Block 3 Flashcards

Click a question to reveal the answer.

1. Why fuse multiple sensors instead of picking the best one?

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.

2. Write the weighted-average estimate of two measurements with the constraint on the weights.

\(\hat{x} = w_1 z_1 + w_2 z_2\) subject to \(w_1 + w_2 = 1\). The constraint keeps the estimate unbiased when both measurements are unbiased.

3. Why is a simple unweighted average a bad choice when sensor uncertainties differ?

An unweighted average gives every sensor equal influence. If \(\sigma_1 \ll \sigma_2\), 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.

4. Write the optimal (inverse-variance) weights for two-sensor fusion.

\(w_1 = \dfrac{1/\sigma_1^2}{1/\sigma_1^2 + 1/\sigma_2^2}\) and \(w_2 = \dfrac{1/\sigma_2^2}{1/\sigma_1^2 + 1/\sigma_2^2}\). Lower-variance sensors get larger weights.

5. Write the optimal fused estimate in compact form.

\(\hat{x}_{\text{opt}} = \dfrac{z_1/\sigma_1^2 + z_2/\sigma_2^2}{1/\sigma_1^2 + 1/\sigma_2^2}\). Multiply each measurement by its information, sum, and normalize by the total information.

6. Write the variance of the optimal fused estimate.

\(\sigma_{\text{opt}}^2 = \left(\dfrac{1}{\sigma_1^2} + \dfrac{1}{\sigma_2^2}\right)^{-1}\). The fused variance is always strictly smaller than the smaller of the two input variances.

7. Define information and write the rule for combining it across sensors.

Information is the inverse of variance: \(I = 1/\sigma^2\). Independent measurements add information: \(I_{\text{total}} = I_1 + I_2 + \cdots + I_n\). The fused variance is the inverse of the total information.

8. Quick numeric: \(z_1 = 100,\ \sigma_1 = 2\), \(z_2 = 110,\ \sigma_2 = 6\). Compute \(\hat{x}_{\text{opt}}\).

\(I_1 = 0.25,\ I_2 = 1/36 \approx 0.0278,\ I_{\text{total}} \approx 0.2778\). \(\hat{x}_{\text{opt}} = (100\cdot 0.25 + 110\cdot 0.0278)/0.2778 \approx 101.0\). The estimate is pulled toward the lower-variance sensor.

9. Same numbers: what is \(\sigma_{\text{opt}}\)?

\(\sigma_{\text{opt}} = 1/\sqrt{I_{\text{total}}} \approx 1/\sqrt{0.2778} \approx 1.897\). Smaller than \(\sigma_1 = 2\), so adding the noisy sensor still bought you a small improvement.

10. What changes if the two sensor errors are correlated rather than independent?

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 \(\boldsymbol{\Sigma}^{-1}\). We will see this exact pattern in the multi-state Kalman filter in Block 5.

11. State the Kalman filter measurement update in scalar form.

\(\hat{x}_k^+ = \hat{x}_k^- + K_k (z_k - \hat{x}_k^-)\) with the Kalman gain \(K_k = P_k^- / (P_k^- + R_k)\). The innovation \(z_k - \hat{x}_k^-\) is the new information; \(K_k\) is the optimal-fusion weight on it.

12. What does the Kalman gain do when the measurement is much more uncertain than the prediction, and vice versa?

If \(R_k \gg P_k^-\), then \(K_k \to 0\): the filter trusts the prediction and largely ignores the measurement. If \(P_k^- \gg R_k\), then \(K_k \to 1\): the filter snaps the estimate onto the measurement. Everywhere in between, it blends the two using the optimal trust factor.

13. Why is recursive fusion (one measurement at a time) equivalent to fusing all measurements at once?

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.

14. In the radar-altimeter / barometer example with \(\sigma_R = 2\) m and \(\sigma_B = 10\) m, what fraction of the fused estimate comes from each sensor?

Information ratio: \(I_R / I_B = (\sigma_B/\sigma_R)^2 = 25\). So \(w_R = 25/26 \approx 0.9615\) and \(w_B = 1/26 \approx 0.0385\). The radar dominates by a factor of 25 to 1.

15. What is the fused variance for the radar/baro example, and how much better is it than the radar alone?

\(\sigma_{\text{opt}}^2 = 1/(1/4 + 1/100) \approx 3.846\), so \(\sigma_{\text{opt}} \approx 1.961\) m. The radar alone is \(\sigma = 2\) 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.