FIRST AND SECOND-ORDERTRANSIENT CIRCUITS

IN CIRCUITS WITH INDUCTORS AND CAPACITORS VOLTAGES AND CURRENTSCANNOT CHANGE INSTANTANEOUSLY. EVEN THE APPLICATION, OR REMOVAL, OF CONSTANT SOURCES CREATES ATRANSIENT BEHAVIOR

LEARNING GOALS

FIRST ORDER CIRCUITSCircuits that contain a single energy storing elements.Either a capacitor or an inductor

SECOND ORDER CIRCUITSCircuits with two energy storing elements in any combination

THE CONVENTIONAL ANALYSIS USING MATHEMATICAL MODELS REQUIRES THE DETERMINATION OF (A SET OF) EQUATIONS THAT REPRESENT THE CIRCUIT.ONCE THE MODEL IS OBTAINED ANALYSIS REQUIRES THE SOLUTION OF THE EQUATIONS FOR THE CASES REQUIRED.

FOR EXAMPLE IN NODE OR LOOP ANALYSIS OF RESISTIVE CIRCUITS ONE REPRESENTS THECIRCUIT BY A SET OF ALGEBRAIC EQUATIONS

WHEN THERE ARE INDUCTORS OR CAPACITORS THE MODELS BECOME LINEAR ORDINARY DIFFERENTIAL EQUATIONS (ODEs). HENCE, IN GENERAL, ONE NEEDS ALL THOSE TOOLS IN ORDER TO BE ABLE TO ANALYZE CIRCUITS WITH ENERGY STORING ELEMENTS.

ANALYSIS OF LINEAR CIRCUITS WITH INDUCTORS AND/OR CAPACITORS

THE GENERAL APPROACH CAN BE SIMPLIFIED IN SOME SPECIAL CASES WHEN THE FORMOF THE SOLUTION CAN BE KNOWN BEFOREHAND. THE ANALYSIS IN THESE CASES BECOMES A SIMPLE MATTER OF DETERMINING SOMEPARAMETERS.TWO SUCH CASES WILL BE DISCUSSED IN DETAIL FOR THE CASE OF CONSTANT SOURCES. ONE THAT ASSUMES THE AVAILABILITY OF THE DIFFERENTIAL EQUATION AND A SECOND THAT IS ENTIRELY BASED ON ELEMENTARY CIRCUIT ANALYSIS… BUT IT IS NORMALLY LONGER

A METHOD BASED ON THEVENIN WILL BE DEVELOPED TO DERIVE MATHEMATICAL MODELSFOR ANY ARBITRARY LINEAR CIRCUIT WITH ONE ENERGY STORING ELEMENT.

WE WILL ALSO DISCUSS THE PERFORMANCE OF LINEAR CIRCUITS TO OTHER SIMPLE INPUTS

THE MODEL

AN INTRODUCTIONINDUCTORS AND CAPACITORS CAN STORE ENERGY. UNDER SUITABLE CONDITIONS THIS ENERGYCAN BE RELEASED. THE RATE AT WHICH IT IS RELEASED WILL DEPEND ON THE PARAMETERSOF THE CIRCUIT CONNECTED TO THE TERMINALS OF THE ENERGY STORING ELEMENT

With the switch on the left the capacitor receivescharge from the battery.

Switch to the rightand the capacitordischarges throughthe lamp

dxxfetxetxet

tTH

xtt

)(1

)()(0

0

0

GENERAL RESPONSE: FIRST ORDER CIRCUITS

0)0(; xxfxdt

dxTH

Including the initial conditionsthe model for the capacitorvoltage or the inductor current will be shown to be of the form

Solving the differential equationusing integrating factors, one tries to convert the LHS into anexact derivative

t

TH efxdt

dx 1/*

TH

ttt

fexedt

dxe

11

TH

tt

fexedt

d

1

t

t0

dxxfetxetxt

t

TH

xttt

)(1

)()(0

0

0

0)0();()()( xxtftaxt

dt

dx

THIS EXPRESSION ALLOWS THE COMPUTATIONOF THE RESPONSE FOR ANY FORCING FUNCTION.WE WILL CONCENTRATE IN THE SPECIAL CASEWHEN THE RIGHT HAND SIDE IS CONSTANT

t

e

/*

circuit

the of speed reaction the on ninformatio

tsignifican provide to shown be willit

constant." time" the called is

.switchings sequentialstudy

to used be can expression general

The arbitrary. is , time, initial The ot

FIRST ORDER CIRCUITS WITHCONSTANT SOURCES

dxxfetxetxt

tTH

xttt

)(1

)()(0

0

0

0)0(; xxfxdt

dxTH

If the RHS is constant

dxef

txetxt

t

xtTH

tt

0

0

)()( 0

xtxt

eee

dxeef

txetxt

t

xtTH

tt

0

0

)()( 0

t

t

xtTH

tt

eef

txetx0

0

)()( 0

00

)()( 0

ttt

TH

tt

eeeftxetx

0

)()( 0

tt

THTH eftxftx

0tt The form of the solution is

021 ;)(0

tteKKtxtt

Any variable in the circuit is ofthe form

021 ;)(0

tteKKtytt

Only the values of the constants K_1, K_2 will change

TRANSIENT

TIMECONSTANT

EVOLUTION OF THE TRANSIENT AND INTERPRETATION OF THE TIME CONSTANT

A QUALITATIVE VIEW:THE SMALLER THE THE TIMECONSTANT THE FASTER THETRANSIENT DISAPPEARS

With less than 2% errortransient is zerobeyond this point

Drops 0.632 of initialvalue in one time constant

Tangent reaches x-axis in one time constant

CRTH

THE TIME CONSTANT

The following example illustratesthe physical meaning of timeconstant

vS

R S a

b

C

+

vc

_

Charging a capacitor

THCC

TH vvdt

dvCR

The model

0)0(, CSS vVv

Assume

The solution can be shown to be

t

SSC eVVtv

)(

CRTH

transientFor practical purposes thecapacitor is charged when thetransient is negligible

0067.0

0183.00498.0135.0

5

432

368.0

t

et

With less than 1%error the transientis negligible afterfive time constants

dt

dvC C

S

SC

R

vv

0

S

SCc

R

vv

dt

dvC

: KCL@a

1. THE CIRCUIT HAS ONLY CONSTANT INDEPENDENT SOURCES

THE DIFFERENTIAL EQUATION APPROACH

CIRCUITS WITH ONE ENERGY STORING ELEMENT

CONDITIONS

2. THE DIFFERENTIAL EQUATION FOR THE VARIABLE OF INTEREST IS SIMPLE TO OBTAIN. NORMALLY USING BASIC ANALYSIS TOOLS; e.g., KCL, KVL. . . OR THEVENIN

3. THE INITIAL CONDITION FOR THE DIFFERENTIAL EQUATION IS KNOWN, OR CAN BE OBTAINED USING STEADY STATE ANALYSIS

SOLUTION STRATEGY: USE THE DIFFERENTIAL EQUATION AND THE INITIAL CONDITIONS TO FIND THE PARAMETERS ,, 21 KK

( )

1 2

FACT: WHEN ALL INDEPENDENT SOURCES ARE CONSTANT

FOR ANY VARIABLE, ( ), IN THE CIRCUIT THE

SOLUTION IS OF THE FORM

( ) ,Ot t

O

y t

y t K K e t t

If the diff eq for y is knownin the form

Use the diff eq to find twomore equations by replacingthe form of solution into thedifferential equation

0

01

)0( yy

fyadt

dya

We can use thisinfo to find the unknowns

feKKaeK

att

2102

1

0110 a

fKfKa

0

120

1 0a

aeKa

at

t

eK

dt

dy 2

0,)( 21 teKKty

t

21)0( KKy

Use the initial condition to getone more equation

12 )0( KyK

SHORTCUT: WRITE DIFFERENTIAL EQ.IN NORMALIZED FORM WITH COEFFICIENTOF VARIABLE = 1.

00

101 a

fy

dt

dy

a

afya

dt

dya

1K

ASSUME FIND 2)0(.0),( SVvttv

)(tv @KCL USE 0.t FORMODEL

0)()(

tdt

dvC

R

Vtv S

2/)0( SVv condition initial

(DIFF. EQ. KNOWN, INITIAL CONDITION KNOWN)

STEP 1 TIME CONSTANT

fydt

dy

Get time constant ascoefficient of derivative

STEP 2 STEADY STATE ANALYSIS

value)state(steady ,t and 0for

IS SOLUTION

1

21 0,)(

Kv(t)

teKKtvt

IN STEADY STATE THE SOLUTION ISA CONSTANT. HENCE ITS DERIVATIVEIS ZERO. FROM DIFF EQ.

SVvdt

dv0 Steady state value

from diff. eq.

SVK

1

values)statesteady (equating

fKfydt

dy 1 THEN ISMODEL THE IF

STEP 3 USE OF INITIAL CONDITION

1221 )0()0(

0

KvKKKv

t

AT

fvK )0(2

2/2/)0( 2 SS VKVv

0,)2/()(

teVVtv RCt

SS :ANSWER

LEARNING EXAMPLE

sVtvtdt

dvRC )()(

R/*

)0();(

0,)(

211

21

xKKxK

teKKtxt

0),( tti FIND

0t FORKVL USE MODEL.

Rv

Lv)(ti

KVL

)()( tdt

diLtRivvV LRS

0)0()0()0(

0)0(0

iii

it

inductor

CONDITIONINITIAL

STEP 1R

Vtit

dt

di

R

L S )()( R

L

STEP 2 STEADY STATER

VKi S 1)(

STEP 3 INITIAL CONDITION

21)0( KKi

RLt

S eR

Vti 1)( :ANS

LEARNING EXAMPLE

)0();(

0,)(

211

21

xKKxK

teKKtxt

1 2( ) , 0t

i t K K e t

0t FORKCL MODEL.

)()(

tiR

tvIS

)(tv

)()( tdt

diLtv )()( tit

dt

di

R

LIS

STEP 1

STEP 2 SS IKIi 1)(

STEP 3 210)0( KKi

RLt

S eIti 1)( :ANS

0)0( i :CONDITIONINITIAL

R

L

LEARNING BY DOING

1 2( ) , 0t

i t K K e t

0t FORMODEL

2

)()(

R

tvti

IT IS SIMPLER TO DETERMINE MODELFOR CAPACITOR VOLTAGE

INITIAL CONDITIONS

VvVkk

kvC 4)0(4)12(

63

3)0(

0)(

)(

||;0)(

)()(

2121

P

P

R

tvt

dt

dvC

RRRR

tvt

dt

dvC

R

tv

kkkRP 26||3

sFCRP 2.0)10100)(102( 63 STEP 1

STEP 2 0)( 1 Kv

STEP 3 VKVKKv 44)0( 221

0],[4)( 2.0

tVetvt

0],[3

4)( 2.0

tmAeti

t

:ANS

0,)( 21 teKKti

t

0t FOR STATESTEADY IN CIRCUIT

Vtvtdt

dvO

O 6)()(5.0

][3)()(5.0

12)(4)(2

Atitdt

di

titdt

di

])[(2)( VtitvO

0),( ttvO FIND

KVL(t>0)

)(ti

KVL USE 0.t FORMODEL

0)()()( 311 tiRtdt

diLtiRVS

5.0 STEP 1

0,)( 21 teKKtv

t

O

LEARNING EXAMPLE

STEP 2: FIND K1 USING STEADY STATEANALYSIS

Vvtvtdt

dvOO

O 6)(6)()(5.0

1)( KvO

VK 61

FOR THE INITIAL CONDITION ONE NEEDSTHE INDUCTOR CURRENT FOR t<0 ANDUSES THE CONTINUITY OF THE INDUCTORCURRENT DURING THE SWITCHING .

THE NEXT STEP REQUIRES THE INITIALVALUE OF THE VARIABLE, )0( Ov

THE STEADY STATE ASSUMPTION FOR t<0SIMPLIFIES THE ANALYSIS

)0();(

0,)(

211

21

xKKxK

teKKtxt

FOR EXAMPLE USE THEVENINASSUMING INDUCTOR IN STEADYSTATE

12||2THR

04412 1 I1I

KVL

KVL ][442 1 VIVV OCTH

][41 AI

][3

4)0()0( AiiL

0,)( 21 teKKtv

t

O

][3

8)0(

3

4)0( Vvi O

0,3

53)( 5.0

teti

t

a

b

0],[3

106)( 5.0

tVetv

t

O

CIRCUIT IN STEADY STATE (t<0)

)(tiL FIND MUST

3

106

3

82221 KKKK

Li

0t

0),( ttvO FIND

C

1R

2R

KCL USE 0.t FORMODEL

0)()(0)( 2121

cCCC vtdt

dvCRR

RR

vt

dt

dvC

sFCRR 6.0)10100)(106()( 6321 STEP 1

)(3

1)(

42

2)( tvtvtv CCO

STEP 2 0,)( 21 teKKtv

t

C 01 K

INITIAL CONDITIONS. CIRCUIT IN STEADY STATE t<0

)0(Cv V)12(9

6

][88)0( 221 VKKKvC STEP 3

0],[8)( 6.0

tVetvt

C

0],[3

8)( 6.0

tVetv

t

O

LEARNING EXTENSION

)(tvc DETERMINE

)0();(

0,)(

1211

21

iKKvK

teKKtv

C

t

C

0),(1 tti FIND

KVL USE 0.t FORMODEL

0)(18 11 tidt

diL

L

0)()(9

11

1 titdt

di

)0();(

0,)(

12111

211

iKKiK

teKKtit

STEP 1 s9

1

STEP 2 01 K

FOR INITIAL CONDITIONS ONE NEEDSINDUCTOR CURRENT FOR t<0

)0(1 i

CIRCUIT IN STEADY STATEPRIOR TO SWITCHING

AV

i 112

12)0(1

STEP 3

][1)0()0( 22111 AKKKii

0],[][)( 991

1

tAeAeti t

t

:ANS

)(1 ti

Lv

LEARNING EXTENSION

USING THEVENIN TO OBTAIN MODELS

Obtain the voltage across the capacitor or the current through the inductor

Circuitw ith

resistancesand

sources

I nductororCapacitor

a

b

R epresentation of an arbitrarycircuit w ith one storage elem ent

Thevenin VTH

R TH

I nductororCapacitor

a

b

VTH

R TH a

b

C

+

vc

_

Case 1.1Voltage across capacitor

VTH

R TH a

b

L

iL

Case 1.2Current through inductor

KCL@ node a

ciRi0 Rc ii

dt

dvCi C

c

TH

THCR R

vvi

0

TH

THCC

R

vv

dt

dvC

THCC

TH vvdt

dvCR

Use KVL

Rv

Lv

THLR vvv

LTHR iRv

dt

diLv L

L

THLTHL viRdt

diL

TH

THL

L

TH R

vi

dt

di

R

LSCi

EXAMPLE

66

6

6

H3

V24

)(tiO

0t

0t;(t)iFind O

The variable of interest is theinductor current. The model is

TH

THO

O

TH R

vi

dt

di

R

L

And the solution is of the form

0;)( 21 teKKti

t

O

66

6

6V24

0t

Thevenin for t>0at inductor terminals

a

b

THv 0 THR ))66(||6(6

sH

R

L

TH

3.010

3

0;03.0 tidt

diO

O

03.0

3.0 3.021

3.02

tt

eKKeK

0;)( 3.02

teKti

t

O01KNext: Initial Condition

66

6

6V24

)0()0( OO ii

0t

Since K1=0 the solution is

0;)( 3.02

teKti

t

O

Evaluating at 0+6

322 K

0;6

32)( 3.0

teti

t

O

66

6

6

H3

V24

)(tiO

0t

Circuit for t<0

1i

2i3i

0)(6)(66 21311 iiiii

0)(6)(624 3212 iiii0)(6)(6 2313 iiii

3)0( iiC

mA632

)(0i:solution C

currentinductor

of continuity and assumption

statesteady Use Determine ).0( Oi

1v

806

24

66 1111

vvvv

66

24)0( 1viO

Loop analysis

Node analysis

+- 0t

k6

k6

k6

k6

F100

V12)(tiO

0t(t),iFind O EXAMPLE

Cv

6kvi0tFor C

O

Hence, if the capacitor voltageis known the problem is solved

Model for v_c

THCC

TH vvdt

dvCR

+- 0t

k6

k6

k6

k6V12)(tiO

a b THv

kkkRTH 36||6

V vTH6

sF 3.010*100*10*3 63

63.0 CC

C

vdt

dv

vforModel

3.021

t

C eKKv

65.1

5.1 3.021

5.12

tt

eKKeK

61 K

Now we need to determinethe initial value v_c(0+)using continuity and thesteady state assumption

+- 0t

k6

k6

k6

k6V12)(tiO

circuit in steady statebefore the switching

)0(Cv

VvC 6)0(

Continuity of capacitor voltage

VvC 6)0(

)0(21 CvKK

06 21 KK

0;6)( tVtvC

0;16

)( tmAk

vti C

O

Diff EqApproach