Image Processing Reference
In-Depth Information
Since Green's function behaves like an impulse response function, we
assume
=
ϕ
()
xGxy
(, )( )
ρ
yy
d
(2.30)
for n integrals where d y = d y 1 , d y 2 ,…, d y n . This can now be substituted back
into Equation 2.28 as follows:
Lx
ϕ
()
=
LGxy
(, )( )
ρ
yy
d
= −
δ
(
xyyy
)( )
ρ
d
=−
ρ
(
x
)
(2.31)
x
x
This way of looking at scattering problems fundamentally changes our per-
spective, whether for a direct or inverse scattering problem. The question now
becomes how we find Green's function or the impulse function, which is not
so easy in general. We note that if G satisfies the boundary conditions, so
does φ.
Let us consider the Fourier transform of G in n -D, that is, n dimensions:
n
1
=
-
(2.32)
Gxy
(, )
g KyeK
(
,)
iK x
,
d

−∞
where Kx
22 and ddd
,
Let us recall the following Fourier transform for a delta function as
,
=
Kx
+
Kx
+
KKK
=
,
.
11
1
2
n
1
2
−=
-
∫∫
Fxy
((
δ
))
e
iK xy
(
)
d
K
(2.33)

π
Using this definition in conjunction with Equation 2.29 yields the follow-
ing relationship:
n
n
1
2
1
2
=− −=
-
=−
-
LGxy
(, )
δ
(
xy
)
gkyLeK
( ,)
iK x
,
d
e
i
KKy iK eK
,
,
d
(2.34)

x
π
x
π
Since the operator L x only operates on x terms, we now have
aaiK
+
+
+
aiK
+
+
0
1
1
nn
Le
iK x
,
=
+
aiK iK
(
)(
)
+
+
a
(
iK
)
n
+
e
iK x
,
(2.35)
x
12
1
2
nn
n
aaiK
+
(
)
α
++
a
(
iK
)
α
n
1
1
n
n
n
We can now simplify this and equate to zero which yields
gkya
(, )[
+
aiK
++
a
(
iK
) ]
α
+
e
iK y
,
e
iKx
,
d
K
= 0
(2.36)
0
1
1
nn nn
,
n
n
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