Fourier Transform in the Form of Dirac-Delta Function

  • #1

Homework Statement


Given [itex]x(t)=8cos(70\pi t)+4sin(132\pi t)+8cos(24\pi t)[/itex], find the Fourier transform [itex]X(f)[/itex] in the form of [itex]\delta[/itex] function.

Homework Equations


[itex]X(f)=\int ^{\infty}_{-\infty}x(t)e^{-j\omega _0t}dt[/itex]
[itex]cos(\omega t)=\frac{e^{j\omega t}+e^{-j\omega t}}{2}[/itex]
[itex]sin(\omega t)=\frac{e^{j\omega t}-e^{-j\omega t}}{2j}[/itex]
[itex]\int ^{\infty}_{-\infty}cos(\omega _0t)e^{-j\omega t}dt=\frac{\pi}{2}(\delta (\omega +\omega _0)+\delta (\omega -\omega _0))[/itex]
[itex]\int ^{\infty}_{-\infty}sin(\omega _0t)e^{-j\omega t}dt=\frac{\pi}{j2}(\delta (\omega +\omega _0)-\delta (\omega -\omega _0))[/itex]

The Attempt at a Solution


[itex]X(f)=\frac{8\pi}{2}(\delta (\omega +70\pi)+\delta (\omega -70\pi))+\frac{4\pi}{j2}(\delta (\omega +132\pi)-\delta (\omega -132\pi))+\frac{8\pi}{2}(\delta (\omega +24\pi)+\delta (\omega -24\pi))[/itex]

Simplifying: [itex]X(f)=4\pi (\delta (\omega +70\pi)+\delta (\omega -70\pi))+\frac{2\pi}{j} (\delta (\omega +132\pi)-\delta (\omega -132\pi))+4\pi (\delta (\omega +24\pi)+\delta (\omega -24\pi))[/itex]
 

Answers and Replies

  • #2
rude man
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All clear, Captain!
(You could changew +1/2j to -j/2 but that would be quibbling!)
Nice work.
 

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