(New page: Let the signal, x(t) be equal to: <math>x(t) = cos(2\pi t) \,</math> The Fourier Transform of a signal in Continuous Time is defined by: <math>X(\omega) = \int_{-\infty}^{\infty}x(t)e^{...)
 
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Let the signal, x(t) be equal to:
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Let the signal x(t) be equal to:
  
 
<math>x(t) = cos(2\pi t) \,</math>
 
<math>x(t) = cos(2\pi t) \,</math>

Revision as of 13:51, 7 October 2008

Let the signal x(t) be equal to:

$ x(t) = cos(2\pi t) \, $

The Fourier Transform of a signal in Continuous Time is defined by:

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

Using this, we obtain:

$ X(\omega) = \int_{-\infty}^{\infty}cos(2\pi t)e^{-j\omega t}dt \, $

Knowing that cos(t) is equal to: $ \frac{e^{jt}+e^{-jt}}{2} $:

$ X(\omega) = \int_{-\infty}^{\infty}\frac{1}{2}(e^{j2\pi t}+e^{-j2\pi t})e^{-j\omega t}dt \, $


$ X(\omega) = \pi \delta(\omega - 2\pi) + \pi \delta(\omega - 2\pi) \, $



Note: This could also be done knowing that the Fourier transform of $ e^{j\omega_0 t} = 2\pi \delta(\omega - \omega_0) \, $.

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva