Through-Wall Imaging With UWB Radar System - KEMT FEI TUKE
Through-Wall Imaging With UWB Radar System - KEMT FEI TUKE
Through-Wall Imaging With UWB Radar System - KEMT FEI TUKE
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4.1 <strong>Through</strong>-<strong>Wall</strong> TOA Estimation 45<br />
computed, what leads to much faster computation. Even if the Tanaka has already<br />
solved this problem in [136], hereunder, we introduce several improvements and<br />
simplifications that speed up computations even more.<br />
One have to mention, that antennas can be placed directly on a wall during<br />
measurements, and two layer model is directly obtained. However, in this case,<br />
the wall is in the antenna’s near field and therefore antenna impulse response in<br />
time domain has been significantly changed. Hereby, the reducing of long antenna<br />
impulse response influence on measured data by deconvolution, as it is described<br />
in Section 2.4.4, is not applicable any more. Therefore, antennas are placed at<br />
least 0.5 m from a wall during SAR scanning in practice, Wdi ≥ 0.5 m.<br />
The true T OAAT can be computed like summation of both times, the time of<br />
flight in the air and the time of flight in the wall:<br />
T OAAT (d) = da<br />
c<br />
+ dw<br />
vw<br />
where da = da1 + da2. The total flight distance dtot is computed as:<br />
where<br />
dtot = da + dw<br />
(4.1.1)<br />
(4.1.2)<br />
�<br />
da = (HZ − Dw) 2 + (HX − d) 2 , dw = � D2 w + d2 (4.1.3)<br />
where HX is the distance between the antenna and target in X direction, HZ is<br />
the distance between the antenna and target in Z direction (Fig. 4.1.1). The true<br />
T OAAT for any antenna position A[xA, zA] and target position T [xT , zT ] that is<br />
represented by the image pixel position HX and HZ is required for formatting<br />
the image during SAR imaging. After substituting da and dw from (4.1.3) to the<br />
(4.1.1), only d will be unknown. In order to estimate d, the minimization process<br />
is computed by derivation:<br />
dT OAAT (d)<br />
= 0 (4.1.4)<br />
dd<br />
After several simple mathematical operations the fourth order polynomial equation<br />
with five coefficients is obtained:<br />
where coefficients:<br />
co1d 4 + co2d 3 + co3d 2 + co4d + co5 = 0 (4.1.5)<br />
co1 =c 2 − v 2 w, co2 = 2HX<br />
co3 =H 2 X<br />
� v 2 w − c 2�<br />
� � 2 2 2<br />
c − vw + c (HZ − Dw) 2 − v 2 wD 2 w<br />
co4 =2v 2 wD 2 wHX, co5 = −v 2 wD 2 wH 2 X<br />
(4.1.6)<br />
The analytical solution of (4.1.5) even with Ferraris or Galois method [139] requires<br />
huge number of divisions and square root operations. Therefore, unlike in [136],