# Lesson 33 Flashcards

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<summary><strong>1. What does a single bearing tell you about a target's position — and what does it not?</strong></summary>
<div class="card-answer"><p>It pins the <em>cross-range</em> position — perpendicular to the line of sight — to about <span class="math notranslate nohighlight">\(R\sigma_\theta\)</span>, and says nothing at all about the along-range position. Slide the target along the line of sight and the bearing does not change.</p></div>
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<summary><strong>2. Write the Jacobian row of one bearing with respect to target position <span class="math notranslate nohighlight">\([N\ E]\)</span>.</strong></summary>
<div class="card-answer"><p>With <span class="math notranslate nohighlight">\(\theta = \mathrm{atan2}(\Delta E, \Delta N)\)</span>, the row is <span class="math notranslate nohighlight">\(\mathbf{h} = [-\Delta E,\ \Delta N]/R^2 = \mathbf{n}^\top/R\)</span>, where <span class="math notranslate nohighlight">\(\mathbf{n}\)</span> is the unit cross-range vector. It is the same row Project 2 called <span class="math notranslate nohighlight">\(\mathbf{H}_k\)</span>.</p></div>
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<summary><strong>3. State the Fisher information matrix for a set of bearing-only measurements.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(\mathbf{J} = \mathbf{H}^\top\mathbf{R}^{-1}\mathbf{H} = \sum_k \mathbf{n}_k\mathbf{n}_k^\top/(R_k^2\sigma_\theta^2)\)</span>, with <span class="math notranslate nohighlight">\(\sigma_\theta\)</span> in radians. Each bearing adds a rank-one piece; information from independent measurements simply adds.</p></div>
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<summary><strong>4. State the Cramér–Rao lower bound and what it means for an estimator.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(\operatorname{Cov}(\hat{\mathbf{p}}) \succeq \mathbf{J}^{-1}\)</span>: no unbiased estimator — however clever — can have a smaller error covariance in any direction. The geometry and the sensor noise set the floor; a good least-squares fix reaches it when the noise is small.</p></div>
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<summary><strong>5. Why can the CRLB be mapped before anyone takes a measurement?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(\mathbf{J}\)</span> depends only on the sensor positions, a hypothetical target position and <span class="math notranslate nohighlight">\(\sigma_\theta\)</span> — not on the measured bearings. Evaluate it at every point and you have a map of how well a network can fix a target anywhere, which is exactly what L14's DOP map and the L33 demo draw.</p></div>
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<summary><strong>6. How do you read the 1σ error ellipse from <span class="math notranslate nohighlight">\(\mathbf{J}\)</span>'s eigenvalues?</strong></summary>
<div class="card-answer"><p>The long semi-axis is <span class="math notranslate nohighlight">\(a = 1/\sqrt{\lambda_{\min}}\)</span> along the eigenvector of <span class="math notranslate nohighlight">\(\lambda_{\min}\)</span>; the short one is <span class="math notranslate nohighlight">\(b = 1/\sqrt{\lambda_{\max}}\)</span> along the other. The <em>small</em> eigenvalue makes the <em>long</em> axis: little information means a large error.</p></div>
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<summary><strong>7. What is DRMS, and how does it come from <span class="math notranslate nohighlight">\(\mathbf{J}\)</span>?</strong></summary>
<div class="card-answer"><p>The 1σ distance root-mean-square error, <span class="math notranslate nohighlight">\(\text{DRMS} = \sqrt{a^2 + b^2} = \sqrt{\operatorname{tr}\mathbf{J}^{-1}}\)</span> — the "Predicted RMS" tile in L14's demo. It adds the squares of the semi-axes, never the semi-axes themselves.</p></div>
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<summary><strong>8. What does the condition number <span class="math notranslate nohighlight">\(\kappa\)</span> of <span class="math notranslate nohighlight">\(\mathbf{J}\)</span> tell you?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(\kappa = \lambda_{\max}/\lambda_{\min} = (a/b)^2\)</span>. A value of 1 is a round error circle; a large value is a long, thin ellipse; <span class="math notranslate nohighlight">\(\kappa \to \infty\)</span> means one direction is unobservable — a seam.</p></div>
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<summary><strong>9. How exactly does L14's demo compute PDOP?</strong></summary>
<div class="card-answer"><p>It builds one row per receiver, scaled by the mean range: <span class="math notranslate nohighlight">\(\tilde{\mathbf{h}}_k = (\bar R/R_k)\,\mathbf{n}_k^\top\)</span>, and takes <span class="math notranslate nohighlight">\(\text{PDOP} = \sqrt{\operatorname{tr}((\tilde{\mathbf{H}}^\top\tilde{\mathbf{H}})^{-1})}\)</span>. The scaling strips out the overall range and leaves a pure geometry factor.</p></div>
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<summary><strong>10. Why does <span class="math notranslate nohighlight">\(\text{DRMS} = \text{PDOP}\cdot\bar R\cdot\sigma_\theta\)</span> hold exactly?</strong></summary>
<div class="card-answer"><p>Because <span class="math notranslate nohighlight">\(\tilde{\mathbf{H}}^\top\tilde{\mathbf{H}} = \bar R^2\sigma_\theta^2\,\mathbf{J}\)</span>: the scaled rows are the information rows times a constant. Inverting and taking the square root of the trace gives <span class="math notranslate nohighlight">\(\text{PDOP} = \sqrt{\operatorname{tr}\mathbf{J}^{-1}}/(\bar R\sigma_\theta)\)</span> — algebra, not an approximation.</p></div>
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<summary><strong>11. What is the PDOP of a symmetric pair at cut angle <span class="math notranslate nohighlight">\(\beta\)</span>, and what is its best value?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(\text{PDOP} = \sqrt{2}/\sin\beta\)</span>, so <span class="math notranslate nohighlight">\(\text{DRMS} = \sqrt{2}\,R\sigma_\theta/\sin\beta\)</span>. The best is <span class="math notranslate nohighlight">\(\sqrt{2} = 1.41\)</span> at a 90° cut; it is infinite at 0° or 180°.</p></div>
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<summary><strong>12. Where is the seam of two bearing-only sites?</strong></summary>
<div class="card-answer"><p>On the whole line through the two sites — beyond both ends (a 0° cut) <em>and</em> on the segment between them (a 180° cut). There both lines of sight are parallel or opposite, <span class="math notranslate nohighlight">\(\mathbf{J}\)</span> is rank one, and position along the line is unobservable.</p></div>
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<summary><strong>13. Why can better bearings not close a seam?</strong></summary>
<div class="card-answer"><p>A smaller <span class="math notranslate nohighlight">\(\sigma_\theta\)</span> scales every eigenvalue of <span class="math notranslate nohighlight">\(\mathbf{J}\)</span> up by the same factor, but zero times any factor is still zero. Only a measurement from another direction — a third site off the line, or a range measurement — adds information where the seam has none.</p></div>
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<summary><strong>14. What does a radar's range measurement add to <span class="math notranslate nohighlight">\(\mathbf{J}\)</span>?</strong></summary>
<div class="card-answer"><p>A term <span class="math notranslate nohighlight">\(\mathbf{u}\mathbf{u}^\top/\sigma_R^2\)</span> along the line of sight <span class="math notranslate nohighlight">\(\mathbf{u}\)</span> — exactly the direction bearings leave empty. Since <span class="math notranslate nohighlight">\(\mathbf{u} \perp \mathbf{n}\)</span>, one range-plus-bearing radar has a full-rank <span class="math notranslate nohighlight">\(\mathbf{J}\)</span> on its own: no seam.</p></div>
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<summary><strong>15. How does noise jamming change a radar network's ability to fix the jammer?</strong></summary>
<div class="card-answer"><p>It denies range and leaves each radar a strobe — a bearing only (L17). Radars crossing strobes form a bearing-only network, with bearing-only seams along the lines joining them; jamming reopens the seams that range had filled.</p></div>
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<summary><strong>16. What do we want from <span class="math notranslate nohighlight">\(\mathbf{J}\)</span> in each direction of the fight?</strong></summary>
<div class="card-answer"><p>Our receivers fixing their emitters (Project 2, Final Project T6): <span class="math notranslate nohighlight">\(\lambda_{\min}\)</span> large — open the cut, shorten the range, and pay for it in exposure. Their ES or IRST network fixing us or our jammer (Final Project T4): their <span class="math notranslate nohighlight">\(\lambda_{\min}\)</span> small along our route — ride their seams and keep the range long.</p></div>
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