EE313 Linear Systems and Signals Fall 2010 Initial conversion of content to PowerPoint by Dr. Wade C. Schwartzkopf Prof. Brian L. Evans Dept. of Electrical and Computer Engineering The University of Texas at Austin 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 Convert complicated rational functions into simpler forms Apply partial fractions decomposition Use lookup tables

Partial Fractions Example #1

Partial Fractions Example #2

Partial Fractions Example #3

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] backquote double quote

Laplace Transform Properties Linearity Time shifting Frequency shifting Differentiation in time

Differentiation in Time Property

Laplace Transform Properties Differentiation in frequency Integration in time Example: f(t) =  (t) Integration in frequency

Laplace Transform Properties Scaling in time/frequency –Under integration, Convolution in time Convolution in frequency t f(t) 2-2 t f(2 t) 1 Area reduced by factor 2

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 Using differentiation in time property we 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  (t), f ’(0 - ) = 0 and f ’(0 + ) = 1 See handout G

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 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 Transfer function Poles at s = 0, s = -1  j2 Zero at s = -3/2