HW 06

Problem

Figure out why $\int _{0}^{\infty }\cos(2\pi \,f\,u)\,du$ 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, $\operatorname {sgn}(t)\,\!$,should map to an imaginary odd function of frequency.

Proof

$F[o(t)]\,\!$	$=\int _{-\infty }^{\infty }\,o(t)\,e^{-j\,2\,\pi \,f\,t}\,dt$
	$=\int _{-\infty }^{\infty }\,o(t)\,\left[\cos(2\,\pi \,f\,t)+j\,\sin(2\,\pi \,f\,t)\right]\,dt$	Euler's identity
	$=\int _{-\infty }^{\infty }\,o(t)\,j\,\sin(2\,\pi \,f\,t)\,dt$	Even function integrates out over symmetric limits
	$=\int _{-\infty }^{\infty }\,\left[{\mbox{Im }}e(t){\mbox{ and an Im }}o(f)\right]\,dt$
	$={\mbox{Im }}o(f)\,\!$	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

$F\left[{\frac {\operatorname {sgn}(t)}{2}}\right]$	$=\int _{-\infty }^{\infty }{\frac {\operatorname {sgn}(t)}{2}}e^{-j\,2\,\pi \,f\,t}\,dt$
	$={\frac {1}{2}}\left[\int _{-\infty }^{0}-1\cdot e^{-j\,2\,\pi \,f\,t}\,dt+\int _{0}^{\infty }1\cdot e^{-j\,2\,\pi \,f\,t}\,dt\right]$
	$=\underbrace {{\frac {1}{2}}\int _{0}^{-\infty }e^{j\,2\,\pi \,f\,u}\,du} _{\begin{matrix}u=-t\\du=-dt\end{matrix}}+\underbrace {{\frac {1}{2}}\int _{0}^{\infty }e^{-j\,2\,\pi \,f\,u}\,du} _{\begin{matrix}u=t\\du=dt\end{matrix}}$
	$=\int _{0}^{-\infty }{\frac {e^{j\,2\,\pi \,f\,u}+e^{-j\,2\,\pi \,f\,u}}{2}}\,du$
	$=\int _{0}^{-\infty }\cos(2\,\pi \,f\,u)\,du$

Correct Solution

$F\left[{\frac {\operatorname {sgn}(t)}{2}}\right]$	$=\int _{-\infty }^{\infty }{\frac {\operatorname {sgn}(t)}{2}}e^{-j\,2\,\pi \,f\,t}\,dt$
	$={\frac {1}{2}}\left[\int _{-\infty }^{0}-1\cdot e^{-j\,2\,\pi \,f\,t}\,dt+\int _{0}^{\infty }1\cdot e^{-j\,2\,\pi \,f\,t}\,dt\right]$
	$={\frac {1}{2}}\left[\int _{0}^{-\infty }1\cdot e^{-j\,2\,\pi \,f\,t}\,dt+\int _{0}^{\infty }1\cdot e^{-j\,2\,\pi \,f\,t}\,dt\right]$
	$=\underbrace {{\frac {1}{2}}\int _{0}^{\infty }-e^{j\,2\,\pi \,f\,u}\,du} _{\begin{matrix}u=-t\\du=-dt\end{matrix}}+\underbrace {{\frac {1}{2}}\int _{0}^{\infty }e^{-j\,2\,\pi \,f\,u}\,du} _{\begin{matrix}u=t\\du=dt\end{matrix}}$
	$=\int _{0}^{\infty }{\frac {-e^{j\,2\,\pi \,f\,u}+e^{-j\,2\,\pi \,f\,u}}{2}}\,du$
	$=\int _{0}^{\infty }-j\,{\frac {e^{j\,2\,\pi \,f\,u}-e^{-j\,2\,\pi \,f\,u}}{2j}}\,du$
	$=\int _{0}^{\infty }-j\,\sin(2\,\pi \,f\,u)\,du$
	$\neq \int _{0}^{\infty }\cos(2\,\pi \,f\,u)\,du$