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The concept of circular convolution in the context of mathematics of image processing. It provides formulas for calculating the convolution of two signals, f and g, with n samples each. The document also includes examples with different values of n and signals defined by vectors. It further discusses the convolution theorem and demonstrates its validity.
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Name: Box #:
Circular convolution Let f and g be two signals with N samples each.
The circular convolution of f and g is the n point sample given by:
f ∗ g(k) =
r=
f (k − r)g(r) =
r=
f (r)g(k − r) (mod N )
The mod N means that we use clock arithmetic in computing k − r, or that we
assume that f and g are defined for all integers , but are periodic with period N ,
i.e., f (k + N ) = f (k), g(k + N ) = g(k). For example, suppose that N = 4 and
that f and g are defined by vectors [x 0 , x 1 , x 2 , x 3 ] and [y 0 , y 1 , y 2 , y 3 ] :
f (r) = xr mod 4, g(r) = yr mod 4.
Then
f ∗ g(0) =
3 ∑
r=
f (−r)g(r) = f (0)g(0) + f (−1)g(1) + f (−2)g(2) + f (−3)g(3)
= f (0)g(0) + f (3)g(1) + f (2)g(2) + f (1)g(3)
= x 0 y 0 + x 3 y 1 + x 2 y 2 + x 1 y 3
f ∗ g(1) = x 1 y 0 + x 0 y 1 + x 3 y 2 + x 2 y 3
f ∗ g(2) = x 2 y 0 + x 1 y 1 + x 0 y 2 + x 3 y 3
f ∗ g(3) = x 3 y 0 + x 2 y 1 + x 1 y 2 + x 0 y 3
1 2
, f (1) =
1 2
f (r) = 0 otherwise. Let g be defined by
an arbitrary vector [y 0 , y 1 ,... , y 7 ]. Find the formulas (for eight samples) for
f ∗ g. y 0 +y 7 2 y 1 +y 0 2 y 2 +y 1 2 y 3 +y 2 2 y 4 +y 3 2 y 5 +y 4 2 y 6 +y 5 2 y 7 +y 6 2
f ∗ g(0)
f ∗ g(1)
. . .
f ∗ g(7)
= Hf Y = Hf
y 0
y 1
. . .
y 7
Hf =
write out the matrix for Hf.
Hf =
x 0 0 0 0 0 0 0 x 1
x 1 x 0 0 0 0 0 0 0
0 x 1 x 0 0 0 0 0 0
0 0 x 1 x 0 0 0 0 0
0 0 0 x 1 x 0 0 0 0
0 0 0 0 x 1 x 0 0 0
0 0 0 0 0 x 1 x 0 0
0 0 0 0 0 0 x 1 x 0
mod 4 and compare to H(z). What do you observe?
F (z) = a 0 +
a 1
z
a 2
z^2
a 3
z^3
G(z) = b 0 +
b 1
z
b 2
z
2
b 3
z
3
H(z) = a 0 b 0 + a 1 b 3 + a 2 b 2 +
a 3 b 1 + a 0 b 1 + a 1 b 0 + a 2 b 3 + a 3 b 2
z
a 0 b 2 + a 1 b 1 + a 2 b 0 + a 3 b 3
z
2
a 0 b 3 + a 1 b 2 + a 2 b 1 + a 3 b 0
z
3
F (z)G(z) =
a 0 +
a 1
z
a 2
z^2
a 3
z^3
b 0 +
b 1
z
b 2
z^2
b 3
z^3
= a 0 b 0 +
a 1 b 0 + a 0 b 1
z
a 2 b 0 + a 1 b 1 + a 0 b 2
z^2
a 3 b 0 + a 2 b 1 + a 1 b 2 + a 0 b 3
z^3
a 3 b 1 + a 2 b 2 + a 1 b 3
z
4
a 3 b 2 + a 2 b 3
z
5
a 3 b 3
z
6
Reduce exponents mod 4 to get
a 0 b 0 +
a 1 b 0 + a 0 b 1
z
a 2 b 0 + a 1 b 1 + a 0 b 2
z
2
a 3 b 0 + a 2 b 1 + a 1 b 2 + a 0 b 3
z
3
a 3 b 1 + a 2 b 2 + a 1 b 3
a 3 b 2 + a 2 b 3
z
a 3 b 3
z^2
= a 0 b 0 + a 2 b 2 + a 1 b 3 + a 3 b 1 +
a 1 b 0 + a 0 b 1 + a 3 b 2 + a 2 b 3
z
a 3 b 3 + a 2 b 0 + a 1 b 1 + a 0 b 2
z^2
a 3 b 0 + a 2 b 1 + a 1 b 2 + a 0 b 3
z^3
= H(z)
f̂ ∗ g = f̂ ̂g
as functions. Since F (z)G(z) is the same as H(z) when the exponentsare
reduced mod 4 then
F (z)G(z) = H(z), if z
4 = 1.
But then
f̂ ∗ g(k) = H(e^2 πi^
k (^4) )
= F (e
2 πi k 4 )G(e
2 πi k 4 )
= f̂ (k)̂g(k).