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 INTRODUCTION

INDUCTORS 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 ∫=− ττττ

GENERAL RESPONSE: FIRST ORDER CIRCUITS

0)0(; xxfxdtdx

TH =+=+τ

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

τ

ττ

t

TH efxdtdx 1/*=+

TH

ttt

fexedtdxe τττ

ττ11

=+

TH

tt

fexedtd ττ

τ1

=⎟⎟⎠

⎞⎜⎜⎝

⎛∫t

t0

dxxfetxetxt

tTH

xttt

)(1)()(0

0

0 ∫−

−−

−+= ττ

τ0)0();()()( xxtftaxt

dtdx

=+=+

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

τt

e−

/*

circuitthe of speed reaction the on ninformatio

tsignifican provide to shown be willit constant."time"thecalled isτ

.switchings sequentialstudy to used be can expression general

Thearbitrary.is,time,initial The ot

FIRST ORDER CIRCUITS WITHCONSTANT SOURCES

dxxfetxetxt

tTH

xttt

)(1)()(0

0

0 ∫−

−−

−+= τττ

0)0(; xxfxdtdx

TH =+=+τ

If the RHS is constant

dxeftxetxt

t

xtTH

tt

∫−

−−

−+=

0

0

)()( 0ττ

τ

⇒=−

−−

τττxtxt

eee

dxeeftxetxt

t

xtTH

tt

∫−

−−

+=0

0

)()( 0τττ

τ

t

t

xtTH

tt

eeftxetx0

0

)()( 0 ⎟⎠⎞

⎜⎝⎛+=

−−

−τττ τ

τ

⎟⎠⎞

⎜⎝⎛ −+=

−−

−ττττ00

)()( 0

ttt

TH

tt

eeeftxetx

( ) τ0

)()( 0

tt

THTH eftxftx−

−−+=

0tt ≥The form of the solution is

021 ;)(0

tteKKtxtt

≥+=−

−τ

Any variable in the circuit is ofthe form

021 ;)(0

tteKKtytt

≥+=−

−τ

Only the values of the constantsK_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=τ

THE TIME CONSTANT

The following example illustratesthe physical meaning of timeconstant

−+vS

RS a

b

C

+

vc

_

Charging a capacitor

THCC

TH vvdt

dvCR =+

The model

0)0(, == CSS vVvAssume

The solution can be shown to be

τt

SSC eVVtv−

−=)(

CRTH=τ

transient

For practical purposes thecapacitor is charged when thetransient is negligible

0067.00183.00498.0135.0

5432

368.0

τττττ

τt

et −

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

dtdvC C

S

SC

Rvv −

0=−

+S

SCc

Rvv

dtdvC

: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 INTERESTIS SIMPLE TO OBTAIN. NORMALLY USING BASIC ANALYSIS TOOLS;e.g., KCL, KVL. . . OR THEVENIN

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

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

( )

1 2

FACT: WHEN ALL INDEPENDENT SOURCES ARE CONSTANTFOR ANY VARIABLE, ( ), IN THE CIRCUIT THESOLUTION IS OF THE FORM

( ) ,Ot t

O

y t

y t K K e t tτ−

−

= + >

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

fyadtdya

=+

=+ We can use thisinfo to find the unknowns

feKKaeKatt

=⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟

⎟⎠

⎞⎜⎜⎝

⎛−

−−ττ

τ 2102

1

0110 a

fKfKa =⇒=

0

120

1 0aaeKaa t

=⇒=⎟⎠⎞

⎜⎝⎛ +−

−τ

ττ

ττ

t

eKdtdy −

−= 2⇒>+=−

0,)( 21 teKKtytτ

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

fydtdy

aafya

dtdya =+⇒=+

τ1K

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

)(tv@KCL USE 0.t FORMODEL >

0)()(=+

− tdtdvC

RVtv S

2/)0( SVv = conditioninitial

(DIFF. EQ. KNOWN, INITIAL CONDITION KNOWN)

STEP 1 TIME CONSTANT

fydtdy

=+τ

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

→∞→>>+=

−

τ

τ

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

SVvdtdv

=⇒= 0 Steady state valuefrom diff. eq.

SVK =∴

1

values)statesteady (equating

fKfydtdy

==+ 1 THEN ISMODEL THE IF τ

STEP 3 USE OF INITIAL CONDITION

1221 )0()0(0

KvKKKvt

−=⇒+== AT

fvK −= )0(22/2/)0( 2 SS VKVv −=⇒=

0,)2/()( >−=−

teVVtv RCt

SS :ANSWER

LEARNING EXAMPLE

sVtvtdtdvRC =+ )()(

R/*

)0();(0,)(

211

21

+=+∞=>+=

−

xKKxKteKKtx

tτ

)(ti

0),( >tti FIND

0tFORKVL USE MODEL. >

−+ Rv

−

+

Lv)(ti

KVL

)()( tdtdiLtRivvV LRS +=+=

0)0()0()0(

0)0(0=+

⎭⎬⎫

+=−⇒=−⇒<

iii

itinductor

CONDITIONINITIAL

STEP 1R

Vtitdtdi

RL S=+ )()( R

L=τ

STEP 2 STEADY STATER

VKi S==∞ 1)(

STEP 3 INITIAL CONDITION

21)0( KKi +=+⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

−R

Lt

S eR

Vti 1)( :ANS

0,)( 21 >+=

−

teKKti

tτ

LEARNING EXAMPLE

)0();(0,)(

211

21

+=+∞=>+=

−

xKKxKteKKtx

tτ

0,)( 21 >+=

−

teKKti

tτ

0t FORKCL MODEL. >

)()( tiRtvIS +=

)(tv

⇒= )()( tdtdiLtv )()( tit

dtdi

RLIS +=

STEP 1

STEP 2 SS IKIi =⇒=∞ 1)(

STEP 3 210)0( KKi +==+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

−R

Lt

S eIti 1)( :ANS

0)0( =+i:CONDITIONINITIAL

RL

=τ

LEARNING BY DOING

0tFORMODEL >

2

)()(R

tvti =

IT IS SIMPLER TO DETERMINE MODELFOR CAPACITOR VOLTAGE

0tFORSTATESTEADY INCIRCUIT <INITIAL CONDITIONS

VvVkk

kvC 4)0(4)12(63

3)0( =+⇒=+

=−

0)()(

||;0)()()(21

21

=+

==++

P

P

Rtvt

dtdvC

RRRR

tvtdtdvC

Rtv

Ω== kkkRP 26||3

sFCRP 2.0)10100)(102( 63 =×Ω×== −τSTEP 1

STEP 2 0)( 1 ==∞ Kv

STEP 3 VKVKKv 44)0( 221 =⇒=+=+

0],[4)( 2.0 >=−

tVetvt

0],[34)( 2.0 >=

−tmAeti

t

:ANS

0,)( 21 >+=−

teKKtitτ

Vtvtdt

dvO

O 6)()(5.0 =+

][3)()(5.0

12)(4)(2

Atitdtdi

titdtdi

=+

=+ ])[(2)( VtitvO =

0),( >ttvO FIND

KVL(t>0)

)(ti

KVLUSE 0.t FORMODEL >

0)()()( 311 =+++− tiRtdtdiLtiRVS

5.0=τ STEP 1

0,)( 21 >+=−

teKKtvt

Oτ

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

+=+∞=>+=

−

xKKxKteKKtx

tτ

0<t

FOR EXAMPLE USE THEVENINASSUMING INDUCTOR IN STEADYSTATE

Ω== 12||2THR

04412 1 =−+− I1I

KVL

KVL ][442 1 VIVV OCTH =−==

][41 AI =

][34)0()0( AiiL =+=−

0,)( 21 >+=−

teKKtvt

Oτ

][38)0(

34)0( Vvi O =+⇒=+

0,353)( 5.0 >−=

−teti

t

a

b

0],[3

106)( 5.0 >−=−

tVetvt

O

CIRCUIT IN STEADY STATE (t<0)

)(tiL

)(tiL FIND MUST

3106

38

2221 =⇒−==+ KKKK

0),( >ttvO FIND

C

1R

2R

KCLUSE 0.t FORMODEL >

0)()(0)( 2121

=++⇒=+

+ cCCC vt

dtdvCRR

RRvt

dtdvC

sFCRR 6.0)10100)(106()( 6321 =×Ω×=+= −τSTEP 1

)(31)(

422)( tvtvtv CCO =+

=

STEP 2 0,)( 21 >+=−

teKKtvt

Cτ 01 =K

INITIAL CONDITIONS. CIRCUIT IN STEADY STATE t<0

−−

+)0(Cv V)12(

96

=

][88)0( 221 VKKKvC =⇒+==+STEP 3

0],[8)( 6.0 >=−

tVetvt

C

0],[38)( 6.0 >=

−tVetv

t

O

LEARNING EXTENSION

)(tvc DETERMINE

)0();(0,)(

1211

21

+=+∞=>+=

−

iKKvKteKKtv

C

t

Cτ

0),(1 >tti FIND

KVLUSE 0.tFORMODEL >

⇒=+ 0)(18 11 ti

dtdiL

L

0)()(91

11 =+ tit

dtdi

)0();(0,)(

12111

211

+=+∞=>+=

−

iKKiKteKKti

tτ

STEP 1 s91

=τ

STEP 2 01 =K

FOR INITIAL CONDITIONS ONE NEEDSINDUCTOR CURRENT FOR t<0

)0(1 −i

CIRCUIT IN STEADY STATEPRIOR TO SWITCHING

AVi 11212)0(1 =Ω

=−

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

Circuitwith

resistancesand

sources

InductororCapacitor

a

b

Representation of an arbitrarycircuit with one storage element

Thevenin −+VTH

RTH

InductororCapacitor

a

b

−+VTH

RTH a

b

C

+

vc_

Case 1.1Voltage across capacitor

−+VTH

RTH a

b

L

iL

Case 1.2Current through inductor

KCL@ node a

ciRi 0=+ Rc ii

dtdvCi C

c =

TH

THCR R

vvi −=

0=−

+TH

THCC

Rvv

dtdvC

THCC

TH vvdt

dvCR =+

Use KVL

−+ Rv

−

+

Lv

THLR vvv =+

LTHR iRv =

dtdiLv L

L =

THLTHL viR

dtdiL =+

==+⎟⎟⎠

⎞⎜⎜⎝

⎛

TH

THL

L

TH Rvi

dtdi

RL

SCi

EXAMPLE

Ω6Ω6

Ω6

Ω6

H3

V24 −+

)(tiO

0=t

0>t;(t)iFind O

The variable of interest is theinductor current. The model is

TH

THO

O

TH Rvi

dtdi

RL

=+

And the solution is of the form

0;)( 21 >+=−

teKKtit

Oτ

Ω6Ω6

Ω6

Ω6V24 −+

0>t

Thevenin for t>0at inductor terminals

a

b

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

sHRL

TH

3.0103

=Ω

==τ

0;03.0 >=+ tidtdi

OO

03.0

3.0 3.021

3.02 =++⎟⎠⎞

⎜⎝⎛−

−−tt

eKKeK

0;)( 3.02 >=

−teKti

t

O⇒= 01KNext: Initial Condition

Ω6Ω6

Ω6

Ω6V24 −+

)0()0( +=− OO ii

0<t

Since K1=0 the solution is

0;)( 3.02 >=

−teKti

t

O

Evaluating at 0+632

2 =K

0;632)( 3.0 >=

−teti

t

O

Ω6Ω6

Ω6

Ω6

H3

V24 −+

)(tiO

0<t

Circuit for t<0

1i

2i3i

0)(6)(66 21311 =−+−+ iiiii0)(6)(624 3212 =−+−+− iiii

0)(6)(6 2313 =−+− iiii3)0( iiC =+

mA632

)(0i:solution C =+

currentinductor of continuity and assumption

statesteady Use Determine ).0( +Oi

1v

806

2466 1

111 =⇒=−

++ vvvv

6624)0( 1viO +=+

Loop analysis

Node analysis

+- 0=t

k6

k6

k6

k6

Fμ100

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 vvdtdvCR =+

+- 0>t

k6

k6

k6

k6V12)(tiO

a b−+ THv

kkkRTH 36||6 ==

VvTH 6=

sF 3.010*100*10*3 63 =Ω= −τ

63.0 =+ CC

C

vdt

dvvforModel

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

+- 0<t

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 +=+ CvKK06 21 =⇒= KK

⇒>= 0;6)( tVtvC

0;16

)( >== tmAk

vti CO

Diff EqApproach