# 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, $\sgn(t)\,\!$,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

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

## Correct Solution

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