HW 06: Difference between revisions

Revision as of 16:12, 3 December 2008

Figure out why $\int_{0}^{\infty} \cos (2 π f u) d u$ seems to equal an imaginary odd function of frequency, but there is no j.

This is the incorrect solution derived in class. Cosine is incorrect, because an odd function of time, $sgn (t)$ ,should map to

$F [\frac{sgn (t)}{2}]$	$= \int_{- \infty}^{\infty} \frac{sgn (t)}{2} e^{- j 2 π f t} d t$
	$= \frac{1}{2} [\int_{- \infty}^{0} - 1 \cdot e^{- j 2 π f t} d t + \int_{0}^{\infty} 1 \cdot e^{- j 2 π f t} d t]$
	$= \underset{\begin{matrix} u = - t \\ d u = - d t \end{matrix}}{\underset{⏟}{\frac{1}{2} \int_{0}^{- \infty} e^{j 2 π f u} d u}} + \underset{\begin{matrix} u = t \\ d u = d t \end{matrix}}{\underset{⏟}{\frac{1}{2} \int_{0}^{\infty} e^{- j 2 π f u} d u}}$
	$= \int_{0}^{- \infty} \frac{e^{j 2 π f u} + e^{- j 2 π f u}}{2} d u$
	$= \int_{0}^{- \infty} \cos (2 π f u) d u$

$F [\frac{sgn (t)}{2}]$	$= \int_{- \infty}^{\infty} \frac{sgn (t)}{2} e^{- j 2 π f t} d t$
	$= \frac{1}{2} [\int_{- \infty}^{0} - 1 \cdot e^{- j 2 π f t} d t + \int_{0}^{\infty} 1 \cdot e^{- j 2 π f t} d t]$
	$= \frac{1}{2} [\int_{0}^{- \infty} 1 \cdot e^{- j 2 π f t} d t + \int_{0}^{\infty} 1 \cdot e^{- j 2 π f t} d t]$
	$= \underset{\begin{matrix} u = - t \\ d u = - d t \end{matrix}}{\underset{⏟}{\frac{1}{2} \int_{0}^{\infty} - e^{j 2 π f u} d u}} + \underset{\begin{matrix} u = t \\ d u = d t \end{matrix}}{\underset{⏟}{\frac{1}{2} \int_{0}^{\infty} e^{- j 2 π f u} d u}}$
	$= \int_{0}^{\infty} \frac{- e^{j 2 π f u} + e^{- j 2 π f u}}{2} d u$
	$= \int_{0}^{\infty} - j \frac{e^{j 2 π f u} - e^{- j 2 π f u}}{2 j} d u$
	$= \int_{0}^{\infty} - j \sin (2 π f u) d u$
	$\neq \int_{0}^{\infty} \cos (2 π f u) d u$

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 This is the incorrect solution derived in class. Cosine is incorrect, because an odd function of time, <math>\sgn(t)\,\!</math>,should map to
-==Solution==
+==Incorrect Solution derived in class==
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