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

Properties of the Fourier Transform

Linearity

$F[ax(t)+bx(t)]$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}[ax(t)+bx(t)]e^{-j2\pi ft}dt$
	$=a\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)e^{-j2\pi ft}dt+b\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)e^{-j2\pi ft}dt$
	$=aF[x(t)]+bF[x(t)]$

Time Invariance (Delay)

$F[x(t-t_0)]$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t-t_0)e^{-j2\pi ft}dt$	Let $u=t-t_0$ and $du=dt$
	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(u)e^{-j2\pi f(u+t_0)}du$
	$=e^{-j2\pi f{t}_0}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(u)e^{-j2\pi fu}du$
	$=e^{-j2\pi f{t}_0}F[x(t)]$	Why isn't this $F[x(u)]$

Frequency Shifting

$F[e^{j2\pi ft}x(t)]$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}[e^{j2\pi f_{0}t}x(t)]e^{-j2\pi ft}dt$
	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)e^{-j2\pi (f-f_0)t}dt$
	$=X(f-f_0)$

Double Sideband Modulation

$F[cos(2\pi f_{0}t)\cdot x(t)]$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}}{2}x(t)e^{-j2\pi ft}dt$
	$=\frac{1}{2}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)[e^{-j2\pi (f-f_0)t}+e^{-j2\pi (f+f_0)t}]dt$
	$=\frac{1}{2}X(f-f_0)+\frac{1}{2}X(f+f_0)$

Differentiation in Time

$x(t)$	$=F^{-1}[X(f)]$
$F\left[\frac{dx}{dt}\right]$	$=F\left[\frac{d}{dt}{F}^{-1}[X(f)]\right]$
	$=F[\frac{d}{dt}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f)e^{j2\pi ft}df]$
	$=F[\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}j2\pi fX(f)e^{j2\pi ft}df]$
	$=F[j2\pi f{F}^{-1}[X(f)]]$
	$=j2\pi fX(f)$	Thus $\frac{dx}{dt}$ is a linear filter with transfer function $j2\pi f$

The Game (frequency domain)

• You can play the game in the frequency or time domain, but it's not advisable to play it in both at same time

Input	LTI System	Output	Reason
$\delta (t)$	$⟹$	$h(t)$	Given
$\delta (t)e^{-j2\pi ft}$	$⟹$	$h(t)e^{-j2\pi ft}$	Proportionality
$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (t)e^{-j2\pi ft}dt=F[\delta (t)]=1$	$⟹$	$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}h(t)e^{-j2\pi ft}dt=F[h(t)]=H(f)$	Superposition
$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (t-\lambda )e^{-j2\pi ft}dt=F[\delta (t-\lambda )]=1\cdot e^{-j2\pi f\lambda }$	$⟹$	$H(f)\cdot e^{-j2\pi f\lambda }$	Time Invariance
$x(\lambda )\cdot 1\cdot e^{-j2\pi f\lambda }$	$⟹$	$x(\lambda )\cdot H(f)\cdot e^{-j2\pi f\lambda }$	Proportionality
$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\lambda )\cdot 1\cdot e^{j2\pi f\lambda }d\lambda =X(f)$	$⟹$	$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\lambda )\cdot H(f)\cdot e^{j2\pi f\lambda }d\lambda =X(f)H(f)$	Superposition

• Having trouble seeing $F[x(t)*h(t)]=X(f)\cdot H(f)$

The Game (Time Domain??)

Input	LTI System	Output	Reason
$1\cdot e^{j2\pi f_{0}t}$	$⟹$	$h(t)*e^{j2\pi f_{0}t}=e^{j2\pi f_{0}t}*h(t)$	Proportionality
		$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}h(\lambda )\cdot e^{j2\pi f_0(t-\lambda )}d\lambda$	$d\lambda$ from 10/3,6 - The Game
		$e^{j2\pi f_0\lambda }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}h(\lambda )\cdot e^{-j2\pi f_0\lambda }d\lambda$
		$e^{j2\pi f_{0}t}H(f_0)$
$X(f_0)\cdot e^{j2\pi f_{0}t}$	$⟹$	$X(f_0)\cdot e^{j2\pi f_{0}t}H(f_0)$	Proportionality
$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f_0)\cdot e^{j2\pi f_{0}t}d{f}_0=x(t)$	$⟹$	$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f_0)H(f_0)\cdot e^{j2\pi f_{0}t}d{f}_0=F^{-1}[X(f)H(f)]$	Superposition

Relation to the Fourier Series

$x(t)$	$=x(t+T)$
	$=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_n{e}^{j2\pi nt/T}$
	$=\underset{Negativefrequencies}{\underbrace{\sum _{m=-\mathrm{\infty }}^{-1}{\alpha }_m{e}^{j2\pi mt/T}}}+{\alpha }_0+\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n{e}^{j2\pi nt/T}$
	$=\sum _{n=1}^{\mathrm{\infty }}{\alpha }_{-n}{e}^{-j2\pi nt/T}+{\alpha }_0+\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n{e}^{j2\pi nt/T}$	Let $n=-m$ and reverse the order of summation
		Note that ${\alpha }_{-n}{e}^{-j2\pi nt/T}$ is the complex conjugate of ${\alpha }_n{e}^{j2\pi nt/T}$
	$={\alpha }_0+2\mathrm{\Re }\left[\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n{e}^{j2\pi nt/T}\right]$	$x(t)+x(t{)}^{*}=2\mathrm{\Re }[x(t)]$
	$={\alpha }_0+\sum _{n=1}^{\mathrm{\infty }}2\mathrm{\Re }\left[\left|{\alpha }_n\right|e^{j(2\pi nt/T+{\theta }_n)}\right]$
	$={\alpha }_0+\sum _{n=1}^{\mathrm{\infty }}2\left|{\alpha }_n\right|\cos (\frac{2\pi nt}{T}+{\theta }_n)$

• How can we assume that the answer exists in the real domain?

Aside: Polar coordinates

Remember from 10/02 - Fourier Series that ${\alpha }_n=\frac{1}{T}\underset{-T/2}{\overset{T/2}{\int }}x(t)e^{-j2\pi nt/T}dt$

• Rectangular coordinates: $a+jb$
• Polar coordinates: $|a+jb|e^{j\theta }$
• $\theta =ta{n}^{-1}\frac{b}{a}$

$a+jb$	$=\sqrt{a^2+b^2}(\cos (\theta )+j\sin (\theta ))$
	$=\sqrt{a^2+b^2}\left(e^{j\theta }\right)$
	$=\sqrt{a^2+b^2}\left(e^{jta{n}^{-1}\left(\frac{b}{a}\right)}\right)$
	$=|a+jb|e^{jta{n}^{-1}\left(\frac{b}{a}\right)}$

Building up to $F[u(t)]$

$e^{i\pi }$	$=\cos \pi +i\sin \pi .$	Euler's Identity
$r_0(t)$	$=-r_0(-t)$	Real odd function of t
$F[r_0(t)]$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{r}_0(t)e^{-j2\pi ft}dt$
	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{r}_0(t)[\cos (-2\pi ft)+j\sin (-2\pi ft)]dt$
	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{r}_0(t)[\cos (2\pi ft)-j\sin (2\pi ft)]dt$	$\cos (-x)=\cos (x)$ & $\sin (-x)=-\sin (x)$
	$=j\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{r}_0(t)\sin (-2\pi ft)dt$	$r_0(t)\cdot j\sin (-2\pi ft)$ = Real odd. Integrates out over symmetric limits.
	$=$Imaginary Odd function of $f$
$r_e(t)$	$=r_e(-t)$	Real even function of t
$F[r_e(t)]$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{r}_e(t)e^{-j2\pi ft}dt$
	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{r}_e(t)[\cos (-2\pi ft)+j\sin (-2\pi ft)]dt$
	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{r}_e(t)[\cos (2\pi ft)-j\sin (2\pi ft)]dt$	$\cos (-x)=\cos (x)$ & $\sin (-x)=-\sin (x)$
	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{r}_e(t)\cos (-2\pi ft)dt$	$r_e(t)\cdot \cos (-2\pi ft)$ = 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? You can break any function down into a Taylor series. There are even and odd powers in the series.
$x_e(t)$	$=\frac{x(t)+x(-t)}{2}$
$x_o(t)$	$=\frac{x(t)-x(-t)}{2}$
$u(t)$	$=\frac{1+\mathrm{sgn}(t)}{2}$	$\mathrm{sgn}(t)=\{\begin{array}{ll}1,& t>0\\ 0,& t=0\\ -1,& t<0\end{array}$
$u_e(t)$	$=\frac{1}{2}$
$u_o(t)$	$=\frac{\mathrm{sgn}(t)}{2}$

$F[u(t)]$

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