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# Inverse Laplace Transform - PowerPoint PPT Presentation

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

• 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

• 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]

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• Linearity

• Time shifting

• Frequency shifting

• Differentiationin time

• Differentiation in frequency

• Integration in time

Example: f(t) = d(t)

• Integration in frequency

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

• 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?

• 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

• 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+

• 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.

• Transfer function

Poles at s = 0, s = -1  j2

Zero at s = -3/2

Initial Value

Final Value