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Lapalician &
Homomorphic Filter
1
Laplacian in Spatial Domain
● Laplacian
–
–
Isotropic
Rotation Invariant
𝑄2
f = f (x +1, y) + f (x - 1, y) +
f (x, y +1) + f (x, y - 1) - 4 f (x, y)
g(x, y) = f (x, y) +c[𝑄2
f (x, y)]
2
2
∂t
+
∂z2
∂2
f ∂2
f
𝑄 f =
2
Laplacian in Frequency Domain
2
2
∂t
+
∂z2
∂2
f ∂2
f
𝑄 f =
3
Laplacian in Frequency
Domain
H(u,v) =-4𝑢2
(u2
+v2
)
With respect to center of frequency rectangle :
H(u,v) =-4π2
[(u - M /2)2
+(v - N /2)2
)]
=-4 π 2
D2
(u,v)
Laplacian of an image:
4
Laplacian in Frequency
Domain
●Enhancement Eq: g(x, y) = f (x, y) +c[▽2
f (x, y)]
c =- 1
● Scales of f (x, y) & ▽
2
f (x, y) as computed by
DFT differ widely due to the DFT process
● Normalize f (x, y) to [0,1] before DFT
● Normalize ▽2
f (x, y) to [-1,1]
5
Laplacian in
Frequency
Domain
6
Comparative Laplacian in
Spatial & Frequency
Domains
7
Unsharp Mask, Highboost Filtering & High-
Frequency-Emphasis Filtering
● In spatial domain:
gmask (x, y) = f (x, y) - f (x, y)
g(x, y) = f (x, y) +k * gmask (x, y)
k =1: Unsharp Masking
k >1: Highboost Filtering
k ∈1: De- emphasized Unsharp Masking
8
Unsharp Mask, Highboost Filtering & High-
Frequency-Emphasis Filtering
9
[ 1+ k*Hhp (u,v)] F(u,v) }
Unsharp Mask, Highboost Filtering & High-
Frequency-Emphasis Filtering
● In frequency domain:
High- Frequency Emphasis Filter
k1>=0: Controlsthe offset from origin
k2 >=0: Controlsthe contribution of high frequencies
10
g(x,y)= {[ k1+k2 * Hhp (u,v)] F(u,v) }
Image:
416x596
D0=40
(5% of
short
side of
padded
image)
k1=0.5
k2=0.75
11
Homomorphic Filtering
● Homomorphic filtering is a FDS that aims at a
simultaneous increase in contrast & dynamic range
compression.
● It is mainly utilized for non-uniformly illuminated
images in medical, sonar images etc. for edge
enhancement that makes the image details clear to the
observer.
● Certain situations where the image is subjected to the
multiplicative interference or noise as depicted
● f(x,y)= i(x,y) . r(x,y)
12
Homomorphic Filtering…
● Illumination-Reflectance Model in FDS
● Illumination Component
–
–
Slow Spatial Variations & Attenuate contributions by
illumination
● Reflectance Component
–
–
Varies abruptly – junctions of dissimilar objects
Amplify contributions by reflectance
● Simultaneous dynamic range compression & contrast
enhancement
● We cannot easily use the product i & r to operate
separately on the frequency components of illumination
& reflection because the FT of f ( x , y) is not separable;
1
3
H F….
• F[f(x,y)) not equal to F[i(x, y)].F[r(x, y)].
14
ln f(x,y) = ln i(x, y) + ln r(x, y).
We can separate them by taking logarithm
F[ln f(x,y)} = F[ln i(x, y)} + F[ln r(x, y)]
F(x,y) = I(x,y) + R(x,y), where F, I & R are the FTs ln f(x,y),ln i(x, y) , &
ln r(x, y). respectively.
F is FT of the sum of 2 images: a low-freq illumination image (suppress) & a
high freq reflectance (enhance)image
0 < i(x,y) < a, It indicate the perfect black body
0 < r(x,y) < 1, It indicate the perfect white body
H.F…
15
 Since i & r combine multiplicatively, they can be added by taking log of the
image intensity,
 so that they can be separated in the FD.
 i variations can be thought as a multiplicative noise & can be reduced by
filtering in the log domain.
 To make the i of an image more even, the HF components are increased and the
LF Components are filtered
 Because the HF are assumed as reflectance in the scene whereas the LF as the
illumination in the scene.i.e.,
 High pass filter is used to suppress LF’s & amplify HF’s in the log intensity
domain.
 i component tends to vary slowly across the image & the reflectance tends to
vary rapidly.
 Therefore, by applying a FD filter the intensity variation across the image can be
reduced while highlighting detail.
H.F….
• Z(x,y) = ln[f(x,y)] = ln[i(x,y)] + ln[r(x,y)] eq-1
• DFT[z(x,y)]
• = DFT{ln[f(x,y)]}
• = DFT{ln[i(x,y)] + ln[r(x,y)]}
• = DFT{ln[i(x,y)]} + DFT{ln[r(x,y)]} eq-2
• Since DFT[f(x,y)] = F(u,v), eq-2 becomes,
• Z(u,v) = Fi(u,v) + Fr(u,v) eq-3
• The function Z represents the FT of the sum of two images: a low
frequency illumination image & a high frequency reflectance
image
16
H.F….
• Thus ,FT of o/p by multiplying the DFT of the i/p with the filter
H(u,v). i.e., S(u,v) = H(u,v) Z(u,v) eq-4
• where S(u,v) is the FT of o/p. Substitute eq-3 in 4,
• we get S(u,v) = H(u,v) [ Fi(u,v) + Fr(u,v) ]
• = H(u,v) Fi(u,v) + H(u,v) Fr(u,v) eq-5
• Applying IDFT to eq-6,
• we get, T -1 [S(u,v)] = T-1 [ H(u,v) Fi(u,v) + H(u,v) Fr(u,v)]
• = T-1 [ H(u,v) Fi(u,v)] + T-1 [H(u,v) Fr(u,v)]
• s(x,y) = i’(x,y) + r’(x,y) eq-6
• The Enhanced image is obtained by taking exponential of the
IDFT s(x,y), i.e.,
17
g(x, y) =es(x,y)
=ei'( x,y)
er'( x,y)
=i (x, y)r (x, y)
0 0
io(x,y) = e i’(x,y) , ro(x,y) = e r’(x,y)
Where, are the I & r components of the enhanced o/p
Homomorphic Filtering
g(x, y) =es(x,y)
=ei'( x,y)
er'( x,y)
=i (x, y)r (x, y)
0 0
18
Homomorphic
Filtering
● Illumination Component
–
–
Slow Spatial Variations
Low Frequencies log of illumination
– attenuate contributions by illumination
–
–
–
Varies abruptly – junctions of dissimilar objects
High frequencies log of reflectance
amplify contributions by reflectance
●Refle
c
Lta
∈
nc
1e Component
● Simultaneous dynamic range compression & contrast
e
n
h
a n
H
c
e
>m
1e
n
t
19
L
L H
2
0
2
-c[D (u,v)/D ]
( - )[
1-e ]+
H(u,v)=
Homomorphic Filtering
20
Image: 1 62x746
γL=0.25, γH=2, c=1, D0=80
21
Band-reject & Band-pass
Filters
HBP (u,v) =1- HBR (u,v)
22
Band-reject &
Band-pass Filters
23
Notch Filters – Narrow Filtering
Q
k=1
HNR (u,v) = Hk (u,v)H- k (u,v)
]
1/2
2
2
1/2
2
2
k
k
k
k
k
N / 2 +v ) ]
M / 2 +u ) +(v -
[(u -
D (u,v) =
+(v - N / 2 - v )
[(u - M / 2 - u )
D (u,v) =
- k
24
- k
k
2n
0k
3
k=1
2n
0k / D (u,v)]
1+[D
1
/ D (u,v)]
1+[D
1
Butterworth Notch Reject Filters
HNR (u,v) =
HNP (u,v) =1- HNR (u,v)
25
D0=80, n=4
26
27
28
Thank you
29

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Digital Image Processing Module 3 Notess

  • 2. Laplacian in Spatial Domain ● Laplacian – – Isotropic Rotation Invariant 𝑄2 f = f (x +1, y) + f (x - 1, y) + f (x, y +1) + f (x, y - 1) - 4 f (x, y) g(x, y) = f (x, y) +c[𝑄2 f (x, y)] 2 2 ∂t + ∂z2 ∂2 f ∂2 f 𝑄 f = 2
  • 3. Laplacian in Frequency Domain 2 2 ∂t + ∂z2 ∂2 f ∂2 f 𝑄 f = 3
  • 4. Laplacian in Frequency Domain H(u,v) =-4𝑢2 (u2 +v2 ) With respect to center of frequency rectangle : H(u,v) =-4π2 [(u - M /2)2 +(v - N /2)2 )] =-4 π 2 D2 (u,v) Laplacian of an image: 4
  • 5. Laplacian in Frequency Domain ●Enhancement Eq: g(x, y) = f (x, y) +c[▽2 f (x, y)] c =- 1 ● Scales of f (x, y) & ▽ 2 f (x, y) as computed by DFT differ widely due to the DFT process ● Normalize f (x, y) to [0,1] before DFT ● Normalize ▽2 f (x, y) to [-1,1] 5
  • 7. Comparative Laplacian in Spatial & Frequency Domains 7
  • 8. Unsharp Mask, Highboost Filtering & High- Frequency-Emphasis Filtering ● In spatial domain: gmask (x, y) = f (x, y) - f (x, y) g(x, y) = f (x, y) +k * gmask (x, y) k =1: Unsharp Masking k >1: Highboost Filtering k ∈1: De- emphasized Unsharp Masking 8
  • 9. Unsharp Mask, Highboost Filtering & High- Frequency-Emphasis Filtering 9 [ 1+ k*Hhp (u,v)] F(u,v) }
  • 10. Unsharp Mask, Highboost Filtering & High- Frequency-Emphasis Filtering ● In frequency domain: High- Frequency Emphasis Filter k1>=0: Controlsthe offset from origin k2 >=0: Controlsthe contribution of high frequencies 10 g(x,y)= {[ k1+k2 * Hhp (u,v)] F(u,v) }
  • 12. Homomorphic Filtering ● Homomorphic filtering is a FDS that aims at a simultaneous increase in contrast & dynamic range compression. ● It is mainly utilized for non-uniformly illuminated images in medical, sonar images etc. for edge enhancement that makes the image details clear to the observer. ● Certain situations where the image is subjected to the multiplicative interference or noise as depicted ● f(x,y)= i(x,y) . r(x,y) 12
  • 13. Homomorphic Filtering… ● Illumination-Reflectance Model in FDS ● Illumination Component – – Slow Spatial Variations & Attenuate contributions by illumination ● Reflectance Component – – Varies abruptly – junctions of dissimilar objects Amplify contributions by reflectance ● Simultaneous dynamic range compression & contrast enhancement ● We cannot easily use the product i & r to operate separately on the frequency components of illumination & reflection because the FT of f ( x , y) is not separable; 1 3
  • 14. H F…. • F[f(x,y)) not equal to F[i(x, y)].F[r(x, y)]. 14 ln f(x,y) = ln i(x, y) + ln r(x, y). We can separate them by taking logarithm F[ln f(x,y)} = F[ln i(x, y)} + F[ln r(x, y)] F(x,y) = I(x,y) + R(x,y), where F, I & R are the FTs ln f(x,y),ln i(x, y) , & ln r(x, y). respectively. F is FT of the sum of 2 images: a low-freq illumination image (suppress) & a high freq reflectance (enhance)image 0 < i(x,y) < a, It indicate the perfect black body 0 < r(x,y) < 1, It indicate the perfect white body
  • 15. H.F… 15  Since i & r combine multiplicatively, they can be added by taking log of the image intensity,  so that they can be separated in the FD.  i variations can be thought as a multiplicative noise & can be reduced by filtering in the log domain.  To make the i of an image more even, the HF components are increased and the LF Components are filtered  Because the HF are assumed as reflectance in the scene whereas the LF as the illumination in the scene.i.e.,  High pass filter is used to suppress LF’s & amplify HF’s in the log intensity domain.  i component tends to vary slowly across the image & the reflectance tends to vary rapidly.  Therefore, by applying a FD filter the intensity variation across the image can be reduced while highlighting detail.
  • 16. H.F…. • Z(x,y) = ln[f(x,y)] = ln[i(x,y)] + ln[r(x,y)] eq-1 • DFT[z(x,y)] • = DFT{ln[f(x,y)]} • = DFT{ln[i(x,y)] + ln[r(x,y)]} • = DFT{ln[i(x,y)]} + DFT{ln[r(x,y)]} eq-2 • Since DFT[f(x,y)] = F(u,v), eq-2 becomes, • Z(u,v) = Fi(u,v) + Fr(u,v) eq-3 • The function Z represents the FT of the sum of two images: a low frequency illumination image & a high frequency reflectance image 16
  • 17. H.F…. • Thus ,FT of o/p by multiplying the DFT of the i/p with the filter H(u,v). i.e., S(u,v) = H(u,v) Z(u,v) eq-4 • where S(u,v) is the FT of o/p. Substitute eq-3 in 4, • we get S(u,v) = H(u,v) [ Fi(u,v) + Fr(u,v) ] • = H(u,v) Fi(u,v) + H(u,v) Fr(u,v) eq-5 • Applying IDFT to eq-6, • we get, T -1 [S(u,v)] = T-1 [ H(u,v) Fi(u,v) + H(u,v) Fr(u,v)] • = T-1 [ H(u,v) Fi(u,v)] + T-1 [H(u,v) Fr(u,v)] • s(x,y) = i’(x,y) + r’(x,y) eq-6 • The Enhanced image is obtained by taking exponential of the IDFT s(x,y), i.e., 17 g(x, y) =es(x,y) =ei'( x,y) er'( x,y) =i (x, y)r (x, y) 0 0 io(x,y) = e i’(x,y) , ro(x,y) = e r’(x,y) Where, are the I & r components of the enhanced o/p
  • 18. Homomorphic Filtering g(x, y) =es(x,y) =ei'( x,y) er'( x,y) =i (x, y)r (x, y) 0 0 18
  • 19. Homomorphic Filtering ● Illumination Component – – Slow Spatial Variations Low Frequencies log of illumination – attenuate contributions by illumination – – – Varies abruptly – junctions of dissimilar objects High frequencies log of reflectance amplify contributions by reflectance ●Refle c Lta ∈ nc 1e Component ● Simultaneous dynamic range compression & contrast e n h a n H c e >m 1e n t 19
  • 20. L L H 2 0 2 -c[D (u,v)/D ] ( - )[ 1-e ]+ H(u,v)= Homomorphic Filtering 20
  • 21. Image: 1 62x746 γL=0.25, γH=2, c=1, D0=80 21
  • 22. Band-reject & Band-pass Filters HBP (u,v) =1- HBR (u,v) 22
  • 24. Notch Filters – Narrow Filtering Q k=1 HNR (u,v) = Hk (u,v)H- k (u,v) ] 1/2 2 2 1/2 2 2 k k k k k N / 2 +v ) ] M / 2 +u ) +(v - [(u - D (u,v) = +(v - N / 2 - v ) [(u - M / 2 - u ) D (u,v) = - k 24
  • 25. - k k 2n 0k 3 k=1 2n 0k / D (u,v)] 1+[D 1 / D (u,v)] 1+[D 1 Butterworth Notch Reject Filters HNR (u,v) = HNP (u,v) =1- HNR (u,v) 25
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