Xử lý ảnh số - Khôi phục ảnh part 2 doc

v`a g l`a c´ac vector cˆo.t trong R6 v`a H l`a ma trˆa.n vuˆong cˆa´p 6:

H =





he(0) he(5) he(4) ··· he(1)

he(1) he(0) he(5) ··· he(2)

he(2) he(1) he(0) ··· he(3)

.

.

.

he(5) he(4) he(3) ··· he(0)





.

Nhu.

ng he(x) = 0 v´o.

i x = 3, 4, 5 v`a he(x) = h(x) v´o.

i x = 0, 1, 2 nˆen

H =





h(0) h(2) h(1)

h(1) h(0) h(2)

h(2) h(1) h(0)

h(2) h(1) h(0)

h(2) h(1) h(0)

h(2) h(1) h(0)





trong d¯´o tˆa´t ca˙’ c´ac phˆa`n tu.

˙’ khˆong d¯u.

o.

. c viˆe

´t ra bˇa`ng khˆong.

Bˆay gi`o. x´et tru.

`o.

ng ho.

. p hai chiˆe` u: hai a˙’nh sˆo´ f(x, y) v`a h(x, y) c´o k´ıch thu.

´o

.

c

A × B v`a C × D tu.

o.

ng ´u.

ng. Cho.n M,N thoa˙’ m˜an

M ≥ A + B − 1, N ≥ C + D − 1.

Ta mo.

˙’ rˆo.ng k´ıch thu.

´o

.

c c´ac a˙’nh bˇa`ng c´ach d¯ˇa.t

fe(x, y) :=







f(x, y) nˆe´u 0 ≤ x ≤ A − 1, v`a 0 ≤ y ≤ B − 1,

0 nˆe´u A ≤ x ≤ M − 1, hoˇa. c B ≤ y ≤ N − 1,

v`a

he(x, y) :=







h(x, y) nˆe´u 0 ≤ x ≤ C − 1, v`a 0 ≤ y ≤ D − 1,

0 nˆe´u C ≤ x ≤ M − 1, hoˇa. c D ≤ y ≤ N − 1,

Ch´ung ta c˜ung gia˙’ thiˆe´t c´ac h`am fe(x, y) v`a he(x, y) tuˆa`n ho`an v´o.

i chu k`y M v`a N

theo hai hu.

´o.

ng x v`a y tu.

o.

ng ´u.

ng. Nhˇa´c la.i

ge(x, y) =

M

X−1

m=0

N

X−1

n=0

fe(m, n)he(x − m, y − n),

v´o.

i x = 0, 1,...,M − 1, v`a y = 0, 1,...,N − 1. H`am ge(x, y) tuˆa`n ho`an v´o.

i c`ung chu

k`y nhu. fe(x, y) v`a he(x, y). D- ˆe˙’ ho`an thiˆe.n mˆo h`ınh suy gia˙’m chˆa´t lu.

o.

. ng r`o.

i ra. c, cˆa`n

thˆem nhiˆe˜u ηe(x, y) v`ao h`am ge(x, y); t´u.

c l`a

ge(x, y) =

M

X−1

m=0

N

X−1

n=0

fe(m, n)he(x − m, y − n) + ηe(x, y), (5.4)

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