18686025 Laplace Transforms

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    S. Boyd EE102

    Lecture 3The Laplace transform

    denition & examples

    properties & formulas

    linearity the inverse Laplace transform time scaling exponential scaling time delay derivative integral multiplication by t convolution

    31

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    Idea

    the Laplace transform converts integral and differential equations intoalgebraic equations

    this is like phasors, but

    applies to general signals, not just sinusoids

    handles non-steady-state conditions

    allows us to analyze

    LCCODEs

    complicated circuits with sources, Ls, Rs, and Cs complicated systems with integrators, differentiators, gains

    The Laplace transform 32

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    Complex numbers

    complex number in Cartesian form: z = x + jy

    x = z, the real part of z

    y = z, the imaginary part of z

    j = 1 (engineering notation); i = 1 is polite term in mixedcompanycomplex number in polar form: z = re j

    r is the modulus or magnitude of z is the angle or phase of z

    exp( j) = cos + j sin

    complex exponential of z = x + jy :

    ez = ex + jy = ex ejy = ex (cos y + j sin y)

    The Laplace transform 33

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    The Laplace transform

    well be interested in signals dened fort 0the Laplace transform of a signal (function) f is the function F = L(f )dened by

    F (s) =

    0f (t)e st dt

    for those sC for which the integral makes sense

    F is a complex-valued function of complex numbers s is called the (complex) frequency variable , with units sec

    1; t is calledthe time variable (in sec); st is unitless

    for now, we assume f contains no impulses at t = 0common notation convention: lower case letter denotes signal; capitalletter denotes its Laplace transform, e.g. , U denotes L(u), V in denotesL(vin ), etc.The Laplace transform 34

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    Example

    lets nd Laplace transform of f (t) = et :

    F (s) =

    0et e st dt =

    0e(1 s ) t dt =

    11 s

    e(1 s ) t

    0=

    1s 1

    provided we can say e(1 s ) t 0 as t , which is true for s > 1:e(1 s ) t = e j ( s ) t

    =1e(1 s ) t = e(1 s ) t

    the integral dening F makes sense for all sC with s > 1 (theregion of convergence of F )

    but the resulting formula for F makes sense for all s

    C except s = 1

    well ignore these (sometimes important) details and just say that

    L(et ) =

    1

    s 1The Laplace transform 35

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    More examples

    constant: (or unit step) f (t) = 1 (for t 0)

    F (s) =

    0e st dt =

    1

    se st

    0=

    1

    s

    provided we can say e st 0 as t , which is true for s > 0 since

    e st = e j ( s ) t

    =1e ( s ) t = e ( s ) t

    the integral dening F makes sense for all s with s > 0

    but the resulting formula for F makes sense for all s except s = 0

    The Laplace transform 36

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    sinusoid: rst express f (t) = cos t as

    f (t) = (1 / 2)ejt + (1 / 2)e jt

    now we can nd F as

    F (s) =

    0e st (1/ 2)ejt + (1 / 2)e jt dt

    = (1 / 2)

    0 e( s + j ) t

    dt + (1 / 2)

    0 e( s j ) t

    dt

    = (1 / 2)1

    s j+ (1 / 2)

    1s + j

    =s

    s2 + 2

    (valid for s > 0; nal formula OK fors = j)

    The Laplace transform 37

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    powers of t : f (t) = tn (n 1)well integrate by parts, i.e. , use

    ba u(t)v (t) dt = u(t)v(t)b

    a b

    av(t)u (t) dt

    with u(t) = tn , v (t) = e st , a = 0 , b = F (s) =

    0tn e st dt = tn e

    st

    s

    0+

    ns

    0tn 1e st dt

    = ns L(t

    n 1)

    provided tn e st 0 if t , which is true for s > 0applying the formula recusively, we obtain

    F (s) =n!

    sn +1

    valid for s > 0; nal formula OK for alls = 0The Laplace transform 38

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    Impulses at t = 0

    if f contains impulses at t = 0 we choose to include them in the integraldening F :

    F (s) =

    0f (t)e st dt

    (you can also choose to not include them, but this changes some formulaswell see & use)

    example: impulse function, f =

    F (s) =

    0(t)e st dt = e st t =0 = 1

    similarly for f = (k ) we have

    F (s) =

    0(k ) (t)e st dt = ( 1)k

    dk

    dtke st

    t =0= sk e st t =0 = s

    k

    The Laplace transform 39

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    Linearity

    the Laplace transform is linear : if f and g are any signals, and a is anyscalar, we have

    L(af ) = aF, L(f + g) = F + Gi.e. , homogeneity & superposition hold

    example:

    L 3(t) 2et = 3L((t)) 2L(et )

    = 3 2

    s 1=

    3s 5s 1

    The Laplace transform 310

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    One-to-one property

    the Laplace transform is one-to-one : if L(f ) = L(g) then f = g(well, almost; see below)

    F determines f inverse Laplace transform L 1 is well dened(not easy to show)example (previous page):

    L 1 3s 5

    s 1= 3 (t) 2et

    in other words, the only function f such that

    F (s) =3s 5s 1

    is f (t) = 3 (t)

    2et

    The Laplace transform 311

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    what almost means: if f and g differ only at a nite number of points(where there arent impulses) then F = G

    examples:

    f dened as f (t) = 1 t = 20 t = 2has F = 0

    f dened asf (t) = 1/ 2 t = 01 t > 0

    has F = 1 /s (same as unit step)

    The Laplace transform 312

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    Inverse Laplace transform

    in principle we can recoverf from F via

    f (t) =1

    2j + j

    j F (s)est ds

    where is large enough that F (s) is dened for s

    surprisingly, this formula isnt really useful!

    The Laplace transform 313

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    Time scaling

    dene signal g by g(t) = f (at ), where a > 0; then

    G(s) = (1 /a )F (s/a )

    makes sense: times are scaled by a, frequencies by 1/a

    lets check:

    G(s) =

    0 f (at )e

    st dt = (1 /a )

    0 f ( )e

    (s/a ) d = (1 /a )F (s/a )

    where = at

    example: L(et) = 1 / (s 1) so

    L(eat ) = (1 /a )1

    (s/a ) 1=

    1s a

    The Laplace transform 314

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    Exponential scaling

    let f be a signal and a a scalar, and dene g(t) = eat f (t); then

    G(s) = F (s a)

    lets check:

    G(s) =

    0 e

    st eat f (t) dt =

    0 e

    ( s

    a ) t f (t) dt = F (s a)

    example:

    L(cos t) = s/ (s2 + 1) , and hence

    L(e t cos t) =

    s + 1(s + 1) 2 + 1

    =s + 1

    s2 + 2 s + 2

    The Laplace transform 315

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    Time delay

    let f be a signal and T > 0; dene the signal g as

    g(t) = 0 0 t < T f (t T ) t T (g is f , delayed by T seconds & zero-padded up to T )

    ttt = T

    f ( t ) g ( t )

    The Laplace transform 316

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    then we have G(s) = e sT F (s)

    derivation:

    G(s) =

    0e st g(t) dt =

    T e st f (t T ) dt

    =

    0e s ( + T ) f ( ) d

    = e sT F (s)

    The Laplace transform 317

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    example: lets nd the Laplace transform of a rectangular pulse signal

    f (t) = 1 if a t b0 otherwisewhere 0 < a < b

    we can write f as f = f 1 f 2 wheref 1(t) =

    1 t a0 t < a f 2(t) =1 t b0 t < b

    i.e. , f is a unit step delayed a seconds, minus a unit step delayed b seconds

    hence

    F (s) = L(f 1) L(f 2)=

    e as e bs

    s

    (can check by direct integration)

    The Laplace transform 318

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    Derivative

    if signal f is continuous at t = 0 , then

    L(f ) = sF (s) f (0)

    time-domain differentiation becomes multiplication by frequencyvariable s (as with phasors) plus a term that includes initial condition ( i.e. , f (0) )

    higher-order derivatives: applying derivative formula twice yields

    L(f ) = sL(f ) f (0)= s(sF (s) f (0)) f (0)= s2F (s) sf (0) f (0)

    similar formulas hold for

    L(f (k ) )

    The Laplace transform 319

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    examples

    f (t) = et , so f (t) = et and

    L(f ) = L(f ) =1

    s 1using the formula, L(f ) = s(

    1s 1

    ) 1, which is the same

    sin t =

    1

    ddt cos t, so

    L(sin t) = 1

    ss

    s2 + 2 1 =

    s2 + 2

    f is unit ramp, so f is unit step

    L(f ) = s1s2 0 = 1/s

    The Laplace transform 320

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    derivation of derivative formula: start from the dening integral

    G(s) =

    0f (t)e st dt

    integration by parts yields

    G(s) = e st f (t)

    0

    0f (t)(se

    st ) dt

    = limt f (t)e st

    f (0) + sF (s)

    for s large enough the limit is zero, and we recover the formula

    G(s) = sF (s) f (0)

    The Laplace transform 321