In previous problems we saw that two PRFs are not in general sufficient to remove ambiguities when there are two targets present. We can remove the two target ambiguity if we add a third PRF.
The third PRF must be carefully chosen to not be harmonically related to the first or the second. Using the set we have been working on over the last several problems, the first PRF is 40 range bins and the second is 41 range bins. We must choose a third PRF that is relatively prime to these two. The 40 and 41 were chosen because any two numbers n and n+1 are relatively prime.
For the third PRF, we cannot choose 42 range bins because 40 and 42 are harmonically related. In part 1 of this problem, you are asked to identify the ambiguity problem in not choosing a PRF carefully.
Identify at least one pair of target ranges that are ambiguous when using two PRFs composed of 40 range bins and 42 range bins.
Limit the range of your possible answers to values of less than 42*40 = 1580 range bins.
Consider a third PRF of 39 range bins. 39 is relatively prime to both 40 and 41, and we already know 40 is relatively prime to 41. A PRF set of 40 range bins for PRF 1, 41 range bins for PRF 2, and 39 range bins for PRF 3 will not have any surprising ambiguity problems. Using a radar with these PRFs, assume detections on the three PRFs as shown below for the region of 5 km to 7 km (range bins 21 to 28).
What is the true range to each of the two targets?