HW 05: Difference between revisions

Latest revision as of 21:27, 23 November 2008

Find the following Fourier Transforms

$F [e^{j ω_{0} t}]$
$F [\cos ω_{0} t]$
$F [\sum_{- \infty}^{\infty} α_{n} e^{j 2 π n t / T}]$
$F [\sin ω_{0} t]$

Solutions

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

@@ Line 1: / Line 1: @@
 Find the following Fourier Transforms
-*<math>F[e^{j \omega_0 t}]</math>
+*<math>F\left[e^{j \omega_0 t}\right]</math>
-*<math>F[\cos {\omega_0 t}]\,\!</math>
+*<math>F\left[\cos {\omega_0 t}\right]\,\!</math>
-*<math>F[\sum_{-\infty}^{\infty}\alpha_n e^{j2\pi nt/T}]</math>
+*<math>F\left[\sum_{-\infty}^{\infty}\alpha_n e^{j2\pi nt/T}\right]</math>
-*<math>F[\sin{\omega_0 t}]\,\!</math>
+*<math>F\left[\sin{\omega_0 t}\right]\,\!</math>
 ==Solutions==
 {| border="0" cellpadding="0" cellspacing="0"
 |-
-|<math>F[e^{j \omega_0 t}]</math>
+|<math>F\left[e^{j \omega_0 t}\right]</math>
 |<math>=\int_{-\infty}^{\infty} e^{j \omega_0 t} e^{-j \omega t}dt</math>
 |-
@@ Line 20: / Line 20: @@
 |<math>=2\pi \delta(\omega_0-\omega)\,\!</math>
 |-
-|<math>F[\cos {\omega_0 t}]\,\!</math>
+|<math>F\left[\cos {\omega_0 t}\right]\,\!</math>
 |<math>=\int_{-\infty}^{\infty}\frac{e^{j\omega_0 t} + e^{-j\omega_0 t}}{2} e^{-j \omega t}dt</math>
 |-
@@ Line 35: / Line 35: @@
 |<math>=\pi\delta(\omega_0-\omega) + \pi\delta(\omega_0+\omega)\,\!</math>
 |-
-|<math>F[\sin{\omega_0 t}]\,\!</math>
+|<math>F\left[\sin{\omega_0 t}\right]\,\!</math>
 |<math>=\int_{-\infty}^{\infty}\frac{e^{j\omega_0 t} - e^{-j\omega_0 t}}{2j} e^{-j \omega t}dt</math>
 |-
@@ Line 50: / Line 50: @@
 |<math>=-j\pi\delta(\omega_0-\omega) + j\pi\delta(\omega_0+\omega)\,\!</math>
 |-
-|<math>F[\sum_{-\infty}^{\infty}\alpha_n e^{j2\pi nt/T}]</math>
+|<math>F\left[\sum_{-\infty}^{\infty}\alpha_n e^{j2\pi nt/T}\right]</math>
 |<math>=\int_{-\infty}^{\infty} \left (\sum_{-\infty}^{\infty}\alpha_n e^{j2\pi nt/T} \right )e^{-j \omega t}dt</math>
 |-
@@ Line 60: / Line 60: @@
 |-
 |
-|<math>=\sum_{-\infty}^{\infty}\alpha_n \delta_{\frac{n}{T}-f} </math>
+|<math>=\sum_{-\infty}^{\infty}\alpha_n \delta\left(\frac{n}{T}-f\right) </math>
-|-
-|
-|<math>=\alpha_{fT}\,\!</math>
 |}
-Is the last problem done correctly?

HW 05: Difference between revisions

Latest revision as of 21:27, 23 November 2008

Solutions

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