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 $du=dt$
	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(u)e^{-j2\pi f(u+t_0)}du$
	$=e^{-j2\pi f{t}_0}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(u)e^{-j2\pi fu}du$
	$=e^{-j2\pi f{t}_0}F[x(t)]$

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 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
$\delta (t)$	$⟹$	$h(t)$	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 $F[x(t)*h(t)]=X(f)\cdot H(f)$
• 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
$1\cdot e^{j2\pi f_{0}t}$	$⟹$	$h(t)*e^{j2\pi f_{0}t}=e^{j2\pi f_{0}t}*h(t)$	Proportionality
		$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}h(\lambda )\cdot e^{j2\pi f_0(t-\lambda )}d\lambda$	Why d lambda instead of dt?
		$e^{j2\pi f_0\lambda }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}h(\lambda )\cdot e^{-j2\pi f_0\lambda }d\lambda$
		$e^{j2\pi f_{0}t}H(f_0)$
$X(f_0)\cdot e^{j2\pi f_{0}t}$	$⟹$	$X(f_0)\cdot e^{j2\pi f_{0}t}H(f_0)$	Proportionality, Why isn't this a convolution?
$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f_0)\cdot e^{j2\pi f_{0}t}d{f}_0=x(t)$	$⟹$	$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f_0)H(f_0)\cdot e^{j2\pi f_{0}t}d{f}_0=F^{-1}[X(f)H(f)]$	Superposition, Not X(f_0)H(f_0)?