10/09 - Fourier Transform

$\left\langle e^{j2\pi nt/T}\mid e^{j2\pi mt/T}\right\rangle$	$=\int _{-\infty }^{\infty }e^{j2\pi nt/T}e^{-j2\pi mt/T}\,dt$
	$=\int _{-\infty }^{\infty }e^{j2\pi (n-m)t/T}\,dt$
	$=\int _{-T/2}^{T/2}e^{j2\pi (n-m)t/T}\,dt$	Assuming the function is perodic with the period T
	$=T\delta _{m,n}\,\!$

Fourier Transform

Remember from 10/02 - Fourier Series

• $\alpha _{m}={\frac {1}{T}}\int _{-T/2}^{T/2}x(t)e^{-j2\pi mt/T}\,dt$
• $x(t)=x(t+T)=\sum _{n=-\infty }^{\infty }\alpha _{m}e^{j2\pi m/T}$

If we let $T\rightarrow \infty$

${\frac {1}{T}}$	$\rightarrow df$
${\frac {n}{T}}$	$\rightarrow f$	Remember $f={\frac {2\pi n}{T}}\,\!$
$T\,\!$	$\rightarrow \infty$
$\sum _{n=-\infty }^{\infty }{\frac {1}{T}}$	$\rightarrow \int _{-\infty }^{\infty }()df$

Definitions

$F[x(t)]\,\!$	$=X(f)\,\!$	$=\int _{-\infty }^{\infty }x(t)e^{-j2\pi ft}dt$	$=\left\langle x(t)\mid e^{j2\pi ft}\right\rangle _{t}$
$F^{-1}[x(t)]\,\!$	$=x(t)\,\!$	$=\int _{-\infty }^{\infty }X(f)e^{j2\pi ft}df$	$=\left\langle X(f)\mid e^{-j2\pi ft}\right\rangle _{f}$

Examples

$\int _{-\infty }^{\infty }e^{j2\pi ft}e^{-j2\pi f\lambda }df$	$=\left\langle e^{j2\pi ft}\mid e^{j2\pi ft}\right\rangle _{f}$	$=\delta (t-\lambda )\,\!$
$\int _{-\infty }^{\infty }e^{j2\pi tf}e^{-j2\pi tf_{0}}dt$	$=\left\langle e^{j2\pi tf}\mid e^{j2\pi tf_{0}}\right\rangle _{t}$	$=\delta (f-f_{0})\,\!$

$F^{-1}[F[x(t)]]\,\!$	$=\int _{-\infty }^{\infty }\left[\int _{-\infty }^{\infty }x(\lambda )e^{-j2\pi f\lambda }d\lambda \right]e^{j2\pi ft}df$	$=\int _{-\infty }^{\infty }X(f)e^{j2\pi ft}df$	$=x(t)\,\!$
		$=\int _{-\infty }^{\infty }x(\lambda )\int _{-\infty }^{\infty }e^{j2\pi f(t-\lambda )}dfd\lambda$	$=\int _{-\infty }^{\infty }x(\lambda )\delta (t-\lambda )d\lambda$	$=x(t)\,\!$
	$=\int _{-\infty }^{\infty }\left[\int _{-\infty }^{\infty }x(\lambda )e^{-j\omega \lambda }d\lambda \right]e^{j\omega t}{\frac {1}{2\pi }}d\omega$	$=\int _{-\infty }^{\infty }x(\lambda )\left[{\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{j(t-\omega )\lambda }d\omega \right]d\lambda$	$=\int _{-\infty }^{\infty }x(\lambda )\delta (t-\omega )d\lambda$	$=x(t)\,\!$

Sifting property of the delta function

The dirac delta function is defined as any function, denoted as $\delta (t-u)\,\!$, that works for all variables that makes the following equation true: $x(t)=\int _{-\infty }^{\infty }x(u)\delta (t-u)du$

• When dealing with $\omega \,\!$, it behaves slightly different than dealing with $f\,\!$. When dealing with $=\int _{-\infty }^{\infty }x(\lambda )\left[{\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{j(t-\omega )\lambda }d\omega \right]d\lambda$, note that the delta function is ${\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{j(t-\omega )\lambda }d\omega$. The ${\frac {1}{2\pi }}$ is tacked onto the front. Thus, when dealing with $\omega \,\!$, you will often need to multiply it by $2\pi \,\!$ to cancel out the ${\frac {1}{2\pi }}$.

More properties of the delta function

$\delta (a\,t)={\frac {1}{\left|a\right|}}$

$\delta (\omega )=\delta (2\pi f)={\frac {1}{2\pi }}\delta (f)$

$\int _{-\infty }^{\infty }\delta (a\,t)\,dt$	$=\int _{-\infty }^{\infty }\delta (u\,t)\,{\frac {du}{\left|a\right|}}$	Let $a\,t=u$ and $du=a\,dt$
	$={\frac {1}{\left|a\right|}}$