HW 05: Difference between revisions

Revision as of 00:25, 18 November 2008

Find the following Fourier Transforms

$F[e^{j\omega _{0}t}]$	$=\int _{-\infty }^{\infty }e^{j\omega _{0}t}e^{-j\omega t}dt$
	$=\int _{-\infty }^{\infty }e^{j(\omega _{0}-\omega )t}dt$
	$=2\pi \left[{\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{j(\omega _{0}-\omega )t}dt\right]$
	$=2\pi \delta (\omega _{0}-\omega )\,\!$
$F[\cos {\omega _{0}t}]\,\!$	$=\int _{-\infty }^{\infty }{\frac {e^{j\omega _{0}t}+e^{-j\omega _{0}t}}{2}}e^{-j\omega t}dt$
	$={\frac {1}{2}}\int _{-\infty }^{\infty }\left(e^{j\omega _{0}t}+e^{-j\omega _{0}t}\right)2e^{-j\omega t}dt$
	$={\frac {1}{2}}\int _{-\infty }^{\infty }\left[2e^{j(\omega _{0}-\omega )t}+2e^{-j(\omega _{0}+\omega )t}\right]dt$
	$=2\pi \left[{\frac {1}{2\pi }}\int _{-\infty }^{\infty }\left(e^{j(\omega _{0}-\omega )t}+e^{-j(\omega _{0}+\omega )t}\right)\,dt\right]$
	$=2\pi \delta (\omega _{0}-\omega )+2\pi \delta (\omega _{0}+\omega )\,\!$
$F[\sin {\omega _{0}t}]\,\!$	$=\int _{-\infty }^{\infty }{\frac {e^{j\omega _{0}t}-e^{-j\omega _{0}t}}{2j}}e^{-j\omega t}dt$

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-|<math>=\frac{1}{2}\int_{-\infty}^{\infty} 2e^{j(\omega_0-\omega) t} + 2e^{-j(\omega_0+\omega) t} dt</math>
+|<math>=\frac{1}{2}\int_{-\infty}^{\infty} \left [2e^{j(\omega_0-\omega) t} + 2e^{-j(\omega_0+\omega) t}\right ] dt</math>
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-|<math>=\int_{-\infty}^{\infty} e^{j(\omega_0-\omega) t} + e^{-j(\omega_0+\omega) t}</math>
+|<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>=\delta(\omega_0-\omega) + \delta(\omega_0+\omega)\,\!</math>
+|<math>=2\pi\delta(\omega_0-\omega) + 2\pi\delta(\omega_0+\omega)\,\!</math>
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 |<math>F[\sin{\omega_0 t}]\,\!</math>