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

Revision as of 18:31, 8 October 2008

$ X(\omega)=\cos{(6\omega + \pi/6)} $

Start by guessing the solution:

$ X(t)=(1/2)e^{-j(\pi/6)}\delta(t-6)+(1/2)e^{j(\pi/6)}\delta(t+6) $

Then take the fourier transform of the guessed solution to make sure it's right...


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


$ \,\mathcal{X}(\omega)=\int_{-\infty}^{+\infty}[(1/2)e^{-j(\pi/6)}\delta(t-6)+(1/2)e^{j(\pi/6)}\delta(t+6)]e^{-j\omega t}\,dt, $


$ \,\mathcal{X}(\omega)=\int_{-\infty}^{+\infty}(1/2)e^{-j(\pi/6)}\delta(t-6)e^{-j\omega t}\,dt + \int_{-\infty}^{+\infty}(1/2)e^{j(\pi/6)}\delta(t+6)e^{-j\omega t}\,dt\, $

$ \,\mathcal{X}(\omega)=(1/2)e^{-j(\pi/6)}\int_{-\infty}^{+\infty}\delta(t-6)e^{-j\omega t}\,dt + (1/2)e^{j(\pi/6)}\int_{-\infty}^{+\infty}\delta(t+6)e^{-j\omega t}\,dt\, $

$ \,\mathcal{X}(\omega)=(1/2)e^{-j(\pi/6)}e^{-6jt} + (1/2)e^{j(\pi/6)}e^{6jt} $


$ \,\mathcal{X}(\omega)=(1/2)e^{-j(6t + \pi/6)} + (1/2)e^{j(6t + \pi/6)} $


$ \,\mathcal{X}(\omega)=\cos{(6\omega + \pi/6)} $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett