Fourier transform.tex

\documentclass[12pt]{article}

\usepackage{mathtools}


\begin{document}
FFT
$$
F(s) 
\equiv
{\int\limits_{-\infty}^{\infty}f(x)\,e^{-2\pi i s x}\,dx~}
$$
DFT
$$
f(x) 
\equiv
{\int\limits_{-\infty}^{\infty}F(s)\,e^{2\pi i s x}\,ds~}
$$
CONVOLUTION
$$
(f*g)(x)
=
{\int\limits_{-\infty}^{\infty} f(y)\,g(x-y)\,dy~} 
=
{\int\limits_{-\infty}^{\infty} f(x-y)\,g(y)\,dy~}  
$$
CORELATION
$$
(f \star g)(x) 
= 
{\int\limits_{-\infty}^{\infty} f(y)\,g(x+y)\,dy~} 
= 
{\int\limits_{-\infty}^{\infty} f(x+y)\,g(y)\,dy~}  
$$
CONVOLUTION THEOREM
$$
{(f*g)(x)}
\Leftrightarrow 
{(F \cdot G )(\xi)}
$$
$$
{(f \cdot g)(x)}
\Leftrightarrow 
{(1/MN)(F*G )(\xi)}
$$
CORELATION THEOREM
$$
{(f \star g)(x)} 
\Leftrightarrow
{(F^* \cdot G)(\xi)}
$$
$$
{(f^*  \cdot g)(x)} 
\Leftrightarrow
{(1/MN)(F \star G)(\xi)}
$$
IMPULSE

$$
\delta(t)=
\begin{cases}
\infty & t = 0 \\
0	 & t \neq 0
\end{cases}
$$

$$
{\int\limits_{-\infty}^{\infty}\delta(t)\,dt~}
=
1
$$

IMPULSE PROPORTIES

$$
{\int\limits_{-\infty}^{\infty}f(t)\delta(t)\,dt~}
=
f(0)
$$

$$
{\int\limits_{-\infty}^{\infty}f(t)\delta(t-t_{0})\,dt~}
=
f(t_{0})
$$

DISKRETE IMPULSE

$$
\delta(t)=
\begin{cases}
1 & t = 0 \\
0	 & t \neq 0
\end{cases}
$$

$$
\sum_{t = - \infty}^{\infty} \delta(t) 
=
1
$$

$$
\sum_{t = - \infty}^{\infty} f(t) 
\delta(t - t_{0}) 
=
f(t_{0})
$$

$$
s_{\varDelta T}(t) 
=
\sum_{k=-\infty}^{\infty}
\delta(t - k\varDelta T)
$$

$$
S(\mu) = \mathcal{F}
\{ s_{\varDelta T}(t)  \}
=
\frac{1}{\varDelta T}
\sum_{n = - \infty}^{\infty}
\delta(
\mu - \frac{n}{\varDelta T}
)
$$

$$
\tilde{f}(t) = 
f(t)s_{\varDelta T}(t)
=
\sum_{n = -\infty}^{\infty}
f(t)
\delta(t - n\varDelta T)
$$

$$
\tilde{F}(\mu) = 
\frac{1}{\varDelta T}
\sum_{n = - \infty}^{\infty}
F(
\mu - \frac{n}{\varDelta T}
)
$$

$$
F(\mu), [-\mu_{max},\mu_{max} ]
$$

$$
\frac{1}{\varDelta T} 
>
2\mu_{max}
$$

$$
F(\mu)
=
H(\mu)
\tilde{F}(\mu)
$$

$$
H(\mu)
=
\begin{cases}
\varDelta T & 
\mu \in [-\mu_{max},\mu_{max}] \\
0 & otherwise
\end{cases}
$$

$$
h(t)
=
\mathcal{F}^{-1}\{H(\mu)\}(t)
=
\int\limits_{-\infty}^{\infty}
H(\mu)
e^{2\pi i \mu t }
d\mu~
=
\int\limits_{-\mu_{max}}^{\mu_{max}}
\varDelta T
e^{2\pi i \mu t }
d\mu~
=
$$
$$
=
\frac{\varDelta T}
{i 2 \pi t} 
\left[
e^{	2\pi i \mu t }
\right]_{-\mu_{max}}^{\mu_{max}}
=
2\varDelta T \mu_{max}
\frac{sin(2\pi t \mu_{max}) }{(2\pi t \mu_{max})}
=
2\varDelta T \mu_{max}
sinc(2t\mu_{max})
$$

$$
f(t)
=
\mathcal{F}^{-1}
\{ F(\mu) \}
$$
$$
=
\mathcal{F}^{-1}
\{
H(\mu)\tilde{F}(\mu)
\}
$$
$$
=
h(t)*\tilde{f}(t)
$$

$$
f(t) 
=
2  \mu_{max} \varDelta T
\sum_{ n = - \infty}^{\infty}
f(n \varDelta T ) 
sinc[ 2  \mu_{max}  (t - n \varDelta T) 	]  
$$

IF $   \varDelta T = \frac{1}{2 \mu_{max}}$
$$
f(t) 
=
\sum_{ n = - \infty}^{\infty}
f(n \varDelta T ) 
sinc[ (t - n \varDelta T) / \varDelta T	]  
$$

2D FFT

$$
F(\mu,\nu)
=
\int\limits_{-\infty}^{\infty}
\int\limits_{-\infty}^{\infty}
f(t,z) \,
e^{-i 2 \pi( \mu t + \nu z)}
dz dt~
$$

2D IFFT

$$
f(t,z)
=
\int\limits_{-\infty}^{\infty}
\int\limits_{-\infty}^{\infty}
F(\mu,\nu) \,
e^{i 2 \pi( \mu t + \nu z)}
d\mu d\nu~
$$

Train of impulses

$$
S_{ \varDelta T \varDelta Z }
(t,z)
=
\sum_
{ m = - \infty}^{\infty}
\sum_
{ n = - \infty}^{\infty}
\delta(
t - m \varDelta T,
z - n \varDelta Z
)
$$

Suggest that $
[- \mu_{max},\mu_{max}] 
\times
[- \nu_{max},\nu_{max}] 
$
$$
\varDelta T 
<
\frac{1}{2 \mu_{max}}
$$
$$
\varDelta Z 
<
\frac{1}{2 \nu_{max}}
$$

$$
\tilde{f}(t,z) 
=
f(t,z)
S_{ \varDelta T \varDelta Z }
=
\sum_
{ m =  0}^{M-1}
\sum_
{ n =  0}^{N-1}
f(t,z)\delta(
t - m \varDelta T,
z - n \varDelta Z
)
$$

where 
$
[-\frac{1}{2},
\frac{1}{2}]
\times
[-\frac{1}{2},
\frac{1}{2}]
$

DISKRET FFT
$$
F(\mu,\nu)
=
\sum_
{ x =  0}^{M-1}
\sum_
{ y =  0}^{N-1}
f(x,y)
e^{
-i2\pi(\mu x / M +
 \nu y / N )
}
$$

$$
F(x,y)
=
\sum_
{ \mu =  0}^{M-1}
\sum_
{ \nu =  0}^{N-1}
f(\mu,\nu)
e^{
 i2\pi(\mu x / M +
 \nu y / N )
}
$$

Cross corelation
$$
\gamma(u,v)
=
\frac{
\sum_{x,y}
[f(x,y) - \overline{f}_{u,v}]
[t(x-u,y-v)-t]
}
{
\sqrt{
\sum_{x,y}
[f(x,y) - \overline{f}_{u,v}]^2
\sum_{x,y}
[t(x-u,y-v)-t]^2
}
} 
$$


$$
t'(x,y)
\equiv
t(x,y) - \overline{t}
$$

$$
f'(x,y)
\equiv
f(x,y) - \overline{f}_{u,v}
$$

$$
\mathcal{F}^{-1}
\{
\mathcal{F}(f')
\mathcal{F}^*(t')
\}
$$
\end{document}