Lesson 35 Flashcards — Pulse-Doppler Radar#
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1. Define pulse width, PRF, and PRI.
Pulse width \(\tau\) is how long the transmitter is on. PRI is the time from the start of one pulse to the start of the next. PRF is pulses per second, with \(PRI = 1/PRF\).
2. How does a radar measure range?
\(R = c\,\Delta t/2\) from the round-trip time. The factor of 2 is because the pulse travels out and back.
3. Give the range resolution.
\(\Delta R = c\tau/2\) — set by pulse width. Two targets closer than \(\Delta R\) merge into one return.
4. A 1 µs pulse gives what range resolution?
\((3\times10^8)(1\times10^{-6})/2 = 150\) m. Two aircraft within 150 m appear as a single target.
5. Give the maximum unambiguous range.
\(R_\text{unamb} = c\,PRI/2\) — set by PRF. The radar must hear the echo before it sends the next pulse. It is the equation sheet's Unambig Range row.
6. What happens to a target beyond the unambiguous range?
Its echo arrives after the next pulse goes out, so the radar times it from the wrong pulse and reports a ghost at the true range minus \(R_\text{unamb}\). With \(R_\text{unamb} = 15\) km, a target at 18 km is reported at 3 km.
7. What does each waveform setting cost?
A short pulse gives fine resolution but carries less energy, so detection range drops. A low PRF gives a long unambiguous range but puts fewer pulses on the target each second. They are two separate trades, and real radars change both by mode.
8. Write the Doppler equation.
\(f_\text{return} = f_\text{original}(1 \pm 2v\cos\theta/c)\) — plus for a closing target, minus for an opening one. \(v\) is the target's speed in m/s, and \(\theta\) is the acute angle between its velocity and the radar's line of sight.
9. Why does the factor of 2 appear in the Doppler equation?
For the same reason as in range: the path is two-way.
10. What Doppler shift does a target flying across the beam produce?
None. At \(\theta = 90^\circ\), \(\cos\theta = 0\) and there is no shift at all.
11. What is the Doppler shift, and how is a return frequency reported?
The Doppler shift is the change in frequency, \(\Delta f = 2v\cos\theta\,f_\text{original}/c\), in Hz. Add it for a closing target, subtract it for an opening one, and report \(f_\text{return}\) to the nearest hertz — the shift is only a few kHz on a GHz carrier.
12. How do you find a target's speed and direction from its return frequency?
A return above \(f_\text{original}\) means the target is closing; below means opening. The speed is \(v = c\,|\Delta f|/(2f_\text{original}\cos\theta)\). The answer is a speed and a direction, such as 125 m/s, opening.
13. A target's speed is given in km/h. What do you do first?
Divide by 3.6 to get m/s, because 1 km/h is 1000 m in 3600 s. A target at 1440 km/h is moving at 400 m/s.