FIRST AND SECOND-ORDER TRANSIENT CIRCUITS

IN CIRCUITS WITH INDUCTORS AND CAPACITORS VOLTAGES AND CURRENTS CANNOT CHANGE INSTANTANEOUSLY. EVEN THE APPLICATION, OR REMOVAL, OF CONSTANT SOURCES CREATES A TRANSIENT BEHAVIOR

LEARNING GOALS

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

SECOND ORDER CIRCUITS Circuits 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 THE CIRCUIT 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 FORM OF THE SOLUTION CAN BE KNOWN BEFOREHAND. THE ANALYSIS IN THESE CASES BECOMES A SIMPLE MATTER OF DETERMINING SOME PARAMETERS. 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 MODELS FOR 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 ENERGY CAN BE RELEASED. THE RATE AT WHICH IT IS RELEASED WILL DEPEND ON THE PARAMETERS OF THE CIRCUIT CONNECTED TO THE TERMINALS OF THE ENERGY STORING ELEMENT

With the switch on the left the capacitor receives charge from the battery.

Switch to the right and the capacitor discharges through the lamp

dxxfetxetxet

tTH

xtt

)(1)()(0

0

0 ∫=− τττ τ

GENERAL RESPONSE: FIRST ORDER CIRCUITS

0)0(; xxfxdtdx

TH =+=+τ

Including the initial conditions the model for the capacitor voltage or the inductor current will be shown to be of the form

Solving the differential equation using integrating factors, one tries to convert the LHS into an exact 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 ∫−

−−

−+= τττ

0)0();()()( xxtftaxtdtdx

=+=+

THIS EXPRESSION ALLOWS THE COMPUTATION OF THE RESPONSE FOR ANY FORCING FUNCTION. WE WILL CONCENTRATE IN THE SPECIAL CASE WHEN THE RIGHT HAND SIDE IS CONSTANT

τt

e−

/*

circuitthe of speed reaction the on ninformatio

tsignifican provide to shown be willit constant." time" the called is τ

.switchings sequentialstudy to used be can expression general The arbitrary. is , time, initial The ot

FIRST ORDER CIRCUITS WITH CONSTANT 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

)()( 0 ττ τ

⇒=−

−−

τττxtxt

eee

dxeeftxetxt

t

xtTH

tt

∫−

−−

+=0

0

)()( 0 τττ τ

t

t

xtTH

tt

eeftxetx0

0

)()( 0

+=

−−

−τττ τ

τ

−+=

−−

−ττττ00

)()( 0ttt

TH

tt

eeeftxetx

( ) τ0

)()( 0tt

THTH eftxftx−

−−+=

0tt ≥The form of the solution is

021 ;)(0

tteKKtxtt

≥+=−

−τ

Any variable in the circuit is of the form

021 ;)(0

tteKKtytt

≥+=−

−τ

Only the values of the constants K_1, K_2 will change

TRANSIENT

TIME CONSTANT

EVOLUTION OF THE TRANSIENT AND INTERPRETATION OF THE TIME CONSTANT

A QUALITATIVE VIEW: THE SMALLER THE THE TIME CONSTANT THE FASTER THE TRANSIENT DISAPPEARS

With less than 2% error transient is zero beyond this point

Drops 0.632 of initial value in one time constant

Tangent reaches x-axis in one time constant

CRTH=τ

THE TIME CONSTANT

The following example illustrates the physical meaning of time constant

−+vS

RS a

b

C

+

vc_

Charging a capacitor

THCC

TH vvdtdvCR =+

The model

0)0(, == CSS vVvAssume

The solution can be shown to be

τt

SSC eVVtv−

−=)(

CRTH=τ

transient

For practical purposes the capacitor is charged when the transient is negligible

0067.00183.00498.0135.0

5432

368.0

τττττ

τt

et−

With less than 1% error the transient is negligible after five 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 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 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 known in the form

Use the diff eq to find two more equations by replacing the form of solution into the differential equation

0

01

)0( yy

fyadtdya

=+

=+ We can use this info 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 get one more equation

12 )0( KyK −+=

SHORTCUT: WRITE DIFFERENTIAL EQ. IN NORMALIZED FORM WITH COEFFICIENT OF 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 = condition initial

(DIFF. EQ. KNOWN, INITIAL CONDITION KNOWN)

STEP 1 TIME CONSTANT

fydtdy

=+τ

Get time constant as coefficient 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 IS A CONSTANT. HENCE ITS DERIVATIVE IS ZERO. FROM DIFF EQ.

SVvdtdv

=⇒= 0 Steady state value from 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

0),( >tti FIND

0t FORKVL USE MODEL. >

−+ Rv

−

+

Lv)(ti

KVL

)()( tdtdiLtRivvV LRS +=+=

0)0()0()0(

0)0(0=+

+=−⇒=−⇒<

iii

itinductor

CONDITIONINITIAL

STEP 1 R

Vtitdtdi

RL S=+ )()( R

L=τ

STEP 2 STEADY STATE R

VKi S==∞ 1)(

STEP 3 INITIAL CONDITION

21)0( KKi +=+

−=

−R

Lt

S eR

Vti 1)( :ANS

LEARNING EXAMPLE

)0();(0,)(

211

21

+=+∞=>+=

−

xKKxKteKKtx

tτ

1 2( ) , 0t

i t K K e 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

1 2( ) , 0t

i t K K e tτ−

= + >

0t FORMODEL >

2

)()(R

tvti =

IT IS SIMPLER TO DETERMINE MODEL FOR CAPACITOR VOLTAGE

INITIAL CONDITIONS

VvVkk

kvC 4)0(4)12(633)0( =+⇒=+

=−

0)()(

||;0)()()( 2121

=+

==++

P

P

Rtvt

dtdvC

RRRR

tvtdtdvC

Rtv

Ω== kkkRP 26||3

sFCRP 2.0)10100)(102(63 =×Ω×== −τ

STEP 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

0t FOR STATESTEADY IN CIRCUIT

Vtvtdt

dvO

O 6)()(5.0 =+

][3)()(5.0

12)(4)(2

Atitdtdi

titdtdi

=+

=+ ])[(2)( VtitvO =

0),( >ttvO FIND

KVL(t>0)

)(ti

KVL USE 0.t FORMODEL >

0)()()( 311 =+++− tiRtdtdiLtiRVS

5.0=τ STEP 1

0,)( 21 >+=−

teKKtvt

Oτ

LEARNING EXAMPLE

STEP 2: FIND K1 USING STEADY STATE ANALYSIS

Vvtvtdt

dvOO

O 6)(6)()(5.0 =∞⇒=+

1)( KvO =∞VK 61 =∴

FOR THE INITIAL CONDITION ONE NEEDS THE INDUCTOR CURRENT FOR t

FOR EXAMPLE USE THEVENIN ASSUMING INDUCTOR IN STEADY STATE

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),( >ttvO FIND

C

1R

2R

KCL USE 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=−

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

KVL USE 0.t FORMODEL >

⇒=+ 0)(18 11 tidtdiL

L

0)()(91

11 =+ tit

dtdi

)0();(0,)(

12111

211

+=+∞=>+=

−

iKKiKteKKti

tτ

STEP 1 s91

=τ

STEP 2 01 =K

FOR INITIAL CONDITIONS ONE NEEDS INDUCTOR CURRENT FOR t==−

−

tAeAeti tt

: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 Cc =

TH

THCR R

vvi −=

0=−+TH

THCC

Rvv

dtdvC

THCC

TH vvdtdvCR =+

Use KVL

−+ Rv

−

+

Lv

THLR vvv =+

LTHR iRv =

dtdiLv LL =

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 the inductor 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 ++

sHRLTH

3.0103

=Ω

==τ

0;03.0 >=+ tidtdi

OO

03.0

3.0 3.0213.02 =++

−

−−tt

eKKeK

0;)( 3.02 >=−

teKtit

O⇒= 01KNext: Initial Condition

Ω6Ω6

Ω6

Ω6V24 −+

)0()0( +=− OO ii

0=−

teKtit

O

Evaluating at 0+ 6

322 =K

0;6

32)( 3.0 >=−

tetit

O

Ω6Ω6

Ω6

Ω6

H3

V24 −+

)(tiO

0

+- 0=t

k6

k6

k6

k6

Fµ100

V12)(tiO

0t(t),iFind O >EXAMPLE

−+ Cv

6kvi0tFor CO =>

Hence, if the capacitor voltage is 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 =+ CCC

vdt

dvvforModel

3.021

t

C eKKv−

+=

65.1

5.1 3.0215.12 =++

−

−−tt

eKKeK

61 =K

Now we need to determine the initial value v_c(0+) using continuity and the steady state assumption

+- 0= 0;6)( tVtvC

0;16

)( >== tmAk

vti CO

ANALYSIS OF CIRCUITS WITH ONE ENERGY STORING ELEMENT CONSTANT INDEPENDENT SOURCES

A STEP-BY-STEP APPROACH

THIS APPROACH RELIES ON THE KNOWN FORM OF THE SOLUTION BUT FINDS THE CONSTANTS USING BASIC CIRCUIT ANALYSIS TOOLS AND FORGOES THE DETERMINATION OF THE DIFFERENTIAL EQUATION MODEL

τ,, 21 KK

0,)( 21 >+=−

teKKtxtτ

statesteady incircuit the analyzing determined be can and variablethe of valuestatesteady the is 1K

21,)0(

KKx

constants the compute to equation second the provides and condition initial the is +

element storingenergy the across Thevenin using determined be can andconstant time the is τ

Obtaining the time constant: A General Approach

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 Cc =

TH

THCR R

vvi −=

0=−+TH

THCC

Rvv

dtdvC

THCC

TH vvdtdvCR =+

Use KVL

−+ Rv

−

+

Lv

THLR vvv =+

LTHR iRv =

dtdiLv LL =

THLTHL viR

dtdiL =+

==+

TH

THL

L

TH Rvi

dtdi

RL

SCi

CIRCUITS WITH ONE ENERGY STORING ELEMENT

TH

TH

RL

CR

=

=

τ

τ

Circuit Inductive

Circuit Capacitive

THE STEPS

)0();(0,)(

211

21

+=+∞=>+=

−

xKKxKteKKtx

tτ

STEP 1. THE FORM OF THE SOLUTION

STEP 2: DRAW THE CIRCUIT IN STEADY STATE PRIOR TO THE SWITCHING AND DETERMINE CAPACITOR VOLTAGE OR INDUCTOR CURRENT

STEP 3: DRAW THE CIRCUIT AT 0+ THE CAPACITOR ACTS AS A VOLTAGE SOURCE. THE INDUCTOR ACTS AS A CURRENT SOURCE. DETERMINE THE VARIABLE AT t=0+

STEP 4: DRAW THE CIRCUIT IN STEADY STATE AFTER THE SWITCHING AND DETERMINE THE VARIABLE IN STEADY STATE.

)0( +x DETERMINE

)(∞x DETERMINE

STEP 5: DETERMINE THE TIME CONSTANT

inductor one withcircuit

capacitor one withcircuit

TH

TH

RL

CR

=

=

τ

τ

STEP 6: DETERMINE THE CONSTANTS K1, K2

)0(),( 211 +=+∞= xKKxK

0),( >tti FIND

0,)( 21 >+=−

teKKtitτ :1 STEP

USE CIRCUIT IN STEADY STATE PRIOR TO THE SWITCHING

mAkVi 2

1224

=Ω

=

][32)2)(2(36)0( VkmAVvc =−=−)0()0( −=+ cC vv

)0( +i Determine :3 STEP

USE A CIRCUIT VALID FOR t=0+. THE CAPACITOR ACTS AS SOURCE

mAkVi

316

632)0( ==+

capacitor across voltageInitial :2 STEP

NOTES FOR INDUCTIVE CIRCUIT (1)DETERMINE INITIAL INDUCTOR CURRENT IN STEP 2 (2)FOR THE t=0+ CIRCUIT REPLACE INDUCTOR BY A CURRENT SOURCE

LEARNING EXAMPLE

)0( −= i

KVL

constant time Determine :5 STEPCRTH=τ :circuit Capacitive

Ω== kkkRTH 5.16||2 FC µ100=

sF 15.0)10100)(105.1( 63 =×Ω×= −τ

21, KK Determine :6 STEP

0,)( 21 >+=−

teKKtitτ 1) (STEP

mAi3

16)0(( =+ 3) STEP 21 KK +=

mAi836)( =∞ 4) (STEP 1K=

)(∞i Determine :4 STEP

USE CIRCUIT IN STEADY STATE AFTER SWITCHING

mAi836)( =∞ 0,

65

836)( 15.0 >+=

−teti

t

WERFINAL ANS

NOTE: FOR INDUCTIVE CIRCUIT

THRL

=τ

ORIGINAL CIRCUIT

65

836

316

2 =−=∴K

USING MATLAB TO DISPLAY FINAL ANSWER

>+

≤= −

0,65

836

02)(

15.0 te

tmAti t » help linspace

LINSPACE Linearly spaced vector. LINSPACE(x1, x2) generates a row vector of 100 linearly equally spaced points between x1 and x2. LINSPACE(x1, x2, N) generates N points between x1 and x2. See also LOGSPACE, :.

%example6p3.m %commands used to display funtion i(t) %this is an example of MATLAB script or M-file %must be stored in a text file with extension ".m” %the commands are executed when the name of the M-file is typed at the %MATLAB prompt (without the extension) tau=0.15; %define time constant tini=-4*tau; %select left starting point tend=10*tau; %define right end point tminus=linspace(tini,0,100); %use 100 points for t=0 iminus=2*ones(size(tminus)); %define i for t=0 plot(tminus,iminus,'ro',tplus,iplus,'bd'), grid; %basic plot command specifying %color and marker title('VARIATION OF CURRENT i(t)'), xlabel('time(s)'), ylabel('i(t)(mA)') legend('prior to switching', 'after switching') axis([-0.5,1.5,1.5,6]);%define scales for axis [xmin,xmax,ymin,ymax]

Command used to define linearly spaced arrays

Script (m-file) with commands used. Prepared with the MATLAB Editor

0),( >ttv FIND

0,)( 21 >+=−

teKKtvtτ :1 STEP

currentinductor Initial :2 STEP

Use circuit in steady state prior to switching

kk 3||6

mAI 4624

1 == 1366)0( IiL +

=−

mAiL 38)0( =−

)0( +v Determine :3 STEP

Use circuit at t=0+. Inductor is replaced by current source

1V

038

126424 111 =+++− VVV ][

320

1 VV =

][3

52][24)0( 1 VVVv =−=+

LEARNING EXAMPLE

)(∞v DETERMINE :4 STEP

USE CIRCUIT IN STEADY STATE AFTER SWITCHING

][24)( Vv =∞

CONSTANT TIME DETERMINE :5 STEP

12||6||4=THR

Ω= 2THR

THRL

=τ :Circuit InductivesH 2

24

=Ω

=τ

21, KK DETERMINE :6 STEP4) (step ][24)(1 VvK =∞=

3) (step 21352)0( KKv +==+

][32024

352

1 VK −=−=

0,24320)( 2 >+−=

−tetv

t

:ANS

ORIGINAL CIRCUIT

0),( >ttvO FIND

0,)( 21 >+=−

teKKtvt

oτ :1 STEP

VOLTAGECAPACITORINITIAL :2 STEP

−−

+)0(Cv

0

)(∞Ov DETERMINE :4 STEP

−

∞

+

)(Ov

][524)12(

52)( VvO ==∞

CONSTANT TIME DETERMINE :5 STEPCRTH=τ :Circuit Capacitive

THRΩ==

544||1THR

sFC582 =⇒= τ

21, KK DETERMINE :6 STEP

][524)(1 VvK O =∞=

][51][5)0( 221 VKKKVvO =⇒+==+

0];[51

524)( 5

8>+=

−

tVetvt

O :ANS

ORIGINAL CIRCUIT

0,)( 21 >+=−

teKKtvt

Oτ :1 STEP

CURRENT INDUCTORINITIAL :2 STEP

)0( −LiOv 022

)4(2

12=+

−−+

− OOO vvv

][34)0()0(

][38

Aii

Vv

LL

O

=+=−

=

)0( +Ov DETERMINE :3 STEP

)0( +Li

][38)0(2)0( Viv LO =+=+

)0( +Ov

0),( >ttvO FIND

)(∞Ov DETERMINE :4 STEP

ORIGINAL CIRCUIT

)(∞Ov

][6)12(22

2)( VvO =+=∞

CONSTANT TIME DETERMINE :5 STEP

THRL

=τ

Circuit Inductive

THR

Ω= 4THR

s5.042==τ

21, KK DETERMINE :6 STEP

4) (step ][6)(1 VvK O =∞=

3) (step 2138)( KKvO +==+

][3

10638

2 VK −=−=

0,3

106)( 5.0 >−=−

tetvt

O :ANS

0),( >ttvO FIND0,)( 21 >+=

−teKKtv

t

Oτ :1 STEP

)0( +Li DETERMINE :2 STEP

AiA 3=

+

−][6 V

][18 V

][3)0()0( Aii LL =+=−

)0( +Ov DETERMINE :3 STEP

][18)0( VvO =+

LEARNING EXAMPLE

)(∞Ov DETERMINE :4 STEP

ORIGINAL CIRCUIT

Bv

06

)2(42

36 '=

−−++

− ABBB ivvv

4' BA

vi =

12/*

636411 ' ×=+ AB iv ][5.4],[18' AiVv AB ==

][9 V

][27)( VvO =∞

KVL

CONSTANT TIME DETERMINE :5 STEP

THRL

=τ

circuit inductive

SC

OCTH i

vR =

sourcesdependent withCircuit

OPEN CIRCUIT VOLTAGE

AiA 6" =

+

−V12

KVL

][361224 VvOC =+=

NOTE: FOR THE INDUCTIVE CASE THE CIRCUIT USED TO COMPUTE THE SHORT CIRCUIT CURRENT IS THE SAME USE TO DETERMINE )(∞Ov

SHORT CIRCUIT CURRENT

ORIGINAL CIRCUIT

1i

'''221

121

26)(236

4)(236

Aiiii

iii

−++=

++=

'''1

2

A

SC

i ii i==

][836 AiSC =

Ω=⇒

==

8][8/36

][36TH

SC

OC RAi

VvsHL

833 =⇒= τ

21, KK DETERMINE :6 STEP4) (step 127)( KvO ==∞

3) (step ][918)0( 221 VKKKvO −=⇒+==+

0,927)( 83

>−=−

tetvt

O :ANS

0),( >ttvo FIND0,)( 21 >+=

−teKKtv

t

Oτ :1 STEP

+= 0t AT VOLTAGECAPACITOR DETERMINE :2 STEP

][12 VvA =

+− ][24 V

KVL ][60122424)0( VvC =++=−

−−

+)0(Cv

)0( +Ov DETERMINE :3 STEP

)0()0( −=+ CC vv

][60)0( Vvvv OCO =+⇒=)(∞Ov DETERMINE :4 STEP

)(∞Ov0=i

0=Av

][24)( VvO =∞

LEARNING EXTENSION

CONSTANT TIME DETERMINE :5 STEP CRTH=⇒τcircuit capacitiveSC

OCTH i

vR =

)(∞Ov][24)( Vvv OOC =∞=

−

+

OCv

ORIGINAL CIRCUIT

OPEN CIRCUIT VOLTAGE

SHORT CIRCUIT CURRENT

SCi

Ω2

KVL

02242 =−− ASC viSCA iv 2−= ][4 AiSC =

Ω== 6424

THR

sF 1226 =×Ω=τ21, KK DETERMINE :6 STEP

4) (step 24)(1 =∞= OvK

3) (step 2160)0( KKvO +==+ ][362 VK =

0,3624)( 12 >+=−

tetvt

O :ANS

Inductor example

STEP 1: Form of the solution

τt

O eKKtv−

+= 21)(

STEP 2: Initial inductor current

Ω2

Ω2 Ω4

V6

0

STEP 4: Find output in steady state after the switching

Ω2

Ω2 Ω4

V6

0>t

_)(∞

+

Ov

−+

0)( =∞Ov

Ω2

Ω2 Ω4

H2V6

0=t

_)(tvO

+

−+

STEP 5: Find time constant after switch

Ω2

Ω2 Ω4

V6

0>t

−+

THR

THRL

=τ

Ω= 8THR

s25.0=τ

STEP 6: Find the solution

VvKK O 6)0(21 −=+=+0)(1 =∞= OvK

0;6)( 25.0 >−=−

tetvt

O

0;6)( 4 >−= − tetv tO

PULSE RESPONSE

WE STUDY THE RESPONSE OF CIRCUITS TO A SPECIAL CLASS OF SINGULARITY FUNCTIONS

><

=0100

)(tt

tu VOLTAGE STEP

CURRENT STEP TIME SHIFTED STEP

PULSE SIGNAL

PULSE AS SUM OF STEPS

))](01.0()([10)( mAtututi −−=

LEARN BY DOING

))](2()1([10)( Vtututv −−−−=

NONZERO INITIAL TIME AND REPEATED SWITCHING

dxxfetxetxt

tTH

xttt

)(1)()(0

0

0 ∫−

−−

−+= τττ

00 )(; xtxfxdtdx

TH =+=+τ

021 ;)(0

tteKKtxtt

≥+=−

−τ

RESPONSE FOR CONSTANT SOURCES

This expression will hold on ANY interval where the sources are constant. The values of the constants may be different and must be evaluated for each interval

The values at the end of one interval will serve as initial conditions for the next interval

LEARNING EXAMPLE 0);( >ttvO VOLTAGEOUTPUT THE FIND

0)(0)(0 =⇒=⇒< tvtvt O 0)0( =+Ov

sFkkCRTH 4.0100)12||6( =×== µτ

Vtvt 9)(0 =⇒>

'1)9(810

8)( Kvo =+=∞

τt

o eKKtv−

+= '2'1)(

0)0( '2`1 =+=+ KKvo

−=

−4.014)(

t

o etv

0)(3.0 =⇒> tvt

')3.0(

"2

"1)( τ

−−

+=t

O eKKtv

3.0=ot )1(4)3.0( 4.03.0

−−=+ evo

4.0' =τ

00)( "1 =⇒=∞ Kvo )(11.2)3.0("2 VvK o =+=

3.0;11.2)( 4.03.0

>=−

−tetv

t

o

+-

Ωk10

Fµ20

−

+

Ov

ab

V12

THE SWITCH IS INITIALLY AT a. AT TIME t=0 IT MOVES TO b AND AT t=0.5 IT MOVES BACK TO a. FIND 0,)( >ttvO

τt

O eKKtv−

+= '2'1)(

'2

'1][12)0( KKVv +==+

'10)( KvO ==∞

5.00,12)( 2.0

%pulse1.m % displays the response to a pulse response tmin=linspace(-0.5,0,50); %negative time segment t1=linspace(0,0.5,50); %first segment t2=linspace(0.5, 1.5,100); %second segment vomin=12*ones(size(tmin)); vo1=12*exp(-t1/0.2); %after first switching vo2=12-11.015*exp(-(t2-0.5)/0.2); %after second switching plot(tmin,vomin,'bo',t1,vo1,'rx',t2,vo2,'md'),grid title('OUTPUT VOLTAGE'), xlabel('t(s)'),ylabel('Vo(V)')

USING MATLAB TO DISPLAY OUTPUT VOLTAGE

슬라이드 번호 1슬라이드 번호 2슬라이드 번호 3슬라이드 번호 4슬라이드 번호 5슬라이드 번호 6슬라이드 번호 7슬라이드 번호 8슬라이드 번호 9슬라이드 번호 10슬라이드 번호 11슬라이드 번호 12슬라이드 번호 13슬라이드 번호 14슬라이드 번호 15슬라이드 번호 16슬라이드 번호 17슬라이드 번호 18슬라이드 번호 19슬라이드 번호 20슬라이드 번호 21슬라이드 번호 22슬라이드 번호 23슬라이드 번호 24슬라이드 번호 25슬라이드 번호 26슬라이드 번호 27슬라이드 번호 28슬라이드 번호 29슬라이드 번호 30슬라이드 번호 31슬라이드 번호 32슬라이드 번호 33슬라이드 번호 34슬라이드 번호 35슬라이드 번호 36슬라이드 번호 37슬라이드 번호 38슬라이드 번호 39슬라이드 번호 40슬라이드 번호 41슬라이드 번호 42슬라이드 번호 43슬라이드 번호 44슬라이드 번호 45슬라이드 번호 46슬라이드 번호 47슬라이드 번호 48