  V0  0
L
  dt
 di(0)
1
  ( RI0  V0 )
or
dt
L

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 critically damped case has the fastest
decay.
Example
Find i(t).
t<0
t>0
Example (Cont’d)
t<0
10
 1 A, v(0)  6i (0)  6 V
46
R
1
1
α
 9, ω0 

 10
2L
LC
0.01
t>0
i (0) 
Init ialcondit ions:
 s1,2  α  α 2  ω02  9  81 100
i( 0 )  1
 di(0)
1
 dt   L Ri(0)  v(0) 


 29(1)  6   6
 9  j 4.359
 i (t )  e 9t  A1 cos 4.359t  A2 sin 4.359t 
 A1  1

 A2  0.6882
The Source-Free Parallel RLC Circuit
Assumed init ial condit ions:
1 0

 i 0   I 0   v(t)dt

L 
v0   V0
Applying KCL gives
(1a)
(1b)
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)  Aest ,
t he charact erist ic equat ion becomes
1
1
2
s 
s
0
RC
LC
s1, 2     2  02
1

  2 RC
where 
1
 0 

LC
Summary
:  > 0
• Overdamped case
v(t )  A1e  A2e
s1t
s 2t
• 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
T o determineA1 and A2 ,
we need v(0) and dv(0) /dt.
 v(0)  V0
V0
dv(0)
0
  I0  C
R
dt

(V0  RI0 )
dv(0)

or
RC
dt

Comparisons
• Series RLC Circuit
• Parallel RLC Circuit
s1, 2     2  02
s1, 2     2  02
R

  2 L
where 
1
0 

LC
Init ial condit ions:
1

  2 RC
where 
1
 0 

LC
Initial conditions:
i (0)  I 0

 di(0)

V0  RI0 
 dt  
L
v(0)  V0

 dv(0)

V0  RI0 
 dt   RC
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
α
 26, ω0 
2 RC
Initialconditions:
1
 10
LC
 s1,2  α  α 2  ω02  2,  50
v(t )  A1e  2t  A2 e 50 t
v(0)  5

 dv(0)   v(0)  Ri(0)  260
 dt
RC
 A1  0.2083

 A2  5.208
Example 1 (Cont’d)
Case 2 : R  5 
1
α
 10, ω0 
2 RC
Initialconditions:
1
 10
LC
 s1,2  α  α 2  ω02  10
v(t )   A1  A2t e 10t
Case 3 : R  6.25 
1
α
 8, ω0 
2 RC
v(0)  5

 dv(0)   v(0)  Ri(0)  100
 dt
RC
 A1  5

 A2  50
Initialconditions:
1
 10
LC
 s1,2  α  α 2  ω02  8  j 6
v(t )   A1 cos6t  A2 sin 6t e 8t
v(0)  5

 dv(0)   v(0)  Ri(0)  80
 dt
RC
 A1  5

 A2  6.667
Example 1 (Cont’d)
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  2 RC  500

ω0  1  354

LC
 s1,2  α  α 2  ω02
 854,  146
v(t )  A1e 854 t  A2 e 146 t
t<0
From t heinit ialcondit ions:
50

v(0)  30  50 (40)  25 V

 i (0)   40  0.5 A

30  50

 dv(0)   v(0)  Ri(0)  25  50 0.5  0
 dt
RC
50 20 106
 A1  5.156

 A2  30.16
Step Response of A Series RLC Circuit
Applying KVL for t  0,
di
Ri  L  v  VS (1)
dt
dv
But i  C
dt
VS
d 2 v R dv v
 2 


(2)
dt
L dt LC LC
v(t )  vt (t )  vss (t )
where
(2) has t hesame formas
in t hesource - free case.
vt : the t ransientresponse

vss : thest eady - st ateresponse
Characteristic Equation
d 2 v R dv v  VS


0
2
dt
L dt
LC
Let v '  v  VS ,
d 2 v ' R dv' v '
 2 

0
dt
L dt LC
T hecharacterist ic equation becomes
R
1
2
 s  s
0
L
LC
Same as in thesource - free case.
Summary
 v(t )  vt (t )  vss (t )
vss (t )  v()  VS
(Overdamped)
 A1e s1t  A2 e s2t

t
(Critically damped)
vt (t )   A1  A2t e
 A cos t  A sin  t e t (Underdamped)
d
2
d
 1
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
Case 1: R = 5 
vss  v()  24 V
R
5

α  2 L  2(1)  2.5

ω  1  2
 0
LC
Initialconditions:
 s1,2  α  α 2  ω02
 1,  4

v(t )  vss  A1e t  A2 e  4t
dv
i (t )  C
dt

24

i (0)  5  1  4 A , v(0)  1i (0)  4 V

dv(0)
dv(0) 4
i (0)  C

  16
dt
dt
C

 A1   64 3

 A2  4 3
Case 2: R = 4 
R
4

α  2 L  2(1)  2

ω  1  2
 0
LC
 s1,2  α  2
v(t )  vss   A1  A2t e  2t
i (t )  C
dv
dt
vss  v()  24 V
Init ialconditions:
24

i (0)  4  1  4.8 A , v(0)  1i (0)  4.8 V

dv(0)
dv(0) 4.8
i (0)  C


 19.2
dt
dt
C

 A1  19.2

 A2  19.2
Case 3: R = 1 
R
1

α


 0.5

2 L 2(1)

ω  1  2
 0
LC
 s1,2  0.5  j1.936
 A1 cos1.936t  0.5t
e
v(t )  vss  
  A2 sin 1.936t 
dv
i (t )  C
dt
vss  v()  24
Initialconditions:
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

 A1  12

 A2  21.694
Example (Cont’d)
Step Response of A Parallel RLC Circuit
Applying KCL for t  0,
v
dv
iC
 IS
(1)
R
dt
di
But v  L
dt
IS
d 2i
1 di
i
 2


(2)
dt
RC dt LC LC
(2) has t hesame formas
in t hesource - free case.
i (t )  it (t )  iss (t )
where
it : t he t ransientresponse

iss : t hesteady - stat eresponse
Characteristic Equation
d 2i
1 di i  I S


0
2
dt
RC dt LC
Let i '  i  I S ,
d 2i '
1 di'
i'
 2 

0
dt
RC dt LC
T hecharacterist ic equation becomes
1
1
2
s 
s
0
RC
LC
Same as in thesource - free case.
Summary
 i (t )  it (t )  iss (t )
iss (t )  i ()  I S
 A1e s1t  A2 e s2t
(Overdamped)

t
(Critically damped)
it (t )    A1  A2t e
 A cos t  A sin  t e t (Underdamped)
d
2
d
 1
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
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
Initialconditions:
v(0  )  v(0  )  12 V (1a)
 
i (0 )  i (0  )  0
(1b)
ApplyingKC at node a (t  0),
v (0  )
i (0 )  iC (0 ) 
2
 iC (0  )  6 A


t>0
dv(0  ) iC (0  )


 12 V/s (1c)
dt
C
Final values for t   :
12

i
(

)

2A

42

v()  2i ()  4 V
Example (Cont’d)
ApplyingKCL at node a gives
v 1 dv
i 
(2)
2 2 dt
ApplyingKVL to theleft mesh gives
dv
 4i  1  v  0
(3)
dt
Substituting (2) into(3) gives
dv 1 dv 1 d 2 v
2v  2 

v 0
2
dt 2 dt 2 dt
d 2v
dv
 2  5  6v  0 (4)
dt
dt
Characteristic equation:
s 2  5s  6  0
t>0
 s  2,  3
v(t )  vss  vt (t )
vss  v()  4
where 
 2t
 3t
v
(
t
)

A
e

A
e
1
2
 t
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
Conductance G
Capacitance C
Voltage v
Voltage source
Node
Series path
Current i
Current source
Mesh
Parallel path
Open circuit
KVL
Thevenin
Short circuit
KCL
Norton
Example 1
• Series RLC Circuit
• Parallel RLC Circuit
R
1
s  s
0
L
LC
1
1
s 
s
0
RC
LC
2
2
Example 2
Application: Smoothing Circuits
Output
from
a D/A
vs v0