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This is a question(2.25) from introduction to radar systems by Merrill Skolnik 3rd edition.

Plot the single-scan probability of detection as function of range, assuming a constant cross-section target of $10m^2$ and free-space propagation. [It is easier to select first the probability of detection (between 0.3 and 0.99) and then find the corresponding signal-to-noise ratio. Also, you may select a single (average) value of the integration improvement factor]

This question has second part which says Plot the probability of detection as function of range for the same situation as above but with decision criteria that target must be found on at least 2 of the 3 scans of the rotating antenna.

I want to ask what is the meaning of "with decision criteria that target must be found on at least 2 of the 3 scans of the rotating antenna" in the 2nd part of the question. Is it just simple binomial distribution use?

Given values are carrier frequency, peak power, pulse width, pulse repetition frequency, receiver noise figure, antenna rotation rate, antenna gain, azimuth beamwidth, system losses, average false alarm time, target cross section.

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  • $\begingroup$ @G.Smith Ok, I removed the plot. $\endgroup$
    – A Q
    Feb 14, 2021 at 7:33

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Yes, you interpretation is correct. For each range you need to convert the single-scan probability of detection $P_D(r)$ to the probability of getting a detection on at least two out of three scans (I'll call it $P_2(r)$). $$P_2(r) = \binom{3}{2}\left[P_D(r)\right]^2\left[1-P_D(r)\right] + \binom{3}{3}\left[P_D(r)\right]^3$$ where the 1st term accounts for the 3 ways of getting exactly 2 detections out of 3 scans and the 2nd term is the probability of getting exactly 3 out of 3 detections.

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