HW 05: Difference between revisions

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|<math>F[\sin{\omega_0 t}]\,\!</math>
|<math>F[\sin{\omega_0 t}]\,\!</math>
|<math>=\int_{-\infty}^{\infty}\frac{e^{j\omega_0 t} - e^{-j\omega_0 t}}{2j} e^{-j \omega t}dt</math>
|<math>=\int_{-\infty}^{\infty}\frac{e^{j\omega_0 t} - e^{-j\omega_0 t}}{2j} e^{-j \omega t}dt</math>
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|<math>=\frac{1}{2j}\int_{-\infty}^{\infty}\left (e^{j\omega_0 t} - e^{-j\omega_0 t}\right )2je^{-j \omega t} dt</math>
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|<math>=\int_{-\infty}^{\infty} \left (e^{j(\omega_0-\omega) t} - e^{-j(\omega_0+\omega) t}\right ) dt</math>
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|<math>=2\pi\left [ \frac{1}{2\pi}\int_{-\infty}^{\infty} \left (e^{j(\omega_0-\omega) t} - e^{-j(\omega_0+\omega) t}\right )\,dt\right]</math>
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|<math>=2\pi\delta(\omega_0-\omega) - 2\pi\delta(\omega_0+\omega)\,\!</math>
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Revision as of 00:37, 18 November 2008

Find the following Fourier Transforms

Solutions