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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]] | ||
| + | ---- | ||
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| + | |||
== The Signal == | == The Signal == | ||
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<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(j \omega)e^{j\omega t}d\omega</math> | <math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(j \omega)e^{j\omega t}d\omega</math> | ||
| + | For this problem I will not be using the above equation but in stead be using duality. | ||
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| + | |||
| + | <math>x(t) = \cos(4 t + \frac{\pi}{3})</math> | ||
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| + | |||
| + | note | ||
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| + | <math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{j k \omega_o t}</math> | ||
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| + | and | ||
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| + | <math>\cos(4 t + \frac{\pi}{3}) = \frac{e^{j(4 t + \frac{\pi}{3})}}{2} + \frac{e^{-j(4 t + \frac{\pi}{3})}}{2}</math> | ||
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| + | <math>\omega_o = 4</math> | ||
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| + | |||
| + | <math>a_1 = e^{j \frac{\pi}{3}}</math> | ||
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| + | |||
| + | <math>a_{-1} = e^{-j \frac{\pi}{3}}</math> | ||
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| + | |||
| + | <math>= 2 \pi e^{j \frac{\pi}{3}} \delta(\omega - 4) + 2 \pi e^{-j \frac{\pi}{3}} \delta(\omega + 4)</math> | ||
| + | |||
| + | |||
| + | duality applied | ||
| − | <math> | + | <math>\frac{1}{2 \pi}( 2 \pi e^{j \frac{\pi}{3}} \delta(-t - 4) + 2 \pi e^{-j \frac{\pi}{3}} \delta(-t + 4))</math> |
| − | <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) $
