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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],

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