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Second-Order Circuits
Instructor: Chia-Ming TsaiElectronics Engineering
National Chiao Tung UniversityHsinchu, Taiwan, R.O.C.
Contents• 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
• Continuity properties
– Capacitor voltage
– Inductor current )0()0(
)0()0(
LL
CC
ii
vv
v
i
+ _
Example
).(,)( (c)
,)0(,)0( (b)
),0(,)0( (a)
Find
vi
dtdvdtdi
vi
V 4)0()0(
A 2)0()0(
V 4)0(2)0(
A 224
12)0(
(a):Sol
vv
ii
iv
i
Example (Cont’d)
04812)0(
0)0()0()0(412
gives KVL applying ,obtain To
,
V/s 201.0
2)0()0(
,
A 2)0()0(
(b):Sol
L
L
L
LL
C
CC
C
v
vvi
vL
v
dt
div
dt
diL
C
i
dt
dv
C
i
dt
dvi
dt
dvC
ii
A/s 025.0
0
)0()0(
L
v
dt
di L
Example (Cont’d)
V 12)(
A 0)(
(c):Sol
v
i
The Source-Free Series RLC Circuit
)4( 1)0(
0)0(
)0(
gives (2) and (1)
(3) 0
(2) 01
gives KVL Applying
(1b)
10
(1a) 0
: conditions initial Assumed
00
0
2
2
0
00
VRILdt
di
Vdt
diLRi
LC
i
dt
di
L
R
dt
id
idtCdt
diLRi
VidtC
v
Ii
t
C
00
0
2
2
1)0( 0
: conditions Initial
0
VRILdt
diIi
LC
i
dt
di
L
R
dt
id
Cont’d
01
01
0
constants are and : Let
1)0( 0
: conditions Initial
0
2
2
2
00
0
2
2
LCs
L
Rs
LCs
L
RsAe
eLC
Ase
L
AReAs
sAAei
VRILdt
diIi
LC
i
dt
di
L
R
dt
id
st
ststst
st
LC
L
R
s
s
LCL
R
L
Rs
LCL
R
L
Rs
1
2 where
1
22
1
22
0
20
22
20
21
2
2
2
1
Characteristicequation
Naturalfrequencies
DampingfactorResonantfrequency
Summary
tsts
tsts
eAeAti
eAieAi
ss
21
21
21
2211
21
)(
:solution generalA
,
:) (if solutions Two
LC
L
R
s
s
1
2 where
0
20
22
20
21
• Three cases discussed– Overdamped case : > 0
– Critically damped case : = 0
– Underdamped case : < 0
Overdamped Case ( > 0)
tsts eAeAti
ss
R
LC
LCL
R
2121
21
2
)(
real. and negative are and Both
41
2
t
i(t)
tse 1
tse 2
Critically damped Case ( = 0)
0
02
equation. aldifferenti original theBack to
!conditions
initial osatisfy twt can'constant Single
)(
2
4
22
2
321
21
2
idt
dii
dt
di
dt
d
idt
di
dt
id
eAeAeAti
RLss
RLC
ttt
t
t
t
tt
t
t
eAtAti
AtAie
Aiedt
d
Aiedt
die
eAidt
di
eAffdt
df
idt
dif
21
21
1
1
1
1
)(
0
Let
Critically damped Case (Cont’d)
t
i(t)
1
te
tte
teAtAti 21)(
Underdamped Case ( < 0)
212
21121
2121
21
21
)(2
)(1
220
2202
2201
2
e whersincos)(
sincos
sincossincos
)(
)(
where
4
BBjA
BBAtAtAeti
tBBjtBBe
tjtBtjtBe
eBeBe
eBeBti
js
js
RLC
ddt
ddt
ddddt
tjtjt
tjtj
d
d
d
dd
dd
Underdamped Case (Cont’d)
tdd etAtAti sincos)( 21
t
i(t)
d2
te
Finding The Constants A1,2
)(1)0(
or
0)0(
)0(
.)0( and )0( need we
, and determine To
00
00
0
21
VRILdt
di
VRIdt
diL
Ii
/dtdii
AA
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 u
nderdamped 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 t > 0
tAtAeti
j
ωααs
.LC, ω
L
Rα
ivi
t
,
359.4sin359.4cos)(
359.49
100819
10010
119
2
V 6)0(6)0( ,A 164
10)0(
219
20
221
0
6882.0
1
66)1(92
)0()0(1)0(
10
:conditions Initial
2
1
A
A
vRiLdt
di)i(
The Source-Free Parallel RLC Circuit
011
becomesequation sticcharacteri the
,Let
(3) 01
(2) 01
gives KCL Applying
(1b) 0
(1a) 1
0
: conditions initial Assumed
2
2
2
0
0
0
LCs
RCs
Aev(t)
LC
v
dt
dv
RCdt
vd
dt
dvCvdt
LR
v
Vv
v(t)dtL
Ii
st
t
LC
RC
s
1
2
1
where
0
20
22,1
Summary• Overdamped case : > 0
• Critically damped case : = 0
• Underdamped case : < 0
tsts eAeAtv 2121)(
tetAAtv 21)(
tAtAetv
js
ddt
d
d
sincos)(
where
21
220
2,1
Finding The Constants A1,2
)()0(
or
0)0(
)0(
.)0( and )0( need we
, and determine To
00
00
0
21
RC
RIV
dt
dvdt
dvCI
R
VVv
/dtdvv
AA
Comparisons
L
RIV
dt
di
Ii
LC
L
R
s
00
0
0
20
22,1
)0(
)0(
:conditions Initial
1
2 where
• Series RLC Circuit • Parallel RLC Circuit
RC
RIV
dt
dv
Vv
LC
RC
s
00
0
0
20
22,1
)0(
)0(
:conditions Initial
1
2
1
where
Example 1
tt
,
eAeAtv
ωααs
LC, ω
RCα
R
502
21
20
221
0
)(
50 ,2
101
262
1
923.1:1 Case
Find v(t) for t > 0.v(0) = 5 V, i(0) = 0Consider three cases: R = 1.923 R = 5 R =6.25
208.5
2083.0
260)0()0()0(
5)0(
:conditions Initial
2
1
A
A
RC
Riv
dt
dvv
Example 1 (Cont’d)
t
,
etAAtv
ωααs
LC, ω
RCα
R
1021
20
221
0
)(
10
101
102
1
5:2 Case
50
5
100)0()0()0(
5)0(
:conditions Initial
2
1
A
A
RC
Riv
dt
dvv
t
,
etAtAtv
jωααs
LC, ω
RCα
R
821
20
221
0
6sin6cos)(
68
101
82
1
25.6:3 Case
667.6
5
80)0()0()0(
5)0(
:conditions Initial
2
1
A
A
RC
Riv
dt
dvv
Example 1 (Cont’d)
Example 2
t < 0 t > 0
Find v(t).
Get x(0). Get x(), dx(0)/dt, s1,2, A1,2.
Example 2 (Cont’d)
tt
,
eAeAtv
ωααs
LCω
RCα
1462
8541
20
221
0
)(
146 ,854
3541
5002
1
t > 0
16.30
156.5
0102050
505025)0()0()0(
505030
40)0(
25)40(5030
50)0(
:conditions initial theFrom
2
1
6
A
A
.
RC
Riv
dt
dv
A.i
V v
t < 0
Step Response of A Series RLC Circuit
case. free-source in the
as form same thehas (2)
(2)
But
(1)
,0for KVL Applying
2
2
LC
V
LC
v
dt
dv
L
R
dt
vd
dt
dvCi
Vvdt
diLRi
t
S
S
response state-steady the:
response transientthe:
where
)()()(
ss
t
sst
v
v
tvtvtv
Characteristic Equation
case. free-source in the as Same
01
becomesequation sticcharacteri The
0
, Let
0
2
''
2
'2
'
2
2
LCs
L
Rs
LC
v
dt
dv
L
R
dt
vd
Vvv
LC
Vv
dt
dv
L
R
dt
vd
S
S
Summary
.)0( and )0( from obtained are where
sincos
)(
)()(
)()()(
2,1
21
21
2121
/dtdvvA
etAtA
etAA
eAeA
tv
Vvtv
tvtvtv
tdd
t
tsts
t
Sss
sst
(Overdamped)
(Critically damped)
(Underdamped)
ExampleFind v(t), i(t) for t > 0.Consider three cases: R = 5 R = 4 R =1
t < 0 t > 0
Get x(0). Get x(), dx(0)/dt, s1,2, A1,2.
Case 1: R = 5
dt
dvCti
eAeAvtv
ωααs
LCω
L
Rα
ttss
,
)(
)(
4 ,1
21
5.2)1(2
5
2
421
20
221
0
3 4
364
164)0()0(
)0(
V 4)0(1)0( ,A 415
24)0(
:conditions Initial
V 24)(
2
1
A
A
Cdt
dv
dt
dvCi
ivi
vvss
Case 2: R = 4
dt
dvCti
etAAvtv
αs
LCω
L
Rα
tss
,
)(
)(
2
21
2)1(2
4
2
221
21
0
2.19
2.19
2.198.4)0()0(
)0(
V 8.4)0(1)0( ,A 8.414
24)0(
:conditions Initial
V 24)(
2
1
A
A
Cdt
dv
dt
dvCi
ivi
vvss
Case 3: R = 1
dt
dvCti
etA
tAvtv
js
LCω
L
Rα
tss
,
)(
936.1sin
936.1cos)(
936.15.0
21
5.0)1(2
1
2
5.0
2
1
21
0
469.21
12
4812)0()0(
)0(
V 12)0(1)0( ,A 1211
24)0(
:conditions Initial
24)(
2
1
A
A
Cdt
dv
dt
dvCi
ivi
vvss
Example (Cont’d)
Step Response of A Parallel RLC Circuit
case. free-source in the
as form same thehas (2)
(2) 1
But
(1)
,0for KCL Applying
2
2
LC
I
LC
i
dt
di
RCdt
id
dt
diLv
Idt
dvCi
R
v
t
S
S
response state-steady the:
response transient the:
where
)()()(
ss
t
sst
i
i
tititi
Characteristic Equation
case. free-source in the as Same
011
becomesequation sticcharacteri The
01
, Let
01
2
''
2
'2
'
2
2
LCs
RCs
LC
i
dt
di
RCdt
id
Iii
LC
Ii
dt
di
RCdt
id
S
S
Summary
.)0( and )0( from obtained are where
sincos
)(
)()(
)()()(
2,1
21
21
2121
/dtdiiA
etAtA
etAA
eAeA
ti
Iiti
tititi
tdd
t
tsts
t
Sss
sst
(Overdamped)
(Critically damped)
(Underdamped)
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.
t < 0 t > 0
Get x(0). Get x(), dx(0)/dt, s1,2, A1,2.
Example (Cont’d)
t > 0t < 0
V 4)(2)(
A 224
12)(
:for valuesFinal
(1c) V/s 12)0()0(
iv
i
tC
i
dt
dv C
A 6)0(
2
)0()0()0(
,)0( nodeat KC Applying
(1b) 0)0()0(
(1a) V 12)0()0(
:conditions Initial
C
C
i
vii
ta
ii
vv
Example (Cont’d)
t > 0
065
:equation sticCharacteri
(4) 065
02
1
2
122
gives (3) into (2) ngSubstituti
(3) 014
givesmesh left the toKVL Applying
(2) 2
1
2
gives nodeat KCL Applying
2
2
2
2
2
ss
vdt
dv
dt
vd
vdt
vd
dt
dv
dt
dvv
vdt
dvi
dt
dvvi
a
(2) usingby obtain becan )(
8 ,12
obtain we(1c) and (1a) From
)(
4)( where
)()(
3 ,2
21
32
21
ti
AA
eAeAtv
vv
tvvtv
s
ttt
ss
tss
Duality• Duality means the same characterizing
equations with dual quantities interchanged.
Resistance R Conductance G
Inductance L Capacitance C
Voltage v Current i
Voltage source Current source
Node Mesh
Series path Parallel path
Open circuit Short circuit
KVL KCL
Thevenin Norton
Table for dual pairs
Example 1
• Series RLC Circuit • Parallel RLC Circuit
012
LCs
L
Rs 0
112 LC
sRC
s
Example 2
Application: Smoothing Circuits
Outputfrom a D/A
v0vs