(The relationship between Fourier and Laplace transform)
 
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<math>X(\sigma+j\omega)=\int_{-\infty}^{\infty}x(t){e^{-(\sigma+j\omega) t}}\, dt</math>
 
<math>X(\sigma+j\omega)=\int_{-\infty}^{\infty}x(t){e^{-(\sigma+j\omega) t}}\, dt</math>
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<math>X(\sigma+j\omega)=\int_{-\infty}^{\infty}[x(t){e^{-\sigma t}}]{e^{-j\omega t}}\, dt</math>
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The Laplace transform of <math>x(t)</math> can be interpreted as the Fourier transform of x(t) after multiplication by a real exponential signal.

Latest revision as of 16:02, 24 November 2008

The relationship between Fourier and Laplace transform

The continuous-time Fourier transform provides us with a representation for signals as linear combinations of complex exponentials of the form $ e^{st} $ with $ s=j\omega $.

For s imaginary (i.e., $ s=jw $), $ X(j\omega)=\int_{-\infty}^{\infty}x(t){e^{-j\omega t}}\, dt $ which corresponds to the Fourier transform of x(t).

For general values of the complex variable s, it is referred to as the Laplace transform of the signal. The complex variable zs can be written as $ s=\sigma+j\omega $, with $ \sigma $ and $ \omega $ the real and imaginary parts, respectively.


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

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

The Laplace transform of $ x(t) $ can be interpreted as the Fourier transform of x(t) after multiplication by a real exponential signal.

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