Through-Wall Imaging With UWB Radar System - KEMT FEI TUKE

2.5 Radar Imaging Methods Overview 26
wavefield, B (kx, f) e −2πikzz . Downgoing means in z direction and upgoing in −z
direction.
It is now apparent that only one boundary condition φ (x, z = 0, t) is not enough
and the solution will be ambiguous. To remove this ambiguity, the two boundary
conditions are required, for example ψ and ∂zψ. But this is not known in our case
so the migration problem is said to be ill-posed and another trick has to be used.
If both conditions were available, A and B would be found as solutions to φ for
z = 0 and it would be specified as φ0 as well as ∂φ/∂z would be specified as φz0:
φ(z = 0) ≡ φ0 = A + B (2.5.36)
and
∂φ
∂z (z = 0) ≡ φz0 = 2πikzA − 2πikzB. (2.5.37)
The solution for this ill-posed problem and ambiguity removing will be done by
assumption of one-way waves. In the receiver position, the wave cannot be downgoing,
only upgoing. This allows the solution:
A(kx, f) = 0 and B(kx, f) = φ0(kx, f) ≡ φ(kx, z = 0, f). (2.5.38)
Then, the wavefield can be expressed as the inverse Fourier transform of φ0:
��
ψ(x, z, t) = φ0(kx, f)e 2πi(kxx−kzz−ft) dkxdf (2.5.39)
and the migrated solution is:
��
ψ(x, z, t = 0) =
φ0(kx, f)e 2πi(kxx−kzz) dkxdf. (2.5.40)
(2.5.40) gives a migration process as a double integration of φ0(kx, f) over f and
kx. Even if it looks like the solution is complete, it has a disadvantage. Only
one of the integrations, that over kx, is a Fourier transform that can be done
by a numerical fast Fourier Transform (FFT). The f integration is not a Fourier
transform because the Fourier kernel e −2πft was lost when the imaging condition
(setting t = 0) was invoked. However, another complex exponential e −2πikzz shows
up in (2.5.40). Stolt suggested a change of variables from (kx, f) to (kx, kz) to
obtain a result in which both integrations are Fourier transforms. The change of
variables is defined by (2.5.34) and can be solved for f to give:
f = v � k 2 x + k 2 z. (2.5.41)
Performing the change of variables from f to kz according to the rules of calculus
transforms (2.5.40) yields into:
��
ψ(x, z, t = 0) = φm(kx, kz)e 2πi(kxx−kzz) dkxdkz (2.5.42)