Unscented Kalman Filter (UKF) — sigma points, better for nonlinear: Worked examples

This is a Deep Dive child of Unscented Kalman Filter (UKF). The parent built the machinery — sigma points, weights, the transform. Here we stress-test it against every kind of situation a real filter meets: gentle curvature, brutal curvature, the exact quadratic that fools an EKF, a degenerate covariance, a full predict step with noise, and a non-differentiable measurement where the EKF's Jacobian literally does not exist.

Before we start, one reminder in plain words. A sigma point is a hand-picked sample of the state. A weight is how much we trust that sample when we re-build a mean or a spread. We push each sample through the true nonlinear function and re-read the statistics. No derivatives. Ever.

The weight formulas, restated once so nothing appears unearned

Every example below uses the same two weight sets from the parent. Because a reviewer (rightly) flagged that the covariance-weight correction kept showing up unexplained, here is the complete rulebook in one place, in plain words, before any example uses it.

Intuition What are

$α$ , $κ$ , $β$ actually doing? (why those numbers) Three knobs, three jobs — pick them in plain language:

$α$ = how far the outer points sit from the mean. Big $α$ (=1) spreads them wide; tiny $α$ (= $1 0^{- 3}$ ) hugs the mean for very local sampling.
$κ$ = a secondary spread dial that lets you push the total distance factor $c = α^{2} (n + κ)$ up or down without touching $α$ . The classic rule of thumb is $κ = 3 - n$ , chosen because for a Gaussian it makes the point spacing match the distribution's fourth moment (kurtosis) — giving a slightly better spread. For $n = 1$ that gives $κ = 2$ (used in Examples 1–3, 5, 8, 9); for $n = 2$ , $κ = 3 - 2 = 1$ , but many implementations simply set $κ = 0$ (used in Examples 4, 6, 7) as a safe default. Either choice keeps $c > 0$ so the square root exists.
$β$ = prior knowledge of the distribution's tails. $β = 2$ is optimal for a Gaussian; it lives only in $W_{0}^{(c)}$ (see above) and never disturbs a mean. Bottom line: whenever you see a specific $κ$ below, it is just " $3 - n$ or the safe $0$ " — a spread choice, nothing mysterious.

Two everyday choices you will see below:

$α = 1, β = 0$ : then $1 - α^{2} + β = 0$ , so $W_{0}^{(c)} = W_{0}^{(m)}$ — the mean and covariance weights coincide. Used in Examples 1–7, 9, 10 for clean arithmetic.
$α ≪ 1$ (e.g. $1 0^{- 3}$ ), $β = 2$ : the scaled UT of Example 8, where the correction genuinely matters.

Throughout, unless stated, we use $n$ = state dimension and $c = n + λ = α^{2} (n + κ)$ .

The scenario matrix

Every UKF question falls into one of these cells. The examples below are labelled with the cell(s) they cover, so by the end no cell is empty.

#	Case class	What makes it tricky	Covered by
A	Linear $f (x) = a x + b$	UT must reduce to the plain [[Kalman Filter (linear)	Kalman filter]] answer exactly
B	Symmetric quadratic $y = x^{2}$ , mean $0$	EKF Jacobian is zero → EKF blind, UT must see it	Ex 2
C	Nonzero-mean nonlinearity	Curvature and offset both matter	Ex 3
D	2D / covariance shape	matrix square root, spread along principal axes	Ex 4 (fig)
E	Full predict step with $Q$	remember to add process noise	Ex 5
F	Non-differentiable $h (x) = ∥ x ∥$	EKF Jacobian undefined at origin; UT is fine	Ex 6 (fig)
G	Degenerate / zero variance	one direction has no uncertainty; points collapse	Ex 7
H	Scaled UT, tiny $α$	limiting behaviour, $W^{(c)}$ correction saves covariance	Ex 8
I	Word problem (real GNC)	translating a physical scenario into $f, x, P$	Ex 9
J	Exam twist: sign/weight trap	using $W^{(m)}$ where $W^{(c)}$ belongs	Ex 10

Example 1 — Cell A: the linear sanity check

Sigma points. Matrix square root: $(n + λ) P = 3 \cdot 4 = 12 = 23$ . $X_{0} = 2, X_{1} = 2 + 23, X_{2} = 2 - 23 .$ Why this step? The three points must reproduce mean $2$ and variance $4$ ; they sit symmetric about $2$ so the mean is untouched and their spread carries $P$ .
Weights. $W_{0}^{(m)} = λ / c = 2/3$ , and $W_{1} = W_{2} = \frac{1}{2 c} = 1/6$ . With $α = 1, β = 0$ the correction $1 - α^{2} + β = 0$ , so $W_{0}^{(c)} = 2/3 + 0 = 2/3$ . Why this step? $α = 1$ makes $W^{(m)} = W^{(c)}$ — no scaled-UT correction, so we can compare cleanly to plain Kalman.
Transform. $Y_{0} = 7, Y_{1} = 7 + 63, Y_{2} = 7 - 63 .$ Why this step? We apply the actual $f$ , not a linearization — but $f$ is linear here, so this is the exact map.
Recover mean (use $W^{(m)}$ ). $\overset{y}{ˉ} = \frac{2}{3} (7) + \frac{1}{6} (7 + 63) + \frac{1}{6} (7 - 63) = 7.$ Why this step? The $\pm 63$ terms cancel; symmetric points give a symmetric mean.
Recover covariance (use $W^{(c)}$ ). Deviations from $7$ : $0, + 63, - 63$ . $P_{y} = \frac{1}{6} (63)^{2} + \frac{1}{6} (63)^{2} = \frac{1}{6} (108) + \frac{1}{6} (108) = 36.$ Why this step? Covariance is the weighted mean of squared deviations; the centre point contributes nothing here because its deviation is $0$ .

Recall Verify

Kalman prediction ::: $\overset{y}{ˉ} = 3 \cdot 2 + 1 = 7$ ✔, $P_{y} = 3^{2} \cdot 4 = 36$ ✔ — the UT reproduces the linear filter exactly. As it must.

Example 2 — Cell B: the quadratic that blinds the EKF

Sigma points. $3 σ^{2} = 3 σ$ , so $X = {0, 3 σ, - 3 σ}$ . Why this step? We refuse to linearize; the two outer points are placed at $\pm c σ$ precisely so their weighted spread reproduces the true variance $σ^{2}$ — we sample $f$ where the uncertainty actually lives.
Transform. $Y = {0, 3 σ^{2}, 3 σ^{2}}$ . Why this step? Squaring both outer points gives the same positive value — curvature shows up as an upward shift the EKF's zero slope never sees.
Mean (use $W^{(m)}$ ). $\overset{y}{ˉ} = \frac{2}{3} (0) + \frac{1}{6} (3 σ^{2}) + \frac{1}{6} (3 σ^{2}) = σ^{2}$ . Why this step? Even though the centre point maps to $0$ , the two lifted outer points drag the weighted average up to $σ^{2}$ — the whole point of not linearizing.

Recall Verify

True mean of $x^{2}$ ::: $E [x^{2}] = Var (x) = σ^{2}$ — the UT is exact, EKF gives $0$ (100% wrong). For $σ = 2$ : UT says $\overset{y}{ˉ} = 4$ , EKF says $0$ .

Example 3 — Cell C: nonzero mean, curvature + offset

Sigma points. $3 \cdot 1 = 3$ , so $X = {1, 1 + 3, 1 - 3}$ . Why this step? The points are centred on the actual mean $1$ (not $0$ ) and stepped out by $\pm c \cdot std = \pm 3$ so their spread equals $P = 1$ — mean and covariance of $x$ are both reproduced before we touch $f$ .
Transform. $Y = {1, (1 + 3)^{2}, (1 - 3)^{2}} = {1, 4 + 23, 4 - 23}$ . Why this step? True $f$ at each shifted sample — the cross-term $23$ is exactly the curvature information EKF discards.
Mean (use $W^{(m)}$ ). $\overset{y}{ˉ} = \frac{2}{3} (1) + \frac{1}{6} (4 + 23) + \frac{1}{6} (4 - 23) = \frac{2}{3} + \frac{1}{6} (8) = \frac{2}{3} + \frac{4}{3} = 2.$ Why this step? The $\pm 23$ terms cancel in the average; what survives is the curvature bias that lifts the mean above $h (\overset{x}{ˉ}) = 1$ to the true $2$ .
Covariance (use $W^{(c)}$ ). Deviations from $2$ : ${- 1, 2 + 23, 2 - 23}$ . $P_{y} = \frac{2}{3} (1) + \frac{1}{6} (2 + 23)^{2} + \frac{1}{6} (2 - 23)^{2} .$ $(2 \pm 23)^{2} = 4 \pm 83 + 12 = 16 \pm 83$ ; their sum is $32$ . $P_{y} = \frac{2}{3} + \frac{1}{6} (32) = \frac{2}{3} + \frac{16}{3} = \frac{18}{3} = 6.$ Why this step? Squaring the deviations and weight-averaging is the definition of variance; here $β = 0$ so $W_{0}^{(c)} = W_{0}^{(m)} = \frac{2}{3}$ , and the centre deviation $(- 1)$ does contribute.

Recall Verify

Both moments ::: $\overset{y}{ˉ} = 2$ (true $\overset{x}{ˉ}^{2} + P = 2$ ) ✔ and $P_{y} = 6$ (true $2 P^{2} + 4 \overset{x}{ˉ}^{2} P = 6$ ) ✔ — the UT captures mean and spread exactly for a quadratic through a Gaussian.

Example 4 — Cell D: 2D covariance, spread along principal axes (figure)

Figure — Unscented Kalman Filter (UKF) — sigma points, better for nonlinear

Figure 1. The lavender curve is the 1-sigma uncertainty ellipse of $P$ . The single coral dot is the centre sigma point $X_{0}$ ; the four mint dots are the outer points. Two straight measuring arrows are drawn: the coral arrow (wide axis, variance $4$ ) reaches $22 \approx 2.83$ ; the slate arrow (narrow axis, variance $1$ ) reaches $2 \approx 1.41$ . Every colour is called out once in the legend so nothing is ambiguous. Axes are the two state components $x_{1}, x_{2}$ . Notice the points sit on the ellipse's axes, not randomly scattered.

Matrix square root. Since $P$ is diagonal, $P = (2001)$ (this is the Cholesky factor). Scale: $(n + λ) P = 2 \cdot P$ , columns $(22, 0)$ and $(0, 2)$ . Why this step? The columns of the scaled square root are the principal directions — how far to step along each axis of the uncertainty ellipse (the lavender curve in Figure 1).
Sigma points. $X_{0} = (0, 0)$ (coral dot); $X_{1, 3} = (\pm 22, 0)$ and $X_{2, 4} = (0, \pm 2)$ (the four mint dots). Why this step? Compare the two measuring arrows in Figure 1: the wide axis (variance $4$ ) pushes points out to $\pm 2.83$ , the narrow axis (variance $1$ ) only to $\pm 1.41$ — the point cloud is stretched exactly like the ellipse, so it carries the correct shape.
Weights. $W_{0}^{(m)} = λ / c = 0/2 = 0$ ; $W_{i} = 1/ (2 \cdot 2) = 1/4$ for the four outer points. Why this step? With $λ = 0$ the centre carries zero mean-weight — the four symmetric points alone rebuild the mean (which is why the coral dot in Figure 1 does not bias anything).

Recall Verify

Reconstruct $P$ ::: $x$ -variance $= \frac{1}{4} [(22)^{2} + (22)^{2}] = \frac{1}{4} (8 + 8) = 4$ ✔; $y$ -variance $= \frac{1}{4} [(2)^{2} + (2)^{2}] = \frac{1}{4} (4) = 1$ ✔; cross term $0$ ✔.

Example 5 — Cell E: a full predict step, remember $Q$

Sigma points. $3 \cdot 1 = 3$ : $X = {0, 3, - 3}$ . Why this step? Outer points at $\pm c \cdot std = \pm 3$ reproduce the prior variance $P = 1$ ; we propagate the prior uncertainty, nothing more yet.
Propagate. $f (x) = x + 0.5 x^{2}$ : $f (0) = 0$ , $f (3) = 3 + 1.5$ , $f (- 3) = - 3 + 1.5$ . Why this step? True dynamics on each point — the $0.5 x^{2}$ bumps both outer points up by $1.5$ , which is the curvature the EKF would miss.
Predicted mean (use $W^{(m)}$ ). $\overset{x}{^}_{k}^{-} = \frac{2}{3} (0) + \frac{1}{6} (3 + 1.5) + \frac{1}{6} (- 3 + 1.5) = \frac{1}{6} (3) = 0.5.$ Why this step? The $\pm 3$ cancel; the surviving $+ 1.5$ on each outer point, weighted, drifts the mean to $0.5$ — the curvature, not any input mean, produced it.
Predicted covariance before $Q$ (use $W^{(c)}$ ). Deviations from $0.5$ : ${- 0.5, 3 + 1, - 3 + 1}$ . $(3 + 1)^{2} = 4 + 23$ , $(- 3 + 1)^{2} = 4 - 23$ , sum $= 8$ . $P_{prop} = \frac{2}{3} (0.25) + \frac{1}{6} (8) = \frac{1}{6} + \frac{8}{6} = \frac{9}{6} = 1.5.$ Why this step? Same variance recipe as before ( $β = 0$ so $W_{0}^{(c)} = \frac{2}{3}$ ); this measures how the propagated points spread — but it is only prior uncertainty carried forward.
Add process noise. $P_{k}^{-} = P_{prop} + Q = 1.5 + 0.2 = 1.7.$ Why this step? Sigma points carry only prior uncertainty; $Q$ is the new uncertainty the world injects each step. Forget it and the filter becomes overconfident and diverges.

Recall Verify

Predict outputs ::: $\overset{x}{^}_{k}^{-} = 0.5$ , $P_{k}^{-} = 1.7$ — confirmed by direct weighted sums.

Example 6 — Cell F: non-differentiable measurement $h (x) = ∥ x ∥$ (figure)

Figure 2. The coral dot at the origin is $X_{0}$ , where $h = 0$ . The four mint dots are the outer sigma points; the mint arrows are the vectors from the origin to each; the butter dashed circle marks radius $2$ . Every mint arrow has the same length $2$ — that is why the averaged range is $2$ . Axes are the state components $x_{1}, x_{2}$ . The EKF's Jacobian would need a slope at the coral dot, where $h$ has a sharp corner — the UT never asks for it.

Sigma points. $(n + λ) P = 2 I$ : $X_{0} = (0, 0)$ , $X_{1, 3} = (\pm 2, 0)$ , $X_{2, 4} = (0, \pm 2)$ . Why this step? $c = 2$ , std $= 1$ in each direction, so outer points land at radius $2$ — exactly the mint dots on the butter circle in Figure 2.
Push through $h$ . $h (0, 0) = 0$ ; each outer point has $∥ \cdot ∥ = 2$ . So $Z = {0, 2, 2, 2, 2}$ . Why this step? $h$ is evaluated, never differentiated — the origin corner (visible as the coral dot in Figure 2) simply never enters the arithmetic.
Weights & mean (use $W^{(m)}$ ). $W_{0} = 0$ , $W_{i} = 1/4$ . $\overset{z}{^} = 0 \cdot 0 + 4 \cdot \frac{1}{4} \cdot 2 = 2 \approx 1.414.$ Why this step? Look at Figure 2: the four mint arrows all have length $2$ , so the measured range averages to $2$ — a finite, sensible, derivative-free estimate exactly where the EKF breaks.

Recall Verify

$\overset{z}{^}$ finite where EKF fails ::: $\overset{z}{^} = 2 \approx 1.41421$ ✔; the EKF Jacobian at origin is $0/0$ , undefined.

Example 7 — Cell G: degenerate / zero-variance direction

Square root. $P = (3000)$ ; scaled columns $(32, 0)$ and $(0, 0)$ . Why this step? A zero eigenvalue produces a zero column, so the transform automatically refuses to spread along a direction with no uncertainty — there is no divide-by-zero because we take a square root, never an inverse, of $P$ .
Sigma points. $X_{0} = (3, 5)$ ; $X_{1, 3} = (3 \pm 32, 5)$ ; $X_{2, 4} = (3, 5)$ . Why this step? The second pair inherits the zero column, so they land exactly on the centre — they still hold weight but sit on $\overset{x}{ˉ}$ , keeping every point on the line $y = 5$ .
Reconstruct. All points have $y = 5$ exactly → reconstructed variance in $y$ is $0$ ✔. In $x$ : $\frac{1}{4} [(32)^{2} + (32)^{2}] = \frac{1}{4} (18 + 18) = 9$ ✔. Why this step? The variance recipe over $y$ -deviations gives $0$ automatically, and over $x$ -deviations gives $9$ — the degenerate case needs no special handling.

Recall Verify

Degenerate handled ::: $x$ -variance rebuilt $= 9$ , $y$ -variance $= 0$ — both exact; sigma points never leave $y = 5$ .

Example 8 — Cell H: scaled UT, tiny $α$ , the $W^{(c)}$ rescue

Compute $λ$ from its definition. Use the actual formula, not a shortcut: $λ = α^{2} (n + κ) - n = (1 0^{- 6}) (3) - 1 = 3 \times 1 0^{- 6} - 1 = - 0.999997.$ So $c = n + λ = 3 \times 1 0^{- 6}$ . Keep $λ$ exact ( $= 3 \times 1 0^{- 6} - 1$ ); do not round it to $- 1$ , because the weights below divide by the tiny $c$ and any rounding of $λ$ corrupts them.
Sigma points. $X_{1} = c = 3 \times 1 0^{- 6} = 1.732 \times 1 0^{- 3}$ , $X_{2} = - 1.732 \times 1 0^{- 3}$ , $X_{0} = 0$ . Why this step? With such small $c$ , the outer points barely leave the mean — we sample $f$ only very locally, which is exactly what the scaled UT is designed to do.
Mean weights (use $W^{(m)}$ ). $W_{0}^{(m)} = λ / c = \frac{3 \times 1 0 ^{- 6} - 1}{3 \times 1 0 ^{- 6}}$ , a large negative number; $W_{1, 2}^{(m)} = \frac{1}{2 c} = \frac{1}{6 \times 1 0 ^{- 6}} \approx 1.667 \times 1 0^{5}$ , huge. Why this step? Points hug the mean, so their weights must blow up (and the centre goes negative) to keep the reconstructed spread of $x$ correct — this is the price of local sampling.
Transform & mean. $Y = {0, 3 \times 1 0^{- 6}, 3 \times 1 0^{- 6}}$ . $\overset{y}{ˉ} = W_{0}^{(m)} \cdot 0 + 2 \cdot (1.667 \times 1 0^{5}) (3 \times 1 0^{- 6}) = 2 \cdot 0.5 = 1.$ Why this step? The tiny transformed values times giant weights recombine to exactly $σ^{2} = 1$ — locality of sampling did not cost accuracy in the mean.
Covariance weight (use $W^{(c)}$ ). $W_{0}^{(c)} = W_{0}^{(m)} + (1 - α^{2} + β) = W_{0}^{(m)} + (1 - 1 0^{- 6} + 2) \approx W_{0}^{(m)} + 2.999999$ . This extra $\approx + 3$ on the centre repairs the covariance that the tight spacing would otherwise underestimate. Why this step? $β = 2$ is the Gaussian-optimal correction; it lives only in $W_{0}^{(c)}$ (per the rulebook at the top), so the mean of step 4 is untouched — that is precisely why we could keep $\overset{y}{ˉ} = 1$ .

Recall Verify

Limit stays exact ::: $\overset{y}{ˉ} = 1$ even for $α = 1 0^{- 3}$ ✔ — proof the scaled UT does not sacrifice the mean.

Example 9 — Cell I: real GNC word problem

Translate the physics into $x, P$ . Altitude is the state; its $\pm 1 σ$ range is $100 \pm 5$ km. Covariance $P = 25$ . Why this step? The UT needs a mean and covariance; the word "std $5$ km" is the covariance once squared. This is the modelling move that turns a physical scenario into filter inputs.
Sigma points. $c P = 3 \cdot 25 = 75 = 53 \approx 8.660$ : $X = {100, 108.66, 91.34}$ . Why this step? Points span $\pm 1.73$ std of altitude — the actual band of altitudes the sensor might be reading, so the transform samples the true operating range.
Push through $h$ . $h (100) = 100$ ; $h (108.66) = 1 0^{6} /108.6 6^{2} \approx 84.70$ ; $h (91.34) = 1 0^{6} /91.3 4^{2} \approx 119.86$ . Why this step? The far (high) point loses more power than the near (low) point gains — that asymmetry is the convex bias the EKF's straight-line slope cannot capture.
Mean (use $W^{(m)}$ ). $\overset{z}{^} = \frac{2}{3} (100) + \frac{1}{6} (84.70) + \frac{1}{6} (119.86) = 66.667 + \frac{1}{6} (204.56) = 66.667 + 34.093 = 100.76.$ Why this step? Weight-averaging the transformed points; the convexity nudges $\overset{z}{^}$ above the naive $h (\overset{x}{^}) = 100$ , which is the physically correct answer an EKF misses.

Recall Verify

Convex bias ::: $\overset{z}{^} \approx 100.76 > 100 = h (\overset{x}{^})$ ✔ — the expected power is higher than the plug-in value, the correct convex (Jensen) effect.

Example 10 — Cell J: exam twist, the weight trap

Correct mean (use $W^{(m)}$ ). $W_{0}^{(m)} = λ / c = 2/3$ regardless of $β$ : reusing Example 3, $\overset{y}{ˉ} = 2$ . Why this step? By the rulebook at the top, $β$ lives only in $W_{0}^{(c)}$ ; the mean must be $β$ -independent by construction.
Wrong mean (misusing $W_{0}^{(c)} = 8/3$ ). The weights no longer sum to $1$ : $\frac{8}{3} + 2 \cdot \frac{1}{6} = \frac{8}{3} + \frac{1}{3} = 3 \neq = 1$ . Plugging in: $\overset{y}{ˉ}_{wrong} = \frac{8}{3} (1) + \frac{1}{6} (4 + 23) + \frac{1}{6} (4 - 23) = \frac{8}{3} + \frac{1}{6} (8) = \frac{8}{3} + \frac{4}{3} = 4.$ Why this step? Weights that sum to $3$ instead of $1$ inflate every contribution — the "mean" balloons to $4$ , carrying the spurious $β$ footprint.
Diagnosis. The correct value is $2$ ; the error is $+ 2$ , precisely the $(1 - α^{2} + β) = 2$ leak from putting a covariance weight where a mean weight belongs. Why this step? Isolating the size of the error to exactly the correction term shows why the rule "means use $W^{(m)}$ " exists — swapping weights injects the covariance correction into the mean.

Recall Verify

Weight trap ::: correct $\overset{y}{ˉ} = 2$ , wrong $\overset{y}{ˉ} = 4$ ; the $+ 2$ error equals the misused $(1 - α^{2} + β)$ term — always use $W^{(m)}$ for means.

One-line recall

Correct-vs-wrong mean weight

Use

W^{(m)}

for the mean,

W^{(c)}

(with

β

) for the covariance — never swap them.

Why add

Q

Sigma points carry only prior uncertainty; process noise

Q

is added after propagation or the filter grows overconfident.

UT vs EKF on

y = x^{2}, \overset{x}{ˉ} = 0

UT gives exact

σ^{2}

; EKF gives

0

because the Jacobian

2 x ∣_{0} = 0

Related: Extended Kalman Filter (EKF) · Kalman Filter (linear) · Particle Filter · Cholesky Decomposition · Taylor Series Expansion · State Estimation in GNC · Nonlinear Systems.