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+ | [[Category:problem solving]] | ||
+ | [[Category:ECE301]] | ||
+ | [[Category:ECE]] | ||
+ | [[Category:Fourier transform]] | ||
+ | [[Category:inverse Fourier transform]] | ||
+ | [[Category:signals and systems]] | ||
+ | == Example of Computation of inverse Fourier transform (CT signals) == | ||
+ | A [[CT_Fourier_transform_practice_problems_list|practice problem on CT Fourier transform]] | ||
+ | ---- | ||
+ | |||
+ | |||
== The Signal == | == The Signal == | ||
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− | <math> | + | <math>e^{j \frac{\pi}{3}} \delta(-t - 4) + e^{-j \frac{\pi}{3}} \delta(-t + 4)</math> |
+ | |||
+ | ---- | ||
+ | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] |
Latest revision as of 12:45, 16 September 2013
Example of Computation of inverse Fourier transform (CT signals)
A practice problem on CT Fourier transform
The Signal
$ X(j \omega) = \cos(4 \omega + \frac{\pi}{3}) $
Taken from 4.22.b from the course book, it looks interesting and I want to try it.
The Inverse Fourier Transform
$ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(j \omega)e^{j\omega t}d\omega $
For this problem I will not be using the above equation but in stead be using duality.
$ x(t) = \cos(4 t + \frac{\pi}{3}) $
note
$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{j k \omega_o t} $
and
$ \cos(4 t + \frac{\pi}{3}) = \frac{e^{j(4 t + \frac{\pi}{3})}}{2} + \frac{e^{-j(4 t + \frac{\pi}{3})}}{2} $
$ \omega_o = 4 $
$ a_1 = e^{j \frac{\pi}{3}} $
$ a_{-1} = e^{-j \frac{\pi}{3}} $
$ = 2 \pi e^{j \frac{\pi}{3}} \delta(\omega - 4) + 2 \pi e^{-j \frac{\pi}{3}} \delta(\omega + 4) $
duality applied
$ \frac{1}{2 \pi}( 2 \pi e^{j \frac{\pi}{3}} \delta(-t - 4) + 2 \pi e^{-j \frac{\pi}{3}} \delta(-t + 4)) $
$ e^{j \frac{\pi}{3}} \delta(-t - 4) + e^{-j \frac{\pi}{3}} \delta(-t + 4) $