Image Processing (Chapter 6 from BKP Horn)
Sampling Theorem
Baba C. Vemuri
Professor, Department of Computer & Information Science and Engineering
University of Florida
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Vemuri
Image Processing/Continuous Images
Image Processing (Chapter 6 from BKP Horn)
Shannon’s Sampling Theorem
1
Image Processing (Chapter 6 from BKP Horn)
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Vemuri
Image Processing/Continuous Images
Image Processing (Chapter 6 from BKP Horn)
Shannon’s Sampling Theorem
Figure: Samples of a function (from ‘Fourier Optics”,Goodman)
Vemuri
Image Processing/Continuous Images
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Image Processing (Chapter 6 from BKP Horn)
How finely must an image function be sampled to preserve
all the information for reconstruction from the samples?
Fine sampling of g(x, y ) ⇒ g can be approximately
reconstrcuted from the samples. How?
For a special class of functions called, bandlimited
functions, the reconstruction is exact.
Shannons Sampling Theorem provides conditions under
which exact reconstruction is possible.
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Vemuri
Image Processing/Continuous Images
Image Processing (Chapter 6 from BKP Horn)
Sampling Theorem
Consider a rectangular lattie of samples:
gs (x, y ) = Ш(
x y
, )g(x, y )
X Y
(1)
Where, impulses are spaced apart X in x-direction and Y
in y-direction.
Taking FT, we get,
Gs (u, v ) = F{Ш(x/X , y /Y )} ⊗ G(u, v )
(2)
Using the table of transforms and the similarity theoreom of
FTs,
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F{Ш(x/X , y /Y )} = XY Ш(Xu, Yv ) and
(3)
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Image Processing/Continuous Images
Image Processing (Chapter 6 from BKP Horn)
XY Ш(Xu, Yv ) =
X∞X m
n
δ u − ,v −
X
Y
n,m=−∞
(4)
Thus, the spectrum of the sampled function is,
Gs (u, v ) =
X∞X
n,m=∞
n
m
G u − ,v −
X
Y
(5)
This implies, the spectrum of gs can be found by simply
erecting the spectum of g about each point (n/X , m/Y ) in
the (u, v ) domain.
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Vemuri
Image Processing/Continuous Images
Image Processing (Chapter 6 from BKP Horn)
Spectrum of the Function
Figure: Sample Spectrum and its Convolution (from ‘Fourier
Optics”,Goodman)
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Image Processing/Continuous Images
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Image Processing (Chapter 6 from BKP Horn)
Since g is assumed to be bandlimited, its spectrum is
nonzero only over a finite region R of the spatial frequency
domain.
For small X and Y , separation in (u, v )-domain is large
enough such that neighboring regions don’t overlap.
Can get G from Gs by passing it through a linear filter that
transmits the term (n = 0, m = 0) in equation 5 and
excludes all others.
Hence, at the output, we have an exact replica of g(x, y ).
Key is to take samples close enough in (x, y )-domain.
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Vemuri
Image Processing/Continuous Images
Image Processing (Chapter 6 from BKP Horn)
Problem: Determine the max. allowable separation
between samples.
Let 2Bx and 2By be the width in the u and v directions
respectively of the smallest rectangle completely enclosing
R.
Since each term in equation 5 is separated by 1/X and
1/Y in u and v directions, separation in spectral region is
assured if,
1
1
andY ≤
(6)
X ≤
2Bx
2By
Problem: What is the transfer function of the filter to apply
to the sampled data?
H(u, v ) = rect(
Vemuri
u
v
,
)
2Bx 2By
Image Processing/Continuous Images
(7)
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Image Processing (Chapter 6 from BKP Horn)
Exact recovery of G from Gs is then got by,
Gs (u, v )rect(
v
u
,
) = G(u, v )
2Bx 2By
Equivalently in the spatial domain,
x y
Ш( , )g(x, y ) ⊗ h(x, y ) = g(x, y )
X Y
(8)
(9)
where, h(x, y ) is the PSF of the filter given by,
ZZ
∞
h(x, y ) =
rect(u/2Bx , v /2By ) exp{j2π(ux + vy )}dudv
−∞
=4Bx By sinc (2Bx x, 2By y )
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Image Processing/Continuous Images
(10)
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Image Processing (Chapter 6 from BKP Horn)
Using the definition of the Sha function, g(x, y ) above can
be written as,
g(x, y ) = 4Bx By XY
X∞X
g(nX , mY )sinc (2Bx (x − nX ), 2By (y − mY ))
−∞
(11)
With X = 1/2Bx and Y = 1/2By , above equation
becomes,
g(x, y ) =
X∞X
−∞
m
n
), 2By (y −
)
g(n/2Bx , m/2By )sinc 2Bx (x −
2Bx
2By
(12)
Equation 12 is the Whitaker-Shannon Sampling Theorem.
Exact recovery of a bandlimited function is possible from
appropriately spaced samples. Reconstruction achieved ufl-logo
using a product of sincs.
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Image Processing/Continuous Images
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