Filtering in Frequency Domain · Filtering in Frequency Domain Basics ... Basics Gaussians Filters...

2D DFT - ReviewFiltering in Frequency Domain

Filtering in Frequency Domain

Dr. Praveen Sankaran

Department of ECE

NIT Calicut

February 4, 2013

Dr. Praveen Sankaran DIP Winter 2013


Outline

1 2D DFT - Review

2 Filtering in Frequency Domain

Basics

Gaussians

Filters in Frequency Domain



2D Sampling

2D Impulse Train

s [t,z ] =∞

∑m=−∞

∞

∑n=−∞

δ [t−m∆T , z−n∆Z ] (1)

f (t,z)s [t,z ]⇒ sampled function.



Another Example



2D DFT

F [u,v ] =M−1

∑m=0

N−1

∑n=0

f [m,n]e−j2π(um/M+vn/N) (2)

f [.] is a digital image of size M×N.

u = 0,1, · · ·M−1 and v = 0,1, · · ·N−1.

Inverse DFT

f [m,n] =1

MN

M−1

∑u=0

N−1

∑v=0

F [u,v ]e j2π(um/M+vn/N) (3)

m = 0,1, · · ·M−1 and n = 0,1, · · ·N−1.



Reconstruction Illustration - Phase Dominates!



2D Convolution Theorem

f [m,n]?h [m,n] =M−1

∑j=0

N−1

∑k=0

f [j ,k]h [m− j ,n−k] (4)

f [m,n]?h [m,n]⇐⇒ F [u,v ]H [u,v ] (5)

f [m,n]h [m,n]⇐⇒ F [u,v ]?H [u,v ] (6)

Equation 5 is of particular importance to us. A spatial convolution

can be expressed as a product in frequency domain. This means

that all the linear �ltering we discussed before as convolutions can

now be easily understood as products.



Zero Padding - Importance



DFT Periodic - Remember!

Remember that we considered just a very speci�c period of the

Fourier Transform, sampled it and obtained the DFT. But

practically we always have to deal with what is present - the

periodic structures when we compute the IDFT. When we �nd a

product of two DFT's, inherently we are convolving two periodic

structures in original domain.

Suppose the function f [.] has A number of samples and function

h [.] has B number of samples, we de�ne a variable,

P ≥ A+B−1 (7)

Append zeros to both functions so that they have the same

length, P → zero padding.



DFT Periodic - Remember!

Remember that we considered just a very speci�c period of the

Fourier Transform, sampled it and obtained the DFT. But

practically we always have to deal with what is present - the

periodic structures when we compute the IDFT. When we �nd a

product of two DFT's, inherently we are convolving two periodic

structures in original domain.

Suppose the function f [.] has A number of samples and function

h [.] has B number of samples, we de�ne a variable,

P ≥ A+B−1 (7)

Append zeros to both functions so that they have the same

length, P → zero padding.



BasicsGaussiansFilters in Frequency Domain

Outline

1 2D DFT - Review


Basics

Gaussians





What does Filtering Involve?

Modify Fourier Transform.

Compute the Inverse Transform.

g [m,n] = ℑ−1 {H [u,v ]F [u,v ]} (8)

H [u,v ]→ �lter function, generally symmetric to simplify

computation.

F [.] , g [.] , H [.] are of size M×N.

pixel-by-pixel multiplication.

Symmetry of the �lter implies that the function F [.] shouldalso be centered.







g [m,n] = ℑ−1 {H [u,v ]F [u,v ]} (8)


computation.

F [.] , g [.] , H [.] are of size M×N.









g [m,n] = ℑ−1 {H [u,v ]F [u,v ]} (8)


computation.

F [.] , g [.] , H [.] are of size M×N.






Zero Padding - Again!

Consider `b', blurring is not uniform. Side edges should have

grayed out.

Zero padding solves the issue in `c'




The Issue with Padding




Filtering Strategy

1 P = 2M, Q = 2N → padding parameters.

2 g [m,n]→ gp [p,q] of size P×Q.

3 gcp [p,q] = gp [p,q]× (−1)m+n.

4 Compute DFT of gcp [p,q]⇒ G [u,v ].

5 Real, symmetric �ltering function, H [u,v ] of size P×Q.

6 Go [u,v ] = H [u,v ]F [u,v ]

7 Obtain processed image,

gop [p,q] ={real

[ℑ−1 [Go [u,v ]]

]}(−1)m+n (9)

8 go [m,n]→ extract M×N region from the top, left quadrant

of gop [p,q].




Outline

1 2D DFT - Review


Basics

Gaussians





Gaussian Example




Options Available

Specify �lter in frequency domain, compute its IDFT, and then

use the resulting, full size spatial �lter as a guide for

constructing smaller spatial �lter mask.

Small spatial �lter is given and we obtain its full size frequency

domain representation.




Gaussian Pair

H [u] = Ae−u2/2σ2 (10)

h [x ] =√2πσAe−2π2σ2x2 (11)

Both are Gaussian

Both Real

σ large → H [u] tends to constant → h [x ] tends to impulse →less �ltering.

More complex �lters can be designed as a mixture of

Gaussians - (e.g. sum or di�erence).




Outline

1 2D DFT - Review


Basics

Gaussians





Ideal Low Pass Filter

2D low pass �lter that passes without attenuation all

frequencies within a circle of radius D0 from the origin.

Cuts o� all frequencies outside this circle.

H [u,v ] =

{1 D [u,v ]≤ D0

0 D [u,v ] > D0

(12)

D [u,v ] =[(u−P/2)2 + (v −Q/2)2

]1/2




Example




Image Blurring Example




Issue with ILPF

Ringing

A box �lter in frequency domain will look like a sinc function

in the spatial domain.

Filtering using this function is equivalent to convolving in the

spatial domain.

The smaller lobes of the sinc causes the ringing.




Butterworth Low-Pass Filter

Idea → smoothen the function.

H [u,v ] =1

1+ [D[u,v ]/D0]2n

(13)

We have seen variable D [.] before.




What about ringing?




Gaussian Low Pass Filtering

H [u,v ] = e−D2[u,v ]/2σ2 = e

−D2[u,v ]/2D20 (14)

No ringing.




Output Example




Highpass Filters

HHP [.] = 1−HLP [.] (15)

Ideal

H [u,v ] =

{0 D [u,v ]≤ D0

1 D [u,v ] > D0

(16)

Butterworth

H [u,v ] =1

1+ [D0/D[u,v ]]2n(17)

Gaussian

H [u,v ] = 1− e−D2[u,v ]/2D20 (18)




Highpass - Illustration




Unsharp Masking

Mask

imask [x ,y ] = f [x ,y ]−fLP [x ,y ] = f [x ,y ]−ℑ−1 {HLP [u,v ]F [u,v ]}

(19)

Final image

f′[x ,y ] = f [x ,y ] +k× imask [x ,y ] (20)

= f [x ,y ] +k× [f [x ,y ]−ℑ−1 {HLP [u,v ]F [u,v ]}]

(21)

= ℑ−1 {[1+k× [1−HLP [u,v ]]]F [u,v ]} (22)

= ℑ−1 {[1+k×HHP [u,v ]]F [u,v ]} (23)

k = 1? k > 1?




High frequency Emphasis Filter

HE [u,v ] = 1+k×HHP [u,v ] (24)

1 in the equation provides an o�set.

Does not eliminate the dc term → preserves tonality.

More generalized form

HE [u,v ] = k1 +k2×HHP [u,v ] (25)

k1 ≥ 0, controls o�set from the zero, thus controlling dc term.

k2 ≥ 0 controls contribution of high frequencies.




Illustration




Questions

4.4, 4.6, 4.7, 4.12, 4.13, 4.15, 4.17(look at 4.3 as reference).

4.21, 4.22, 4.23, 4.26, 4.28, 4.33

4.31, 4.32


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