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| + | [[Category:problem solving]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:Fourier transform]] | ||
| + | [[Category:signals and systems]] | ||
| + | == Example of Computation of Fourier transform of a CT SIGNAL == | ||
| + | A [[CT_Fourier_transform_practice_problems_list|practice problem on CT Fourier transform]] | ||
| + | ---- | ||
== The Signal == | == The Signal == | ||
| − | <math>( | + | <math>(t e^{-4t} \sin{6 \pi t}) u(t) \ </math> |
== The Fourier Transform == | == The Fourier Transform == | ||
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| + | <math>X(\omega)=\int_{-\infty}^{\infty} x(t) e^{-j\omega t}dt</math> | ||
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| + | |||
| + | <math>X(\omega)=\int_{-\infty}^{\infty} (te^{-4t}\sin{6\pi t})u(t) e^{-j\omega t}dt</math> | ||
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| + | |||
| + | <math>X(\omega)=\int_{0}^{\infty} (te^{-4t}\sin{6\pi t}) e^{-j\omega t}dt</math> | ||
| + | |||
| + | |||
| + | <math>X(\omega)=\int_{0}^{\infty} (te^{-4t})(\frac {e^{j 6 \pi t} - e^{-j 6 \pi t}}{2 j}) e^{-j\omega t}dt</math> | ||
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| + | |||
| + | <math>X(\omega)=\int_{0}^{\infty} \frac {t e^{-4t} e^{j 6 \pi t} e^{-j\omega t}}{2 j} - \frac {t e^{-4t} e^{-j 6 \pi t} e^{-j\omega t}}{2 j}dt</math> | ||
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| + | <math>X(\omega)=\int_{0}^{\infty} \frac {t e^{t(j(6 \pi - \omega)-4)}}{2 j} - \frac {t e^{t(-j(6 \pi + \omega)-4)}}{2 j}dt</math> | ||
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| + | <math>X(\omega)= \frac{(t (j(6 \pi - \omega)-4) - 1) e^{t(j(6 \pi - \omega)-4)}}{2 j (j(6 \pi - \omega)-4)} - \frac{(t (-j(6 \pi + \omega)-4) - 1) e^{t(-j(6 \pi + \omega)-4)}}{2 j (-j(6 \pi + \omega)-4)}\bigg]_0^\infty</math> | ||
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| + | <math>X(\omega)= \frac{-1}{2 j (j(6 \pi - \omega)-4)} + \frac{1}{2 j (-j(6 \pi + \omega)-4)}</math> | ||
| + | |||
| + | |||
| + | ---- | ||
| + | ==Comments/questions== | ||
| + | *A faster/easier way to solve this problem would be to use the Multiplication Property | ||
| + | |||
| + | ---- | ||
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 11:27, 16 September 2013
Contents
Example of Computation of Fourier transform of a CT SIGNAL
A practice problem on CT Fourier transform
The Signal
$ (t e^{-4t} \sin{6 \pi t}) u(t) \ $
The Fourier Transform
$ X(\omega)=\int_{-\infty}^{\infty} x(t) e^{-j\omega t}dt $
$ X(\omega)=\int_{-\infty}^{\infty} (te^{-4t}\sin{6\pi t})u(t) e^{-j\omega t}dt $
$ X(\omega)=\int_{0}^{\infty} (te^{-4t}\sin{6\pi t}) e^{-j\omega t}dt $
$ X(\omega)=\int_{0}^{\infty} (te^{-4t})(\frac {e^{j 6 \pi t} - e^{-j 6 \pi t}}{2 j}) e^{-j\omega t}dt $
$ X(\omega)=\int_{0}^{\infty} \frac {t e^{-4t} e^{j 6 \pi t} e^{-j\omega t}}{2 j} - \frac {t e^{-4t} e^{-j 6 \pi t} e^{-j\omega t}}{2 j}dt $
$ X(\omega)=\int_{0}^{\infty} \frac {t e^{t(j(6 \pi - \omega)-4)}}{2 j} - \frac {t e^{t(-j(6 \pi + \omega)-4)}}{2 j}dt $
$ X(\omega)= \frac{(t (j(6 \pi - \omega)-4) - 1) e^{t(j(6 \pi - \omega)-4)}}{2 j (j(6 \pi - \omega)-4)} - \frac{(t (-j(6 \pi + \omega)-4) - 1) e^{t(-j(6 \pi + \omega)-4)}}{2 j (-j(6 \pi + \omega)-4)}\bigg]_0^\infty $
$ X(\omega)= \frac{-1}{2 j (j(6 \pi - \omega)-4)} + \frac{1}{2 j (-j(6 \pi + \omega)-4)} $
Comments/questions
- A faster/easier way to solve this problem would be to use the Multiplication Property
