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Laplace Transforms – recap for ccts

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  • MAE140 Linear Circuits107

    Laplace Transforms recap for ccts

    Whats the big idea?1. Look at initial condition responses of ccts due to capacitor

    voltages and inductor currents at time t=0Mesh or nodal analysis with s-domain impedances

    (resistances) or admittances (conductances)Solution of ODEs driven by their initial conditions

    Done in the s-domain using Laplace Transforms2. Look at forced response of ccts due to input ICSs and IVSs

    as functions of time

    Input and output signals IO(s)=Y(s)VS(s) or VO(s)=Z(s)IS(s)The cct is a system which converts input signal to

    output signal3. Linearity says we add up parts 1 and 2

    The same as with ODEs

  • MAE140 Linear Circuits108

    Laplace transforms

    The diagram commutesSame answer whichever way you go

    Linearcct

    Differentialequation

    Classicaltechniques

    Responsesignal

    Laplacetransform L

    Inverse Laplacetransform L-1

    Algebraicequation

    Algebraictechniques

    Responsetransform

    Tim

    e do

    mai

    n (t

    dom

    ain) Complex frequency domain(s domain)

  • MAE140 Linear Circuits109

    Laplace Transform - definition

    Function f(t) of timePiecewise continuous and exponential order

    0- limit is used to capture transients and discontinuities at t=0s is a complex variable (+j)

    There is a need to worry about regions of convergence ofthe integral

    Units of s are sec-1=HzA frequency

    If f(t) is volts (amps) then F(s) is volt-seconds (amp-seconds)

    btKetf

  • MAE140 Linear Circuits110

    Laplace transform examplesStep function unit Heavyside Function

    After Oliver Heavyside (1850-1925)

    Exponential functionAfter Oliver Exponential (1176 BC- 1066 BC)

    Delta (impulse) function (t)

    0if1

    )()(

    0

    )(

    000

    >=+

    !=!===

    "+!"!"

    !

    !"

    !

    !## $

    %$

    %$

    sj

    e

    s

    edtedtetusF

    tjststst

    =!! ! asFeatuatf asL

  • MAE140 Linear Circuits116

    Laplace Transform Properties

    Initial Value Property

    Final Value Property

    Caveats:Laplace transform pairs do not always handle

    discontinuities properlyOften get the average value

    Initial value property no good with impulsesFinal value property no good with cos, sin etc

    )(lim)(lim0

    ssFtfst !"+"

    =

    )(lim)(lim0

    ssFtfst !"!

    =

  • MAE140 Linear Circuits117

    Rational FunctionsWe shall mostly be dealing with LTs which are

    rational functions ratios of polynomials in s

    pi are the poles and zi are the zeros of the functionK is the scale factor or (sometimes) gain

    A proper rational function has nmA strictly proper rational function has n>mAn improper rational function has n

  • MAE140 Linear Circuits118

    A Little Complex Analysis

    We are dealing with linear cctsOur Laplace Transforms will consist of rational function (ratios

    of polynomials in s) and exponentials like e-sThese arise from discrete component relations of capacitors and inductors the kinds of input signals we apply

    Steps, impulses, exponentials, sinusoids, delayedversions of functions

    Rational functions have a finite set of discrete polese-s is an entire function and has no poles anywhere

    To understand linear cct responses you need to look atthe poles they determine the exponential modes inthe response circuit variables.Two sources of poles: the cct seen in the response to Ics

    the input signal LT poles seen in the forced response

  • MAE140 Linear Circuits119

    A Little More Complex Analysis

    A complex function is analytic in regions where it hasno polesRational functions are analytic everywhere except at a finite

    number of isolated points, where they have poles of finiteorderRational functions can be expanded in a Taylor Series

    about a point of analyticity

    They can also be expanded in a Laurent Series about anisolated pole

    General functions do not have N necessarily finite

    ...)(2)(!2

    1)()()()( +!!"+!"+= afazafazafzf

    ! !"

    "=

    #

    =

    "+"=1

    0

    )()()(

    Nn n

    nn

    nn azcazczf

  • MAE140 Linear Circuits120

    Residues at poles

    Functions of a complex variable with isolated, finiteorder poles have residues at the polesSimple pole: residue =

    Multiple pole: residue =

    The residue is the c-1 term in the Laurent Series

    Cauchy Residue TheoremThe integral around a simple closed rectifiable positively

    oriented curve (scroc) is given by 2j times the sum ofresidues at the poles inside

    )()(lim sFasas

    !"

    [ ])()(lim)!1(

    11

    1

    sFas

    ds

    d

    m

    m

    m

    m

    as

    !! !

    !

    "

  • MAE140 Linear Circuits121

    Inverse Laplace Transforms the Bromwich Integral

    This is a contour integral in the complex s-plane is chosen so that all singularities of F(s) are to the left of

    Re(s)=It yields f(t) for t0

    The inverse Laplace transform is always a causal functionFor t

  • MAE140 Linear Circuits122

    Inverse Laplace Transform Examples

    Bromwich integral of

    On curve C1

    For given there is r such that

    Integral disappears on C1 for positive t

    x

    R

    s-plane

    pole a

    t0 t

  • MAE140 Linear Circuits123

    Inverting Laplace Transforms

    Compute residues at the poles

    Example

    )()(lim sFasas

    !"

    !"#

    $%& '

    '

    '

    (')()(

    1

    1lim

    )!1(

    1sF

    mas

    mds

    md

    asm

    ( ) ( ) ( ) ( )31

    3

    21

    1

    1

    2

    31

    3)1(2)1(2

    31

    522

    +!

    ++

    +=

    +

    !+++=

    +

    +

    ssss

    ss

    s

    ss

    3)1(

    )52()1(lim 323

    1=

    +

    ++

    ssss

    s1

    )1()52()1(lim 3

    23

    1=

    +

    ++

    ssss

    dsd

    s

    2)1(

    )52()1(lim!2

    13

    23

    2

    2

    1=

    +

    ++

    ssss

    dsd

    s

    ( ) )(32)1(52 23

    21 tutte

    sss t +=

    +

    + L

  • MAE140 Linear Circuits124

    Inverting Laplace Transforms

    Compute residues at the poles

    Bundle complex conjugate pole pairs into second-order terms if you want

    but you will need to be careful

    Inverse Laplace Transform is a sum of complexexponentials

    For circuits the answers will be real

    ( )[ ]222 2))(( !""!"!" ++#=+### ssjsjs

    )()(lim sFasas

    !"

    !

    1

    (m "1)!lims#a

    dm"1

    dsm"1

    (s" a)mF(s)[ ]

  • MAE140 Linear Circuits125

    Inverting Laplace Transforms in Practice

    We have a table of inverse LTsWrite F(s) as a partial fraction expansion

    Now appeal to linearity to invert via the tableSurprise!Nastiness: computing the partial fraction expansion is best

    done by calculating the residues

    !

    F(s) =bms

    m + bm"1sm"1 +L+ b1s+ b0

    ansn + an"1s

    n"1 +L+ a1s+ a0

    = K(s" z1)(s" z2)L(s" zm)

    (s" p1)(s" p2)L(s" pn )

    =#1

    s" p1( )+

    #2

    s" p2( )+

    #31

    (s" p3)+

    #32

    s" p3( )2

    +#33

    s" p3( )3

    + ...+#q

    s" pq( )

  • MAE140 Linear Circuits126

    Example 9-12

    Find the inverse LT of)52)(1(

    )3(20)( 2 ++++

    =sss

    ssF

    21211)(

    *221

    js

    k

    js

    k

    s

    ksF

    +++

    !++

    +=

    !4

    5

    2555

    21)21)(1(

    )3(20)()21(

    21lim2

    10

    1522

    )3(20)()1(

    1lim1

    jej

    jsjss

    ssFjs

    jsk

    sss

    ssFs

    sk

    =""=

    +"=+++

    +="+

    +"#=

    =

    "=++

    +=+

    "#=

    )()4

    52cos(21010

    )(252510)( 4

    5)21(

    4

    5)21(

    tutee

    tueeetf

    tt

    jtjjtjt

    !"

    #$%

    &++=

    !!

    "

    #

    $$

    %

    &

    ++=

    ''

    '''++''

    (

    ((

  • MAE140 Linear Circuits127

    Not Strictly Proper Laplace Transforms

    Find the inverse LT of

    Convert to polynomial plus strictly proper rational functionUse polynomial division

    Invert as normal

    348126)( 2

    23

    ++

    +++=

    ssssssF

    3

    5.0

    1

    5.02

    34

    22)(

    2

    ++

    +++=

    ++

    +++=

    sss

    ss

    sssF

    )(5.05.0)(2)(

    )( 3 tueetdt

    tdtf tt !"

    #$%

    &+++=

    ''((

  • MAE140 Linear Circuits128

    Multiple Poles

    Look for partial fraction decomposition

    Equate like powers of s to find coefficients

    Solve

    )())(()(

    )())((

    )()(

    12221212

    211

    22

    22

    2

    21

    1

    12

    21

    1

    pskpspskpskKzKs

    ps

    k

    ps

    k

    ps

    k

    psps

    zsKsF

    !+!!+!=!

    !+

    !+

    !=

    !!

    !=

    112221122

    1

    22212121

    211

    2

    )(22

    0

    Kzpkppkpk

    Kkppkpk

    kk

    =!+

    =++!!

    =+

  • MAE140 Linear Circuits129

    Introductory s-Domain Cct AnalysisFirst-order RC cct

    KVL

    instantaneous for each tSubstitute element relations

    Ordinary differential equation in terms of capacitor voltage

    Laplace transform

    Solve

    Invert LT

    +_

    R

    VA

    i(t)t=0

    CvcvSvR+ + +

    -

    -

    -0)()()( =!! tvtvtv CRS

    dt

    tdvCtitRitvtuVtv

    CRAS

    )()(),()(),()( ===

    )()()(

    tuVtvdt

    tdvRC AC

    C =+

    ACCC Vs

    sVvssVRC1

    )()]0()([ =+!

    RCs

    v

    RCss

    RCVsV

    CAC

    /1

    )0(

    )/1(

    /)(

    ++

    +=

    Volts)()0(1)( tueveVtv RCt

    CRCt

    AC !"

    #$%

    &+''(

    )**+

    ,-=

    --

  • MAE140 Linear Circuits130

    An Alternative s-Domain Approach

    Transform the cct element relationsWork in s-domain directly OK since L is linear

    KVL in s-Domain

    +_

    R

    VA

    i(t)t=0

    CvcvSvR+ + +

    -

    -

    -

    +_

    R

    VA

    I(s)

    Vc(s)

    +

    -

    sC

    1

    s

    1

    +_s

    Cv )0(

    )0()()(

    )0()(

    1)(

    CCC

    CCC

    CvssCVsI

    s

    vsI

    CssV

    !=

    += Impedance + source

    Admittance + source

    ACCC Vs

    sVCRvssCRV1

    )()0()( =+!

  • MAE140 Linear Circuits131

    Time-varying inputsSuppose vS(t)=Vacos(t), what happens?

    KVL as before

    Solve

    +_

    R

    VA

    i(t)t=0

    CvcvSvR+ + +

    -

    -

    -

    +_

    R I(s)

    Vc(s)

    +

    -

    sC

    1

    22 !+s

    AsV

    +_s

    Cv )0(

    RCs

    v

    RCss

    RC

    sV

    sV

    s

    sVRCvsVRCs

    C

    A

    C

    ACC

    1

    )0(

    )1)(()(

    )0()()1(

    22

    22

    ++

    ++=

    +=!+

    "

    "

    )()0(2)(1

    )cos(2)(1

    )( tuRCt

    eCvRCt

    e

    RC

    AVt

    RC

    AVtCv !

    "

    #$%

    & '+

    '

    +'+

    +

    =(

    )((