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based on slides by J. Yan Slide 1.1 Laplace Transform Part 1: Introduction (I&N Chap 13) Definition of the L.T. L.T. of Singularity Functions L.T. Pairs Properties of the L.T. Inverse L.T. • Convolution IVT(initial value theorem) & FVT (final value theorem)

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  • based on slides by J. Yan Slide 1.1

    Laplace Transform Part 1: Introduction (I&N Chap 13)

    • Definition of the L.T.• L.T. of Singularity Functions• L.T. Pairs • Properties of the L.T. • Inverse L.T. • Convolution• IVT(initial value theorem) & FVT (final value

    theorem)

  • Slide 1.2

    Lessons from Phasor Analysis What good are phasors?

    dt

    tdiLtRitvtvtv LR

    )()()()()( +=+= IIVVV LjRLR ω+=+=

    ( )RLM tLR

    Vti ωω

    ω1

    22tancos

    )()( −−

    += ( )RLM

    LR

    V

    LjR

    ω

    ω

    ω1

    22tan0

    )(

    −−°∠+

    =

    +=⇒

    I

    VI

    Time domain

    How might we solve i(t)?

    Phasor domain

    ⇐We’d much rather solve algebraic equations in the phasor domain than DEs in the time domain.

    based on slides by J. Yan

  • Slide 1.3

    Motivation for Laplace Transforms Laplace transforms: generalized tool for circuit analysis(transient

    response, steady state,…).

    )( :Input tf

    )( :Output tg

    Time domain (t-domain)

    Laplace/complex frequency domain (s-domain)

    ⇐Laplace Xforms:

    • Change linear DEs into algebraic equations (easier to solve)

    • Handle a wider variety of inputs than just sinuosoids

    • Incorporate ICs in the solution automatically

    • Provide the total response (natural+forced) in one operation

    Circuit

    )( :Input sF

    )( :Output sG

    based on slides by J. Yan

  • Laplace Transform

    Slide 1.4based on slides by J. Yan

  • Slide 1.5

    Definition of Laplace Transform Given f(t), its (one-sided) Laplace transform (if it exists) is given by:

    ∫∞ −==0

    dte)t(f)s(F)]t(f[ stL

    Examples: Recall singularity functionsare discontinuous or have discontinuous derivatives (useful for modeling switching in signals).

    ≥<

    =0 if 1

    0 if 0)( :)Function" Heaviside" (a.k.a. step Unit :1 E.g.

    t

    ttu

    ≥<

    =−=0

    00 if 1

    if 0 :step Unit Shifted :2 E.g.

    0 tt

    tt)tt(u)t(ut

    )()()(1)(but 0for 0)(

    :)Function" Delta Dirac" (a.k.a. impulse Unit :3 E.g.

    00 tfdttfttdtttt =−⇒=≠= ∫∫∞

    ∞−

    ∞−δδδ

    based on slides by J. Yan

  • Slide 1.6

    More Examples: ramp, exp & sine t)t(u)t(r ⋅= :ramp Unit :4 E.g.

    )t(ue at− :lexponentia Decaying :5 E.g.

    )sin()t(u)tsin()t(u Ttπω 2 :Sinusoid :6 E.g. ⋅=⋅

    based on slides by J. Yan

  • Slide 1.7

    • Not all functions have a LT.

    • For most circuits (including any you’re required to analyse for EECE 253), the LT will exist in some region of convergence.

    • (f1(t)=f2(t))⇒(F1(s)=F2(s)). Is the converse true?

    • f(t): R→R. What about F(s)?

    Comments about the LT

    based on slides by J. Yan

  • Slide 1.8

    Laplace Transform Pairs

    *Defined for t≥0; f(t)=0 for t

  • Slide 1.9

    Properties of the L.T.

    )(lim)(lim :(FVT) Theorem Value Final .12

    )(lim)(lim :(IVT) Theorem Value Initial .11

    0

    0

    ssFtf

    ssFtf

    st

    st

    →∞→

    ∞→→

    =

    =

    based on slides by J. Yan

  • Slide 1.10

    • You won’t be asked to compute the LT from the definition. Instead, use the look-up tables of the preceding two slides (on an exam, these will be provided so no need to memorise them now).

    • The LT is interesting mathematically but also takes much time to understand. For now, focus more on using this tool rather than understanding why/how it works.

    • The power of this tool largely depends on the properties of time differentiation/integration. Observe what happens in the s-domain.

    • We’ll increasingly see that the poles of a LT (i.e., roots of the denominator) are quite important. You’ve already used this fact earlier this term…where?

    Comments about Using the LT in 253

    based on slides by J. Yan

  • Slide 1.11

    Examplestetuttf 23)(2)()( : E.g. −−+= δ

    t)u(t)(ttf 2sin)( : E.g. 2=

    ≤≤

    = otherwise 0

    32for 10)( : E.g.

    ttg

    based on slides by J. Yan

  • Slide 1.12

    Inverse Laplace Transform If the region of convergence for F(s) is Re(s)>σc, then the inverse Laplace transform is given by:

    ∫∞+

    ∞−==

    j

    j

    stdsesFj

    tfsF1

    1

    )(2

    1)()]([

    σ

    σπ1-

    L

    Fortunately, in 253, this computation isn’t required but you’ll need to generate a partial fraction expansion (PFE) and use look-up tables.

    Algorithm to find inverse LT:

    1. Find all poles of F(s). ID them as simple vs. repeated vs. complex.

    2. Find partial fraction expansion (PFE) in basic terms.

    3. Look up inverse of each basic term in tables.

    Consider F(s)=N(s)/D(s) where N(s) & D(s) are polynomials in s with degree (N(s))

  • Slide 1.13

    Examples ).( find ,

    16

    5

    3

    41)(Given

    2tf

    s

    s

    ssF

    +−

    ++=

    ).( find ,)4)(3)(1(

    )2(6)(Given tf

    sss

    ssF

    ++++=

    based on slides by J. Yan

  • Slide 1.14

    Poles of F(s)There are 3 relatively distinct types of poles that F(s) may have:

    Simple: pi is real and negative (pi

  • Slide 1.15

    F(s) Partial Fraction Expansion• Given F(s)=N(s)/D(s) and the poles of D(s), you often need to find the

    coefficients in the PFE. The text demonstrates the Residue Method for all coefficients but I recommend using this only for a pole’s highest degree (i.e., if simple, if repeated). For the others (complex poles and lower degrees of a pole), I recommend a form of the Algebraic Method (examples on next two slides).

    • Note subtle differences in my choice of notation compared to I&N (and other textbooks). Consider what reasons I might have for these differences. – textbook uses (s+pi) as a factor of D(s) whereas I prefer (s-pi).

    – textbook uses {(s+α)2+β2} as a factor but I prefer {(s-σ)2+ω2}.

    • I specified the poles must be in the LHP. Why?

    based on slides by J. Yan

  • Slide 1.16

    Example

    ( ) ( ) ).( find ,3162

    )(Given 2

    3

    tgsss

    sssG

    ++++=

    based on slides by J. Yan

  • Slide 1.17

    Example

    ( )( ) ).( find ,134110

    )(Given 2

    tgsss

    sG+++

    =

    based on slides by J. Yan

  • Slide 1.18

    Consider a linear time-invariant (LTI) system having impulse response h(t). If the system excitation (or input) is x(t), the response (or output) y(t) can be computed from the convolution integral:

    Convolution Integral

    )()()()()(0

    thtxdthxtyt

    ⊗=−= ∫ λλλThis formula is explained at great length in the text. However, I mainly require that you know the following:

    { } { } )()()()()()( sYsHsXthtxty =⋅=⊗= LL

    based on slides by J. Yan

  • Slide 1.19

    Convolution Integral NotesThe convolution integral applies to systems which are causal, linear

    and time-invariant. Suppose you have a system with zero ICs and you know its impulse (Dirac Delta) response is h(t).

    • Causal ⇒ h(t)=0 for t

  • Slide 1.20

    Linear Integrodifferential Equations

    1)0()0( where44 Solve : Example ===++ − vvevvv t &&&&

    0)0( where2)(2)(3 Solve :Example 30

    ==++ −∫ yedytyytt ττ&

    Linear integroDEs can be Xformed by the LT into s-domain, solved algebraically (include any ICs) and Xformed back into t-domain.

    based on slides by J. Yan

  • Slide 1.21

    ExampleIf the network is in steady state prior to t=0, find i(t) for t>0. (NB: Already solved this type of question before but now can solve using L.T.)

    based on slides by J. Yan