Inverse laplace transform
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Inverse Laplace Transform. Inverse Laplace Transform. Definition has integration in complex plane We will use lookup tables instead Roberts, Appendix F Many Laplace transform expressions are ratios of two polynomials, a.k.a. rational functions

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Inverse Laplace Transform

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Inverse laplace transform

Inverse Laplace Transform


Inverse laplace transform1

Inverse Laplace Transform

  • Definition has integration in complex plane

    We will use lookup tables instead

    Roberts, Appendix F

  • Many Laplace transform expressions are ratios of two polynomials, a.k.a. rational functions

  • Convert complicated rational functions into simpler forms

    Apply partial fractions decomposition

    Use lookup tables


Partial fractions example 1

Partial Fractions Example #1


Partial fractions example 2

Partial Fractions Example #2


Partial fractions example 3

Partial Fractions Example #3


Mathematica

Mathematica

  • Function Apart performs partial fractions but returns conjugate poles in quadratic form

    Apart[(2 s^2 + 5)/(s^2 + 3 s + 2), s]

  • Laplace transform is an add-on package

    Needs[ “Calculus`Master`” ]

  • Forward Laplace Transform

    LaplaceTransform[Exp[-a*t], t, s]

    Note that u(t), which is UnitStep[t], is implied.

  • Inverse Laplace Transform

    InverseLaplaceTransform[1/(s+a),s,t]

double quote

backquote


Laplace transform properties

Laplace Transform Properties

  • Linearity

  • Time shifting

  • Frequency shifting

  • Differentiationin time


Differentiation in time property

Differentiation in Time Property


Laplace transform properties1

Laplace Transform Properties

  • Differentiation in frequency

  • Integration in time

    Example: f(t) = d(t)

  • Integration in frequency


Laplace transform properties2

f(2 t)

f(t)

t

t

-1

1

-2

2

Laplace Transform Properties

  • Scaling in time/frequency

    • Under integration,

  • Convolution in time

  • Convolution in frequency

Area reduced by factor 2


Example

Example

  • Compute y(t) = e a t u(t) * e b t u(t) , where a b

  • If a = b, then we would have resonance

  • What form would the resonant solution take?


Linear differential equations

Linear Differential Equations

  • Using differentiation in time propertywe can solve differential equations (including initial conditions) using Laplace transforms

  • Example: y”(t) +5 y’(t) + 6 y(t) = f ’(t) + f(t)

    With y(0-) = 2, y’(0-) =1, and f(t) = e- 4 t u(t)

    So f ’(t) = -4 e-4 t u(t) + e-4 t d(t), f ’(0-) = 0 and f ’(0+) = 1

    See handout G


Mathematica solution

Mathematica Solution

  • Define DSolve

    Needs[ "Calculus`Master`" ];

  • Define f(t) and solve for y(t)

    f[t_] := Exp[-4 t];

    DSolve[ { y''[t] + 5 y'[t] + 6 y[t] == D[ f[t], t ] + f[t],y[0] == 2, y'[0] == 1 },y[t], t ]

  • Does not distinguish between 0- and 0+


Initial and final values

Initial and Final Values

  • Values of f(t) as t 0 and t   may be computed from its Laplace transform F(s)

  • Initial value theorem

    If f(t) and its derivative df/dt have Laplace transforms, then provided that the limit on the right-hand side of the equation exists.

  • Final value theorem

    If both f(t) and df/dt have Laplace transforms, then provided that s F(s) has no poles in right-hand plane or on imaginary axis.


Final and initial values example

Final and Initial Values Example

  • Transfer function

    Poles at s = 0, s = -1  j2

    Zero at s = -3/2

Initial Value

Final Value


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