Laplace Transform

Forward Laplace Transform Decompose a signal f(t) into complex sinusoids of the form es t where s is complex: s = s + j2pf Forward (bilateral) Laplace transform f(t): complex-valued function of a real variable t F(s): complex-valued function of a complex variable s Bilateral means that the extent of f(t) can be infinite in both the positive t and negative t direction (a.k.a. two-sided)

Inverse (Bilateral) Transform is a contour integral which represents integration over a complex region– recall that s is complex c is a real constant chosen to ensure convergence of the integral Notation F(s) = L{f(t)} variable t implied for L f(t) = L-1{F(s)} variable s implied for L-1

Laplace Transform Properties Linear or nonlinear? Linear operator L F(s) f(t)

Laplace Transform Properties Time-varying or time-invariant? This is an odd question to ask because the output is in a different domain than the input.

Example

Convergence The condition Re{s} > -Re{a} is the region of convergence, which is the region of s for which the Laplace transform integral converges Re{s} = -Re{a} is not allowed (see next slide)

Regions of Convergence What happens to F(s) = 1/(s+a) at s = -a? (1/0) -e-a t u(-t) and e-a t u(t) have same transform function but different regions of convergence Im{s} Re{s} = -Re{a} Re{s} f(t) 1 t f(t) f(t) = e-a t u(t) causal t -1 f(t) = -e- a t u(-t) anti-causal

Review of 0- and 0+ d(t) not defined at t = 0 but has unit area at t = 0 0- refers to an infinitesimally small time before 0 0+ refers to an infinitesimally small time after 0

Unilateral Laplace Transform Forward transform: lower limit of integration is 0- (i.e. just before 0) to avoid ambiguity that may arise if f(t) contains an impulse at origin Unilateral Laplace transform has no ambiguity in inverse transforms because causal inverse is always taken: No need to specify a region of convergence Disadvantage is that it cannot be used to analyze noncausal systems or noncausal inputs

Existence of Laplace Transform As long as e-s t decays at a faster rate than rate f(t) explodes, Laplace transform converges for some M and s0, there exists s0 > s to make the Laplace transform integral finite We cannot always do this, e.g. does not have a Laplace transform

Key Transform Pairs

Key Transform Pairs

Fourier vs. Laplace Transform Pairs Assuming that Re{a} > 0