Two Boats on the Same Bearing at Six Miles
A 12-inch X-band radar on a harbor tug is set to the 6 NM range scale. The transmitter uses a 0.5 microsecond pulse and a PRF of 1200 Hz. Two small boats are ahead on nearly the same bearing, one at 1.200 NM and the other at 1.227 NM. The display shows a single elongated echo. The operator increases gain and sea clutter control, but the echo remains merged. The radar has a manual tuning control and a receiver bandwidth matched to the pulse. The operator asks whether the boats are too close to resolve and whether a different range scale will help. The tide is slack, sea state 1, and the boats are crossing slowly. The radar is otherwise performing normally, with good echoes on buoys and the harbor breakwater.
1. Calculate the theoretical range resolution and decide whether the two boats can be separated on this display.
Read the worked answer
Range resolution is set mainly by pulse length: ΔR = c × τ / 2, where c is about 3 × 10^8 m/s and τ is the pulse length in seconds.
For τ = 0.5 microseconds = 0.5 × 10^-6 s: ΔR = (3 × 10^8 × 0.5 × 10^-6) / 2 = 75 m.
The boat separation is 1.227 − 1.200 = 0.027 NM. Since 1 NM = 1852 m, that is 0.027 × 1852 ≈ 50 m. The separation is less than the 75 m resolution cell, so the two echoes overlap in range. Increasing gain will brighten both echoes and the noise, but it cannot undo the pulse-length limitation. The single elongated echo is consistent with unresolved targets, not necessarily a single boat.
2. What change would actually separate the echoes, and what tradeoffs would that change create?
Read the worked answer
The direct fix is a shorter transmitted pulse. For example, a 0.05 microsecond pulse would give a theoretical resolution of about 7.5 m, enough to separate 50 m targets in range. The receiver bandwidth must be increased to match the shorter pulse; otherwise the receiver will smear the echo and give back much of the gain.
The tradeoffs are significant. A shorter pulse carries less energy, so detection range and echo strength on weak or distant targets fall. A wider receiver bandwidth admits more noise, reducing the signal-to-noise ratio. The display spot size may still make very close echoes hard to read, and if the boats are also close in bearing, antenna beamwidth can merge them even when range resolution is adequate.
The operator should not expect a different range scale to solve the problem by itself. Changing from 6 NM to 3 NM may enlarge the display and make the merged echo easier to see, but it does not change the 75 m pulse-length limit. The correct operational answer is to use a shorter pulse if the radar provides one, accept the detection tradeoff, and treat the merged echo as a possible two-target situation until visual or AIS data resolves it.
3. Calculate the maximum unambiguous range for the 1200 Hz PRF and explain whether second-time-around echoes are a concern on this 6 NM scale.
Read the worked answer
The maximum unambiguous range is Rmax = c / (2 × PRF).
Rmax = (3 × 10^8) / (2 × 1200) = 125,000 m.
Convert to nautical miles: 125,000 / 1852 ≈ 67.5 NM.
The radar is operating on a 6 NM scale, far below 67.5 NM, so second-time-around echoes are not a practical concern here. A target at 6 NM returns well before the next pulse is transmitted. The merged echo is therefore a range-resolution problem, not a PRF ambiguity problem. The calculation also shows that this PRF is relatively low; it favors unambiguous long-range performance rather than a very high pulse repetition rate.