HW 05: Difference between revisions

Revision as of 00:25, 18 November 2008

Find the following Fourier Transforms

$F [e^{j ω_{0} t}]$	$= \int_{- \infty}^{\infty} e^{j ω_{0} t} e^{- j ω t} d t$
	$= \int_{- \infty}^{\infty} e^{j (ω_{0} - ω) t} d t$
	$= 2 π [\frac{1}{2 π} \int_{- \infty}^{\infty} e^{j (ω_{0} - ω) t} d t]$
	$= 2 π δ (ω_{0} - ω)$
$F [\cos ω_{0} t]$	$= \int_{- \infty}^{\infty} \frac{e^{j ω_{0} t} + e^{- j ω_{0} t}}{2} e^{- j ω t} d t$
	$= \frac{1}{2} \int_{- \infty}^{\infty} (e^{j ω_{0} t} + e^{- j ω_{0} t}) 2 e^{- j ω t} d t$
	$= \frac{1}{2} \int_{- \infty}^{\infty} [2 e^{j (ω_{0} - ω) t} + 2 e^{- j (ω_{0} + ω) t}] d t$
	$= 2 π [\frac{1}{2 π} \int_{- \infty}^{\infty} (e^{j (ω_{0} - ω) t} + e^{- j (ω_{0} + ω) t}) d t]$
	$= 2 π δ (ω_{0} - ω) + 2 π δ (ω_{0} + ω)$
$F [\sin ω_{0} t]$	$= \int_{- \infty}^{\infty} \frac{e^{j ω_{0} t} - e^{- j ω_{0} t}}{2 j} e^{- j ω t} d t$

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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>
 |-
 |
-|<math>=\delta(\omega_0-\omega) + \delta(\omega_0+\omega)\,\!</math>
+|<math>=2\pi\delta(\omega_0-\omega) + 2\pi\delta(\omega_0+\omega)\,\!</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>
 |}