Exercises — Gyroscope — angular velocity measurement, bias, noise

A constant offset integrated over time is a ramp — the classic $\int bd\tau =bt$. So the error grows linearly in time. (Noise, not bias, is the one that grows as $\sqrt{t}$.) Answer: linearly, $\propto t$.

• ARW is zero-mean high-frequency noise. Averaging more samples lets the random shoves cancel → you can reduce it (this is the slope $-1/2$ region of the Allan curve, see Allan Variance Analysis).
• Bias is a low-frequency offset with a non-zero mean; averaging a constant just returns the constant. You cannot average it away. Answer: ARW is averageable; bias is not.

Step 1 (WHAT): use the bias law $\Delta \theta =b\cdot t$. WHY: a constant offset integrated is a ramp. Step 2: $\Delta \theta =0.04°/\text{s}\times 90\text{s}=3.6°$. Answer: $\boxed{3.6°}$. A "tiny" $0.04°/\text{s}$ becomes a real error in a minute and a half — this is why an Inertial Measurement Unit (IMU) fuses in GPS or a magnetometer.

Step 1 (WHAT): use ${\sigma }_{\theta }=\text{ARW}\cdot \sqrt{t}$ with $t$ in hours. WHY: integrated white noise is a random walk whose spread grows as $\sqrt{t}$. (a) ${\sigma }_{\theta }=0.6\times \sqrt{1}=0.6°$. (b) ${\sigma }_{\theta }=0.6\times \sqrt{9}=0.6\times 3=1.8°$. Answer: $0.6°$ and $1.8°$. Note it took $9\times$ the time to only triple the error — the signature of $\sqrt{t}$.

Step 1 (WHAT + WHY): we use ${\sigma }_{\theta }=\text{ARW}\cdot \sqrt{t}$. WHY the $\sqrt{t}$: the noise is a stream of independent little random kicks; integrating them is a random walk, and a random walk's spread grows as the square root of the number of steps (equivalently, of elapsed time) because the kicks partly cancel rather than all pushing the same way. So the same $\sqrt{t}$ law from 2.2 applies here — the only new work is a unit conversion. Step 2: convert time to hours: $100\text{s}=100/3600\text{h}=0.02\overline{7}\text{h}$. Step 3: ${\sigma }_{\theta }=0.6\times \sqrt{0.02\overline{7}}=0.6\times 0.1667\approx 0.10°$. Answer: $\approx 0.10°$. Small, because $\sqrt{t}$ is gentle over short times.

Step 1 (set them equal, WHAT): we want $bt=\text{ARW}\sqrt{t/3600}$, with $t$ in seconds on the left and the right side's $t$ converted to hours. Write $\text{ARW}$-in-per-root-second: $0.5°/\sqrt{\text{h}}=0.5/\sqrt{3600}°/\sqrt{\text{s}}=0.5/60=0.008\overline{3}°/\sqrt{\text{s}}$. Step 2 (solve): $bt=k\sqrt{t}$ with $k=0.0083\overline{3}$. Divide by $\sqrt{t}$: $b\sqrt{t}=k\Rightarrow \sqrt{t}=\frac{k}{b}=\frac{0.008333}{0.005}=1.667\Rightarrow t_c=1.667^2\approx 2.78\text{s}.$ Step 3 (interpret, WHY): below $\sim 2.8\text{s}$ the $\sqrt{t}$ noise term is bigger (it dominates instantly, then flattens); above it the linear bias ramp overtakes and grows without bound. Answer: $t_c\approx 2.8\text{s}$; for $t<t_c$ noise dominates, for $t>t_c$ bias dominates. See the crossover figure at the top.

• Slope $-1/2$: Angle Random Walk (white noise) — averaging longer helps, error falls.
• Flat bottom (slope $0$): bias instability — the floor you cannot average past.
• Slope $+1/2$: rate random walk / slowly drifting bias — averaging hurts, drift wins. This bathtub shape (see Allan Variance Analysis) is how one plot separates the two enemies from the parent note. Link to the underlying stochastic model: Random Walk & Wiener Process.

(a) Bias offset: $bt=0.01\times 60=0.6°$. (b) ARW spread: $t=60/3600=0.0167\text{h}$, so ${\sigma }_n=0.4\times \sqrt{0.0167}=0.4\times 0.1291\approx 0.0516°$. (c) Quadrature sum (WHY quadrature): two independent error sources have variances that add, so standard deviations add in quadrature: ${\sigma }_{\text{tot}}=\sqrt{(0.6{)}^2+(0.0516{)}^2}=\sqrt{0.36+0.00266}\approx \sqrt{0.3627}\approx 0.602°.$ Answer: $(a)0.6°$, $(b)0.052°$, $(c)0.602°$. The bias utterly dominates the budget — a Kalman Filter would spend its effort estimating $b$, not the noise.

Step 1 (WHAT): the bias is now $b(t)=0.01+\frac{0.02-0.01}{60}t=0.01+0.0001667t$ (°/s). We integrate it: $\Delta \theta ={\int }_0^{60}b(t)dt.$ Step 2 (WHY integrate): heading is the time-integral of rate, and the false rate here is $b(t)$. Step 3 (evaluate): for a linearly varying bias the integral equals the average bias times the time: $\bar{b}=\frac{0.01+0.02}{2}=0.015°/\text{s},\Delta \theta =0.015\times 60=0.9°.$ Answer: $0.9°$, versus $0.6°$ for the constant $0.01°/\text{s}$ bias — the drift added $0.3°$. This is precisely why a "measure bias once and subtract" strategy fails: bias moves during the run.

Step 1 (which error dominates?): compare the two growth rates near the deadline. Bias grows linearly and, as L3 showed, quickly overtakes ARW. Estimate the crossover: convert ARW to $°/\sqrt{\text{s}}$: $0.5/60=0.00833°/\sqrt{\text{s}}$. Crossover $\sqrt{t_c}=k/b=0.00833/0.02=0.4167\Rightarrow t_c=0.174\text{s}$. So beyond a fraction of a second bias dominates completely — we can size $T$ from bias alone and treat ARW as a small correction. Step 2 (bias-only bound, WHAT): require $bT\le 1.0°$: $T\le \frac{1.0}{0.02}=50\text{s}.$ Step 3 (check with quadrature, WHY): at $T=50\text{s}$, ARW spread $=0.5\sqrt{50/3600}=0.5\times 0.1179=0.0589°$. Combine with bias in quadrature: ${\sigma }_{\text{tot}}=\sqrt{(0.02\times 50{)}^2+0.0589^2}=\sqrt{1.0+0.00347}=1.0017°>1.0°.$ That just exceeds the spec. Back off slightly: solve $\sqrt{(0.02T{)}^2+(0.5\sqrt{T/3600}{)}^2}=1.0$. The bias term overwhelmingly sets the scale, so $T\approx 49.9\text{s}$. Answer: reset at least every ~50 s (49.9 s to include the noise margin). Practically you'd reset well inside this — say every 10 s — for safety headroom.

First, what is $Q$? $n(\tau )$ is the gyro's random noise in the rate reading (units $\text{rad/s}$). The symbol $\mathbb{E}[\cdot ]$ means "average over many runs." The quantity $Q$ is the noise power spectral density — physically, how much noise power sits in each hertz of bandwidth, with units $(\text{rad/s}{)}^2/\text{Hz}$. A bigger $Q$ means a shakier gyro. The white-noise property $\mathbb{E}[n(\tau )n({\tau }^{\prime })]=Q\delta (\tau -{\tau }^{\prime })$ says two noise samples at different instants are uncorrelated (the spike $\delta$ is non-zero only when $\tau ={\tau }^{\prime }$), and the height of that correlation is exactly $Q$. Step 1 (WHAT): the variance of an integral is a double integral of the covariance: $\mathrm{Var}[\theta ]=\mathbb{E}\left[{\int }_0^{t}nd\tau {\int }_0^{t}nd{\tau }^{\prime }\right]={\int }_0^t{\int }_0^t\mathbb{E}[n(\tau )n({\tau }^{\prime })]d\tau d{\tau }^{\prime }.$ Step 2 (use white noise, WHY the delta): substitute $\mathbb{E}[n(\tau )n({\tau }^{\prime })]=Q\delta (\tau -{\tau }^{\prime })$. The delta "fires" only when $\tau ={\tau }^{\prime }$, collapsing one integral: $\mathrm{Var}[\theta ]={\int }_0^{t}Qd\tau =Qt.$ Step 3 (finish, and connect to ARW): ${\sigma }_{\theta }=\sqrt{Qt}=\sqrt{Q}\cdot \sqrt{t}$. Compare this with the growth law ${\sigma }_{\theta }=\text{ARW}\cdot \sqrt{t}$ used all over this page: matching the two term-by-term shows the ARW coefficient is $\sqrt{Q}$. So the datasheet's ARW number is just the square root of the gyro's noise power density, re-expressed in $°/\sqrt{\text{h}}$. That is where the whole $\sqrt{t}$ law — and the ARW spec — comes from. See Random Walk & Wiener Process.

Bias error grows as? ::: linearly in time, $bt$ ARW spread grows as? ::: $\sqrt{t}$, i.e. $\text{ARW}\cdot \sqrt{t}$ with $t$ in hours Convert ARW from $°/\sqrt{\text{h}}$ to $°/\sqrt{\text{s}}$? ::: divide by $\sqrt{3600}=60$ Two independent errors combine by? ::: quadrature (variances add), $\sqrt{{\sigma }_1^2+{\sigma }_2^2}$ Which error do you calibrate first? ::: bias — it dominates over seconds and up, growing linearly without bound, so removing it gives the biggest accuracy win before you touch the noise