2012/10/24

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Second-Order Circuits•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 oftwo energy storage elements.

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Finding Initial and Final Values•v(0), i(0), dv(0)/dt, di(0)/dt, v(), and i()

•Two key points:–v and i are defined according to the passive sign

convention.

–Continuity properties:

•Capacitor voltage: (VS-like)

•Inductor current: (IS-like))0()0(

)0()0(

LL

CC

ii

vv

v

i

+ _

Example

).(,)((c),)0(,)0((b)

),0(,)0((a)

Find:Q

vidtdvdtdi

vi

V4)0()0(

A2)0()0(

V4)0(2)0(

A224

12)0(

0.for tanalysisdcApply(a):Sol

vv

ii

iv

i

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Cont’d

V12)(A0)(

.0foranalysisdcApply

(c):Sol

vi

t

Cont’d

V/s2000

A200

findeasilycanWecase.in thissourcecurrenta

astreatedbecaninductorTheabruptly.change

cannotcurrentinductortheSince

:0

findTo(b):Sol

Ci

dtdv

ii

Ci

dtdv

idtdv

C

dtdv

C

C

CC

)()(

)()(

)(

t = 0+2 A

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Cont’d

t = 0+

A/s0250000

haveweThus

048120

0000412

givesKVLapplying,0obtainTo

case.in thissourcevoltageaastreatedbecancapacitorThe

abruptly.changecannotvoltagecapacitortheSince

0findTo(b):Sol

.)()(

)(

)()()(

)(

:)(

Lv

dtdi

v

vvi

v

Lv

dtdi

vdtdi

L

dtdi

L

L

CL

L

LL

The Source-Free Series RLC Circuit

)4(1)0(

0)0(

)0(

gives(2)and(1)required.is)0((3),solveTo

(3)0

(2)01

givesKVLApplying

(1b)1

0

(1a)0

:conditionsinitialAssumed

00

0

2

2

0

00

VRILdt

di

Vdt

diLRi

dtdiLCi

dtdi

LR

dtid

idtCdt

diLRi

VidtC

v

Ii

t

C

00

0

2

2

1)0(0

:conditionsInitial

0

VRILdt

diIi

LCi

dtdi

LR

dtid

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Cont’d

01

01

0

constants.areand:Let

1)0(0

:conditionsInitial

0

2

2

2

00

0

2

2

LCs

LR

s

LCs

LR

sAe

eLCA

seL

AReAs

sAAei

VRILdt

diIi

LCi

dtdi

LR

dtid

st

ststst

st

LC

LR

s

s

LCLR

LR

s

LCLR

LR

s

12where

122

122

0

20

22

20

21

2

2

2

1

Characteristicequation

Naturalfrequencies

DampingfactorResonantfrequency

(or undamped naturalfrequency)

Summary

.conditionsinitialthefromdeterminedareandwhere

)(

:solutiongeneralA,

:)(ifsolutionsTwo

21

21

2211

21

21

21

AAeAeAti

eAieAi

ss

tsts

tsts

LC

LR

s

s

ss

12where

02

:equationsticCharacteri

0

20

22

20

21

20

2

•Three cases discussed–Overdamped case (distinct real roots) : > 0

–Critically damped case (repeated real root) : = 0

–Underdamped case (complex-conjugate roots): < 0

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Overdamped Case (> 0)

tsts eAeAti

ssR

LC

LCLR

2121

21

2

)(

real.andnegativeareandBoth

412

t

i(t)tse 1

tse 2

Critically damped Case (= 0)

0

02

equation.aldifferentioriginaltheBack to

!conditionsinitialosatisfy twtcan'constantSingle

)(

24Let

22

2

321

21

2

idtdi

idtdi

dtd

idtdi

dtid

eAeAeAti

RLssRLC

ttt

t

t

t

tt

t

t

eAtAti

AtAie

Aiedtd

Aiedtdi

e

eAidtdi

eAffdtdf

idtdi

f

21

21

1

1

1

1

)(

0

Let

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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

ewhersincos)(

sincos

sincossincos

)(

)(

where

4Let

BBjABBA

tAtAeti

tBBjtBBe

tjtBtjtBe

eBeBe

eBeBti

js

js

RLC

ddt

ddt

ddddt

tjtjt

tjtj

d

d

d

dd

dd

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Underdamped Case (Cont’d)

L

RαetAtAti t

dd 2,sincos)( 21

t

i(t)

d2

te

Finding The Constants A1,2

)(1)0(

or

0)0(

gives0atKVL2.

)0(1..)0(and)0(needwe

,anddetermineTo

00

00

0

21

VRILdt

di

VRIdt

diL

t

Ii/dtdii

AA

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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, theoverdamped case has the longest settling time.The underdamped case has the fastest decay.(If a constant 0 is assumed.)

Example

Find i(t).

t < 0 t > 0

(6+3)

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Example (Cont’d)

t < 0 t > 0

tAtAeti

j

ωααs

.LC, ω

LR

αb

ivia

t

,

359.4sin359.4cos)(

359.49

100819

10010

119

2)(

V6)0(6)0(,A164

10)0()(

219

20

221

0

6882.01

66)1(92

)0()0(1)0(

10

:conditionsInitial

2

1

AA

vRiLdt

di)i(

-

The Source-Free Parallel RLC Circuit

011

becomesequationsticcharacterithe,)(Let

(3)01

(2)01

givesKCLApplying

(1b)0

(1a))(1

0

:conditionsinitialAssumed

2

2

2

0

0

0

LCs

RCs

AetvLCv

dtdv

RCdtvd

dtdv

CvdtLR

v

Vv

dttvL

Ii

st

t

LC

RC

s

12

1

where0

20

22,1

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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

RCRIV

dtdv

dtdv

CIRV

t

Vv/dtdvv

AA

)()0(or

0)0(

gives0atKCL2.

)0(1..)0(and)0(needwe

,anddetermineTo

00

00

0

21

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Comparisons

LRIV

dtdi

Ii

LC

LR

s

00

0

0

20

22,1

)0(

)0(

:conditionsInitial

12where

•Series RLC Circuit •Parallel RLC Circuit

RCRIV

dtdv

Vv

LC

RC

s

00

0

0

20

22,1

)0(

)0(

:conditionsInitial

12

1

where

Example 1

tt

,

eAeAtv

ωααs

LC, ω

RCα

R

502

21

20

221

0

)(

50,2

101

262

1923.1:1Case

Find v(t) for t > 0.v(0) = 5 V, i(0) = 0Consider three cases:R = 1.923 R = 5 R =6.25

208.52083.0

260)0()0()0(

5)0(

:conditionsInitial

2

1

AA

RCRiv

dtdvv

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Example 1 (Cont’d)

t

,

etAAtv

ωααs

LC, ω

RCα

R

1021

20

221

0

)(

10

101

102

15:2Case

505

100)0()0()0(

5)0(

:conditionsInitial

2

1

AA

RCRiv

dtdvv

t

,

etAtAtv

jωααs

LC, ω

RCα

R

821

20

221

0

6sin6cos)(

68

101

82

125.6:3Case

667.65

80)0()0()0(

5)0(

:conditionsInitial

2

1

AA

RCRiv

dtdv

v

Example 1 (Cont’d)

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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(

:conditionsinitialtheFrom

2

1

6

A

A

.RC

Rivdt

dv

A.i

Vv

t < 0

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Step Response of A Series RLC Circuit

case.free-sourcein theasformsamethehas(2)

(2)

But

(1)

,0forKVLApplying

2

2

LCV

LCv

dtdv

LR

dtvd

dtdv

Ci

Vvdtdi

LRi

t

S

S

responsestate-steadythe:responsetransientthe:

where

)()()(

ss

t

sst

vv

tvtvtv

Characteristic Equation

case.free-sourcein theasSame

01

becomesequationsticcharacteriThe

0

,Let

0

2

''

2

'2

'

2

2

LCs

LR

s

LCv

dtdv

LR

dtvd

VvvLC

Vvdtdv

LR

dtvd

S

S

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Summary

.)0(and)0(fromobtainedarewhere

sincos)(

)()(

0)(where

)()()(

2,1

21

21

2121

/dtdvvA

etAtAetAAeAeA

tv

Vvv

v

tvtvtv

tdd

t

tsts

t

Sss

t

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.

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Case 1: R = 5

dtdv

Cti

eAeAvtv

ωααs

LCω

LR

α

ttss

,

)(

)(

4,1

21

5.2)1(2

52

421

20

221

0

1

2

( ) 24 V

Initial conditions:

24(0) 4 A , (0) 1 (0) 4 V

5 1(0) (0) 4

(0) 16

64 34 3

ssv v

i v i

dv dvi C

dt dt CAA

t < 0 t > 0

Case 2: R = 4

dtdv

Cti

etAAvtv

αsLC

ω

LR

α

tss

,

)(

)(

2

21

2)1(2

42

221

21

0

2.192.19

2.198.4)0()0(

)0(

V8.4)0(1)0(,A8.414

24)0(

:conditionsInitial

V24)(

2

1

AA

Cdtdv

dtdv

Ci

ivi

vvss

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Case 3: R = 1

dtdv

Cti

etA

tAvtv

jsLC

ω

LR

α

tss

,

)(

936.1sin936.1cos

)(

936.15.0

21

5.0)1(2

12

5.0

2

1

21

0

469.2112

4812)0()0(

)0(

V12)0(1)0(,A1211

24)0(

:conditionsInitial

24)(

2

1

AA

Cdtdv

dtdv

Ci

ivi

vvss

Example (Cont’d)

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Step Response of A Parallel RLC Circuit

case.free-sourcein theasformsamethehas(2)

(2)1

But

(1)

,0forKCLApplying

2

2

LCI

LCi

dtdi

RCdtid

dtdi

Lv

Idtdv

CiRv

t

S

S

responsestate-steadythe:responsetransientthe:

where

)()()(

ss

t

sst

ii

tititi

Characteristic Equation

case.free-sourcein theasSame

011

becomesequationsticcharacteriThe

01

,Let

01

2

''

2

'2

'

2

2

LCs

RCs

LCi

dtdi

RCdtid

IiiLC

Iidtdi

RCdtid

S

S

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Summary

.)0(and)0(fromobtainedarewhere

sincos)(

)()(

0)(where

)()()(

2,1

21

21

2121

/dtdiiA

etAtAetAAeAeA

ti

Iiti

i

tititi

tdd

t

tsts

t

Sss

t

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 thetwo initial conditions x(0) and dx(0)/dt.

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Example

Find v, ifor t > 0.

t < 0 t > 0

Get x(0). Get x(), dx(0)/dt, s1,2, A1,2.

Example (Cont’d)

t > 0t < 0

V4)(2)(

A224

12)(

:forvaluesFinal

(1c)V/s12)0()0(

iv

i

tC

idt

dv C

A6)0(2

)0()0()0(

,)0(nodeatKCLApplying

(1b)0)0()0(

(1a)V12)0()0(

:conditionsInitial

C

C

i

vii

ta

ii

vv

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Example (Cont’d)

t > 0

065

:equationsticCharacteri

(4)2465

1221

21

22

gives(3)into(2)ngSubstituti

(3)1214

givesmeshleftthetoKVLApplying

(2)21

2

givesnodeatKCLApplying

2

2

2

2

2

ss

vdtdv

dtvd

vdt

vddtdv

dtdv

v

vdtdi

i

dtdvv

i

a

(2)usingbyobtainbecan)(8,12

obtainwe(1c)and(1a)From

)(

4)(where

)()(3,2

21

32

21

tiAA

eAeAtv

vv

tvvtvs

ttt

ss

tss

Duality•Duality means the same characterizing equations with

dual quantities interchanged.

Resistance R Conductance GInductance L Capacitance CVoltage v Current iVoltage source Current sourceNode MeshSeries path Parallel pathOpen circuit Short circuitKVL KCLThevenin Norton

Table for dual pairs

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A Case Study

0:KVL)(

)()(

21

22

11

n

nn

vvvifv

ifvifv

0:KCL)(

)()(

21

22

11

n

nn

iiivfi

vfivfi

….+ v1 - + v2 - + vn -

i

….

i1 i2 in +v_

Element Transformations

)(Voltage

)(Current

)ce(Capacitan

)e(Inductanc

)ce(Conductan

SSS

SSS

IIvIi

VViVv

Ldtdv

Lidtdi

Lv

Cdtdi

Cvdtdv

Ci

RRviRiv

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Example 1

•Series RLC Circuit •Parallel RLC Circuit

012

LCs

LR

s 012

LCs

LR

s

Example 2

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Application: Smoothing CircuitsOutputfrom aD/Aconverter

v0vs