Fundamentals of Digital Image Processing - Control of Color Imaging Systems: Analysis and Design - page 50

If n ¼ 0 and m ¼ 0, then

y(0, 0) ¼ x(0, 0) þx(0, 1) þx( 1, 0) þx( 1, 1)

¼ x(0, 0) þx(0, 1) þx(1, 0) þx(1, 1) ¼ 2 þ 1 1 þ 3 ¼ 5

If n ¼ 0 and m ¼ 1, then

y(0, 1) ¼ x(0, 1) þx(0, 0) þx( 1, 1) þx( 1, 0)

¼ x(0, 1) þx(0, 0) þx(1, 1) þx(1, 0) ¼ 1 þ 2 þ 3 1 ¼ 5

If n ¼ 1 and m ¼ 0, then

y(1, 0) ¼ x(1, 0) þx(1, 1) þx(0, 0) þx(0, 1)

¼ x(1, 0) þx(1, 1) þx(0, 0) þx(0, 1) ¼ 1 þ 3 þ 2 þ 1 ¼ 5

Finally, if n ¼ 1 and m ¼ 1, then

y(1, 1) ¼ x(1, 1) þx(0, 1) þx(0, 1) þx(0, 0) ¼ 3 þ 1 1 þ 2 ¼ 5

Therefore,

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55

y(n, m) ¼ x(n, m) h(n, m) ¼

b. Linear convolution. In this case, we have

y(n, m) ¼ x(n, m) * h(n, m)

X X

1 1

¼ h(u, v)x(nu, y v ) n ¼ 0, 1, 2 m ¼ 0, 1, 2

v¼ 0 u¼ 0

Expanding the above sum results in

y(n,m) ¼h(0,0)x(n,m) þh(0, 1)x(n,m 1) þh(1, 0)x(n 1,m)

þh(1, 1)x(n 1,m 1)

¼x(n,m) þx(n,m 1) þx(n 1,m) þx(n 1,m 1)

Computing y(n, m) for n ¼ 0, 1, 2 and m ¼ 0, 1, 2, we would have

y(0, 0) ¼ x(0, 0) þx(0, 1) þx( 1, 0) þx( 1, 1) ¼ 2 þ 0 þ 0 þ 0 ¼ 2

y(0, 1) ¼ x(0, 1) þx(0, 0) þx( 1, 1) þx( 1, 0) ¼ 1 þ 2 þ 0 þ 0 ¼ 3

y(0, 2) ¼ x(0, 2) þx(0, 1) þx( 1, 2) þx( 1, 1) ¼ 0 þ 1 þ 0 þ 0 ¼ 1

y(1, 0) ¼ x(1, 0) þx(1, 1) þx(0, 0) þ x(0, 1) ¼ 1 þ 0 þ 2 þ 0 ¼ 1

y(1, 1) ¼ x(1, 1) þx(0, 1) þx(0, 1) þx(0, 0) ¼ 3 þ 1 1 þ 2 ¼ 5