Lesson 35 Flashcards

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1. How does fusing bearings over time differ from L14's fusion?

L14 fused many receivers at one instant, in one batch least-squares solve; the baseline was the distance between receivers. A single aircraft fuses its own bearings as they arrive, one at a time; the baseline is the distance the own-ship flies between them.

2. State the scalar fusion rule for two independent estimates.

Information adds: \(1/\sigma^2 = 1/\sigma_1^2 + 1/\sigma_2^2\). The fused estimate is the inverse-variance-weighted average, and its \(\sigma\) is smaller than either input's.

3. What is the Kalman gain in the scalar case, and what does it do?

\(K = \sigma_1^2/(\sigma_1^2 + \sigma_2^2)\), and the update is \(\hat x = x_1 + K(x_2 - x_1)\): move a fraction \(K\) of the way from the prior toward the measurement. \(K \to 1\) trusts the measurement; \(K \to 0\) keeps the prior.

4. What makes an estimator "recursive"?

Each fused estimate and its variance become the prior for the next measurement, so old measurements never need to be revisited — everything they said is already folded in. For a linear problem this reproduces the batch least-squares answer one measurement at a time.

5. Write the Kalman filter's predict step.

\(\hat{\mathbf x}^- = F\hat{\mathbf x}\) and \(P^- = FPF^\top + Q\): carry the estimate and its covariance forward with the motion model \(F\), adding process noise \(Q\) for motion the model does not describe.

6. Write the update step for one bearing.

\(S = HP^-H^\top + \sigma_\theta^2\), \(K = P^-H^\top/S\), \(\hat{\mathbf x} = \hat{\mathbf x}^- + K\nu\), \(P = (I - KH)P^-\), with the innovation \(\nu = \operatorname{wrap}(z - h(\hat{\mathbf x}^-))\). Each line is the scalar rule in matrix form.

7. What are the state, F and Q for a stationary emitter and for a moving one?

Stationary: \(\mathbf x = [N\;E]^\top\), \(F = I\), \(Q = 0\) — recursive nonlinear least squares. Constant velocity: \(\mathbf x = [N\;E\;v_N\;v_E]^\top\), with \(F\) moving the position by \(\mathbf v\,\Delta t\); this lesson keeps \(Q = 0\) because its emitter truly moves at constant velocity.

8. What is the "extended" part of an EKF for a bearing?

The bearing \(z = \text{atan2}(\Delta E, \Delta N)\) is nonlinear, so the filter uses its slope at the current estimate: \(H = [-\Delta E\;\;\Delta N]/R^2\), L33's row, zero-padded for velocity. It is evaluated at the estimate, never the truth.

9. Why must the bearing innovation be wrapped?

A predicted bearing of −179° and a measured 179° disagree by 2°, not 358°. Wrapping the innovation into [−180°, 180°) keeps a bearing near due south from producing a huge false correction.

10. Which way does one bearing shrink the error ellipse?

Only across the line of sight. \(H\), and so \(K\), point across the LOS, so the along-LOS axis is untouched: a round prior becomes a sliver along the line of sight. A second bearing from the same spot only makes the sliver thinner.

11. When is a stationary emitter's position observable from bearings alone?

From any own-ship leg that is not flown along the line of sight: each bearing from a new point is a new line of position, and as the bearing swings the lines cross. Flown straight at the emitter, every bearing is the same line — the seam — and range is unobservable.

12. Why is a constant-velocity emitter's range unobservable from a constant-velocity own-ship?

For any \(\alpha > 0\), the ghost track \(\mathbf p_o + \alpha(\mathbf p_e - \mathbf p_o)\) gives exactly the same bearings and still moves at constant velocity. Every ghost fits every bearing, so no accuracy and no update rate can pick the range.

13. What restores observability for a constant-velocity emitter?

An own-ship maneuver — a dogleg, or any change of heading or speed. After the turn the scaled ghosts are no longer straight, steady tracks, so only the true range fits every bearing. The bigger the change in bearing rate, the faster the ellipse collapses.

14. Why does maneuvering to observe cost exposure?

The turns that collapse the ellipse fastest swing the line of sight and close the range — usually a turn toward the threat, which spends time inside its ring and lowers \(P_S\). In the companion script the dogleg buys 4.3 km → 0.6 km for 143 s inside a 40 km ring.

15. Name three ways to get range on a pop-up without your own turn.

A wingman's simultaneous bearing from another position (an instant baseline, L14); an active range measurement (L33's radar row), at the cost of radiating; or accept the long ellipse and route around its whole length (L34).

16. How can an EKF fool itself on a bearings-only track?

It linearizes at its own estimate, so when that estimate is badly wrong its \(P\) can shrink along a direction the geometry never observed. The check is to compare the claimed \(\sqrt{\operatorname{tr} P}\) with the actual error, and to distrust a small ellipse on a geometry that should not produce one.