View source for HW10 Daniel Barjum: Example illustration relationship between Fourier and Laplace transform ECE301Fall2008mboutin - Rhea

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[[Category:Fourier transform]]
[[Category:Laplace transform]]
[[Category:relation between Laplace transform and Fourier transform]]
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[[Category:ECE301Fall2008mboutin]]

=Comparing the Fourier transform and the Laplace transform of <math>x(t) = e^{-at}u(t)</math>=

Let the signal '''<math>x(t) = e^{-at}u(t)</math>'''.  The Fourier transform '''<math>X(\omega)</math>''' converges for '''<math>a > 0</math>''' and is given by:

<math>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  </math>         (1)

the Laplace transform is:


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

with <math>s = \sigma + j\omega</math>

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

By comparing with (1) above, this last equation is the Fourier transform of '''<math>e^{-(\sigma + a)t}u(t)</math>'''

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

since '''<math>^{s = \sigma + j\omega}</math>''' and <math>\sigma = \mathbb{R} (s)</math>

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

that is,

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

for example, for '''<math>a = 0, x(t)</math>''' is the unit step with Laplace transform 

<math>X(s) = \frac{1}{s}, \mathbb{R} (s) > 0 </math>
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