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 an odd function of time, $\operatorname {sgn}(t)\,\!$,should map to

Solution

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

Solution

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