Second-Order Transient Circuits

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=++


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


• 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=++


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


)()()(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


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


02 20

2 =++ ωαss


2

2 0: =++ cbxaxEquationQaudratic2

24−±−

=a

acbbx

22

20

2 2: ωα ++ ssEquationsticCharacteri

20

220

2

24)2(2

ωααωαα

−±−=−±−

=s



22

Roots of the characteristic equation:

20

21 ωαα −+−=s

22 20

22 ωαα −−−=s

ωαζ = Dampening ratio


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.


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.


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


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.


Analysis of a Second-Order Circuit with a DC Source

Apply KVL around the loop:

SC VtvtRitdiL =++ )()()(


SC VtvtRidt

L ++ )()(


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


LCLCdtLdt 2


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)( ==


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



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)(

:


tstsC eKeKtv 21

2110)( ++=


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 +=


22110

220

110)0(

KsKseKseKsdt

dvc +=+==


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 −=

−−−−

=−

=−

−−==


tstsc eetv 21 708.11708.110)( −+=


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


ss 021

:

10==+== αωαα

tstsC teKeKtv 11

2110)( ++=



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 ++=


2110

20

210

11 )0(0)0(

KKseKeKseKsdt

dvc +=++==


sKsKKKs

KK54

1112211

11

10)10(10100

10010

−=−==−=→=+

−=→=+

tsts teetv 21 5

1112211

101010)(

)(

=c teetv 21 101010)( −−=



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

−−−−

−−

+−−−=

++=



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 →+


)sin(774.5)cos(1010)( tetetv nt

nt

c ωω αα −− −−=

O d d


Overdamped

C iti ll d d


Critically damped

U d d d


Underdamped


Normalized Step Response of a Second Order System

01)(00)(

>=<=

ttuttu

)(


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 =


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?


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:


∫ =+++ 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)(

=++


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 =


)()()(2)( 20 tftv

dtdt=++ ωα


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


L2

Example Parallel LC Circuit: RF-ID Tag

L

C



Find v(t) for R=25Ω, 50Ω and 250Ω



Find v(t) for R=25Ω, 50Ω and 250Ω

radxLC

sec/101))((

11 5730 ===ω

FxHxLC

1)101)(101( 73

=

−−

α


RC2


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.


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


21)(

:

tstsT

P

tstsC

eKeKtv

tveKeKtv

21

21

21

21

)(

0)()(

+=

=+=



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


62211

022

011 10)0(

=+=+=+ sKsKesKesK

dtdv


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

−−−− −=−=



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


1121)(

:

+=

tstsT

P

C

teKeKtv

tv11

21

21

)(

0)(

+=

=



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( +

=→==


ttetv510610)( −=


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


xjxjs nωα −−=−−=

)i ()()(

0)()sin()cos()( 21

tKtKt

tvteKteKtv

ttP

nt

nt

C ωω

αα

αα

−−

−−

=+=


)sin()cos()( 21 teKteKtv nt

nt

T ωω αα +=


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−=


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Second-Order Transient Circuits