Instantaneous Velocity and Speed — Question 7

Question 7

A shorter sampling interval can amplify position error

To estimate velocity at $t=1\ \mathrm{s}$, an instrument measures positions at $1-h$ and $1+h$ seconds and forms $$\widehat v_h=\frac{x_{\mathrm{meas}}(1+h)-x_{\mathrm{meas}}(1-h)}{2h}.$$ In all formulas here, $t,h$ are numerical seconds and $x$ numerical metres. For this design exercise the exact reference motion is $x(t)=2t+t^3$. Each position measurement equals its true value plus an independent unknown error in $[-0.01,0.01]\ \mathrm{m}$. Times are exact. These are hard bounds, not statistical errors.

Available half-widths are $h=0.50,0.20,0.10\ \mathrm{s}$ only. The design must guarantee absolute velocity error at most $0.10\ \mathrm{m/s}$ for every allowed pair of position errors.

1. Find the true velocity at 1 s by differentiation. Expand the centered quotient using exact positions and show its difference from the true velocity.

2. Bound the measurement-error contribution to $\widehat v_h$. Combine it with the exact model difference to obtain the worst possible absolute error for each available $h$.

3. Which available choices satisfy the guarantee, and which has the smallest worst-case error? Explain why choosing the shortest interval is not automatically best.

4. Does reducing $h$ to zero with this fixed position-error bound guarantee a more accurate experimental estimate? Distinguish the exact mathematical limit from the noisy measurement calculation.

Question 7 - Solution

A shorter sampling interval can amplify position error

(a) Compare the estimator with a known reference derivative.

Differentiation gives $v(t)=2+3t^2\ \mathrm{m/s}$, so $\boxed{v(1)=5\ \mathrm{m/s}}$. Expanding the two positions gives $$\frac{x(1+h)-x(1-h)}{2h}=\frac{10h+2h^3}{2h}=5+h^2\ \mathrm{m/s}.$$ The exact-position estimator therefore exceeds the true velocity by $h^2\ \mathrm{m/s}$ under the stated numerical-unit convention.

(b) Propagate hard bounds through the subtraction.

If the two errors are $e_+$ and $e_-$, their contribution is $(e_+-e_-)/(2h)$. Since $h>0$ and $e_+-e_-\in[-0.02,0.02]\ \mathrm{m}$, its magnitude is at most $0.01/h\ \mathrm{m/s}$. Hence the largest possible absolute total error is $$\boxed{E(h)=h^2+0.01/h\ \mathrm{m/s}}.$$ This bound is attained by choosing $e_+=+0.01$ and $e_-=-0.01\ \mathrm{m}$, reinforcing the positive model difference. $$\begin{aligned}E(0.50)&=0.25+0.02=0.27\ \mathrm{m/s},\\E(0.20)&=0.04+0.05=0.09\ \mathrm{m/s},\\E(0.10)&=0.01+0.10=0.11\ \mathrm{m/s}.\end{aligned}$$ Each error component and total have units m/s.

(c) Apply the guarantee to the allowed choices.

Only $\boxed{h=0.20\ \mathrm{s}}$ meets the 0.10 m/s requirement, and it has the smallest worst-case error among the three choices. Shortening the interval reduces the model difference but divides the position error by a smaller time separation.

(d) Exact limits and bounded measurement errors behave differently.

$$\lim_{h\to0^+}h^2=0,\qquad\lim_{h\to0^+}\frac{0.01}{h}=\infty.$$

With exact positions the model difference tends to zero as $h\to0$. With the fixed nonzero error bounds, the worst-case allowance $0.01/h$ grows without bound. A derivative’s mathematical existence does not make arbitrarily close noisy position samples arbitrarily accurate velocity measurements.