Page 416 - Fundamentals of Radar Signal Processing

(5.128)

To make the procedure clearer, consider the case of r = 3 PRFs. Then n t

satisfies

(5.129)

where

(5.130)

(for example, α = β N N ) and the β are the smallest integers such that
i
2
1
1
0
(5.131)

To illustrate the procedure, suppose that the true range of a target,
normalized to the range bin size, is n = 19. Further suppose that the three PRFs
t
are chosen such that the number of range bins in the unambiguous range for each
PRF are N = 11, N = 12, and N = 13. On the first PRF the target will be
2
1
0
detected in the apparent range bin . Similarly, and
. From Eq. (5.131), β is the smallest integer that satisfies ((β × 12
0
0
× 13)) = 1; that is, β satisfies β × 12 × 13 = 156β = 11k + 1 for some integer
0
11
0
0
k. The solution is β = 6. In the same manner it is found that β = 11 and β = 7.
1
2
0
Equation (5.130) then gives α = 6 × 12 × 13 = 936, α = 1573, and α2 = 924.
1
0
Finally, Eq. (5.129) gives the estimate of the true range bin as
(5.132)

which is the correct result.
A serious problem with the CRT is its extreme sensitivity to errors induced

by noise and range quantization. There is no guarantee that the actual range R t
will be an integer multiple of the range bin spacing ΔR as assumed previously;
the target may in fact straddle range bins. In addition, noise in the measurements
may cause the target to be located in an incorrect range bin. To illustrate the
effect of such errors, repeat the previous example but suppose that is for
some reason measured to be 7 instead of the correct value of 6. Carrying out the

previous calculations will give instead of 19, a huge error.
A “robust CRT” that controls the maximum error is given in Li et al.