State and prove the property of kernel separating and linearity for 2D-DFT.
1 Answer

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


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.

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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}} $

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


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.

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