Lesson 13 Flashcards#
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1. Why is AoA the only "spatial" field in the PDW?
Every other field — frequency, pulse width, TOA, power — is a timing measurement on a single antenna and says what the emitter is. AoA depends on geometry and says where it is. It is also the only field that needs more than one antenna (or one receiver) to resolve.
2. What does AoA do for the receiver that the other fields cannot?
It deinterleaves — splitting emitters that share a band when their pulses overlap, since pulses from different bearings arrive at different angles — and it cues the aircrew and steers countermeasures at the right threat.
3. Why is AoA the hardest PDW field to measure well?
It is geometry, not timing, measured from a maneuvering aircraft with no scanning dish. And a single bearing is only a line, not a fix — a position needs two bearings crossed (L14).
4. What are the two physical handles a passive receiver has on bearing?
Amplitude — how loud the signal is in beams pointing different ways — and phase — how the wavefront arrives at antennas spaced apart. These give amplitude comparison and phase interferometry.
5. How does amplitude comparison find a bearing?
Several broad antennas are squinted in different directions so their patterns overlap; the classic RWR uses four, one per quadrant. A signal off boresight is louder in the nearer beam, and the power ratio between adjacent beams gives bearing within the overlap.
6. What are the strengths and the accuracy limit of amplitude comparison?
It is single-pulse, wideband, cheap, and degrades gracefully. Its accuracy is limited by how well each antenna pattern is known (calibration), landing within roughly \(10\text{–}30^\circ\) — enough to label a quadrant, not to target.
7. Why does an off-boresight wave produce a phase difference across an interferometer?
A plane wave arriving at angle \(\theta\) reaches the far element after traveling an extra path \(d\sin\theta\), where \(d\) is the baseline. That extra path is a measurable phase shift between the two channels.
8. State the interferometer relation and its inversion for bearing.
\(\Delta\phi = \frac{2\pi d}{\lambda}\sin\theta\), inverted to \(\theta = \arcsin\!\left(\frac{\lambda\,\Delta\phi}{2\pi d}\right)\). Measure one number, \(\Delta\phi\), and solve for the angle \(\theta\).
9. Why does a longer baseline give a finer bearing?
Angle accuracy scales like \(\lambda/(2\pi d)\): a fixed phase-measurement error maps to a smaller angle error as \(d/\lambda\) grows. Long baselines reach sub-degree accuracy, far past amplitude comparison.
10. What is the phase-wrap (ambiguity) problem?
A phase meter cannot tell \(\Delta\phi\) from \(\Delta\phi + 2\pi\) — phase is known only mod \(2\pi\), folding into \([-\pi,\pi]\). So several arrival angles share one reading, and the inversion can pick the wrong one.
11. What baseline keeps the bearing unambiguous over \(\pm 90^\circ\), and why?
\(d \le \lambda/2\). Unambiguity requires \(|\Delta\phi| \le \pi\); setting \(\sin\theta = 1\) in the relation gives \(d \le \lambda/2\). But that is a short baseline, so it is coarse — accuracy and unambiguity pull opposite ways.
12. How does a multi-baseline array resolve the ambiguity?
The short pair (\(d \le \lambda/2\)) gives a coarse, unambiguous bearing that predicts which \(2\pi\) cycle the long pair is in; unwrap the long pair with that cycle and invert for a fine bearing. Short pair picks the cycle; long pair sets the precision.
13. Compare amplitude comparison and interferometry on accuracy and failure mode.
Amplitude comparison reads a power ratio — coarse (\(\sim 10\text{–}30^\circ\)), fails on pattern-calibration error. Interferometry reads phase across a baseline — fine (\(<1^\circ\)), fails on phase wrap. Both are passive and single-pulse; systems often amplitude-compare for the quadrant, then interferometer-refine.