Difference between revisions of "HW5.3 Tyler Johnson ECE301Fall2008mboutin" - Rhea

Latest revision as of 11:39, 16 September 2013

Example of Computation of inverse Fourier transform (CT signals)

A practice problem on CT Fourier transform

$ 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)} $

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