0
15kviews
State and prove the property of kernel separating and linearity for 2D-DFT.
1
584views

1-Dimensional Fourier transform:

Let f(x) be a continuous function of x. The Fourier transform of f(x) is

$$F(u) =\int_{-∞}^∞ f(x) e^{-j2πux} dx \\ \hspace{3.4cm}=\int_{-∞}^∞ f(x)[cos⁡2πux - j sin⁡ 2πux] dx$$

Similarly, the inverse fourier transform is

$$F(x)= \int_(-∞)^∞ f(x) e^{+j2πux} dx$$

Because f(u)is a complex

$$F(u)= R(u)+j I(u)$$

Where R is real and j is imaginary.

F(u) has a magnitude plot as well as phase plot

$Mag =|f(u)| =[R^2(u)+I^2(u)]^{1/2}$

$Phase= \angle F(u) = tan^{-1}\left[\frac{I(u)}{R(u)}\right]$

|f(u)|, when plotted is called magnitude plot of the Fourier spectrum.

The power spectrum, $P(u) =|f(u)|^2= R^2(u)+I^2(u)$

2-Dimensional fourier transform:

The 1-D Fourier transform can be easily extended to a function(x,y)of two variables. Hence the images being 2-Dimensional functions.

$F{f(x,y)}= F(u,v) = ∬_{-∞}^∞ f(x,y) e^{-j2π(ux+vy)} dxdy$

$F-1{F(u,v)}= ∬_{-∞}^∞ f(u,v;) e^{-j2π(ux+vy)} dudv$

In 1-D Fourier transform the magnitude plot is given by

$|f(u,v)|=[R^2(u,v)+I^2(u,v)]^{1/2}$

And the phase plot is given by

$F(u,v)= tan^{-1} \left[\frac{I(u,v)}{R(u,v)}\right]$

The power spectrum is given by,

$P(u,v) = | F(u,v)|^2= R^2(u,v) + I^2(u,v)$

Properties of discrete Fourier Transform:

1. Seperability property:

This property states that a 2D-DFT can be separated into two 1D DFt’s

$F(u, v)= \sum_{x=0}^{N-1} \sum_{y=0}^{N-1} f(x, y) e^{-j2π \left(\frac{ux+vy}{n}\right)}$

This can be split up as:

$F(u,v) = \sum_{x=0}^{N-1} e^{-j2π \frac{ux}{N}} \sum{y=0}^{N-1} f(x,y) e^{-j2π (\frac{vy}{N})}$

Let $F(u,v) = \sum{y=0}^{N-1} f(x,y) e^{-j2π(\frac{vy}{N})}$

$\therefore F(u, v)= \sum{y=0}^{N-1} f(x,y) e^{-j2π (\frac{ux}{N}) }$

The principle advantage of the separability property is that the 2D DFT can be obtained in two steps by successive application of 1D DFT.

F(u,v) can be obtained by applying 1D-DFT along the rows and then along the columns.

2. Translation Property (Shifting Property):

$f(x, y) e^{+j2π \left(\frac{uax+vay}{N}\right)}$

If f(x,y) is multiplied by an exponential, the original Fourier transform f(u,v) gets shifted in frequency by the F(u-u0, v-v0)

We know,

$F{f(x,y)}= F(u,v) = \sum{x=0}^{N-1} \sum_{y=0}^{N-1} f(x,y) e^{-j2π(\frac{ux+vy}{N})}$

Multiplying f(x,y) by $e^{j2π (\frac{u0x+v0y}{N})}$

$F{f(x,y) e^{j2π (\frac{u0x+v0y}{N})}} = \sum_{x=0}^{N-1}\sum_{y=0}^{N-1} f(x,y) e^{-j2π(\frac{ux+vy}{N})} e^{j2π(\frac{u0x+v0y}{N})}$

$F{f(x,y) e^{j2π (\frac{u0x+v0y}{N})}} = \sum_{x=0}^{N-1} \sum_{y=0}^{N-1} f(x,y) e^{(\frac{-j2πux}{N}+\frac{j2πu0x}{N})} e^{\frac{-j2πvx+j2πv0y}{N}}$

$F{f(x,y) e^{j2π (\frac{u0x+v0y}{N})}} = F(u-u0, v- v0)$

Hence,

If $F{f(x, y)} → F (u, v)$

If $F{f(x, y) e^{j2π \frac{uax+vay}{N}}}$

3. Rotational property:

Polar coordinates is

X=rcosθ ; Y=rsinθ

U= wcosΦ ; Y=wsinΦ

Then f(x,y) and F(u,v) become f(r,θ) and F(w,Φ)

Put this in the continuous or discrete Fourier transform pair, get

F(r,θ+θ0) = F(w,Φ+θ0)

That if the rotated f(x,y) by an angle θ0, the fourier spectrum F(u,v) also rotates the same angle.