Second-Order Circuit..
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Transcript Second-Order Circuit..
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
42
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
eti 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 jd
where d 02 2
i (t ) B1e ( jd ) t B2 e ( jd )t
e t ( B1e jd t B2 e jd t )
e t B1 cosd t j sin d t B2 cosd t j sin d t
e t B1 B2 cosd t j B1 B2 sin d t
i (t ) e
t
A1 cosd t A2 sin d t
A1 B1 B2
where
A2 j B1 B2
Underdamped Case (Cont’d)
i(t ) A1 cosd 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
46
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)
29(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
v0 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 et
• Underdamped case
s1, 2 jd
: = 0
: < 0
where d 02 2
v(t ) e t A1 cosd 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 20 106
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
11
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
iC
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
42
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