# Block 3 Flashcards

Click a question to reveal the answer.

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<summary><strong>1. Why fuse multiple sensors instead of picking the best one?</strong></summary>
<div class="card-answer"><p>Two independent measurements of the same quantity always carry more information than one. By combining them with the right weights you produce an estimate with smaller variance than either input. Picking only the best sensor throws away the information in the other.</p></div>
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<summary><strong>2. Write the weighted-average estimate of two measurements with the constraint on the weights.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(\hat{x} = w_1 z_1 + w_2 z_2\)</span> subject to <span class="math notranslate nohighlight">\(w_1 + w_2 = 1\)</span>. The constraint keeps the estimate unbiased when both measurements are unbiased.</p></div>
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<summary><strong>3. Why is a simple unweighted average a bad choice when sensor uncertainties differ?</strong></summary>
<div class="card-answer"><p>An unweighted average gives every sensor equal influence. If <span class="math notranslate nohighlight">\(\sigma_1 \ll \sigma_2\)</span>, letting the noisy sensor pull on the estimate as hard as the quiet one inflates the fused variance unnecessarily. Optimal weights respect the relative quality of the measurements.</p></div>
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<summary><strong>4. Write the optimal (inverse-variance) weights for two-sensor fusion.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(w_1 = \dfrac{1/\sigma_1^2}{1/\sigma_1^2 + 1/\sigma_2^2}\)</span> and <span class="math notranslate nohighlight">\(w_2 = \dfrac{1/\sigma_2^2}{1/\sigma_1^2 + 1/\sigma_2^2}\)</span>. Lower-variance sensors get larger weights.</p></div>
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<summary><strong>5. Write the optimal fused estimate in compact form.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(\hat{x}_{\text{opt}} = \dfrac{z_1/\sigma_1^2 + z_2/\sigma_2^2}{1/\sigma_1^2 + 1/\sigma_2^2}\)</span>. Multiply each measurement by its information, sum, and normalize by the total information.</p></div>
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<summary><strong>6. Write the variance of the optimal fused estimate.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(\sigma_{\text{opt}}^2 = \left(\dfrac{1}{\sigma_1^2} + \dfrac{1}{\sigma_2^2}\right)^{-1}\)</span>. The fused variance is always strictly smaller than the smaller of the two input variances.</p></div>
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<summary><strong>7. Define information and write the rule for combining it across sensors.</strong></summary>
<div class="card-answer"><p>Information is the inverse of variance: <span class="math notranslate nohighlight">\(I = 1/\sigma^2\)</span>. Independent measurements add information: <span class="math notranslate nohighlight">\(I_{\text{total}} = I_1 + I_2 + \cdots + I_n\)</span>. The fused variance is the inverse of the total information.</p></div>
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<summary><strong>8. Quick numeric: <span class="math notranslate nohighlight">\(z_1 = 100,\ \sigma_1 = 2\)</span>, <span class="math notranslate nohighlight">\(z_2 = 110,\ \sigma_2 = 6\)</span>. Compute <span class="math notranslate nohighlight">\(\hat{x}_{\text{opt}}\)</span>.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(I_1 = 0.25,\ I_2 = 1/36 \approx 0.0278,\ I_{\text{total}} \approx 0.2778\)</span>. <span class="math notranslate nohighlight">\(\hat{x}_{\text{opt}} = (100\cdot 0.25 + 110\cdot 0.0278)/0.2778 \approx 101.0\)</span>. The estimate is pulled toward the lower-variance sensor.</p></div>
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<summary><strong>9. Same numbers: what is <span class="math notranslate nohighlight">\(\sigma_{\text{opt}}\)</span>?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(\sigma_{\text{opt}} = 1/\sqrt{I_{\text{total}}} \approx 1/\sqrt{0.2778} \approx 1.897\)</span>. Smaller than <span class="math notranslate nohighlight">\(\sigma_1 = 2\)</span>, so adding the noisy sensor still bought you a small improvement.</p></div>
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<summary><strong>10. What changes if the two sensor errors are correlated rather than independent?</strong></summary>
<div class="card-answer"><p>Correlation reduces the effective information in the second measurement. The optimal fusion rule still exists, but the inverse-variance form is replaced by a more general inverse-covariance form using <span class="math notranslate nohighlight">\(\boldsymbol{\Sigma}^{-1}\)</span>. We will see this exact pattern in the multi-state Kalman filter in Block 5.</p></div>
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<summary><strong>11. State the Kalman filter measurement update in scalar form.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(\hat{x}_k^+ = \hat{x}_k^- + K_k (z_k - \hat{x}_k^-)\)</span> with the Kalman gain <span class="math notranslate nohighlight">\(K_k = P_k^- / (P_k^- + R_k)\)</span>. The innovation <span class="math notranslate nohighlight">\(z_k - \hat{x}_k^-\)</span> is the new information; <span class="math notranslate nohighlight">\(K_k\)</span> is the optimal-fusion weight on it.</p></div>
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<summary><strong>12. What does the Kalman gain do when the measurement is much more uncertain than the prediction, and vice versa?</strong></summary>
<div class="card-answer"><p>If <span class="math notranslate nohighlight">\(R_k \gg P_k^-\)</span>, then <span class="math notranslate nohighlight">\(K_k \to 0\)</span>: the filter trusts the prediction and largely ignores the measurement. If <span class="math notranslate nohighlight">\(P_k^- \gg R_k\)</span>, then <span class="math notranslate nohighlight">\(K_k \to 1\)</span>: the filter snaps the estimate onto the measurement. Everywhere in between, it blends the two using the optimal trust factor.</p></div>
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<summary><strong>13. Why is recursive fusion (one measurement at a time) equivalent to fusing all measurements at once?</strong></summary>
<div class="card-answer"><p>Because for independent measurements, information adds. Adding two measurements simultaneously and adding them one-after-the-other produce the same total information and therefore the same fused variance. Recursion is just an implementation that uses constant memory, which is what makes the Kalman filter practical for real-time navigation.</p></div>
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<summary><strong>14. In the radar-altimeter / barometer example with <span class="math notranslate nohighlight">\(\sigma_R = 2\)</span> m and <span class="math notranslate nohighlight">\(\sigma_B = 10\)</span> m, what fraction of the fused estimate comes from each sensor?</strong></summary>
<div class="card-answer"><p>Information ratio: <span class="math notranslate nohighlight">\(I_R / I_B = (\sigma_B/\sigma_R)^2 = 25\)</span>. So <span class="math notranslate nohighlight">\(w_R = 25/26 \approx 0.9615\)</span> and <span class="math notranslate nohighlight">\(w_B = 1/26 \approx 0.0385\)</span>. The radar dominates by a factor of 25 to 1.</p></div>
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<summary><strong>15. What is the fused variance for the radar/baro example, and how much better is it than the radar alone?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(\sigma_{\text{opt}}^2 = 1/(1/4 + 1/100) \approx 3.846\)</span>, so <span class="math notranslate nohighlight">\(\sigma_{\text{opt}} \approx 1.961\)</span> m. The radar alone is <span class="math notranslate nohighlight">\(\sigma = 2\)</span> m, so fusion buys about 0.04 m of additional precision. When one sensor is much better than the other, the marginal value of the worse sensor is small but non-zero.</p></div>
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