Latest revision as of 07:33, 18 January 2013

Comparing the Fourier transform and the Laplace transform of $ x(t) = e^{-at}u(t) $

Let the signal $ x(t) = e^{-at}u(t) $. The Fourier transform $ X(\omega) $ converges for $ a > 0 $ and is given by:

$ X(\omega) = \int_{-\infty}^{\infty} e^{-at}u(t)e^{-j\omega t} \, dt = \int_{0}^{\infty} e^{-at}e^{-j\omega t} \, dt = \frac{1}{j\omega + a}, a > 0 $ (1)

the Laplace transform is:

$ X(s) = \int_{-\infty}^{\infty} e^{-at}u(t)e^{-st} \, dt = \int_{0}^{\infty} e^{-(s + a)t} \, dt $

with $ s = \sigma + j\omega $

$ X(\sigma + j\omega) = \int_{0}^{\infty} e^{-(\sigma + a)t} e^{-j\omega t} \, dt $

By comparing with (1) above, this last equation is the Fourier transform of $ e^{-(\sigma + a)t}u(t) $

$ \Rightarrow X(\sigma + j\omega) = \frac{1}{(\sigma + a) + j\omega}, \sigma + a > 0 $

since $ ^{s = \sigma + j\omega} $ and $ \sigma = \mathbb{R} (s) $

$ X(s) = frac{1}{s + a}, \mathbb{R} (S) > -a $

that is,

$ e^{-at}u(t) \xrightarrow{\mathcal{L}} \frac{1}{s + a}, \mathbb{R}(s) > -a $

for example, for $ a = 0, x(t) $ is the unit step with Laplace transform

$ X(s) = \frac{1}{s}, \mathbb{R} (s) > 0 $

Back to HW10

Back to ECE301 Fall 2008, Prof. Boutin