Lecture 14.pdf
Transcript of Lecture 14.pdf
Lecture 14Lecture 14Second Order Transient Response
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Damped Harmonic Motion
2 dxxd
oscillatorMechanical
)(2
ill tEl t i l
tfkxdtdxb
dtxdm m=++
)(12tvqdqRqdL
oscillatorElectrical
++ )(2 tvqCdt
Rdt
L s=++
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
The Series RLC Circuit and its Mechanical Analog
• Drive voltage v (t) Applied force f(t)• Drive voltage vS(t) Applied force f(t)
• Charge q Displacement x
• Current i Velocity v
• Inductance L Mass m
• Inductive Energy U=(1/2)Li2 Inertial Energy (1/2)mv2
• Capacitance C Spring constant 1/k
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
• Capac. Energy U=(1/2)(1/C)q2 Spring energy U=(1/2)kx2
• Resistance R Dampening constant b
Second –Order Circuits
( ) ( ) ( ) ( ) ( )tvvdttiC
tRidt
tdiL sC
t
=+++ ∫ 01
Differentiating with respect to time:
Cdt ∫0
tdvtitdiRtidL s )()(1)()(2=++
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
dtti
CdtR
dtL )(2 =++
Second –Order Circuits
dttdv
Lti
LCdttdi
LR
dttid s )(1)(1)()(
2
2=++
dtLLCdtLdt 2
LR
2=α Dampening
coefficientDefine:L2
LC1
0 =ω
coefficient
Undamped resonant frequencyLC frequency
tdvtf s )(1)( = Forcing function
dtLtf )( =
)()(2 tditid
Forcing function
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
)()()(2)( 202 tfti
dttdi
dttid
=++ ωα
Solution of the Second-Order Equation
solutionParticular
( ) ( ) ( ) ( )2 22
=++ tftitditid
solutionParticular
ωα ( ) ( )2 02 =++
solutionaryComplement
tftidtdt
ωα
( ) ( ) ( )22 tditid
solutionaryComplement
( ) ( ) ( ) 02 202 =++ ti
dttdi
dttid ωα
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Solution of the Complementary Equation
( ) ( )2 tditid ( ) ( ) ( ) 02 202
2=++ ti
dttdi
dttid ωα
02
:)(22 ++
=
ωα KesKeKes
KetiTryststst
stC
:02
22
0 =++ ωαFactoring
KesKeKes
t
:0)2( 2
02 =++ ωα
equationsticCharacteriKess st
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
02 20
2 =++ ωαss
Solution of the Complementary Equation
2
2 0: =++ cbxaxEquationQaudratic2
24−±−
=a
acbbx
22
20
2 2: ωα ++ ssEquationsticCharacteri
20
220
2
24)2(2
ωααωαα
−±−=−±−
=s
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Solution of the Complementary Equation
22
Roots of the characteristic equation:
20
21 ωαα −+−=s
22 20
22 ωαα −−−=s
ωαζ = Dampening ratio
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
0ω
1. Overdamped case (ζ > 1). If ζ > 1 (or equivalently, if α > ω0), the roots of the characteristic equation are real and distinct. Then the complementary solution is:
( ) tstsc eKeKti 21
21 +=
In this case, we say that the circuit is overdamped.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
2. Critically damped case (ζ = 1). If ζ = 1 (or equivalently, if α = ω0 ), the roots are real and equal. Then the complementary solution is
( ) tsts( ) tstsc teKeKti 11
21 +=
In this case, we say that the circuit is iti ll d dcritically damped.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
3. Underdamped case (ζ < 1). Finally, if ζ < 1 (or equivalently if α < ω ) the roots are(or equivalently, if α < ω0), the roots are complex. (By the term complex, we mean that the roots involve the square root of 1 ) In otherthe roots involve the square root of –1.) In other words, the roots are of the form:
nn jsjs ωαωα −−=+−= 21 and
in which j is the square root of -1 and the natural frequency is given by:f q y g y
220 αωω −=n
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
I l t i l i i j th th iIn electrical engineering, we use j rather than i to stand for square root of -1,because we use i f t F l t thfor current. For complex roots, the complementary solution is of the form:
( ) ( ) ( )teKteKti nt
nt
c ωω αα sincos 21−− +=
In this case, we say that the circuit is underdamped.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Analysis of a Second-Order Circuit with a DC Source
Apply KVL around the loop:
SC VtvtRitdiL =++ )()()(
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
SC VtvtRidt
L ++ )()(
Analysis of a Second-Order Circuit with a DC Source
SC VtvtRidt
tdiL =++ )()()(
dttdv
CtiSubstitute C=)(
)(
LCbDi id
Vtvdt
tdvRC
dttvd
LC sCCC =++ )(
)()(2
2
LCV
tvLCdt
tdvLR
dtvd
LCbyDivide
sC
CC =++ )(1)()(
:
2
2
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
LCLCdtLdt 2
Analysis of a Second-Order Circuit with a DC Source
To find the particular solution we consider the steadyTo find the particular solution, we consider the steady state. For DC steady state, we can replace the inductors by a short circuit and the capacitor by an open circuit:
VVt 10)(ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
VVtv sCp 10)( ==
Analysis of a Second-Order Circuit with a DC Source
fi d h h l i id hTo find the homogeneous solution, we consider three cases; R=300Ω, 200Ω and 100Ω. These values will correspond to over-damped, critically damped and under-damped cases.p , y p p
sec/10)1)(10(
11 40 rads
FmHLC===
μω
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Analysis of a Second-Order Circuit with a DC Source
RICase 300: Ω=
OverdampedxxmHL
R4
4
0
4 5.110
105.1105.1)10(2
3002
→====Ω
==ωαζα
xxxs 42424420
21 10618.2)10()105.1(105.1 −=−−−=−+−= ωαα
xs 420
22 10382.0−=−−−= ωαα
tsts KK
solutionsarycomplementandparticulartheAdding
10)(
:
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
tstsC eKeKtv 21
2110)( ++=
Analysis of a Second-Order Circuit with a DC Source
To find K1 and K2 we use the initial conditions:To find K1 and K2 we use the initial conditions:
0021
10100)0(
10)( 21
KKeKeKv
eKeKtv tstsc
++=++==
++=
2121
0)0(
10100)0(
i
KKeKeKvc
=
++=++==
0)0(
0)0()(
)(
0)0(
dtdv
idt
tdvCti
i
cc =→==
2211
)0(
)(21
dv
eKseKsdt
tdv tstsc +=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
22110
220
110)0(
KsKseKseKsdt
dvc +=+==
Analysis of a Second-Order Circuit with a DC Source
KsKsKK
0010
2211
21
=+=++
KsKsKK
010
2211
21
2211
=+−=+
xsssK 7111082.310)1)(0())(10(0
1104
222 ==−
=−−
=
−
=xxssss
ss
K
101
71.1)10618.2(103820.0)1)(())(1(11 44
1212
21
1
−
=−−−
=−
=−
==
xxx
sss
sss
ss
sK 7.11
)10618.2(103820.01018.2610
)1)(())(1()10)(()0)(1(
110
44
4
12
1
12
1
21
12 −=
−−−−
=−
=−
−−==
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
tstsc eetv 21 708.11708.110)( −+=
Analysis of a Second-Order Circuit with a DC Source
dampedCriticallymHL
R
RIICase
4
44 1
101010
)10(2200
2
200:
→====Ω
==
Ω=
ωαζα
ss
mHL
420
221
0
10
10)10(22
−=−=−+−== αωαα
ω
solutionsarycomplementandparticulartheAdding
ss 021
:
10==+== αωαα
tstsC teKeKtv 11
2110)( ++=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Analysis of a Second-Order Circuit with a DC Source
To find K1 and K2 we use the initial conditions:To find K1 and K2 we use the initial conditions:
0021
)()(
10)( 11 teKeKtv tstsc ++=
10
20
1
0)0(
10)0(100)0(
i
KeKeKvc
=
+=++==
0)0(
0)0()(
)(
0)0(
dtdv
idt
tdvCti
i
cc =→==
=
22111
)0(
)(111
dv
eKteKseKsdt
tdv tststsc ++=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
2110
20
210
11 )0(0)0(
KKseKeKseKsdt
dvc +=++==
Analysis of a Second-Order Circuit with a DC Source
sKsKKKs
KK54
1112211
11
10)10(10100
10010
−=−==−=→=+
−=→=+
tsts teetv 21 5
1112211
101010)(
)(
=c teetv 21 101010)( −−=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Analysis of a Second-Order Circuit with a DC Source
100: RIIICase Ω=
5.010
105.0105.0)10(2
1002 4
4
0
4 dUnderdampexxmHL
Rωαζα →====
Ω==
8660)105.0()10( 2424220 xn αωω =−=−=
:solutionsarycomplementandparticulartheAdding
tt αα
)cos()sin()sin()cos()(
)sin()cos(10)(
2211
21
teKteKteKteKdt
tdvteKteKtv
nnt
nt
nnt
ntC
nt
nt
C
ωωωαωωωα
ωω
αααα
αα
−−−−
−−
+−−−=
++=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Analysis of a Second-Order Circuit with a DC Source
To find K1 and K2 we use the initial conditions:
)0(0)0(
dvvc =
To find K1 and K2 we use the initial conditions:
0)0(
dtdvc =
774.5)10(50000
10010
1221
11
KKKK
KKαωα −=−==→=+−
−=→=+
774.5)10(8660
0 1221 KKKK
tt
nn ω
ωα →+
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
)sin(774.5)cos(1010)( tetetv nt
nt
c ωω αα −− −−=
O d d
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Overdamped
C iti ll d d
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Critically damped
U d d d
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Underdamped
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Normalized Step Response of a Second Order System
01)(00)(
>=<=
ttuttu
)(
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Normalized Step Response of a Second
Example: Closing a switch at t 0 to apply a DC voltage A
Order SystemExample: Closing a switch at t=0 to apply a DC voltage A
)()( tAutv =
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Normalized Step Response of a Second Order System
What value of ζshould be used to getshould be used to get to the steady state position quickly without overshooting?
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Circuits with Parallel L and CdtdvC
dtdqiivdt
Li
RVi
t
cLLR ==+== ∫0
)0(1
We can replace the circuit with it’s Norton equivalent and then analyze the circuit by writing KCL at the top
∫ =+++t
tiidttvtvtdvC )()0()(1)(1)(
y y g pnode:
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
∫ =+++ nL tiidttvL
tvRdt
C0
)()0()()(
Circuits with Parallel L and C
dt
)( tiidttvL
tvRdt
tdvC nL )()0()(1)(1)(
0
=+++ ∫
tditdvtvdC
atingdifferenti
n )()(1)(1)(
:2
tditdvtvddt
tditv
Ldttdv
RdttvdC n
)(11)(1)(
)()(1)(1)(
2
=++
dttdi
Ctv
LCdttdv
RCdttvd n )(1)(1)(1)(
=++
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Circuits with Parallel L and C)(1)(1)(1)(2 tdi
tvtdvtvd n=++
1
)(
tcoefficienDampening
dtCtv
LCdtRCdt
=
++
α
12
0frequencyresonantUndamped
RCtcoefficienDampening
=ω
α
)(1)(
0
tditffunctionForcing
LCfrequencyresonantUndamped
n
ω
)()()(2)(
)(
22
ftdvtvddtC
tffunctionForcing =
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
)()()(2)( 20 tftv
dtdt=++ ωα
Circuits with Parallel L and C
tdvtvd )()( 22
tftvdt
tdvdt
tvd )()()(2)( 20 =++ ωα
This is the same equation as we found for the series LC circuit with the following changes for α:
RCcircuitParallel
21
=α
LRcircuitSeries
2=α
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
L2
Example Parallel LC Circuit: RF-ID Tag
L
C
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Circuits with Parallel L and C
Find v(t) for R=25Ω, 50Ω and 250Ω
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Circuits with Parallel L and C
Find v(t) for R=25Ω, 50Ω and 250Ω
radxLC
sec/101))((
11 5730 ===ω
FxHxLC
1)101)(101( 73
=
−−
α
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RC2
Circuits with Parallel L and C
Vp(t) = 0
To find the particular solution vP(t) (steady state response) replace C with an open circuit and L with a short circuitshort circuit.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Circuits with Parallel L and CRICase
15
25: Ω=
Overdampedsrad
sxsxFxRC 5
15
0
157 2
/10102102
)101)(25(21
21
→====Ω
==−
−− ω
αζα
xxxs
xxxs
52525520
22
52525520
21
1073.3)10()102(102
10268.0)10()102(102
−=−−−=−−−=
−=−+−=−+−=
ωαα
ωαα
tsts KK
solutionsarycomplementandparticulartheAdding
21)(
:
tstsT
P
tstsC
eKeKtv
tveKeKtv
21
21
21
21
)(
0)()(
+=
=+=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Circuits with Parallel L and C
0)0(0)0(:conditionsInitial
Since the voltage across a capacitor
0)0(0)0(0)0(0)0(=+→=−
=+→=−ii
vv
LL
Since the voltage across a capacitor and the current through an inductor cannot change instantaneously
)0()0()0(1.0
0+
++++
=
= +
dvCivtatKCL
L
66 /10
101.01.01.0)0()0(1.0
)0(1.0
===+
→+
=
+++
−sv
xCdtdv
dtdvC
dtCi
R L
210
20
1
)(0)0( =+→+=+
dKKeKeKv
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
62211
022
011 10)0(
=+=+=+ sKsKesKesK
dtdv
Circuits with Parallel L and C
0KK +
10
06
2211
21
KsKs
KK
=+
=+
89.21010)1)(10())(0(1010
55
66622
6
1ss
K =−
=−
=−
==
892
)1073.3(10268.0)1)(())(1(11
12
551212
21
1
KK
xxssssss−=−=
−−−−−
)(89.289.289.2)(
89.2
5555 1073.310268.01073.310268.0
12
txtxtxtx eeeetv
KK
−−−− −=−=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Circuits with Parallel L and C
RIICase 50: Ω
dampedCriticallysrad
sxsxFxRC
RIICase
5
15
0
157 1
/10101101
)101)(50(21
21
50:
→====Ω
==
Ω=−
−− ω
αζα
xs 51 101−=−= α
tstsC teKeKtv
solutionsarycomplementandparticulartheAdding
1121)(
:
+=
tstsT
P
C
teKeKtv
tv11
21
21
)(
0)(
+=
=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Circuits with Parallel L and C
tsts teKeKtv 11 21)( +=
KeKeKv 10
20
1
21
0)0()0(
)(
==+=
tsteKtv 12)( =
tsts teKseKdt
tdv 11 212)(
+
+=
KeKdt
dv
5
62
602 1010)0( +
=→==
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
ttetv510610)( −=
Circuits with Parallel L and C
250: RIIICase Ω=
2.0/10
102.0102.0)101)(250(2
12
1
250:
5
15
0
157 dUnderdampe
sradsxsx
FxRC
RIIICase
ωαζα
−−
−→====
Ω==
Ω
1098.0102.0
1098.0)102.0()101(
55
551
52225220
xjxjs
xxx
n
n
ωα
αωω
+−=+−=
=−=−=
:
1098.0102.0 552
solutionsarycomplementandparticulartheAdding
xjxjs nωα −−=−−=
)i ()()(
0)()sin()cos()( 21
tKtKt
tvteKteKtv
ttP
nt
nt
C ωω
αα
αα
−−
−−
=+=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
)sin()cos()( 21 teKteKtv nt
nt
T ωω αα +=
Circuits with Parallel L and C
tt αα
)cos()sin()sin()cos()()sin()cos()(
2211
21
teKteKteKteKdt
tdvteKteKtv
nt
nnt
nt
nnt
nt
nt
−−−−
−−
+−−−=
+=
ωωωαωωωα
ωω
αααα
αα
10)0(
00)0(6
11
dv
KeKv t− =→== α
2.101098.0
1010)0(52221
6
5
xKKKK
dtdv
nn ==→=+−== ωωα
)1098.0sin()2.10()( 5102.0 5txetv tx−=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.