Chapter 05

Steady State, the ARE, and When the Theorems Fail

The time-varying Kalman gain is awkward to certify and expensive to run, so in practice everyone uses its limit. Finding that limit means solving the Riccati equation at equilibrium, choosing between its roots, and proving the recursion actually gets there. Two theorems settle all of that at a glance — and the exam has spent nine sittings asking about systems where they do not apply.

Reading: ~42 min Interactive: 2 widgets Source: Polimi MIDA2 2025/26 — Lecture 7 (Ch. 2.3 — extensions and the asymptotic theorems) · Polimi MIDA2 2025/26 — Lecture 14 (Kalman filter exercises, part 1) · Polimi MIDA2 2025/26 — Lecture 15 (Kalman filter exercises, part 2) · Polimi MIDA2 2025/26 — Lecture 18 (Kalman filter exercises, part 3) · Polimi MIDA2 2025/26 — Lecture 19, slides 1–3 (the ARE in MATLAB)

01 · Motivation

From a recursion to a constant gain

Chapter 04 built a filter whose gain changes at every step. That is optimal and almost nobody uses it, for two reasons that have nothing to do with optimality.

Start from an uncomfortable observation. Even when the plant is perfectly time-invariant, the Kalman filter is not: $K(t)$ moves, so the filter is a linear time-varying system. Two problems follow.

Certifying stability becomes hard

For an LTI system you check the eigenvalues of one matrix. For an LTV system that test is not available — and worse, it is not even a valid heuristic: a system whose $F(t)$ has all its eigenvalues strictly inside the unit circle at every single instant can still be unstable. Proving LTV stability directly is genuinely difficult, especially for large $n$ .

Running it costs a matrix inverse per step

Every sampling instant needs a fresh $K(t)$ , hence a fresh $P(t)$ , hence an inversion of the $p\times p$ matrix $\big(HP(t)H^{\top}+V_2\big)$ . On a fast loop with several sensors that is a real budget item.

The fix, in one line

The DRE is autonomous (chapter 04, §03). So if the recursion settles,

If P settles, so does K

$P(t)\longrightarrow\bar P\quad\Longrightarrow\quad K(t)\longrightarrow\bar K$

and we can simply use the constant $\bar K$ from the start. Both problems evaporate at once: the filter becomes LTI, so its stability is an eigenvalue test again, and there is no per-step inversion because the gain was computed once, offline. This is the asymptotic Kalman filter, and it is what ships.

The filter’s own state matrix

Eliminating $e(t)$ from the filter’s state equation shows what that LTI system actually is:

The asymptotic Kalman filter, in closed-loop form

$\hat x(t+1|t)=\big(F-\bar KH\big)\hat x(t|t-1)+\bar K\,y(t)$

So $(F-\bar KH)$ is the state matrix of the filter, and the criterion is the ordinary one:

The asymptotic Kalman filter is asymptotically stable if and only if every eigenvalue of $(F-\bar KH)$ lies strictly inside the unit circle.

	stability depends on
the plant	$F$
the filter	$F-\bar KH$

key

Two different matrices, two different questions

Those are separate objects, so their answers can differ — and the useful direction is the surprising one: the filter can be asymptotically stable even when the plant is unstable. Estimating a diverging state is not the same as stabilising it. The estimate diverges too; what stays bounded is the difference. §07 makes this concrete with a plant whose mode runs away at $F=2$ and whose estimation error still converges.

Three questions that have to be answered

Using $\bar K$ presupposes a lot. Written out, there are three separate things to establish:

Existence — does the equation defining $\bar P$ have any positive-semidefinite solution?
Convergence — supposing $\bar P$ exists, does $P(t)$ actually reach it? An equilibrium is a ball in a bowl or a ball on a dome; only the first attracts.
Stability — supposing both, is the resulting $(F-\bar KH)$ stable?

Answering all three from scratch is hard. §02 gives two theorems that answer all three at once when they apply, and §03–§06 give the machinery for when they do not — which, on this exam, is most of the time.

02 · The headline result

The two asymptotic theorems, and why failing them proves nothing

Two theorems settle existence, convergence and stability together. Both are sufficient conditions, and the single most repeated instruction in this course’s exam archive is to check them on a system that does not satisfy them.

9 questions, 57 marks, 9 of the 12 sittings

“Can you apply the asymptotic theorems?” is the most-asked stem in the whole of chapter 2.3 — nine questions across nine different sittings (15/07/2023, 08/01/2024, 29/01/2024, 25/06/2024, 26/07/2024, 10/02/2025, 14/07/2025, 12/02/2026, 26/06/2026), worth 57 marks between them.

In seven of the nine the answer is “no, neither applies.” In six of those seven the reason is the same one: $V_{12}\ne0$ , because the paper wrote a single noise source into both equations. The other two are parametric — “for which values of $\alpha$ and $\beta$ …” — and want the conditions turned into inequalities.

A full-credit answer never stops at “the theorems fail”. It says which hypothesis fails, why, and then adds the sentence that earns the rest of the marks: these are sufficient conditions, so their failure says nothing about whether a steady-state filter exists — it only means you must analyse the DRE by hand, which is §03–§05.

Reconciling two counts: the frequency analysis lists a narrower topic row — “Asymptotic KF theorems”, 4 questions and 29 marks over 4 sittings — because it classifies each question by its whole subject. The nine above are every question that asks the applicability check as a part, counted from the prompts directly. Both numbers are right; the larger one is the one to revise for.

The first theorem

✓

1st asymptotic Kalman filter theorem

If $V_{12}=0$ and the system is asymptotically stable, then

the ARE has one and only one positive-semidefinite solution $\bar P\ge0$ ;
the DRE converges to it for every $P_0\ge0$ ;
the resulting $\bar K$ makes the Kalman filter asymptotically stable.

Short hypotheses, and a strong conclusion — but it needs the plant to be stable, which throws away exactly the case §01 advertised as the interesting one.

Ingredients for the second theorem

The second theorem drops the stability requirement and pays for it with two structural conditions. The first is observability from chapter 01: $\mathcal O=[H;HF;\dots;HF^{n-1}]$ full rank. The second is unusual enough to deserve its own build-up.

The state equation has two inputs — the control $u(t)$ and the noise $v_1(t)$ — and the condition concerns reachability from the noise, not from the control. To state it, normalise the noise so it has identity covariance:

Normalising the process noise

$x(t+1)=Fx(t)+Gu(t)+\Gamma\,\omega(t),\qquad \omega(t)\sim\mathrm{WN}(0,I),\qquad V_1=\Gamma\Gamma^{\top}$

$\Gamma$ is a factorisation of $V_1$ — a square root, in the matrix sense. Concretely, for $x(t+1)=\tfrac12x(t)+v_1(t)$ with $v_1\sim\mathrm{WN}(0,4)$ : rewrite as $x(t+1)=\tfrac12x(t)+2\,\omega(t)$ with $\omega\sim\mathrm{WN}(0,1)$ , so $\Gamma=2$ and indeed $\Gamma\Gamma^{\top}=4=V_1$ .

The state is then reachable from the noise when

Reachability of the noise channel

$\mathcal R=\big[\Gamma\ \ F\Gamma\ \ F^{2}\Gamma\ \cdots\ F^{\,n-1}\Gamma\big]\ \text{is full rank}$

why

The condition says the noise must reach every state — and that is good news

Read literally: there must be no noise-free state equation. A state direction that no disturbance ever touches breaks the hypothesis.

That is genuinely counter-intuitive, because noise is normally the enemy. Here it is the carrier of information: a state that nothing excites also produces no signature in the data for the filter to lock onto. Chapter 07 pushes this all the way — gray-box identification works by adding a fictitious noise to a parameter so that the filter can estimate it.

The second theorem

✓

2nd asymptotic Kalman filter theorem

If $V_{12}=0$ , and $(F,H)$ is observable, and $(F,\Gamma)$ is reachable with $\Gamma\Gamma^{\top}=V_1$ , then

the ARE has one and only one positive-definite solution $\bar P>0$ ;
the DRE converges to it for every $P_0\ge0$ ;
the resulting $\bar K$ makes the Kalman filter asymptotically stable.

Comparing them

	1st theorem	2nd theorem
needs $V_{12}=0$	yes	yes
needs the plant stable	yes	no
extra hypotheses	—	observability and reachability from the noise
conclusion on $\bar P$	unique semidefinite, $\bar P\ge0$	unique definite, $\bar P>0$

The second is the powerful one precisely because it drops plant stability — which is what makes the Kalman filter usable on unstable plants, consistent with §01’s $(F-\bar KH)$ result. When they apply, both let you skip the direct convergence analysis entirely, which is a large saving.

The three ways this question is failed

Concluding that no asymptotic filter exists. Sufficient conditions that fail tell you nothing. Six of the archive’s nine questions go on to construct a perfectly good steady-state filter after the theorems have failed.
Checking reachability with $G$ or with $V_1$ instead of $\Gamma$ . The pair is $(F,\Gamma)$ with $\Gamma\Gamma^{\top}=V_1$ . In the scalar case $\Gamma=\sqrt{V_1}$ , so using $V_1$ gives the same yes/no answer by luck; in the multivariable case it is simply the wrong matrix.
Reading “asymptotically stable” as “stable”. $F=1$ is simply stable, not asymptotically stable, and theorem 1 fails on it — that is exactly the trap in the 29/01/2024 paper, where $V_1=0$ then knocks out theorem 2 as well.

2025-07-q2a2025Q02Asymptotic KF theoremsmedium6 pts

Consider the following system: $$\begin{cases}x(t+1)=2x(t)+v_1(t)\\ y(t)=h\cdot x(t)+v_2(t)\end{cases}$$ where $v_1(t)\sim\mathrm{WN}(0,q)$, $v_2(t)\sim\mathrm{WN}\!\left(0,\tfrac34\right)$ and $v_1(t)\perp v_2(t)$. a. For which values of $h$ and $q$ can you apply the asymptotic theorems? b. Set $h=\tfrac12$ and $q=5$. Compute the DRE and the ARE asymptotic solution(s). c. Check the convergence to the asymptotic solution(s) of point (b) using the graphical method.

2026-06-q2a2026Q02Asymptotic KF theoremshard7 pts

Consider the system $$\begin{cases}x(t+1)=(\alpha-1)x(t)+(1-\beta)v(t)\\ y(t)=x(t)+\beta v(t)+w(t)\end{cases}$$ with $v(t)\sim\mathrm{WN}(0,1)$, $w(t)\sim\mathrm{WN}(0,2)$, $v(t)\perp w(t)$, $\alpha\in\mathbb R$, $\beta\in\mathbb R$. a. Find $F$, $H$, $V_1$, $V_2$ and $V_{12}$. b. For which values of $\alpha$ and $\beta$ can you apply the first and second asymptotic theorems, respectively? c. Set $\alpha=0$ and $\beta=0$. Analyze graphically the convergence of the DRE to the unique steady-state solution $\bar P=2$ (no need to check the stability of the filter).

03 · The equation

The algebraic Riccati equation and its roots

A constant gain needs a constant $P$ . For a discrete-time recursion $P(t+1)=f\big(P(t)\big)$ , the constants are the fixed points — solutions of $\bar P=f(\bar P)$ .

4 questions, 23 marks, 4 sittings — and it opens most of Q2

The topic “DRE and ARE” is set as its own question in four sittings (08/01/2024, 25/06/2024, 26/07/2024, 12/02/2026) for 23 marks. Counting more broadly, deriving the DRE and solving the ARE is step one of nearly every Q2 in the archive, which is why an error here is expensive: everything after it inherits the wrong $\bar P$ .

Setting $P(t+1)=P(t)=\bar P$ in the difference Riccati equation removes the time index and leaves an algebraic equation:

The algebraic Riccati equation (ARE)

$\bar P=\big(F\bar PF^{\top}+V_1\big)-\big(F\bar PH^{\top}+V_{12}\big)\big(H\bar PH^{\top}+V_2\big)^{-1}\big(F\bar PH^{\top}+V_{12}\big)^{\top}$

Same three blocks as chapter 04’s mnemonic — STATE, OUTPUT, MIX — with $\bar P$ in place of $P(t)$ . It is a system of nonlinear algebraic matrix equations, and $\bar P$ , if it exists, is an equilibrium of the DRE.

why

Discrete time changes what an equilibrium looks like

For a continuous-time system $\dot x=f(x)$ the equilibria solve $0=f(\bar x)$ . For a discrete-time system $x(t+1)=f\big(x(t)\big)$ they solve $\bar x=f(\bar x)$ — not zero, unchanged. Writing $0=f(\bar P)$ here is a common and instantly fatal slip.

Which root is the answer

The ARE is quadratic in the scalar case, so it generally has two roots, and two separate filters are applied to them in order:

Order matters Selecting the steady-state covariance

Filter 1 — admissibility

$\bar P$ is a variance, so any negative root is discarded immediately. This is a statement about what $P$ is, not about stability — and it is often enough on its own. In the 08/01/2024 paper the roots are $-1$ and $2$ , and the negative one is simply not a covariance.

Filter 2 — stability, when admissibility is not decisive

Sometimes both roots are non-negative and filter 1 separates nothing. Then compute $\bar K$ for each and keep the one with $\lvert F-\bar KH\rvert<1$ .

The archetypal case: $F=3$ , $H=2$ , $V_1=4$ , $V_2=1$ , $V_{12}=-2$ gives $P(t+1)=\frac{49P(t)}{4P(t)+1}$ , whose ARE $4\bar P(\bar P-12)=0$ has roots $\bar P=0$ and $\bar P=12$ — both perfectly non-negative.

\bar K=\frac{6\bar P-2}{4\bar P+1}=\begin{cases}-2&\bar P=0\\ \tfrac{10}{7}&\bar P=12\end{cases} \qquad F-\bar KH=\begin{cases}3+4=7&\text{unstable}\\ 3-\tfrac{20}{7}=\tfrac17&\text{stable}\end{cases}

So $\bar P=12$ is the answer, selected by stability alone.

What happens to the rejected root

It is a genuine fixed point of the DRE — just not an attracting one. Starting exactly at $P_0=0$ above, the recursion sits on $\bar P=0$ forever. Any $P_0>0$ climbs away from it and converges to $12$ . That is why the answer to “for which $P_0$ does the asymptotic predictor exist?” is $P_0>0$ strictly, and it is what attracting and repelling equilibria means in practice.

Non-negativity does not always decide

Two habits are worth separating. Rejecting a negative root is about $P$ being a variance. Choosing between two non-negative roots is about stability, and there is no shortcut — you have to compute both gains. Papers are built both ways: 08/01/2024 and 14/07/2025 are decided by sign, 26/07/2024 and the archetypal exercise above are decided by stability.

2024-01-08-q22024Q02DRE and AREhard8 pts

Given the system described by the following equations: $$\begin{cases}x(t+1)=-2x(t)+v_1(t)\\ y(t)=\tfrac12 x(t)+v_2(t)\end{cases}$$ where $v_1(t)\sim\mathrm{WN}(0,3)$ is uncorrelated with the state, $v_2(t)\sim\mathrm{WN}\!\left(0,\tfrac12\right)$ and $\mathbb E\!\left[v_1(t)\,v_2^\top(t)\right]=-1$. a. Can you apply the asymptotic theorems? b. Analyze the convergence of the DRE to the ARE solution. Compute the asymptotic Kalman gain. Analyze the stability of the resulting asymptotic predictor. c. Compute the transfer function from $y(t)$ to $\hat x(t|t)$ (state **filter**, not predictor).

04 · In full generality

The scalar DRE is always a hyperbola

Before drawing anything, it is worth knowing what shape you are drawing. For a first-order system the answer is always the same shape, and its four coefficients are available in closed form — which turns a page of algebra into four multiplications.

Write the scalar DRE over a common denominator:

Expanding the scalar DRE

$P(t+1)=\frac{\big(F^2P(t)+V_1\big)\big(H^2P(t)+V_2\big)-\big(FHP(t)+V_{12}\big)^2}{H^2P(t)+V_2}$

The numerator looks quadratic. It is not: the $P^2$ terms are $F^2H^2P^2$ from the first product and $F^2H^2P^2$ from the square, and they cancel exactly. What survives is linear over linear:

The DRE as a Möbius map

P(t+1)=\frac{a\,P(t)+b}{c\,P(t)+d},\qquad \begin{aligned} a&=H^2V_1+F^2V_2-2FHV_{12}\\ b&=V_1V_2-V_{12}^{2}\\ c&=H^2\\ d&=V_2 \end{aligned}

A Möbius map — a hyperbola. Never anything worse, for any first-order system.

key

This formula computes every DRE in the archive

Four products and you are done. Checked against every scalar exam question in the bank whose covariances are given, it reproduces the published DRE exactly:

paper	$F,H,V_1,V_2,V_{12}$	$P(t+1)$
08/01/2024	$-2,\ \tfrac12,\ 3,\ \tfrac12,\ -1$	$\dfrac{\tfrac34P+\tfrac12}{\tfrac14P+\tfrac12}=\dfrac{3P+2}{P+2}$
14/07/2025	$2,\ \tfrac12,\ 5,\ \tfrac34,\ 0$	$\dfrac{\tfrac{17}4P+\tfrac{15}4}{\tfrac14P+\tfrac34}$
04/09/2025	$3,\ -2,\ 2,\ \tfrac12,\ -\tfrac34$	$\dfrac{\tfrac72P+\tfrac7{16}}{4P+\tfrac12}=\tfrac78$
12/02/2026	$2,\ 1,\ \tfrac34,\ \tfrac12,\ \tfrac12$	$\dfrac{\tfrac34P+\tfrac18}{P+\tfrac12}$
16/07/2026	$1,\ 2,\ \tfrac34,\ \tfrac13,\ \tfrac12$	$\dfrac{\tfrac43P}{4P+\tfrac13}$

Expanding by hand is still legitimate and still what the marking scheme shows. But the shortcut is a free cross-check, and — as the last two rows hint — the coefficients themselves diagnose the degenerate cases before you draw anything.

The four features you need in order to draw it

Here they are on the chapter’s own worked example. Notice where the two awkward ones landed — the vertical asymptote and the $x$ -intercept are both off in the shaded half-plane, the one a variance can never visit. That is not luck, and it is the whole reason the method works:

feature	value	sign
horizontal asymptote	$\dfrac{a}{c}=\dfrac{H^2V_1+F^2V_2-2FHV_{12}}{H^2}$	$\ge0$
vertical asymptote	$-\dfrac{d}{c}=-\dfrac{V_2}{H^2}$	$\le0$
$y$ -intercept, at $P(t)=0$	$\dfrac{b}{d}=\dfrac{V_1V_2-V_{12}^{2}}{V_2}$	$\ge0$
$x$ -intercept, at $P(t+1)=0$	$-\dfrac{b}{a}$	$\le0$

Those signs are not coincidences, and they are the reason the graphical method in §05 always works. The $y$ -intercept’s sign is the covariance inequality: correlation is bounded by $1$ , so $V_{12}^2\le V_1V_2$ , hence $b\ge0$ . The vertical asymptote is negative because $V_2>0$ and $H^2\ge0$ .

key

Why the cobweb always converges

Both asymptote-and-intercept pairs that could cause trouble sit in the negative quadrant. So on the physically meaningful domain $P\ge0$ the curve is increasing, concave and bounded above by $a/c$ . An increasing bounded concave map on a half-line crosses the $45°$ line at most once from above — which is precisely the picture in which a cobweb converges from any starting point. The covariance inequality is doing all the work.

The degenerate cases, and how the coefficients announce them

b = 0 — the map passes through the origin

$V_1V_2=V_{12}^2$ : the two noises are perfectly correlated. $\bar P=0$ becomes a fixed point, and the question is whether it attracts. Two sub-cases: the $45°$ line cuts the curve again at a positive $\bar P$ (then $P_0=0$ stays at zero, every $P_0>0$ climbs to the positive root), or it does not (then everything falls into zero). The commonest degenerate form in the archive.

a/c = b/d — the map is constant

The two ratios coincide, the fraction collapses, and $P(t+1)=\alpha$ for every $P(t)$ . The DRE reaches $\bar P=\alpha$ in one step from any $P_0$ . Both the 04/09/2025 paper and subsystem B of the 15/07/2023 paper are this case, and neither announces itself.

$H=0$ — the map degenerates to a straight line. With no measurement there is no correction, and $c=0$ leaves

The H = 0 case

$P(t+1)=F^{2}\,P(t)+\frac{V_1V_2-V_{12}^{2}}{V_2}$

whose behaviour is decided entirely by the slope $F^2$ against the $45°$ line:

	condition	behaviour
plant asymptotically stable	$\lvert F\rvert<1$	one intersection ⟹ a single $\bar P$ , reached from every $P_0$
plant unstable	$F^2>1$	$b\ne0$ : the DRE diverges for every $P_0$ — there is genuinely no asymptotic filter. $b=0$ : $\bar P=0$ if $P_0=0$ , diverges otherwise
$F=1$	$F^2=1$	$b\ne0$ : parallel lines ⟹ no asymptotic solution. $b=0$ : $P(t)=P_0$ forever ⟹ infinitely many

✓

H = 0 is the no-measurement case

Filter stability normally depends on $F-\bar KH$ ; with $H=0$ that collapses to $F$ , so the filter simply inherits the plant’s stability. This is not an exotic corner — §07 manufactures it deliberately, by splitting off a subsystem that the output does not measure.

Deep dive Proof that the horizontal asymptote is never negative

The claim is $a=H^2V_1+F^2V_2-2FHV_{12}\ge0$ , which is what keeps the curve above the axis for large $P$ . It follows from the covariance inequality $-\sqrt{V_1V_2}\le V_{12}\le\sqrt{V_1V_2}$ by completing the square, in two cases split on the sign of $FH$ .

Case $FH\ge0$ . Complete the square with a minus sign:

H^2V_1+F^2V_2-2FHV_{12} =\underbrace{\big(H\sqrt{V_1}-F\sqrt{V_2}\big)^2}_{\ge0} -\underbrace{2FH}_{\ge0}\underbrace{\big(V_{12}-\sqrt{V_1V_2}\big)}_{\le0}\ \ge0

Case $FH<0$ . The same identity with a plus sign:

H^2V_1+F^2V_2-2FHV_{12} =\underbrace{\big(H\sqrt{V_1}+F\sqrt{V_2}\big)^2}_{\ge0} -\underbrace{2FH}_{\le0}\underbrace{\big(V_{12}+\sqrt{V_1V_2}\big)}_{\ge0}\ \ge0

In both cases a non-negative square minus a non-positive product. ∎

This is the only place in the course where the covariance inequality does structural work rather than serving as an arithmetic check, and it is worth seeing once: the geometry that makes every exam cobweb converge is a consequence of $V_{12}$ being a legitimate cross-covariance and nothing more.

One caveat on scope. The taxonomy above is stated for $V_{12}\ge0$ ; several exam systems have $V_{12}<0$ . The graphical method is unaffected — only the sign bookkeeping inside this proof changes, and the conclusion $a\ge0$ holds either way, which is exactly what the two-case split establishes.

05 · The method exams ask for

Convergence by drawing the map

Having a fixed point is not having convergence. The papers ask you to establish it by drawing — and they ask by name, in seven of the last eight sittings.

7 questions, 41 marks, 7 of the last 8 sittings

“Graphically check the convergence of the DRE to the ARE solution(s)” appears verbatim in seven papers — 26/07/2024, 04/09/2024, 14/07/2025, 04/09/2025, 12/02/2026, 26/06/2026 and 16/07/2026, worth 41 marks. That is every sitting from July 2024 onwards except 10/02/2025. Two earlier papers ask for convergence analysis without naming the method.

Graders want the picture and the reasoning, not a limit computation: the two asymptotes, the intercepts, the $45°$ line, the intersection(s), and a sentence about which initial conditions go where. A correct $\bar P$ with no diagram does not answer the question that was asked.

Three of the seven (04/09/2024, 04/09/2025, 16/07/2026 — the “noise covariances” topic, 18 marks between them) pair the graphical check with building $V_1$ , $V_2$ and $V_{12}$ from a shared noise source first. That step is chapter 04, §02; everything after it is this section.

Hands-on

The DRE is a hyperbola — cobweb it

Pick a system from the archive, then move P₀ across the fixed point and watch the staircase turn around. Up to the curve, across to the 45° line, repeat — exactly the construction the papers ask you to draw.

P₀ — 1/2

steps drawn — 6

a, b, c, d

9/4, 7/4, 9/16, 2

ARE roots

Limit from this P₀

§05 worked example · From P₀ = 1/2 the DRE converges to P̄ = 2. Hyperbola — the generic shape. Increasing, concave, bounded above by a/c. V₁₂ ≠ 0, so neither theorem applies and the picture is the only route. P̄ = 2, chosen by sign alone.

Try thisDrag P₀ to exactly 0 on the 16/07/2026 system. Because b = V₁V₂ − V₁₂² = 0 the curve passes through the origin, so P₀ = 0 is a fixed point and stays there forever — while every P₀ > 0 climbs to the other root. That is why the marked answer is “converges for every P₀ > 0”, and why “converges” on its own does not earn it.

TakeawayThe graphical check is not a weaker substitute for algebra — for the systems these papers set it is the only route, because V₁₂ ≠ 0 kills both asymptotic theorems. Four coefficients give you the shape, the 45° line gives you the roots as a free check on your quadratic, and the staircase turns a limit computation into something you can see. Write the conclusion as a case list over P₀, never as one word.

The procedure

Do this every time The graphical convergence check

Get the map into hyperbola form

Either expand the DRE by hand or read $a,b,c,d$ off §04. Either way you end with $P(t+1)=\frac{aP+b}{cP+d}$ — and if the coefficients collapse (constant map, straight line, through the origin), notice that now, because it changes the picture entirely.

Locate the four features

Horizontal asymptote $a/c$ ; vertical asymptote $-d/c$ ; $y$ -intercept $b/d$ ; $x$ -intercept $-b/a$ . Only the first two matter for the shape, but the intercepts anchor the sketch on the positive side, which is the side that exists.

Draw the 45° line and mark the intersections

$P(t+1)=P(t)$ is where the recursion stands still. Its intersections with the hyperbola are exactly the ARE roots you computed in §03 — which is a free consistency check on your algebra: if the picture shows two crossings on $P\ge0$ and the quadratic gave one, something is wrong.

Cobweb from a few starting points

From $P_0$ , go vertically to the curve (that is $P_1$ ), horizontally to the $45°$ line (that carries $P_1$ back to the input axis), and repeat. Do it from below a fixed point and from above it. On the standard picture the staircase climbs from small $P_0$ and descends from large $P_0$ , both landing on the same $\bar P$ .

State the conclusion by basin, not in general

Write the answer as a case list over $P_0$ . The archetypal one:

$P_0$	limit
$P_0=0$	stays at $\bar P=0$
$0<P_0<12$	climbs to $\bar P=12$
$P_0=12$	stays at $\bar P=12$
$P_0>12$	descends to $\bar P=12$

“Converges for every $P_0>0$ ” is the sentence that earns the mark. “Converges” on its own does not, because $P_0=0$ is a counterexample sitting in plain sight.

A full worked example, in the exam’s format

Do this in 10 minutes From a system to a convergence verdict

The system as given

$\begin{cases}x(t+1)=\tfrac34x(t)+w(t)\\ y(t)=\tfrac34x(t)-\tfrac12w(t)+\tfrac{\sqrt7}{2}v(t)\end{cases}$

with $v\sim\mathrm{WN}(0,1)$ , $w\sim\mathrm{WN}(0,1)$ , $v\perp w$ . One shared noise $w$ , as usual.

Covariances, then the theorems

$F=\tfrac34$ , $H=\tfrac34$ , $v_1=w$ , $v_2=-\tfrac12w+\tfrac{\sqrt7}{2}v$ :

$V_1=1,\qquad V_2=\tfrac14+\tfrac74=2,\qquad V_{12}=\mathbb E\big[w\big(-\tfrac12w\big)\big]=-\tfrac12$

$V_{12}\ne0$ , so neither theorem applies — the graphical method is the only route left. Say this explicitly; it is a marked step.

The DRE, via the coefficients

From §04, with $F=H=\tfrac34$ :

a=\tfrac9{16}+\tfrac9{16}\cdot2-2\cdot\tfrac9{16}\cdot\big(-\tfrac12\big)=\tfrac94, \qquad b=1\cdot2-\tfrac14=\tfrac74,\qquad c=\tfrac9{16},\qquad d=2

$P(t+1)=\frac{\tfrac94P(t)+\tfrac74}{\tfrac9{16}P(t)+2}$

Expanding the full Riccati expression by hand gives the same thing, and takes about ten times longer.

Solve the ARE

\tfrac9{16}\bar P^{2}+2\bar P=\tfrac94\bar P+\tfrac74 \ \Longrightarrow\ 9\bar P^{2}-4\bar P-28=0 \ \Longrightarrow\ \bar P=\frac{4\pm32}{18}=\begin{cases}2&\text{admissible}\\ -\tfrac{14}9&\text{rejected, negative}\end{cases}

So $\bar P=2$ , decided by sign alone — no stability argument needed to pick the root here.

Draw it

feature	value
horizontal asymptote	$a/c=\tfrac{9/4}{9/16}=4$
vertical asymptote	$-d/c=-\tfrac{32}9$
$y$ -intercept	$b/d=\tfrac{7/8}{1}=\tfrac78$
$x$ -intercept	$-b/a=-\tfrac79$

The branch on $P\ge0$ starts at $\tfrac78$ (above the bisector, since $\tfrac78>0$ ), rises towards the asymptote at $4$ , and crosses the $45°$ line once, at $\bar P=2$ . Cobweb from $P_0<2$ and it climbs; from $P_0>2$ it descends. $P(t)\to2$ for every $P_0\ge0$ .

Gain, stability, and the predictors

\bar K=\frac{\tfrac9{16}\bar P-\tfrac12}{\tfrac9{16}\bar P+2}=\frac{5/8}{25/8}=\tfrac15, \qquad F-\bar KH=\tfrac34-\tfrac15\cdot\tfrac34=\tfrac35

$\lvert\tfrac35\rvert<1$ , so the filter is stable and the asymptotic prediction error variance is bounded. The one-step predictor is then

$\hat x(t+1|t)=\tfrac35\hat x(t|t-1)+\tfrac15\,y(t)$

and — using chapter 04’s multi-step rule, with no new gain and no new Riccati equation — the two-step predictor is just $F$ applied to it:

$\hat x(t+2|t)=F\hat x(t+1|t)=\tfrac9{20}\hat x(t|t-1)+\tfrac3{20}\,y(t)$

Why this is the only available route so often

The two theorems of §02 would answer convergence instantly — when they apply. They mostly do not, and the graphical method has no hypotheses at all: it works with $V_{12}\ne0$ , with an unstable plant, with an unobservable mode. That is the whole reason it dominates the archive.

Four ways the graphical part loses marks

Answering “it converges” without the basins. The origin case has a fixed point that only attracts itself; a blanket statement is wrong on a set of measure zero that the examiner chose deliberately.
Not noticing a collapsed map. The 04/09/2025 paper’s DRE is a constant: numerator and denominator share the factor $\left(P+\tfrac18\right)$ . Grinding the quadratic still gets the right $\bar P$ , but the far more interesting one-step convergence is invisible unless you factor.
Drawing only the positive quadrant. The vertical asymptote is at $-V_2/H^2$ , on the negative side. Leaving it out is fine for the answer and costs you the argument for why the branch on $P\ge0$ is well behaved.
Assuming a fixed point at the origin attracts. Passing through the origin says nothing about stability of that point. In the 16/07/2026 paper $\bar P=0$ repels and the physical limit is $\tfrac14$ .

2024-07-q2a2024Q02DRE and AREmedium4 pts

Consider a system whose DRE is: $$P(t+1)=\frac{4P(t)}{P(t)+\tfrac14}$$ a. Compute the asymptotic solution(s) of the ARE. b. Check graphically the convergence of the DRE to the ARE solution(s).

2025-09-q2a2025Q02Noise covariances V1, V2, V12hard6 pts

Consider the following system: $$\begin{cases}x(t+1)=3x(t)+w(t)\\ y(t)=-2x(t)-\tfrac38 w(t)+\sqrt{\tfrac7{32}}\,v(t)\end{cases}$$ where $w(t)\sim\mathrm{WN}(0,2)$, $v(t)\sim\mathrm{WN}(0,1)$, $w(t)\perp v(t)$, and the state is uncorrelated with the noises. a. Compute $V_1$, $V_2$ and $V_{12}$. Are the asymptotic Kalman theorems applicable? b. Compute the ARE solution(s) and discuss their existence. c. Graphically check the convergence of the DRE to the ARE solution(s).

You compute a scalar DRE and find that its numerator and denominator have a common linear factor, so the map reduces to a constant. What does that tell you about convergence?

06 · Choosing the root

The asymptotic gain and filter stability

With $\bar P$ settled, the gain is one substitution and the stability verdict is one subtraction. Both are short — which is why they are usually bolted onto the end of a longer question rather than asked alone.

Asked in its own right once, and as a sub-part constantly

“Compute the asymptotic Kalman gain and assess the filter stability” is a topic of its own in one paper (10/02/2025, 4 marks), but the same two steps appear as a sub-part of nearly every Q2 in the archive. Treat them as reflexes rather than as a section to revise.

The asymptotic gain and the stability test

\bar K=\big(F\bar PH^{\top}+V_{12}\big)\big(H\bar PH^{\top}+V_2\big)^{-1}, \qquad \text{filter stable}\iff \text{all eigenvalues of }\big(F-\bar KH\big)\text{ inside the unit circle}

Three things are worth stating explicitly, because each is a place where the exam probes.

The gain re-uses the same blocks. MIX over OUTPUT, with $\bar P$ substituted for $P(t)$ . Nothing new is being computed; chapter 04’s mnemonic still rebuilds it.

The test is on the filter, not on the plant. $F$ can sit far outside the unit circle. In the 14/07/2025 paper $F=2$ and the filter is fine; in the 15/07/2023 paper’s second subsystem $F=1$ and $F-\bar KH=0$ , which is as stable as it gets.

A stable filter on an unstable plant is not a paradox. The state runs away, the estimate runs away with it, and the estimation error — the only thing the filter controls — stays bounded. What the filter stabilises is the error dynamics:

The error dynamics

$e_x(t+1)=\big(F-\bar KH\big)e_x(t)+v_1(t)-\bar K\,v_2(t)$

which is a stable linear system driven by white noise, hence a stationary process with finite variance, regardless of what $x(t)$ itself is doing.

key

What the measurement update is worth, numerically

$\bar P$ is the predictor’s error variance. Fold in the measurement at time $t$ and you get the filter’s, which chapter 04 showed is $F^{-1}$ applied to the whole recursion:

$x(t)-\hat x(t|t)=\big(I-F^{-1}\bar KH\big)e_x(t)-F^{-1}\bar K\,v_2(t)$

For the archetypal system ( $F=3$ , $H=2$ , $\bar K=\tfrac{10}7$ , $V_2=1$ , $\bar P=12$ ) the two coefficients are $\tfrac1{21}$ and $\tfrac{10}{21}$ , and since $e_x(t)\perp v_2(t)$ the variances add:

$\operatorname{var}\big[x(t)-\hat x(t|t)\big]=\tfrac1{441}\cdot12+\tfrac{100}{441}\cdot1=\tfrac{16}{63}\approx0.25$

Against $\bar P=12$ that is roughly a 47-fold reduction, bought by nothing more than using $y(t)$ instead of stopping at $y(t-1)$ . The predictor/filter distinction stops being notational at this point.

Deep dive Deriving the gain from scratch, by minimising the error variance

The Riccati route produces $\bar K$ without ever saying what it optimises. The direct route does, and the two answers agree — including on which root to throw away. (This derivation is flagged in the exercise sessions as not required for the exam; it is here because it explains the rest.)

Take the stationary error equation above. Its variance must reproduce itself:

$\bar P=\big(F-\bar KH\big)\bar P\big(F-\bar KH\big)^{\top}+V_1+\bar KV_2\bar K^{\top}-V_{12}\bar K^{\top}-\bar KV_{12}^{\top}$

For the archetypal system $F=3$ , $H=2$ , $V_1=4$ , $V_2=1$ , $V_{12}=-2$ , solving that for $\bar P$ as a function of the still-unknown $\bar K$ gives

$\bar P(\bar K)=\frac{\bar K^{2}+4\bar K+4}{-8+12\bar K-4\bar K^{2}}$

and minimising it — $\frac{d\bar P}{d\bar K}=0$ — yields the stationary points $\bar K=\tfrac{10}{7}$ and $\bar K=-2$ : the same two candidates the ARE produced. The second derivative separates them: positive at $\tfrac{10}{7}$ , negative at $-2$ .

So the rejected root is not merely non-stabilising, it is the gain that maximises the error variance. Two independent arguments — Riccati fixed point plus stability test, and direct variance minimisation — pick the same winner and each explains why the other’s loser lost.

The optimality statement itself is worth recording: $K(t)$ minimises $\gamma^{\top}P(t)\gamma$ for every direction $\gamma$ in the state space. Not the trace, not one component — the error covariance along every direction simultaneously. That is a strong claim, and it is what “optimal” means here.

07 · Divide and conquer

Decoupling into independent subsystems

When the theorems fail on a second-order system, there is one more move before resorting to a $2\times2$ Riccati equation by hand: check whether the system is really two first-order problems wearing a single matrix.

2 questions, 9 marks — and always after the theorems have failed

“Can the system be decoupled into two independent subsystems?” is set in two sittings (26/07/2024 and 10/02/2025) for 9 marks, and always as part (b) of a question whose part (a) established that the theorems do not apply to the system as a whole. The 15/07/2023 paper asks the same thing under a different heading. The pattern is fixed: block-diagonal $F$ , one mode measured and one not, a cross-covariance that breaks the theorems globally but vanishes on one of the halves.

When it is legal

Decoupling requires that the states never mix: $F$ block-diagonal and the noise channels uncorrelated across blocks. Then the two modes evolve independently, the output picks up some of them and not others, and a single Kalman filter is genuinely two filters running side by side.

The pay-off is large. Hypotheses that fail globally can hold locally — a plant that is unstable overall may be unstable only in one block, and a plant that is unobservable overall is usually perfectly observable in the block the output measures.

The fictitious-output trick

A subsystem that the output does not measure has $H_A=0$ , and the gain’s denominator $H_A\bar P_AH_A^{\top}+V_{2A}$ collapses to $V_{2A}$ — which does not exist, because the subsystem has no output equation and therefore no measurement noise.

The repair is to invent one:

A measurement that measures nothing

$y_A(t)=0\cdot x_A(t)+v_{2A}(t),\qquad v_{2A}\sim\mathrm{WN}(0,\varepsilon),\qquad \varepsilon>0$

A fictitious output that is pure noise. It carries zero information, which is exactly right — the real system tells you nothing about $x_A$ — but it makes the subsystem a well-posed Kalman problem. And then:

$\bar K_A=\frac{F_A\bar P_AH_A+V_{12A}}{H_A\bar P_AH_A+V_{2A}}=\frac{0}{\varepsilon}=0$

key

Zero gain is the correct answer, and ε > 0 is what makes it computable

A zero gain means no correction: the estimator for $x_A$ is an open-loop simulation of its own dynamics, $\hat x_A(t+1|t)=F_A\hat x_A(t|t-1)$ . That is the right response to having no information — the filter runs the model and applies no measurement update, because there is no measurement.

Note where $\varepsilon$ goes: it cancels out of the answer entirely, but it must be strictly positive, or the gain is $0/0$ rather than $0$ . It is scaffolding that has to be present and then disappears.

The two routes, and why they agree

Method I — split, solve, append

Separate the subsystems, run the theorems on each (they often apply now), solve each scalar ARE, and append the two predictors. Fast, and it makes the structure visible. Only legal when the plant is block-diagonal — do not present it as general.

Method II — the full matrix ARE

Write $\bar P=\begin{bmatrix}\ell&\beta\\ \beta&\gamma\end{bmatrix}$ — symmetric, because it is a covariance — impose $\ell\ge0$ , $\gamma\ge0$ , $\ell\gamma-\beta^2\ge0$ , and grind out the three scalar equations. Always legal, more work.

Method II is worth doing once, because of what falls out. The off-diagonal equation reduces to $\beta\gamma=0$ ; the branch $\gamma=0$ dies on a contradiction; so $\beta=0$ . The algebra discovers the decoupling that Method I assumed — the steady-state covariance is block-diagonal because the system is, and nothing had to be imposed by hand.

The result worth remembering

Take a plant with a stable unobservable mode and an unstable observable one — $F=\mathrm{diag}(\tfrac12,2)$ , $H=[0\ \ 1]$ , unit noise covariances. Both theorems fail globally: the plant is unstable and unobservable. Split it, and both halves are fine — subsystem A by theorem 1 (it is stable), subsystem B by theorem 2 (it is observable and noise-reachable). Appending two asymptotically stable predictors gives an asymptotically stable predictor, so the error is bounded.

Then compare the predictor and filter error variances state by state:

state	$\bar P$ (predictor)	filter	change
$x_1$ — unobservable	$\tfrac43\approx1.33$	$\tfrac43\approx1.33$	none
$x_2$ — observable	$2+\sqrt5\approx4.24$	$\tfrac{1+\sqrt5}{4}\approx0.81$	↓ about 5×

✓

Two things this table is saying

The measurement update does nothing at all for the unobservable mode. Conditioning on $y(t)$ cannot improve an estimate that $y$ carries no information about — so its filter variance equals its predictor variance exactly. Students expect at least a little improvement; there is none, and the zero gain from the fictitious-output argument already predicted it.

The error stays bounded although $x_2$ diverges. $F_B=2$ , so the true state runs away and so does its estimate. The gap between them converges. This is §01’s point, made numerically.

2023-07-q22023Q02Asymptotic KF theoremshard8 pts

Given the system described by the following equations: $$\begin{cases}x(t+1)=\begin{bmatrix}\tfrac12&0\\0&1\end{bmatrix}x(t)+v_1(t)\\ y(t)=[0\;\;1]\,x(t)+v_2(t)\end{cases}$$ where $v_1(t)=\begin{bmatrix}v_{11}(t)\\v_{12}(t)\end{bmatrix}\sim \mathrm{WN}\!\left(0,\begin{bmatrix}1&0\\0&1\end{bmatrix}\right)$ is uncorrelated with the state, $v_2(t)\sim\mathrm{WN}\!\left(0,\tfrac12\right)$, $v_{11}(t)\perp v_2(t)$, and $\mathbb E\!\left[v_{12}(t)v_2(t)^\top\right]=\tfrac12$. a. Can you apply the asymptotic K.F. theorems to the given system? b. Can the system be decoupled into two independent subsystems? If so, evaluate for each subsystem: (i) convergence of the DRE to the ARE solution, (ii) the asymptotic Kalman gain, (iii) stability of the asymptotic predictor.

2025-02-q2a2025Q02Subsystem decouplingmedium5 pts

Given the system described by the following equations: $$\begin{cases}x(t+1)=\begin{bmatrix}-\tfrac35&0\\0&2\end{bmatrix}x(t)+v_1(t)\\ y(t)=[0\;\;2]\,x(t)+v_2(t)\end{cases}$$ where $v_1(t)=\begin{bmatrix}v_{11}(t)\\v_{12}(t)\end{bmatrix}\sim\mathrm{WN}\!\left(0,\begin{bmatrix}1/4&0\\0&1/2\end{bmatrix}\right)$ is uncorrelated with the state, $v_2(t)\sim\mathrm{WN}\!\left(0,\tfrac35\right)$, $v_{11}(t)\perp v_2(t)$, and $\mathbb E\!\left[v_{12}(t)v_2(t)^\top\right]=\tfrac15$. a. Can you apply the asymptotic K.F. theorems to the given system? b. Can the system be decoupled into two independent subsystems? If so, explain how and state whether you can apply the theorems to each of the subsystems.

08 · In practice

Solving the ARE in MATLAB

One function, stated for the dual problem, with two transposes to get right.

idare — implicit discrete algebraic Riccati equation — solves

What idare actually solves

$E^{\top}\bar PE=A^{\top}\bar PA+V_1-\big(A^{\top}\bar PB+V_{12}\big)\big(B^{\top}\bar PB+V_2\big)^{-1}\big(A^{\top}\bar PB+V_{12}\big)^{\top}$

Compare that against §03’s ARE: same three blocks, but with $A^{\top}$ where we write $F$ and $B^{\top}$ where we write $H$ . It is the control-theoretic dual. Setting $E=I$ , $A=F^{\top}$ and $B=H^{\top}$ recovers the estimation form exactly, at the cost of transposing on the way in — and the returned gain on the way out:

[P, Ktilde, EIGS] = idare(F', H', V1, V2, V12, eye(n));
K = Ktilde';        % <-- the transpose that is easy to forget
% EIGS is already eig(F - K*H): the §06 stability test, for free

The second transpose is the silent one

Transposing $F$ and $H$ going in is self-correcting — get it wrong and dimensions usually complain. Transposing Ktilde coming out is not: in a scalar or square problem the untransposed gain has a perfectly valid shape and simply holds the wrong numbers. Use the returned EIGS as a cross-check — if it does not match the $F-\bar KH$ you compute from your K, the transpose is missing.

The companion function kalman, its 'current'/'delayed' flag, and the $\tilde G=I,\ \tilde H=0$ trick that lets the course’s covariances pass through unchanged are all in chapter 04, §08.

Load-bearing ideas

Why steady state at all: a time-varying $K(t)$ makes the filter LTV, which is hard to certify (eigenvalues inside the unit circle at every instant do not imply LTV stability) and expensive to run (a $p\times p$ inverse per step). Freezing the gain at $\bar K$ makes the filter LTI.
The filter’s state matrix is $F-\bar KH$ , not $F$ . Stability of the plant and of the filter are different questions, and a stable filter on an unstable plant is normal — what converges is the error, not the state.
The ARE is the DRE at a fixed point: $\bar P=f(\bar P)$ , not $0=f(\bar P)$ . Same STATE/OUTPUT/MIX blocks, no time index.
Choosing the root: discard negative roots because $P$ is a variance; when two roots are non-negative, only $\lvert F-\bar KH\rvert<1$ separates them. The rejected root is still a fixed point — it just repels, which is why $P_0>0$ is required strictly.
The two theorems both need $V_{12}=0$ . Theorem 1 adds plant stability; theorem 2 replaces it with observability plus reachability from the noise, $(F,\Gamma)$ with $\Gamma\Gamma^{\top}=V_1$ . Both are sufficient only.
The scalar DRE is always a Möbius map $\frac{aP+b}{cP+d}$ with $a=H^2V_1+F^2V_2-2FHV_{12}$ , $b=V_1V_2-V_{12}^2$ , $c=H^2$ , $d=V_2$ — the $P^2$ terms cancel exactly. $b=0$ puts it through the origin; $a/c=b/d$ collapses it to a constant; $H=0$ makes it a straight line.
The graphical method needs the horizontal asymptote $a/c$ , the vertical asymptote $-d/c$ , the intercepts, the $45°$ line, and a cobweb — then a conclusion stated by basin of attraction.
Decoupling works when $F$ is block-diagonal with uncorrelated noise channels; an unmeasured subsystem gets a fictitious output $\mathrm{WN}(0,\varepsilon)$ with $\varepsilon>0$ strictly, which gives $\bar K=0$ — an open-loop estimator, correct because there is no information. The unobservable mode’s filter variance equals its predictor variance.
MATLAB: idare(F', H', V1, V2, V12, eye(n)), and transpose the returned gain.

Exam radar. This is the heaviest chapter in the course, and its weight is concentrated in two reflexes. First, §02: nine questions and 57 marks turn on “can you apply the theorems?”, and the answer is “no” seven times out of nine — usually because a shared noise source makes $V_{12}\ne0$ . Practise saying which hypothesis fails and then adding that sufficient conditions failing proves nothing. Second, §05: seven questions and 41 marks ask for the graphical check by name, in every sitting since July 2024 bar one. Draw the asymptotes, draw the bisector, cobweb it, and answer by basin.

Between them sits the pipeline every Q2 walks: covariances → DRE → ARE → root selection → gain → stability → transfer function or variance. Time yourself on it end to end. §04’s coefficient formula is the cheapest speed-up available — four products instead of a page of expansion, and it doubles as a check on the expansion you were going to do anyway.