# Block 9 Flashcards

Click a question to reveal the answer.

<div data-flashcards data-deck="block-9"></div>

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<summary><strong>1. Why is treating <span class="math notranslate nohighlight">\(N\)</span> raw samples as <span class="math notranslate nohighlight">\(N\)</span> independent samples wrong for navigation error?</strong></summary>
<div class="card-answer"><p>Consecutive navigation-error samples are not independent: the filter has memory measured in seconds, and a 0.1-s sampling interval gives you essentially the same number twice. Treating them as independent inflates your apparent confidence by orders of magnitude and produces test plans that fail audit immediately.</p></div>
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<summary><strong>2. Define the autocorrelation function and write its form for an AR(1)-like navigation error.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(R(\tau) = \mathbb{E}[e(t)\,e(t+\tau)]\)</span>. For exponentially-correlated errors: <span class="math notranslate nohighlight">\(R(\tau) = \sigma^2 e^{-|\tau|/T_{\rm corr}}\)</span>. At <span class="math notranslate nohighlight">\(\tau = 0\)</span> it equals the variance; at <span class="math notranslate nohighlight">\(\tau = T_{\rm corr}\)</span> it has fallen to <span class="math notranslate nohighlight">\(\sigma^2/e \approx 0.37\sigma^2\)</span>.</p></div>
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<summary><strong>3. What is the F-47 ANS HITL correlation time, and why does it matter for test planning?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(T_{\rm corr} \approx 15\)</span> s for horizontal navigation error in steady-state operation. It sets the floor on how long an accuracy-assessment run must be: <span class="math notranslate nohighlight">\(T_{\rm total} \geq 2\,T_{\rm corr}\,N_{\rm eff,\,req} = 2 \cdot 15 \cdot 300 = 9{,}000\)</span> s, or about 150 minutes.</p></div>
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<summary><strong>4. State the standard formula for <span class="math notranslate nohighlight">\(N_{\rm eff}\)</span> for an exponentially-correlated time series.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(N_{\rm eff} = T_{\rm total} / (2\, T_{\rm corr})\)</span>. The factor of 2 comes from the fact that each sample is correlated with neighbors both forward and backward in time, so each "independent look" occupies a window of <span class="math notranslate nohighlight">\(2 T_{\rm corr}\)</span> on the timeline.</p></div>
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<summary><strong>5. What is the 10% rule of thumb, and how does it compare to the standard formula?</strong></summary>
<div class="card-answer"><p>Take the ACF down to 10% of its peak; that lag <span class="math notranslate nohighlight">\(L \cdot dt = T_{\rm corr} \ln 10 \approx 2.30\, T_{\rm corr}\)</span> is the decimation interval that yields effectively-independent samples. The standard formula uses an implicit threshold of <span class="math notranslate nohighlight">\(\rho = e^{-2} \approx 13.5\%\)</span>, so the 10% rule yields an <span class="math notranslate nohighlight">\(N_{\rm eff}\)</span> about 13% smaller — a conservative estimate.</p></div>
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<summary><strong>6. Quick numeric: at 10 Hz with <span class="math notranslate nohighlight">\(T_{\rm corr} = 15\)</span> s, how long must an accuracy run be for <span class="math notranslate nohighlight">\(N_{\rm eff} \geq 300\)</span>?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(T_{\rm total} \geq 2 \cdot 15 \cdot 300 = 9{,}000\)</span> s = 150 min. The raw sample count at 10 Hz would be 90,000 — vastly more than 300 — but only 300 of those are independent. Sampling faster does not help; only longer runs increase <span class="math notranslate nohighlight">\(N_{\rm eff}\)</span>.</p></div>
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<summary><strong>7. Why is sampling faster (e.g., raising from 10 Hz to 100 Hz) ineffective for increasing <span class="math notranslate nohighlight">\(N_{\rm eff}\)</span>?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(N_{\rm eff}\)</span> depends only on <span class="math notranslate nohighlight">\(T_{\rm total}\)</span> and <span class="math notranslate nohighlight">\(T_{\rm corr}\)</span>; the sample rate cancels out. Faster sampling adds redundant measurements that all sit inside the same correlation window. To grow <span class="math notranslate nohighlight">\(N_{\rm eff}\)</span> you must run longer or reduce <span class="math notranslate nohighlight">\(T_{\rm corr}\)</span>; faster sampling alone gives you no new statistical information.</p></div>
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<summary><strong>8. What kind of statistic is "horizontal error <span class="math notranslate nohighlight">\(\leq 1\)</span> m at 95% confidence" — a tail or a mean?</strong></summary>
<div class="card-answer"><p>A tail. The right tool is the 95th-percentile of the empirical CDF of <span class="math notranslate nohighlight">\(e_H\)</span>: <span class="math notranslate nohighlight">\(e_{H,95} = \mathrm{prctile}(e_H, 95)\)</span>. A 95% CI on the mean tests a different question (where the mean lives) and is the wrong tool for this requirement, even if the language sounds similar.</p></div>
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<summary><strong>9. Why does the requirement use a tail rather than a mean?</strong></summary>
<div class="card-answer"><p>Because operational performance is dominated by worst-case events, not average events. A system with zero mean error and a fat tail can still violate operational requirements; a system with a small bias but a tight distribution can be perfectly acceptable. The 95th-percentile threshold says "bad days happen at most 5% of the time", which is operationally meaningful in a way the mean is not.</p></div>
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<summary><strong>10. Write the inertial drift error model and the OLS estimator for the slope.</strong></summary>
<div class="card-answer"><p>Model: <span class="math notranslate nohighlight">\(e_H(t) = a + b t + \varepsilon(t)\)</span>. OLS: stack <span class="math notranslate nohighlight">\(\mathbf{X} = [\mathbf{1}\;\mathbf{t}]\)</span> and solve <span class="math notranslate nohighlight">\(\hat{\boldsymbol{\beta}} = \mathbf{X} \backslash \mathbf{e}_H\)</span>. The slope <span class="math notranslate nohighlight">\(\hat{b}\)</span> in m/s converts to NM/hr by <span class="math notranslate nohighlight">\(\hat{b}_{\rm NM/hr} = \hat{b}_{\rm m/s} \cdot 3600/1852\)</span>.</p></div>
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<summary><strong>11. For Requirement 3, what statistic do you compare to the 1.0 NM/hr threshold — point estimate or CI bound?</strong></summary>
<div class="card-answer"><p>The <strong>upper 95% confidence bound</strong> on the slope: <span class="math notranslate nohighlight">\(b_{\rm upper} = \hat{b} + t_{N-2,\,0.975} \cdot \hat{\sigma}_b\)</span>. Even if the point estimate is below 1.0 NM/hr, a wide CI can leave the upper bound above the requirement, in which case the system fails. The CI is the test's honest statement of "what the data has ruled out".</p></div>
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<summary><strong>12. Why is the upper-bound rule conservative, and how does <span class="math notranslate nohighlight">\(N_{\rm eff}\)</span> tighten it?</strong></summary>
<div class="card-answer"><p>The upper-bound rule rejects systems whose drift could plausibly be above the requirement. As <span class="math notranslate nohighlight">\(N_{\rm eff}\)</span> grows, the OLS standard error <span class="math notranslate nohighlight">\(\hat{\sigma}_b\)</span> shrinks, the CI tightens, and the upper bound moves toward the point estimate. With enough data, the upper bound and the point estimate converge — and that is the only honest path to declaring a marginal system compliant.</p></div>
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<summary><strong>13. Why is detection time <span class="math notranslate nohighlight">\(t_D\)</span> not directly observable in the F-47 capstone?</strong></summary>
<div class="card-answer"><p>The SPO has directed that DT use only production-representative sensors and avionics-bus data. There is no Flight Test Instrumentation tap into the integrity-monitoring algorithm, so the internal "fault detected" flag is not available. <span class="math notranslate nohighlight">\(t_D\)</span> must be inferred from the externally-observable navigation-error time history.</p></div>
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<summary><strong>14. State the four steps of the peak-and-snapback algorithm.</strong></summary>
<div class="card-answer"><p>(1) Compute pre-fault baseline <span class="math notranslate nohighlight">\(e_{\rm base}\)</span> from a window ending just before <span class="math notranslate nohighlight">\(t_0\)</span>. (2) Find peak error <span class="math notranslate nohighlight">\(e_{\rm peak}\)</span> in the post-fault search window. (3) Set snapback threshold <span class="math notranslate nohighlight">\(e_{\rm thr} = e_{\rm base} + \alpha(e_{\rm peak} - e_{\rm base})\)</span> with <span class="math notranslate nohighlight">\(\alpha = 0.25\)</span>. (4) <span class="math notranslate nohighlight">\(t_D\)</span> is the first time after the peak at which <span class="math notranslate nohighlight">\(e_H \leq e_{\rm thr}\)</span> for at least 0.5 s.</p></div>
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<summary><strong>15. What is the role of the 0.5-second debounce in the snapback rule?</strong></summary>
<div class="card-answer"><p>It immunizes the algorithm against single-sample noise dips that happen to fall below threshold. Requiring the error to stay below threshold for at least 0.5 s ensures the system has actually snapped back, not just briefly oscillated through the threshold value. Without it, very short coincidental dips would produce false-low <span class="math notranslate nohighlight">\(t_D\)</span> readings.</p></div>
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<summary><strong>16. State the 27/30 rule and what it implies about per-event reliability.</strong></summary>
<div class="card-answer"><p>For 30 independent spoof events, at least 27 must satisfy <span class="math notranslate nohighlight">\(T_D \leq 5\)</span> s, <span class="math notranslate nohighlight">\(\mathrm{HMI}_H \leq 1\)</span> s, and <span class="math notranslate nohighlight">\(\mathrm{HMI}_V \leq 1\)</span> s. Three failures are tolerated. The implied per-event success probability is in the 0.90 to 0.95 range, computed from the binomial distribution: with <span class="math notranslate nohighlight">\(p = 0.95\)</span>, <span class="math notranslate nohighlight">\(P(\geq 27\,\mathrm{of}\,30) \approx 0.94\)</span>.</p></div>
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<summary><strong>17. Which dataset is used for which requirement?</strong></summary>
<div class="card-answer"><p><code>data_AllSource</code>: Requirement 1 (AllSource accuracy). <code>data_AltNav</code>: Requirement 2 (AltNav accuracy). <code>data_Inertial</code>: Requirement 3 (inertial drift). <code>data_Spoof</code>: Requirement 4 (fault detection / integrity). Critical: <code>data_Spoof</code> must NOT be used for accuracy assessment because the fault-injection events produce episodic high errors that inflate the 95th-percentile readout artificially.</p></div>
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