Chapter 13 Laplace Transform Analysis -...

68
Chapter 13 Laplace Transform Analysis

Transcript of Chapter 13 Laplace Transform Analysis -...

  • Chapter 13 Laplace Transform Analysis

  • Chapter 13: Outline

  • Laplace transforms

  • Laplace Transform

    l s-domain phasor analysis:

    l Laplace transform:)cos()( xt

    temXtx φ+ωσ= xmXX φ∠=

    +≤≤−∞

  • Laplace Transform (Three conditions)

    l Unilateral (one-sided Laplace transform): Laplacetransform holds for t=0. Previous effects are included in the initial conditions at t=0-.

    l Existence condition: l The resulting transform is a function of s.

    . ,0lim )( ctetft σσ

    σ >=−∞→

  • One-Sided Waveform

    [ ] [ ] +− ≤≤∞<

    =⇒

    ><

    =

    00,)( if ,

    0),(0,0

    )()(0,10,0

    )( :stepunit

    ttfLL

    ttft

    tutftt

    tu

    f(t)f(t)u(t)

  • Sufficient Existence Condition

    econvergenc of abscissa :order lexponentia of is )(

    ,0)( if finite remains integral The

    )()()()(

    finite is )(

    c

    0t

    000 0

    0

    lim

    σ

    σσσ

    σ

    ⇒⇒

    >=⇒

    ==≤

    ∫∫∫∫ ∫

    ∞ −

    ∞→

    ∞ −∞ −∞ ∞ −−

    ∞ −

    −−− −

    tf

    dtetf

    dtetfdtetfdtetfdtetf

    dtetf

    ct

    tststst

    st

  • Inverse Laplace Transform

    l Inverse Laplace transform for t=0:

    l Inverse Laplace transform is usually done by partial fraction expansion (will be covered in section 13.2).

    ccjc dsstesF

    jsFtf jc σ

    ωπ ω

    >+≡−= ∫ − ,)(21)]([1L)(

  • Example 13.1

    [ ]

    [ ]

    [ ] [ ][ ] )0Re complex, becan :(Note

    1)(1 ,0 if

    when ,1

    0 if ,0lim)(lim

    0

    )(

    0

    )(

    >+

    ===

    −>+

    ===

    >+==

    ∫∫∞ +−∞ −−−

    +−

    ∞→

    ∞→

    −−

    σ

    σ

    σσ

    aas

    tuLLa

    aas

    dtedteeeL

    aeetf

    eL

    ctsastatat

    ta

    t

    at

    t

    at

  • Example 13.1

  • Example 13.2

    [ ] ( )

    0

    112

    1

    21

    sin

    22

    =

    +=

    +

    −−

    =

    −= −

    c

    tjtj

    sjsjsj

    eej

    LtL

    σ

    ββ

    ββ

    β ββ

  • Transform Properties

    – Linear combination:

    [ ] [ ][ ][ ] )()(

    )()()()()()( ),()(

    sAFtAfLsBGsAFtBgtAfL

    sGtgLsFtfL

    =+=+

    ==

  • Transform Properties

    – Multiplication by e-at:

    [ ][ ] )()(

    )()(

    asFtfeL

    sFtfLat +=

    =−

  • Transform Properties

    – Multiplication by t:

    [ ])()(

    )()(

    0

    0

    ttfLdtettf

    dtetfdsd

    sFdsd

    st

    st

    −=−=

    =

    ∫∞ −

    ∞ −

  • Transform Properties

    – Time delay:

    [ ] )()(),(

    ,0)()()(

    0

    00

    000

    sFetgL

    ttttf

    ttttuttftg

    st−=

    >−<

    =−−=

  • Transform Properties

    – Differentiation:

    [ ]

    [ ] )0()0()()(

    )0()()(

    )(

    2 −−

    ′−−=′′

    −=

    =′

    fsfsFstfL

    fssFdt

    tdfLtfL

  • Transform Properties

    – Integration:

    [ ] )(1)()(

    ,0)0( ,)()(

    )()(

    0

    0

    sFs

    dfLtgL

    se

    ddtegdttftdg

    dftg

    t

    stst

    t

    =

    =

    ===

    =

    −−−

    λλ

    λλ

  • Example 13.3

    se

    se

    ssF

    DtutuDt

    tf

    sDsD

    −− −=−+=

    −−=

  • Example 13.4

    [ ]

    [ ] [ ]

    [ ]

    [ ] [ ] 22

    22

    2222

    22

    sincossinsincoscos)cos(

    cos

    )0sin(cos)(

    )(sin)(

    βφβφ

    βφβφφβ

    ββ

    ββ

    ββ

    ββ

    ββ

    β

    +−

    =−=+

    +=

    +=−

    +==′

    +=⇒=

    ss

    ttLtL

    ss

    tL

    ss

    ss

    tLtfL

    stfLttf

  • Table 13.1

  • Table 13.2

  • Solving Differential Equations– The transformation automatically incorporates

    the initial conditions (zero-input response).– Transformation converts linear differential

    equations to s-domain algebraic equations.– Transformation is similar to the s-domain

    phasor analysis. Denominator of the s-domain function includes the characteristic polynomial.

    – Inverse transformation is required to obtain the resultant time domain function.

  • First-Order Example

    Vv 6)0( =−

    [ ]

    )5(75

    56

    )5(756

    )(

    15)()0()(2.0

    15)()(2.0

    ++

    +=

    ++

    =

    =+−

    =+′

    ssssss

    sV

    ssVvssV

    tvtv

    [ ]

    ( ) 0,9151)2.0(756)5(

    755

    6)()(

    555

    111

    ≥−=−−+=

    +

    +

    +==

    −−−

    −−−

    teee

    ssL

    sLsVLtv

    ttt

  • Second-Order Example

    11

    2

    1

    )0( ,)0(

    0 t,0 t,

    RIvIi

    II

    iin

    ==

    ><

    =

    −−

  • Second Order Example (Cont.)

    [ ][ ]

    [ ]

    0for ,)()(

    switching) (no if/1)/(

    /)/()(

    /)()(

    )()()()()(

    )()()(

    0

    11

    12

    221

    21

    21

    1

    2

    ≥=⇒=

    =++++

    =

    =+−=+−

    =+′=+′

    >

    tItisI

    sI

    IILCsLRss

    LCIsILRsIsI

    sIsIRIssVC

    sVsRIIssILItItvC

    tvtRitiL

    t

    Characteristic polynomial

  • Transform Inversion

  • Partial-Fraction Expansion

    l Partial-fraction expansion of a strictly proper rational function:

    1,

    011

    1

    011

    1)()()( −≤

    +++−−+

    +++−−+== nmasansna

    ns

    bsbmsmbmsmb

    sDsNsF

    L

    L

    l Three cases will be considered: distinct real poles, complex poles and repeated poles.

  • Case 1: Distinct Real Poles

    l (Heaviside's theorem, cover-up rule.)

    K

    L

    L

    ,2,1 ,)()(

    0 ,)(

    )()(

    )(

    2121

    2

    2

    1

    1

    =−=

    ≥+++=

    −++

    −+

    −==

    = isFssA

    teAeAeAtf

    ssA

    ssA

    ssA

    sDsN

    sF

    i

    n

    ssii

    tsn

    tsts

    n

    n

  • Example 13.5: Inversion of a Third-Order Function (Heaviside’s theorem)

    =−=

    ==Ω=

    AIAI

    FCHLR

    2,2201

    ,1,12

    21

    ++

    ++=

    LCsLRss

    LCIsILRsIsI

    /1)/(2

    /21)/(2

    1)(

    0,52)(10

    12

    52)(

    1)()10(

    5)()2(

    2)(

    102)10)(2(40242

    )(

    102

    103

    22

    01

    3212

    ≥+−=⇒+

    ++

    −+=

    =+=

    −=+=

    ==

    ++

    ++=

    +++−−

    =

    −−

    −=

    −=

    =

    teetisss

    sI

    sIsA

    sIsA

    ssIA

    sA

    sA

    sA

    sssss

    sI

    tt

    s

    s

    s

  • Example 13.5: (Method of Undetermined Coefficients)

    133

    311113

    52

    3314

    )(

    10252

    )10)(2(40242

    )(

    3

    331

    32

    =⇒

    +=+−==

    ++

    +−=

    +++−−

    =

    =

    A

    AAsI

    sA

    sssssss

    sI

    s

  • Case 2: Complex Poles

    )()3)(202

    2()( nsssssssD −−ω+α+= L

    )cos(21

    21

    )(

    )()(2

    21

    21

    )(

    )()(

    , , if

    )()(

    3

    22021

    20

    2

    φβ

    βα

    βαβα

    αωαβαωα

    αβαβα

    φβα

    +=+=

    =−+=

    +++

    −+=

    −+=

    −±−=±−=<

    −+−∗−−

    +−=

    =∑

    teKeKKetg

    eKsFjsK

    js

    K

    js

    KsG

    ssA

    sGsF

    jjss

    tm

    tjtj

    jmjs

    n

    i i

    i

  • Case 2: Complex Poles (Cont.)

    [ ]

    KjKKCBjB

    KCB

    sCBBs

    ssCBs

    sG

    sKKs

    tgL

    teKtg

    mm

    mm

    tm

    =+=−+⇒⇒

    −++−−+

    =++

    +=

    ++−+

    =

    += −

    φφβαωβα

    αωααα

    ωα

    βαφβφα

    φβα

    sincos/)(

    . determine then ts,coefficien compare , and , , , Find

    )()()(

    2)(

    )(sincos)(

    )(

    )cos()(

    Or,...

    20

    220

    220

    2

    22

    Undetermined coefficients

  • Example 13.6: Inversion with Complex Poles

    432

    2

    21

    2

    1

    3212

    2

    )256)(2(71615

    2

    5)()2(

    2564321

    4321

    )(

    )(2

    )(

    43, ,2)256)(2(

    71615)(

    js

    s

    sssss

    K

    sFsA

    ssCBs

    js

    K

    js

    KsG

    sGs

    AsF

    jjssssss

    sssF

    +−=

    −=

    +++−−

    =

    =+=

    +++

    =++

    +−+

    =

    ++

    =

    ±−=±−=−=⇒+++

    −−= βα

  • Example 13.6: (Cont., method of undetermined coefficients)

    0 ),4.674cos(265)(5)(

    4.67262410

    66252

    52527

    )0(

    25610

    25

    )256)(2(71615

    )(

    10515 ,2562

    515)(

    25625

    )256)(2(71615

    )(

    0322

    0

    22

    2

    2

    2

    3

    3

    22

    2

    ≥++=+=

    ∠=+=−

    +=

    −=⇒+=⋅−

    =

    +++

    ++

    =+++

    −−=

    =⇒+=∞→++

    ++

    +=

    ++

    =⇒

    +++

    ++

    =+++

    −−=

    −−− tteetgetf

    jCB

    jBK

    CC

    F

    ssCs

    ssssss

    sF

    BBsss

    CsBss

    sss

    ssF

    ssCBs

    ssssss

    sF

    ttt

    βα

    LL

    )cos()( φβα += − teKtg tm

  • Case 3: Repeated Poles

    [ ]i

    i

    ssii

    ssii

    n

    j j

    jiiiii

    ii

    ii

    i

    i

    i

    i

    n

    n

    sFssdsdA

    sFssA

    ssA

    sssssFss AA

    tsteAtseAtg

    ssA

    ssAsG

    ssA

    ssAsGsF

    =

    =

    =

    −=

    −=⇒

    −−−=− ∑++

    +=

    −+

    −=

    −++

    −+=

    )()(

    )()(

    )()()()(

    21

    22

    3

    221

    2

    21

    221

    3

    3

    )(

    poles) (double )(

    )(

    )()( L

  • Case 3: Repeated Poles (Cont.)

    [ ][ ]

    [ ]i

    i

    i

    ssii

    ssii

    ssii

    ii

    ii

    ii

    i

    i

    i

    i

    i

    i

    sFssdsd

    A

    sFssdsd

    A

    sFssA

    tsetAtsteAtseAtg

    ssA

    ssA

    ssAsG

    =

    =

    =

    −=

    −=

    −=⇒

    ++=

    −+

    −+

    −=

    )()(!2

    1

    )()(

    )()(21

    32

    2

    1

    32

    33

    2321

    33

    221

    )(

    )()()(

  • Example 13.7: Inversion with a Triple Pole

    0 ,26

    )(

    )4(6

    )4(0

    41

    051

    646

    1641

    32014

    )0(

    10)(lim

    )4(6

    )4(4)5()4(

    142

    6)()4( ,1)()5(

    )4()4(4)5()4(

    142

    5424

    32

    1212

    11

    321211

    3

    2

    43

    1354

    43

    132

    12113

    2

    51)(

    51)(

    5)(

    ≥++−=

    +++

    =⇒+++−==

    −=⇒=

    +++++

    +−−

    =+==+=

    +++++

    +−−

    −−−

    ∞→

    −=−=

    ++++−=

    ++++==

    ++++==

    teetetf

    sss

    AA

    F

    AssF

    ssA

    sA

    ss

    ss

    sFsAsFsA

    As

    As

    AsA

    ss

    ss

    ttt

    s

    ss

    ssF

    ssF

    ssF

  • Time DelayInitial-Value TheoremFinal-Value Theorem

  • Time Delay

    [ ] [ ]

    0 ),()()()(

    )()()(

    )()()(

    )()()(

    0021

    2121

    00

    00

    0

    ≥−−+=

    +=+

    =

    =−−

    −−

    tttuttftftf

    esFsFsD

    esNsNsF

    etgLttuttgL

    stst

    st

  • Example 13.8: Inversion with Time Delay

    [ ] 0 ),3(8844)3()3(2)()(44)(

    544

    )5(20

    )(

    )(2)()5(

    4020)(

    4020)(5)(

    0)0( ),3(40)(20)()(5)(

    )3(40)(20)( :excitation

    )3(5511

    51

    1

    311

    3

    3

    ≥−−−−=−−−=

    −=

    −−+=

    −−=

    −=−

    +−=

    +−=−

    =−+−=−=−′

    −−=

    −−

    ttueetutftfty

    etf

    sssssF

    esFsFss

    esY

    se

    sYssY

    ytututxtyty

    tututx

    tt

    t

    st

    s

  • Initial-Value Theorem

    ( )properstrictly is )()(lim)0(

    sF

    ssFfs ∞→

    + =

  • Initial-Value Theorem (Cont.)

    [ ]( )0 ),0()()0()( i.e.,

    )()0()0()()(

    >−=−

    −−=+−

    −+

    tftfftf

    tufftftf

    c

    c

  • Initial-Value Theorem (Cont.)

    [ ][ ]

    ( )[ ]

    0)0()(lim

    0)0()0()0()(

    00over finite is )(

    )0()(lim0)(lim

    )()0()()(

    /)0()0()()(

    0

    0

    =−

    =−−−⇒

    ≤≤′

    −==′

    ′=−=′

    −−=

    +

    ∞→

    −−+

    +−

    ∞→

    ∞ −

    ∞→

    ∞ −−

    −+

    ∫∫

    fssF

    fffssF

    ttf

    fssFdtetf

    dtetffssFtfL

    sffsFsF

    s

    c

    ccs

    stcs

    stcccc

    c

  • Initial Slope

    [ ]

    )()()0()(

    lim)0(

    )0()()(

    )0()()(

    known is)0( assuming )0( slope initial Find

    sDsDfssN

    sf

    fsDssN

    fssFtfL

    ff

    s

    ∞→

    +

    −−

    −+

    −=′

    −=−=′

  • Final-Value Theorem

    =∞→

    axisimaginary on the poles no origin, at the poles multiple no

    RHP, in the poles no

    )(lim)(0

    ssFfs

    [ ]

    )0()()()0()(lim

    )()0()()(

    00

    0

    −∞−

    ∞ −−

    −∞=′=−

    ′=−=′

    ∫∫

    ffdttffssF

    dtetffssFtfL

    s

    st

  • Example 13.9: Calculating Initial and Final Values

    5)0( =−f

    proper)(strictly )8009018(

    16005)()(

    )( 233

    +++−

    ==ssss

    ssDsN

    sF

    9090

    lim)0(

    proper)(strictly )8009018(

    560045090)(

    )()0()( :)0(

    )0(5)(lim)0(

    4

    3

    23

    23

    −=+

    −−=′

    +++−−−

    =−′

    ===

    ∞→

    +

    −+

    ∞→

    +

    LL

    ss

    sf

    sssssss

    sDsDfssN

    f

    fssFf

    s

    s

  • Example 13.9: (Cont.)

    5)0( =−f

    proper)(strictly )8009018(

    16005)()(

    )( 233

    +++−

    ==ssss

    ssDsN

    sF

    2)(lim)(locations polecheck values,finalFor

    0−==∞

    →ssFf

    s

  • Transform circuit analysis

  • Transform Circuit Analysis

    l Given a circuit with some initial state at t=0-and an excitation x(t) starting at t=0, find the resulting behavior of any voltage or current y(t) for t=0.

    l Zero-state response, natural response, forced response, zero-input response and complete response.

  • Zero-State Responsel Zero-state response: a circuit with no stored energy at t=0-.

    ( )functionnetwork domain - theas same

    )(

    )state" initial zero(" )(

    )]([)()],([)(

    network,order th -nan For

    011

    1

    011

    1

    0101

    s

    asasasabsbsbsb

    XY

    sH

    sYsdt

    ydL

    tyLsYtyLsY

    xbdtdx

    bdt

    xdbya

    dtdy

    adt

    yda

    nn

    nn

    mm

    mm

    kk

    k

    m

    m

    mn

    n

    n

    ++++++++

    =≡

    =

    ==⇓

    +++=+++

    −−

    −−

    LL

    LL

  • Zero-State Response

    [ ]

    polynomial sticcharacteri :)()()(

    )( ,)()(

    )(

    )()( (3).

    )()()( ),()( (2).)()(

    )(obtain diagram,domain - Draw (1).

    :Steps

    1

    sPsDsN

    sXsPsN

    sH

    sYLty

    sXsHsYsXtxsXsY

    sHs

    X

    XH ==

    =

    =→

    =

  • Step Response (zero initial state, by definition)

    0,1)()( >== ttutx

    ssX /1)( =

    )()(1

    )()(ssPsN

    ssHsY H=×=

    A pole at the origin

  • Example 13.10: Step Response

    Z(s)

    ssVsssZ

    sH

    sHsVsI

    s

    s

    1)( ,)16/2()(

    )(1

    )()()(

    =+==

    =

  • Example 13.10: (Cont.)

    5.0)(lim)(

    0)(lim)0(

    0 ,5.05.0)(

    )4(1

    45.05.0

    )4(81

    )()(

    0

    44

    22

    ==∞

    ==

    ≥−−=

    +−

    +−=

    ++

    =×=

    ∞→

    +

    −−

    ssIi

    ssIi

    tteeti

    ssssss

    ssHsI

    s

    s

    tt

    Steady state response natural response

  • Zero-state AC response

    )cos()( xm tXtx φβ +=

    22)( β+= ssDX

    ymymF

    FN

    FNFN

    XH

    YXjHYtYtytytyty

    ssN

    sPsN

    sYsYssP

    sNsNsY

    φβφβ

    ββ

    ∠==+=+=

    ++=+≡

    +=

    )( where)cos()()()()(

    )()()(

    )()())(()()(

    )( 2222

    Natural responsefrom Y(s)

    Forced responseFrom phasor analysis

  • Example 13.11: Zero-State AC Response (from Fig. 13.9)

    ttN

    FNF

    s

    ss

    teeti

    sIsIs

    sNss

    ssss

    sVsHsI

    ss

    sVtttv

    44

    22

    22

    2

    2

    10)(

    )()(64

    )()4(

    104

    1)64()4(

    40050)()()(

    6450

    )(0 ,8cos50)(

    −− −−=

    +≡+

    ++

    −+−

    =

    +++

    ==

    +=⇒≥=

    Phasor analysis

  • Example 13.11: (Cont.)

    0)9.81cos(07.71)0(

    0 ),9.818cos(07.7)101()()()(

    )9.818cos(07.7)(

    9.8107.7)8(

    9.81141.0)8( ,050

    0

    04

    0

    0

    00

    =−+−=

    ≥−++−=+=

    −=

    −∠=⋅=

    −∠=∠=

    +

    i

    ttettititi

    tti

    VjHI

    jHV

    tFN

    F

    s

    s

  • Natural Response and Forced Response

    response) forced thefind toanalysisphasor use(or 0 ,5105)(

    35

    210

    15

    )3)(2)(1(10

    )()()(

    )2,1( 3 :(1) case10)(

    )( ,)2)(1(

    1)(Let

    32

    000

    221

    0

    ≥+−=

    ++

    +−

    ++

    =+++

    ==

    −≠−≠−==

    +=++

    =

    −−−

    −−

    teeety

    sssssssXsHsY

    sssetx

    eAeAtyss

    sH

    ttt

    ts

    ttN

  • Natural Response and Forced Response(Cont.)

    0 ,101010)(

    )2(10

    210

    110

    )2)(1(10

    )(

    analysis transformuse

    frequency natural a as same theisfrequency excitation when applicablenot is analysisphasor

    )2(

    2 :(2) case

    22

    22

    0

    ≥−−=

    +−

    ++

    −+

    +=

    ++=

    ⇒∞→−

    −=

    −−− tteeety

    ssssssY

    H

    s

    ttt

    natural mixed forced

  • Zero-Input Response

    l The excitation equals zero for t=0 but the circuit contains stored energy at t=0-.

    l Thévenin/Norton equivalent circuits can be established.

  • Zero-Input Response (Cont.)

    [ ]

    −=

    +=⇒

    −=

    ′=

    )0()()( :Norton

    )(1)0(

    )( :Thevenin

    )0()()(

    )()(

    CCC

    CC

    C

    CCC

    CC

    CvssCVsI

    sIsCs

    vsV

    vssVCsI

    tvCti

  • Zero-Input Response (Cont.)

    [ ]

    +=

    −=⇒

    −=

    ′=

    )(1)0(

    )( :Norton

    )0()()( :Thevenin

    )0()()(

    )()(

    sVsLs

    isI

    LissLIsV

    issILsV

    tiLtv

    LL

    L

    LLL

    LLL

    LL

  • Example 13.12: Calculating a Zero-Input Response

    VvAi

    tAti

    CL

    s

    120206)0( ,6)0(

    :analysis statesteady DC0 ,6)(

    =⋅==

  • Example 13.12: (Cont.)

    0 ,)455cos(26

    )cos()(

    452666 ,5 ,5

    25010606

    )(

    12012)(

    100202

    :responseinput -zero

    05

    0

    20

    22

    ≥−=

    +=

    −∠=−===⇒

    +++

    =++

    +=

    +=

    ++

    tAte

    teKti

    jK

    ssCBs

    sss

    sI

    ssI

    ss

    t

    tmL

    L

    L

    φβ

    βα

    ωα

    α

  • Complete Response

    l Complete response: complete response=zero-input response + zero-state response (with fictitious sources)

  • Example 13.13: Calculating a Complete Response

    ssVVvAi

    t

    tvtV

    tVtv

    SCL

    Cs

    20)( ,20)0( ,4)0(

    0for analysis statesteady DC

    )( find 0 ,20

    0 ,20)(

    −===

    <

    ≥−<

    =

    −−

  • Example 13.13: (Cont.)

    0 ,)6.268cos(52020)(808

    )(

    5.05.0

    /202)(

    5.01

    51

    40

    04

    21

    ≥−+−=++

    ++=

    +−

    =

    ++

    − tVtetvss

    CBssA

    sV

    ss

    sVs

    s

    tC

    C

    C

    Overshoot due to underdamped

  • Chapter 13: Problem Set

    l 6, 10, 16, 21, 25, 28, 31, 35, 38, 42, 46, 51, 57, 58, 63, 65