10/10,13,16,17 - Fourier Transform Properties

Properties of the Fourier Transform

Linearity

$F\left[a\,x(t)+b\,x(t)\right]$	$=\int _{-\infty }^{\infty }\left[a\,x(t)+b\,x(t)\right]e^{-j\,2\pi f\,t}\,dt$
	$=a\int _{-\infty }^{\infty }\,x(t)\,e^{-j\,2\pi f\,t}\,dt+b\int _{-\infty }^{\infty }\,x(t)\,e^{-j\,2\pi f\,t}\,dt$
	$=a\,F[x(t)]+b\,F[x(t)]$

Time Invariance (Delay)

$F[x(t-t_{0})]\,\!$	$=\int _{-\infty }^{\infty }x(t-t_{0})\,e^{-j\,2\pi f\,t}\,dt$	Let $u=t-t_{0}\,\!$ and $du=dt\,\!$
	$=\int _{-\infty }^{\infty }x(u)\,e^{-j\,2\pi f\,(u+t_{0})}\,du$
	$=e^{-j\,2\pi f\,t_{0}}\int _{-\infty }^{\infty }x(u)\,e^{-j\,2\pi f\,u}\,du$
	$=e^{-j\,2\pi f\,t_{0}}\,F[x(t)]$

Frequency Shifting

$F\left[e^{j\,2\pi f\,t}x(t)\right]$	$=\int _{-\infty }^{\infty }\left[e^{j\,2\pi f_{0}\,t}x(t)\right]e^{-j\,2\pi f\,t}\,dt$
	$=\int _{-\infty }^{\infty }x(t)\,e^{-j\,2\pi (f-f_{0})\,t}\,dt$
	$=X(f-f_{0})\,\!$

Double Sideband Modulation

$F[cos(2\pi f_{0}\,t)\cdot x(t)]$	$=\int _{-\infty }^{\infty }{\frac {e^{j\,2\pi f_{0}\,t}+e^{-j\,2\pi f_{0}\,t}}{2}}x(t)\,e^{-j\,2\pi f\,t}\,dt$
	$={\frac {1}{2}}\int _{-\infty }^{\infty }x(t)\left[e^{-j\,2\pi (f-f_{0})\,t}+e^{-j\,2\pi (f+f_{0})\,t}\right]\,dt$
	$={\frac {1}{2}}X(f-f_{0})+{\frac {1}{2}}X(f+f_{0})$

Differentiation in Time

$x(t)\,\!$	$=F^{-1}\left[X(f)\right]$
$F\left[{\frac {dx}{dt}}\right]$	$=F\left[{\frac {d}{dt}}F^{-1}\left[X(f)\right]\right]$
	$=F\left[{\frac {d}{dt}}\int _{-\infty }^{\infty }X(f)\,e^{j\,2\pi f\,t}\,df\right]$
	$=F\left[\int _{-\infty }^{\infty }j\,2\pi fX(f)e^{j\,2\pi f\,t}\,df\right]$
	$=F\left[j\,2\pi f\,F^{-1}[X(f)]\right]$
	$=j\,2\pi f\,X(f)$	Thus ${\frac {dx}{dt}}$ is a linear filter with transfer function $j\,2\pi f$

The Game (frequency domain)

• You can play the game in the frequency or time domain, but not both at the same time
  • Then how can you use the Fourier Transform, but can't build up to it?

Input	LTI System	Output	Reason
$\delta (t)\,\!$	$\Longrightarrow$	$h(t)\,\!$	Given
$\delta (t)\,e^{-j\,2\,\pi f\,t}$	$\Longrightarrow$	$h(t)\,e^{-j\,2\,\pi f\,t}$	Proportionality
$\int _{-\infty }^{\infty }\delta (t)\,e^{-j\,2\,\pi f\,t}\,dt=F[\delta (t)]=1$	$\Longrightarrow$	$\int _{-\infty }^{\infty }h(t)\,e^{-j\,2\,\pi f\,t}\,dt=F[h(t)]=H(f)$	Superposition
$\int _{-\infty }^{\infty }\delta (t-\lambda )\,e^{-j\,2\,\pi f\,t}\,dt=F[\delta (t-\lambda )]=1\cdot e^{-j\,2\,\pi f\,\lambda }$	$\Longrightarrow$	$H(f)\cdot e^{-j\,2\,\pi f\,\lambda }$	Time Invariance
$x(\lambda )\cdot 1\cdot e^{-j\,2\,\pi f\,\lambda }$	$\Longrightarrow$	$x(\lambda )\cdot H(f)\cdot e^{-j\,2\,\pi f\,\lambda }$	Proportionality
$\int _{-\infty }^{\infty }x(\lambda )\cdot 1\cdot e^{j\,2\,\pi f\,\lambda }\,d\lambda =X(F)$	$\Longrightarrow$	$\int _{-\infty }^{\infty }x(\lambda )\cdot H(f)\cdot e^{j\,2\,\pi f\,\lambda }\,d\lambda =X(F)\,H(f)$	Superposition

• Having trouble seeing $F\left[x(t)*h(t)\right]=X(f)\cdot H(f)$
• Since we were dealing in the frequency domain, is that the reason why multiplying one side did not result in a convolution on the other?

The Game (Time Domain??)

Input	LTI System	Output	Reason
$1\cdot e^{j\,2\,\pi f_{0}\,t}$	$\Longrightarrow$	$h(t)*e^{j\,2\,\pi f_{0}\,t}=e^{j\,2\,\pi f_{0}\,t}*h(t)$	Proportionality
		$\int _{-\infty }^{\infty }h(\lambda )\cdot e^{j\,2\,\pi f_{0}\,(t-\lambda )}\,d\lambda$	Why d lambda instead of dt?
		$e^{j\,2\,\pi f_{0}\,\lambda }\int _{-\infty }^{\infty }h(\lambda )\cdot e^{-j\,2\,\pi f_{0}\,\lambda }\,d\lambda$
		$e^{j\,2\,\pi f_{0}\,t}\,H(f_{0})$
$X(f_{0})\cdot e^{j\,2\,\pi f_{0}\,t}$	$\Longrightarrow$	$X(f_{0})\cdot e^{j\,2\,\pi f_{0}\,t}\,H(f_{0})$	Proportionality, Why isn't this a convolution?
$\int _{-\infty }^{\infty }X(f_{0})\cdot e^{j\,2\,\pi f_{0}\,t}\,df_{0}=x(t)$	$\Longrightarrow$	$\int _{-\infty }^{\infty }X(f_{0})H(f_{0})\cdot e^{j\,2\,\pi f_{0}\,t}\,df_{0}=F^{-1}\left[X(f)H(f)\right]$	Superposition, Not X(f_0)H(f_0)?

Relation to the Fourier Series

$x(t)\,\!$	$=x(t+T)\,\!$
	$=\sum _{n=-\infty }^{\infty }\alpha _{n}e^{j\,2\pi n\,t/T}$
	$=\underbrace {\sum _{m=-\infty }^{-1}\alpha _{m}e^{j\,2\pi m\,t/T}} _{Negativefrequencies}+\alpha _{0}+\sum _{n=1}^{\infty }\alpha _{n}e^{j\,2\pi n\,t/T}$
	$=\sum _{n=1}^{\infty }\alpha _{-n}e^{-j\,2\pi n\,t/T}+\alpha _{0}+\sum _{n=1}^{\infty }\alpha _{n}e^{j\,2\pi n\,t/T}$	Let $n=-m\,\!$ and reverse the order of summation
	$=\alpha _{0}+2\Re \left[\sum _{n=1}^{\infty }\alpha _{n}e^{j\,2\pi n\,t/T}\right]$	Assume that $x(t)\in \Re$

• Does the server reset every hour?
• How can we assume that the answer exists in the real domain?

Remember from 10/02 - Fourier Series that $\alpha _{n}={\frac {1}{T}}\int _{-T/2}^{T/2}x(t)e^{-j\,2\,\pi \,n\,t/T}\,dt$

• $\alpha _{n}=\left|\alpha _{n}\right|\,e^{j\theta }$?
• Rest of page

Building up to $F\left[u(t)\right]$

$e^{i\pi }\,\!$	$=\cos \pi +i\sin \pi .\,\!$	Euler's Identity
$r_{0}(t)\,\!$	$=-r_{0}(-t)\,\!$	Real odd function of t
$F\left[r_{0}(t)\right]\,\!$	$=\int _{-\infty }^{\infty }r_{0}(t)\,e^{-j\,2\pi \,f\,t}\,dt$
	$=\int _{-\infty }^{\infty }r_{0}(t)\,\left[\cos(-2\pi \,f\,t)+j\sin(-2\pi \,f\,t)\right]\,dt$
	$=\int _{-\infty }^{\infty }r_{0}(t)\,\left[\cos(2\pi \,f\,t)-j\sin(2\pi \,f\,t)\right]\,dt$	$\cos(-x)=\cos(x)\,\!$ & $\sin(-x)=-\sin(x)\,\!$
	$=j\int _{-\infty }^{\infty }r_{0}(t)\,\sin(-2\pi \,f\,t)\,dt$	$r_{0}(t)\cdot j\sin(-2\pi \,f\,t)$ = Real odd. Integrates out over symmetric limits.
	$=\,\!$Imaginary Odd function of $f\,\!$
$r_{e}(t)\,\!$	$=r_{e}(-t)\,\!$	Real even function of t
$F\left[r_{e}(t)\right]\,\!$	$=\int _{-\infty }^{\infty }r_{e}(t)\,e^{-j\,2\pi \,f\,t}\,dt$
	$=\int _{-\infty }^{\infty }r_{e}(t)\,\left[\cos(-2\pi \,f\,t)+j\sin(-2\pi \,f\,t)\right]\,dt$
	$=\int _{-\infty }^{\infty }r_{e}(t)\,\left[\cos(2\pi \,f\,t)-j\sin(2\pi \,f\,t)\right]\,dt$	$\cos(-x)=\cos(x)\,\!$ & $\sin(-x)=-\sin(x)\,\!$
	$=\int _{-\infty }^{\infty }r_{e}(t)\,\cos(-2\pi \,f\,t)\,dt$	$r_{e}(t)\cdot \cos(-2\pi \,f\,t)$ = Real odd. Integrates out over symmetric limits.
	$=\,\!$Real Even function of $f\,\!$

Definitions

$x(t)\,\!$	$=x_{e}(t)+x_{(}o)t\,\!$	Can't x(t) have parts that aren't even or odd?
$x_{e}(t)\,\!$	$={\frac {x(t)+x(-t)}{2}}$
$x_{o}(t)\,\!$	$={\frac {x(t)-x(-t)}{2}}$
$u(t)\,\!$	$={\frac {1+\operatorname {sgn}(t)}{2}}$	$\operatorname {sgn}(t)={\begin{cases}1,&t>0\\0,&t=0\\-1,&t<0\end{cases}}$
$u_{e}(t)\,\!$	$={\frac {1}{2}}$
$u_{o}(t)\,\!$	$={\frac {\operatorname {sgn}(t)}{2}}$

$F\left[u(t)\right]$

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