HW 06

Problem

Figure out why $\underset{0}{\overset{\mathrm{\infty }}{\int }}\cos (2\pi fu)du$ seems to equal an imaginary odd function of frequency, but there is no j.

Background

This is the incorrect solution derived in class. Cosine is incorrect, because a real odd function of time, $\mathrm{sgn}(t)$,should map to an imaginary odd function of frequency.

Proof

$F[o(t)]$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}o(t)e^{-j2\pi ft}dt$
	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}o(t)[\cos (2\pi ft)+j\sin (2\pi ft)]dt$	Euler's identity
	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}o(t)j\sin (2\pi ft)dt$	Even function integrates out over symmetric limits
	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}[\text{Im }e(t)\text{ and an Im }o(f)]dt$
	$=\text{Im }o(f)$	Time integrates out

• The odd function of time has no component (ie. 0) of frequency. Thus it is an even function in frequency.

Functions

• Even*Even=Even
• Odd*Odd=Even
• Odd*Even=Odd

Incorrect Solution derived in class

$F\left[\frac{\mathrm{sgn}(t)}{2}\right]$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{\mathrm{sgn}(t)}{2}{e}^{-j2\pi ft}dt$
	$=\frac{1}{2}[\underset{-\mathrm{\infty }}{\overset{0}{\int }}-1\cdot e^{-j2\pi ft}dt+\underset{0}{\overset{\mathrm{\infty }}{\int }}1\cdot e^{-j2\pi ft}dt]$
	$=\underset{\begin{array}{c}u=-t\\ du=-dt\end{array}}{\underbrace{\frac{1}{2}\underset{0}{\overset{-\mathrm{\infty }}{\int }}{e}^{j2\pi fu}du}}+\underset{\begin{array}{c}u=t\\ du=dt\end{array}}{\underbrace{\frac{1}{2}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-j2\pi fu}du}}$
	$=\underset{0}{\overset{-\mathrm{\infty }}{\int }}\frac{e^{j2\pi fu}+e^{-j2\pi fu}}{2}du$
	$=\underset{0}{\overset{-\mathrm{\infty }}{\int }}\cos (2\pi fu)du$

Correct Solution

$F\left[\frac{\mathrm{sgn}(t)}{2}\right]$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{\mathrm{sgn}(t)}{2}{e}^{-j2\pi ft}dt$
	$=\frac{1}{2}[\underset{-\mathrm{\infty }}{\overset{0}{\int }}-1\cdot e^{-j2\pi ft}dt+\underset{0}{\overset{\mathrm{\infty }}{\int }}1\cdot e^{-j2\pi ft}dt]$
	$=\frac{1}{2}[\underset{0}{\overset{-\mathrm{\infty }}{\int }}1\cdot e^{-j2\pi ft}dt+\underset{0}{\overset{\mathrm{\infty }}{\int }}1\cdot e^{-j2\pi ft}dt]$
	$=\underset{\begin{array}{c}u=-t\\ du=-dt\end{array}}{\underbrace{\frac{1}{2}\underset{0}{\overset{\mathrm{\infty }}{\int }}-e^{j2\pi fu}du}}+\underset{\begin{array}{c}u=t\\ du=dt\end{array}}{\underbrace{\frac{1}{2}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-j2\pi fu}du}}$
	$=\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{-e^{j2\pi fu}+e^{-j2\pi fu}}{2}du$
	$=\underset{0}{\overset{\mathrm{\infty }}{\int }}-j\frac{e^{j2\pi fu}-e^{-j2\pi fu}}{2j}du$
	$=\underset{0}{\overset{\mathrm{\infty }}{\int }}-j\sin (2\pi fu)du$
	$\ne \underset{0}{\overset{\mathrm{\infty }}{\int }}\cos (2\pi fu)du$