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IT523: Digital Image Processing Chapter 4: Frequency Domain Enhancement

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Page 1: IT523: Digital Image Processingcourses.daiict.ac.in/pluginfile.php/28248/mod_resource/... · 2018-02-16 · IT523: Digital Image Processing Chapter 4: Frequency Domain Enhancement

IT523: Digital Image Processing

Chapter 4: Frequency Domain Enhancement

Page 2: IT523: Digital Image Processingcourses.daiict.ac.in/pluginfile.php/28248/mod_resource/... · 2018-02-16 · IT523: Digital Image Processing Chapter 4: Frequency Domain Enhancement

Frequency domain filtering methodology

Figure: Frequency domain filtering

IT523 DIP: Lecture 8 2/28

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Padding during convolution

Figure: Ciruclar and linear convolution

IT523 DIP: Lecture 8 3/28

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Notch filtering

Figure: Periodic noise

IT523 DIP: Lecture 8 4/28

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Notch filtering

H(u, v) =0 (u, v) = (p, q)

=1 otherwise.

Figure: Notch filteringIT523 DIP: Lecture 8 5/28

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Low pass and High pass filtering

Figure: Low and High pass filtering

IT523 DIP: Lecture 8 6/28

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Ideal low pass filter

Figure: Ideal low pass fitler. (top) Frequency response (bottom) Impulseresponse

IT523 DIP: Lecture 8 7/28

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Ideal low pass filter

Figure: Ideal Low pass filter output

IT523 DIP: Lecture 8 8/28

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Butterworth filter

Hn(u, v) = 11+(D(u,v)/D0)2n

, where D(u, v) is the Euclidean

distance from origin and D0 is the cut-off frequency.

Figure: Butterworth low pass fitler. (top) Frequency response (bottom)Impulse response

IT523 DIP: Lecture 8 9/28

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Butterworth low pass filter

Figure: Butterworth Low pass filter output

IT523 DIP: Lecture 8 10/28

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Gaussian low pass filter

Hn(u, v) = exp(−D2

u0,v0(u, v)/2D20

), where

Du0,v0(u, v) = (u − u0)2 + (v − v0)2 is the Euclidean distancefrom (u0, v0) and D0 is the cut-off frequency.

Figure: Gaussian low pass fitler. (top) Frequency response (bottom)Impulse response

IT523 DIP: Lecture 8 11/28

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Gaussian low pass filter

Figure: Gaussian Low pass filter output

IT523 DIP: Lecture 8 12/28

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Applications of Low pass filters

Figure: Applications of low pass filter (top) Character recognition(bottom) Removing horizontal scan lines.

IT523 DIP: Lecture 8 13/28

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High pass filters

Hhp(u, v) = 1− Hlp(u, v).

Figure: High pass filter frequency responses.

IT523 DIP: Lecture 8 14/28

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High pass filters

Figure: High pass filter impulse responses.

IT523 DIP: Lecture 8 15/28

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Ideal High pass filters

Figure: Ideal High pass filter outputs.

IT523 DIP: Lecture 8 16/28

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Gaussian High pass filters

Figure: Gaussian High pass filter outputs.

IT523 DIP: Lecture 8 17/28

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Laplacian filter

One-dimensional signals:

F(

d2

dx2f (x)

)= −ω2F (ω).

Two-dimensional signals:

F(

∂2

∂x2f (x , y) + ∂2

∂y2 f (x , y))

= −(u2 + v2)F (u, v).

Therefore the Laplacian filter’s frequency response is

Hlap(u, v) = −(u2 + v2)

IT523 DIP: Lecture 8 18/28

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Laplacian filter

One-dimensional signals:

F(

d2

dx2f (x)

)= −ω2F (ω).

Two-dimensional signals:

F(

∂2

∂x2f (x , y) + ∂2

∂y2 f (x , y))

= −(u2 + v2)F (u, v).

Therefore the Laplacian filter’s frequency response is

Hlap(u, v) = −(u2 + v2)

IT523 DIP: Lecture 8 18/28

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Laplacian filter

One-dimensional signals:

F(

d2

dx2f (x)

)= −ω2F (ω).

Two-dimensional signals:

F(

∂2

∂x2f (x , y) + ∂2

∂y2 f (x , y))

= −(u2 + v2)F (u, v).

Therefore the Laplacian filter’s frequency response is

Hlap(u, v) = −(u2 + v2)

IT523 DIP: Lecture 8 18/28

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Laplacian filter

Figure: Laplacian frequency response and impulse response

IT523 DIP: Lecture 8 19/28

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Laplacian filter

Figure: Laplacian filter output

IT523 DIP: Lecture 8 20/28

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Phase of Image DFT

Fourier transform is a complex valued function:

F (f (x , y)) =F (u, v)

F (u, v) =R(u, v) + jI (u, v)

Phase: φ(u, v) = tan−1(

I (u,v)R(u,v)

).

IT523 DIP: Lecture 8 21/28

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Phase of Image DFT

Fourier transform is a complex valued function:

F (f (x , y)) =F (u, v)

F (u, v) =R(u, v) + jI (u, v)

Phase: φ(u, v) = tan−1(

I (u,v)R(u,v)

).

IT523 DIP: Lecture 8 21/28

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Phase

Figure: Phase is more crucial than DFT magnitude.

IT523 DIP: Lecture 8 22/28

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Effect of Translation and rotation

Figure: Effect of translation and rotation on DFT magnitude and phase.

IT523 DIP: Lecture 8 23/28

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Homomorphic filtering

Image formation model: f (x , y) = i(x , y)r(x , y)

We are interested in r(x , y), not in the illumination i(x , y).

Unfortunately f is a multiplicative combination of i and r . SoF (u, v) 6= I (u, v)R(u, v).

IT523 DIP: Lecture 8 24/28

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Homomorphic filtering

Image formation model: f (x , y) = i(x , y)r(x , y)

We are interested in r(x , y), not in the illumination i(x , y).

Unfortunately f is a multiplicative combination of i and r . SoF (u, v) 6= I (u, v)R(u, v).

IT523 DIP: Lecture 8 24/28

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Homomorphic filtering

Image formation model: f (x , y) = i(x , y)r(x , y)

We are interested in r(x , y), not in the illumination i(x , y).

Unfortunately f is a multiplicative combination of i and r . SoF (u, v) 6= I (u, v)R(u, v).

IT523 DIP: Lecture 8 24/28

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Homomorphic filtering

Use Logarithm:z(x , y) = ln(f (x , y)) = ln(i(x , y)) + ln(r(x , y)).

Take Fourier Transform: Z (u, v) = Fi (u, v) + Fr (u, v), whereFi (u, v) = F{ln(i(x , y))} and Fr (u, v) = F{ln(r(x , y))}.

Apply a filter H(u, v) to eliminate Fi (u, v):

S(u, v) = H(u, v)Z (u, v) = H(u, v)Fi (u, v) + H(u, v)Fr (u, v)

Output image:

s(x , y) =F−1{S(u, v)}=F−1{H(u, v)Fi (u, v) + H(u, v)Fr (u, v)}=F−1{H(u, v)Fi (u, v)}+ F−1{H(u, v)Fr (u, v)}=i1(x , y) + r1(x , y)

IT523 DIP: Lecture 8 25/28

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Homomorphic filtering

Use Logarithm:z(x , y) = ln(f (x , y)) = ln(i(x , y)) + ln(r(x , y)).

Take Fourier Transform: Z (u, v) = Fi (u, v) + Fr (u, v), whereFi (u, v) = F{ln(i(x , y))} and Fr (u, v) = F{ln(r(x , y))}.

Apply a filter H(u, v) to eliminate Fi (u, v):

S(u, v) = H(u, v)Z (u, v) = H(u, v)Fi (u, v) + H(u, v)Fr (u, v)

Output image:

s(x , y) =F−1{S(u, v)}=F−1{H(u, v)Fi (u, v) + H(u, v)Fr (u, v)}=F−1{H(u, v)Fi (u, v)}+ F−1{H(u, v)Fr (u, v)}=i1(x , y) + r1(x , y)

IT523 DIP: Lecture 8 25/28

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Homomorphic filtering

Use Logarithm:z(x , y) = ln(f (x , y)) = ln(i(x , y)) + ln(r(x , y)).

Take Fourier Transform: Z (u, v) = Fi (u, v) + Fr (u, v), whereFi (u, v) = F{ln(i(x , y))} and Fr (u, v) = F{ln(r(x , y))}.

Apply a filter H(u, v) to eliminate Fi (u, v):

S(u, v) = H(u, v)Z (u, v) = H(u, v)Fi (u, v) + H(u, v)Fr (u, v)

Output image:

s(x , y) =F−1{S(u, v)}=F−1{H(u, v)Fi (u, v) + H(u, v)Fr (u, v)}=F−1{H(u, v)Fi (u, v)}+ F−1{H(u, v)Fr (u, v)}=i1(x , y) + r1(x , y)

IT523 DIP: Lecture 8 25/28

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Homomorphic filtering

Use Logarithm:z(x , y) = ln(f (x , y)) = ln(i(x , y)) + ln(r(x , y)).

Take Fourier Transform: Z (u, v) = Fi (u, v) + Fr (u, v), whereFi (u, v) = F{ln(i(x , y))} and Fr (u, v) = F{ln(r(x , y))}.

Apply a filter H(u, v) to eliminate Fi (u, v):

S(u, v) = H(u, v)Z (u, v) = H(u, v)Fi (u, v) + H(u, v)Fr (u, v)

Output image:

s(x , y) =F−1{S(u, v)}=F−1{H(u, v)Fi (u, v) + H(u, v)Fr (u, v)}=F−1{H(u, v)Fi (u, v)}+ F−1{H(u, v)Fr (u, v)}=i1(x , y) + r1(x , y)

IT523 DIP: Lecture 8 25/28

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Inverting the logarithm using exponential:

g(x , y) = exp(s(x , y))

= exp(i1(x , y) + r1(x , y))

= exp(i1(x , y)) exp(r1(x , y))

g(x , y) =i0(x , y)r0(x , y)

IT523 DIP: Lecture 8 26/28

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Assumption about Illumination

Illumination component typically has low spatial variationscompared to reflectance component, for example aroundedges.

Figure: Homomorphic filter response.

IT523 DIP: Lecture 8 27/28

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Homomorphic filtering output

Figure: Eliminating Illuminance component.

IT523 DIP: Lecture 8 28/28