
How Laplace Transform related to Fourier Transform?  
asked  920 views  1 answers.  
Difference between Laplace Transform and Fourier Transform and how they are related.  
LaplaceTransformFourier  
 
0
 This brief note is about Fourier transform and how related to Laplace Transform.
To begin, we first state the definition of the Fourier transform [$$:]\hat{f}(\omega) = \int_{\infty}^\infty f(x) e^{j\omega x} dx[/:$$] and its inverse [$$:]f(x) = \frac{1}{2\pi}\int_{\infty}^\infty \hat{f}(\omega) e^{j\omega x} d\omega[/:$$] There are two notable features. One, with Laplace transform, the lower integration limit is zero, not negative infinity. Second, Laplace transform uses a transform variable in the complex plane. The transform variable in Fourier transform is a pure imaginary number, restricted to [$:]s = j\omega[/:$]. Generally, Laplace transform is for functions that are semiinfinite or piecewise continuous, as in the step or rectangular pulse functions. We also impose the condition that the function is zero at negative times: [$$:]f(t)=0,\;t<0[/:$$] More formally, we say the function must be of exponential order as [$:]t[/:$] approaches infinity so that the transform integral converges. A function is of exponential order if there exists(real) constants [$:]K[/:$],[$:]c[/:$] and [$:]T[/:$] such that [$$:]f(t) < Ke^{ct}\; \text{ for }\; t > T[/:$$] or in other words, the quantity [$:]e^{ct} f(t)[/:$] is bounded. If [$:]c[/:$] is chosen sufficiently large, the socalled abscissa of convergence, then [$:]e^{ct} f(t)[/:$] should approach zero as [$:]t[/:$] approaches infinity. In terms of the Laplace transform integral[$:]\int_0^{\infty}f(t) e^{st} dt[/:$], it means that the real part of [$:]s[/:$] must be larger than the real part of all the poles of [$:]f(t)[/:$] in order the integral to converge. Otherwise, we can force a function to be transformable with [$:]e^{\gamma t} f(t)[/:$] if we can choose [$:]\gamma > c[/:$] such that [$:]Ke^{(\gamma c)t}[/:$] approaches zero as [$:]t[/:$] goes to infinity. We now do a quick twostep to see how the definition of Laplace transform may arise from that of Fourier transform. First, we write the inverse transform of the Fourier transform of the function [$:] e^{\gamma t} f(t)[/:$], which of course, should recover the function itself: [$$:]e^{\gamma t} f(t) = \frac{1}{2 \pi}\int_{\infty}^\infty e^{j\omega t} d \omega \int_0^\infty e^{\gamma \tau} f(\tau) e^{j\omega \tau} d\tau[/:$$] where we have changed the lower integration limit of the Fourier transform from[$:]\infty[/:$] to [$:]0[/:$] because [$:] f(t) = 0[/:$] when [$:]t < 0[/:$]. Next, we move the exponential function [$:]e^{\gamma t}[/:$] to the RHS to go with the inverse integral and then combine the two exponential functions in the transform integral to give [$$:] f(t) = \frac{1}{2 \pi}\int_{\infty}^\infty e^{(\gamma + j\omega)t} d \omega \int_0^\infty f(\tau) e^{(\gamma + j\omega) \tau} d\tau[/:$$] Now we define [$$:] s = \gamma + j\omega[/:$$] So, [$$:] f(t) = \frac{1}{2 \pi j}\int_{\gammaj\infty}^{\gamma+j\infty} e^{st} ds \int_0^\infty f(\tau) e^{s \tau} d\tau [/:$$] From this form, we can extract the definitions of Laplace transform and its inverse: [$$:]F(s) = \int_0^\infty f(t) e^{st} dt [/:$$] and [$$:]f(t) = \frac{1}{2 \pi j}\int_{\gammaj\infty}^{\gamma+j\infty} F(s) e^{st} ds[/:$$] where again, [$:]\gamma[/:$] must be chosen to be larger than the real parts of all the poles of [$:]F(s)[/:$]. Thus the path of integration of the inverse is the imaginary axis shift by the quantity [$:]\gamma[/:$] to the right.  
 
Add comment 
Post Your Answer Here :
Add comment