written 8.1 years ago by | • modified 8.1 years ago |
1-Dimensional Fourier transform:
Let f(x) be a continuous function of x. The Fourier transform of f(x) is
$$F(u) =\int_{-∞}^∞ f(x) e^{-j2πux} dx \\ \hspace{3.4cm}=\int_{-∞}^∞ f(x)[cos2πux - j sin 2πux] dx$$
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.
![enter image description here](https://i.imgur.com/evbB3kl.jpg)
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.