Exercises — GPS — pseudorange, trilateration, dilution of precision

It is the pseudorange $\rho$. The receiver clock is off by an unknown bias $b$, so its $t_{rx}$ is wrong by the same amount for every satellite. The separator is the term $cb$: $\rho =r+cb.$ The true range $r$ is the honest geometric distance $\sqrt{(x-x_i{)}^2+(y-y_i{)}^2+(z-z_i{)}^2}$; the pseudorange adds the clock error converted to metres.

Four unknowns: $x,y,z$ (position) and $b$ (clock bias). Four unknowns need four independent equations, so the minimum is 4 satellites. Three would pin the geometry only if the clock were perfect — it is not.

$cb=(3\times 10^8)(2\times 10^{-6})=600\text{m}.$ Why this tool? $b$ is a time; distance = speed × time, so multiplying by $c$ converts the clock error into the metres it corrupts every measurement by. The same $600$ m sits on every satellite's pseudorange.

Clock offset in metres: $cb=(3\times 10^8)(-1.5\times 10^{-6})=-450\text{m}$. $r_i={\rho }_i-cb={\rho }_i-(-450)={\rho }_i+450.$

• $r_1=22,000,000+450=22,000,450\text{m}$.
• $r_2=20,500,000+450=20,500,450\text{m}$. Why add here? A slow clock makes $t_{rx}$ read too small ⇒ underestimated distance ⇒ we must add back the missing $450$ m. The same correction hits both — the shared-bias insight.

Between the satellites, $|x-0|=x$ and $|x-100|=100-x$. So: ${\rho }_A=cb+x=340,{\rho }_B=cb+(100-x)=280.$ Subtract: ${\rho }_A-{\rho }_B=x-(100-x)=2x-100=60$, giving $x=80\text{m}$. Back-substitute: $cb=340-80=260\text{m}$, i.e. $b=260/(3\times 10^8)\approx 8.67\times 10^{-7}\text{s}$. Why two equations? Two unknowns ($x$ and $b$) need two satellites — the 1D echo of "4 satellites for 4 unknowns."

${\sigma }_{\text{pos}}=\text{PDOP}\times {\sigma }_{\text{UERE}}=2.5\times 4=10\text{m}.$ Why multiply? DOP is the geometric amplification factor (from the satellite directions); UERE is the raw per-measurement noise (from the signals). Final error = geometry × noise, and they are independent factors.

$G=\left[\begin{array}{cc}1& 0\\ 0.5& 0.866\end{array}\right],G^{\top }G=\left[\begin{array}{cc}{1}^2+0.5^2& 1\cdot 0+0.5\cdot 0.866\\ \cdot & 0^2+0.866^2\end{array}\right]=\left[\begin{array}{cc}1.25& 0.433\\ 0.433& 0.75\end{array}\right].$ Determinant: $\det =1.25\cdot 0.75-0.433^2=0.9375-0.1875=0.75$. Inverse: $Q=\frac{1}{0.75}\left[\begin{array}{cc}0.75& -0.433\\ -0.433& 1.25\end{array}\right]=\left[\begin{array}{cc}1.0& -0.577\\ -0.577& 1.667\end{array}\right]$. Trace of $Q$: $1.0+1.667=2.667$. So $\text{DOP}=\sqrt{2.667}\approx 1.63$. Why the trace? The diagonal entries of $Q$ are the per-axis variance multipliers; summing them and taking a square root gives the combined position DOP.

$\cos 10^{\circ }\approx 0.9848,\sin 10^{\circ }\approx 0.1736$. $G^{\top }G=\left[\begin{array}{cc}1+0.9848^2& 0.9848\cdot 0.1736\\ 0.1710& 0.1736^2\end{array}\right]=\left[\begin{array}{cc}1.9698& 0.1710\\ 0.1710& 0.0301\end{array}\right].$ $\det =1.9698\cdot 0.0301-0.1710^2=0.05929-0.02924=0.03015={\sin }^{2}10^{\circ }$. Trace shortcut — why $\text{tr}Q=\frac{\text{tr}(G^{\top }G)}{\det }$ for a $2\times 2$: For any $2\times 2$ matrix $M=\left[\begin{array}{cc}a& b\\ b& d\end{array}\right]$, the inverse is $M^{-1}=\frac{1}{\det }\left[\begin{array}{cc}d& -b\\ -b& a\end{array}\right]$. Its diagonal entries are $d/\det$ and $a/\det$, so their sum is $\frac{a+d}{\det }=\frac{\text{tr}(M)}{\det }$. That is only a $2\times 2$ convenience (the swap-diagonal inverse); in higher dimensions you must actually invert. Applying it here: $\text{tr}Q=\frac{\text{tr}(G^{\top }G)}{\det }=\frac{1.9698+0.0301}{0.03015}=\frac{1.9999}{0.03015}\approx \mathrm{66.3.}$ $\text{DOP}=\sqrt{66.3}\approx 8.14$ — versus $1.63$ at $60^{\circ }$. Why? $\det (G^{\top }G)={\sin }^2(\theta )$ for two unit directions $\theta$ apart. As $\theta \to 0$ the directions become parallel, $\det \to 0$, $Q$ entries blow up — the spheres cut at a glancing angle and errors smear. Wide spread = small DOP = good.

Sign convention first (flag it explicitly, 3B1B-style). We follow the parent's definitions exactly, so no sign is left to guess:

• Residual: $\Delta {\rho }_i={\rho }_i^{\text{measured}}-{\rho }_i^{(0)}$ (measured minus predicted). A positive residual means the measured pseudorange was longer than the guess predicted.
• Line-of-sight entries: $e_{ix}=(x-x_i)/r_i$ — the parent's convention, pointing from satellite to receiver. Evaluated at the guess $(0,0)$.
• Clock column entry $=+1$, because $\partial {\rho }_i/\partial (cb)=+1$; a positive $c\Delta b$ lengthens every predicted range.

Compute the unit vectors at $(0,0)$. For $S_1=(0,10)$: ${\mathbf{e}}_1=(0-0,0-10)/10=(0,-1)$. For $S_2=(10,0)$: $(0-10,0-0)/10=(-1,0)$. For $S_3=(-8,-6)$, $r_3=\sqrt{64+36}=10$: ${\mathbf{e}}_3=(0-(-8),0-(-6))/10=(0.8,0.6)$. $G=\left[\begin{array}{ccc}0& -1& 1\\ -1& 0& 1\\ 0.8& 0.6& 1\end{array}\right].$ Key insight: an identical positive residual on every satellite looks exactly like a pure clock error — the column of 1's (all $+1$) absorbs it. The model $\Delta \rho =G\Delta \mathbf{x}$ with $\Delta \mathbf{x}=(0,0,0.30)$ gives $G\Delta \mathbf{x}=(0.30,0.30,0.30)$, matching the data with zero position change. So $\Delta x=0,\Delta y=0,c\Delta b=+0.30\text{km}=+300\text{m}$. The positive sign says "our clock guess made predictions too short by 300 m ⇒ correct the bias upward." Why is this the unique least-squares answer? $G$ is $3\times 3$ and invertible (satellites well spread), so the system is exact: $\Delta \mathbf{x}=G^{-1}\Delta \rho$ is unique, and we exhibited a solution — it must be the solution. A common signal on all channels is diagnosed as clock, not position — precisely the job of the 1's column.

Horizontal: ${\sigma }_H=\text{HDOP}\times {\sigma }_{\text{UERE}}=1.2\times 5=6\text{m}$. Vertical: ${\sigma }_V=\text{VDOP}\times {\sigma }_{\text{UERE}}=2.1\times 5=10.5\text{m}$. PDOP combines them: $\text{PDOP}=\sqrt{{\text{HDOP}}^2+{\text{VDOP}}^2}=\sqrt{1.44+4.41}=\sqrt{5.85}\approx 2.42$. Total: ${\sigma }_{\text{pos}}=\text{PDOP}\times {\sigma }_{\text{UERE}}=2.42\times 5\approx 12.1\text{m}$. Why does VDOP dominate? Satellites are all above you — none below the horizon — so the vertical direction is always poorly bracketed. That is the structural reason GPS altitude is worse than horizontal position.

If the four unit directions ${\mathbf{e}}_i$ all lie in one plane, they span only a 2D subspace of the 3D direction space. The columns of the position part of $G$ are then linearly dependent (rank $<3$), so $G^{\top }G$ is singular: $\det (G^{\top }G)=0$. Consequently $(G^{\top }G{)}^{-1}$ does not exist — formally the DOP $\to \infty$ along the direction perpendicular to the plane. That perpendicular coordinate is unobservable: no combination of these ranges constrains motion out of the plane. Physical picture: if every satellite lies on your horizon plane, nudging your altitude changes all four ranges by the same second-order amount only — first order it's invisible. You need at least one satellite well out of the common plane (e.g. overhead) to observe the third dimension. This is the continuous version of the L3 "angle $\to 0$" blow-up, now in 3D.

No. Two identical satellites contribute two identical rows to $G$. Identical rows are linearly dependent, so $G$ has rank at most $3$, and $G^{\top }G$ is singular ($\det =0$). You effectively have only 3 independent equations for 4 unknowns — underdetermined, exactly the failure mode of "3 satellites" from the parent note. Why "4 measurements" isn't enough: what matters is independent rows (distinct geometry), not the raw count. A duplicated measurement adds no new constraint. You must have four geometrically distinct line-of-sight directions.

Statement of the mechanism. DOP blows up in a coordinate when the corresponding column of $G$ is nearly parallel to (nearly a linear combination of) the other columns — that is the multi-column version of the L3 "two directions parallel ⇒ $\det \to 0$" story. The clock column is all ones. TDOP is large precisely when a shift in receiver position mimics a clock shift, i.e. when the all-ones column is nearly reproducible from the position columns.

Why common elevation triggers it. Each line-of-sight unit vector makes the same angle with the local vertical (same elevation). Its coupling to a vertical position change is therefore identical across all satellites — and a vertical move then changes every predicted range by nearly the same amount, which is exactly what a clock bias does (adds the same $c\Delta b$ to all ranges). So "you moved up/down" and "your clock drifted" produce almost the same residual pattern; the estimator cannot separate them, and the clock variance $q_{tt}$ (hence TDOP $=\sqrt{q_{tt}}$) inflates.

The cure. Mix high and low elevation satellites. A satellite directly overhead couples strongly to vertical position but weakly reproduces the flat common-mode clock signature; a satellite near the horizon does the opposite. That contrast makes the all-ones clock column no longer expressible from the position columns, breaks the position–clock ambiguity, and shrinks TDOP.

Why this is the right diagnosis (and not "just add satellites"): adding more satellites at the same elevation keeps the columns near-parallel and does nothing — the fix is angular diversity in elevation, matching the L3 lesson that spread, not count, controls DOP. See Covariance Propagation for how this ambiguity appears as an ill-conditioned $(G^{\top }G{)}^{-1}$.