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Electrical & Computer Engineering Dr. D. J. Jackson Lecture 9-
Computer Vision &
Digital Image Processing
Fourier Transform Properties, the Laplacian, Convolution and Correlation
Periodicity of the Fourier transform
- The discrete Fourier transform (and its inverse) are periodic with period N. F (u,v) = F (u+N,v) = F (u,v+N) = F (u+N,v+N)
- Although F(u,v) repeats itself infinitely for many values of u and v , only N values of each variable are required to obtain f(x,y) from F(u,v) - i.e. Only one period of the transform is necessary to specify F(u,v) in the frequency domain. - Similar comments may be made for f(x,y) in the spatial domain
Electrical & Computer Engineering Dr. D. J. Jackson Lecture 9-
Conjugate symmetry of the Fourier
transform
- If f(x,y) is real (true for all of our cases), the Fourier transform exhibits conjugate symmetry
F( u,v )=F*(- u ,- v )
or, the more interesting
|F( u,v )| = |F(- u ,- v )|
where F*( u,v ) is the complex conjugate of F( u,v )
Implications of periodicity & symmetry
- Consider a 1-D case:
- F(u) = F(u+N) indicates F(u) has a period of length N
- |F(u)| = |F(-u)| shows the magnitude is centered about the origin
- Because the Fourier transform is formulated for values in the range from [0,N-1], the result is two back-to-back half periods in this range
- To display one full period in the range, move (shift) the origin of the transform to the point u=N/
Electrical & Computer Engineering Dr. D. J. Jackson Lecture 9-
Distributivity & Scaling
- The Fourier transform (and its inverse) are distributive over addition but not over multiplication
- So,
- For two scalars a and b ,
ℑ{ f 1 (^) ( x , y )+ f 2 ( x , y )}=ℑ{ f 1 ( x , y )}+ℑ{ f 2 ( x , y )}
ℑ{ f 1 (^) ( x , y )× f 2 ( x , y )}≠ℑ{ f 1 ( x , y )}×ℑ{ f 2 ( x , y )}
( , )^1 ( / , / )
( , ) (, ) f axby abFu av b
af x y aFuv ⇔
⇔
Average Value
- A widely used expression for the average value of a 2-D discrete function is:
- From the definition of F(u,v), for u=v=0,
- Therefore,
−
−
=
1 0
1 (^2 ) (, )^1 (, )
N x
N y
f xy N f xy
−
−
=
1 0
1 0
( 0 , 0 )^1 (, )
N x
N y
F N f xy
f ( x , y )= N^1 F ( 0 , 0 )
Electrical & Computer Engineering Dr. D. J. Jackson Lecture 9-
The Laplacian
- The Laplacian of a two variable function f(x,y) is given as:
- From the definition of the 2-D Fourier transform,
- The Laplacian operator is useful for outlining edges in an image
2
2 2 2 2 ∇ f ( x , y )=∂∂ xf +∂∂ yf
ℑ{∇ 2 f ( x , y )} ⇔−( 2 π)^2 ( u^2 + v^2 ) F ( u , v )
The Laplacian: Matlab example
% Given F(u,v), use the Laplacian % to construct an edge outlined % representation of the f(x,y) [f,fmap]=bmpread('lena128.bmp'); F=fft2(f); Fedge=zeros(128); for u=1: for v=1: Fedge(u,v)=- (2pi).^2(u.^2+v.^2)*F(u,v); end end fedge=ifft2(Fedge); image(real(fedge));colormap(gray(256);
Electrical & Computer Engineering Dr. D. J. Jackson Lecture 9-
1-D convolution example (continued)
- Then, for any value x , we multiply g( x - α) and f(α) and integrate from -∞ to +∞
- For 0≤x ≤ 1 we have For 1 ≤ x ≤ 2 we have
1
1
α
f(α)g(x- α)
1
1
α
f(α)g(x- α)
1-D convolution example (continued)
- Thus we have
- Graphically,
.
1 2
0 1
0
1 / 2
/ 2 ()* () elsewhere
x
x x
x f x gx ≤ ≤
≤ ≤
⎪⎩
⎪⎨
⎧ = −
1
1/
x
f(x)*g(x)
2
Electrical & Computer Engineering Dr. D. J. Jackson Lecture 9-
Convolution and impulse functions
- Of particular interest will be the convolution of a function f(x) with an impulse function δ (x-x 0 )
- The function δ (x-x 0 ) may be viewed as having an area of unity in an infinitesimal neighborhood around x 0 and 0 elsewhere. That is
+∞
−∞
f ( x ) δ( x − x 0 ) dx = f ( x 0 )
+∞ −∞
−
− = − =
0 0
( 0 ) ( 0 ) 1
x x
δ x x dx δ x x dx
Convolution and impulse functions
(continued)
- We usually say that δ (x-x 0 ) is located at x=x 0 and the strength of the impulse is given by the value of f(x) at x=x 0
- If f(x)=A then, A δ (x-x 0 ) is impulse of strength A at x=x 0.
- Graphically this is:
x 0 x A δ (x-x 0 )
A
Electrical & Computer Engineering Dr. D. J. Jackson Lecture 9-
Convolution and the Fourier transform
- f(x)*g(x) and F(u)G(u) form a Fourier transform pair
- If f(x) has transform F(u) and g(x) has transform G(u) then f(x)*g(x) has transform F(u)G(u)
- These two results are commonly referred to as the convolution theorem
f x g x F u G u
f x g x F uG u ⇔
Frequency domain filtering
- Enhancement in the frequency domain is straightforward
- Compute the Fourier transform
- Multiply the result by a filter transform function
- Take the inverse transform to produce the enhanced image
- In practice, small spatial masks are used considerably more than the Fourier transform because of their simplicity of implementation and speed of operation
- However, some problems are not easily addressable by spatial techniques - Such as homomorphic filtering and some image restoration techniques
Electrical & Computer Engineering Dr. D. J. Jackson Lecture 9-
Lowpass frequency domain filtering
- Given the following relationship
- where F(u,v) is the Fourier transform of an image to be smoothed
- The problem is to select an H(u,v) that yields an appropriate G(u,v)
- We will consider zero-phase-shift filters that do not alter the phase of the transform (i.e. they affect the real and imaginary parts of F(u,v) in exactly the same manner)
G ( u , v )= H ( u , v ) F ( u , v )
Ideal lowpass filter (ILPF)
- A transfer function for a 2-D ideal lowpass filter (ILPF) is given as
- where D 0 is a stated nonnegative quantity (the cutoff frequency) and D(u,v) is the distance from the point (u,v) to the center of the frequency plane
⎩
⎨
⎧ > = ≤ 0
0 0 ifD(u,v) D H ( u , v )^1 ifD(u,v) D
D ( u , v )= u^2 + v^2
v u
H(u,v)
