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Sampling and Quantization
Lecture #5January 21, 2002
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Jan 21, 2003 Sampling 2
Sampling and Quantization
� Spatial Resolution (Sampling)� Determines the smallest perceivable image
detail.� What is the best sampling rate?
� Gray-level resolution (Quantization)� Smallest discernible change in the gray level
value.� Is there an optimal quantizer?
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Jan 21, 2003 Sampling 3
Image sampling and quantization
In 1-D
f(x,y)
(Continuous image)
Samplerfs(m,n)
Quantizeru(m,n)
To Computer
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Jan 21, 2003 Sampling 4
1-D1−D
x(t)
Time domain
X(u)
Frequency
Ts(t)
xs(t) = x(t) s(t) = Σ x(kt) δ (t-kT)
s(t)1/T
1/TXs(f)
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Jan 21, 2003 Sampling 5
2-D: Comb function
ycomb(x,y;∆x,∆y)
x∆x
∆y
Comb( , ; , ) ( , )x y x y x m x y n ynm
∆ ∆ ∆ ∆≅ − −=−∞
∞
=−∞
∞
∑∑ δ
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Jan 21, 2003 Sampling 6
Sampled Image
f x y f x y x y x y
f m x n y x m x y n y
x y x y u v
x yu v
x y
s
nm
( , ) ( , ) ( , ; , )
( , ) ( , )
( , ; , ) ( , )
, ; , )
=
= − −
← → ==−∞
∞
=−∞
∞
ℑ
∑∑
comb
comb COMB
comb(
∆ ∆
∆ ∆ ∆ ∆
∆ ∆
∆ ∆ ∆ ∆
δ
1 1 1
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Jan 21, 2003 Sampling 7
Sampled Spectrum
F u v F u v u v
x yF u v u k
xv l
y
x yF u k
xv l
y
s
k l
k l
( , ) ( , ) ( , )
( , ) ,
,
,
,
= ∗
= ∗ − −FHG
IKJ
= − −FHG
IKJ
∑∑
∑∑
=−∞
∞
=−∞
∞
COMB
1
1
∆ ∆ ∆ ∆
∆ ∆ ∆ ∆
δ
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Jan 21, 2003 Sampling 8
Sampled Spectrum: Example
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Jan 21, 2003 Sampling 9
Bandlimited Images
A function f(x,y) is said to be band limited if the Fourier transform
F(u,v) = 0 for | u | > u0, | v | > v0
u0, v0 Band width of the image in the x- and y- directions
v0
u0region of support
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Jan 21, 2003 Sampling 10
Foldover Frequencies
Sampling frequencies:Let us and vs be the sampling frequencies
Then us > 2u0 ; vs > 2v0
or ∆ x <1/2u0 ; ∆ y <1/2v0
Frequencies above half the sampling frequencies are called fold over frequencies.
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Jan 21, 2003 Sampling 11
Sampling Theorem
A band limited image f(x,y) with F(u,v) as its Fourier transform; and F(u,v) = 0 |u| > u0 |v| > v0 ; and sampled uniformly on a rectangular grid with spacing ∆∆∆∆x and ∆∆∆∆y, can be recovered without error from the sample values f(m ∆∆∆∆x, n ∆∆∆∆y) provided the sampling rate is greater than the nyquist rate.
i.e 1/ ∆∆∆∆x = us > 2 u0, 1/ ∆∆∆∆y = vs > 2 v0
The reconstructed image is given by the interpolation formula:
f x y f m x n y xu mxu m
y v ny v nm n
s
s
s
s
( , ) ( , ) ( )( )
( )( ),
= −−
−−∑∑
=−∞
∞
∆ ∆ sin sinππ
ππ
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Jan 21, 2003 Sampling 12
Reconstruction
0
R1
R2
δR
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Jan 21, 2003 Sampling 13
Reconstruction via LPF
F(u,v) can be recovered by a LPF with
H u vx y u v R
R RR
R
F u v H u v F u v F u vf x y F u v
s
( , )( , )
~( , ) ( , ) ( , ) ( , )
( , ) [ ( , )]
=∈RST
= =
= ℑ −
∆ ∆0
1
2
1
Other wise is any region whose boundary is contained
within the annular ring between the rectangles and in the figure. Reconstructed signal is
∂
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Jan 21, 2003 Sampling 14
Aliasing
Note: If us and vs are below the Nyquist rate, the periodicreplications will overlap, resulting in a distorted spectrum.
This overlapping of successive periods of the spectrumcauses the foldover frequencies in the original image toappear as frequencies below us/2, vs/2 in the sampled image. This is called aliasing.
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Jan 21, 2003 Sampling 15
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Jan 21, 2003 Sampling 16
Example
there will be aliasing.
f x y x yF u v u v u v
u v
x y u v u vs s
( , ) ( ( ))( , ) ( , ) ( , )
,
. ,.
= += - - + + +
fi = =
= = fi = = = <
2 2 3 43 4 3 4
3 4
0 2 10 2
5 2
0 0
0 0
cos
Let ,< 2
pd d
D D
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Jan 21, 2003 Sampling 17
Example:(contd.)
F
Let
Otherwise
cos
s ( , ) ( , )
[ ( , ) ( , )]
( , ). . , . .
( , ) ( , ) ( , )( , ) ( , )
~( , ) (2 (2 ))
,
,
u v F u ku v lv
u k v l u k v l
H u vu u
F u v H u v F u vu v u v
f x y x y
s sk l
k l
s
= − −
= − − − − + + − + −
=− ≤ ≤ − ≤ ≤RST
∴ == + + + − −
∴ = +
∑∑
∑∑
=−∞
∞
=−∞
∞
25
25 3 5 4 5 3 5 4 5
2 5 2 5 2 5 2 50
2 1 2 1
2
125
δ δ
δ δπ
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Jan 21, 2003 Sampling 18
Examples
Original and the reconstructed image from samples.
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Jan 21, 2003 Sampling 19
Another example
Sampling filter
sampled image
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Jan 21, 2003 Sampling 20
Aliasing Problems (real images!)