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Inverse Laplace TransformPowerPoint Presentation

Inverse Laplace Transform

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Inverse Laplace Transform

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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(2 t)

f(t)

t

t

-1

1

-2

2

- 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