Second-Order Circuits Instructor: Chia-Ming Tsai Electronics Engineering National Chiao Tung University Hsinchu, Taiwan, R.O.C.

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