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<math> x(t) = \frac{d}{dt} {u(-2-t) + u(t-2)} </math>
 
<math> x(t) = \frac{d}{dt} {u(-2-t) + u(t-2)} </math>
  
<math> x(t) = \int_{-\infty}^{infty} \frac{d}{dt} {u(-2-t) + u(t-2)} e^{-j \omega t} dt </math>
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<math> X(\omega) = \int_{-\infty}^{\infty} \frac{d}{dt} {u(-2-t) + u(t-2)} e^{-j \omega t} dt </math>
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<math> X(\omega) = \int_{-\infty}^{\infty} \delta (-2-t) e^{-j \omega t} dt + \int_{-\infty}^{\infty} \delta (t-2) e^{-j \omega t} dt</math>
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<math> X(\omega) = -e^{2j \omega} + e^{-2j \omega} </math>
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<math> X(\omega) = -2j (\frac{e^{2j \omega} - e^{-2j \omega}}{2j}) </math>
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<math> X(\omega) = -2j sin(2 \omega) </math>

Latest revision as of 17:26, 24 October 2008

Fourier Transform of delta functions

1.

$ x(t) = \delta (t+1) + \delta (t-1) $

$ X(\omega) = \int_{-\infty}^{\infty} \delta (t+1)e^{-j \omega t} + \int_{-\infty}^{\infty} \delta (t-1)e^{-j \omega t} dt $

$ X(\omega) = e^{j \omega}+ e^{-j \omega} = \frac{1}{2} (e^ {j \omega} + e^ {-j \omega})^2 $

$ X(\omega) = 2cos(\omega) $

2.

$ x(t) = \frac{d}{dt} {u(-2-t) + u(t-2)} $

$ X(\omega) = \int_{-\infty}^{\infty} \frac{d}{dt} {u(-2-t) + u(t-2)} e^{-j \omega t} dt $

$ X(\omega) = \int_{-\infty}^{\infty} \delta (-2-t) e^{-j \omega t} dt + \int_{-\infty}^{\infty} \delta (t-2) e^{-j \omega t} dt $

$ X(\omega) = -e^{2j \omega} + e^{-2j \omega} $

$ X(\omega) = -2j (\frac{e^{2j \omega} - e^{-2j \omega}}{2j}) $

$ X(\omega) = -2j sin(2 \omega) $

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EISL lab graduate

Mu Qiao