Worked examples — LQG — LQR + Kalman filter, separation principle

Before any number: three tiny reminders so no symbol is unearned.

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The scenario matrix

Every cell below is hit by at least one worked example. Read it as a checklist.

#	Case class	What makes it special	Hit by
A	Stable scalar plant ($A<0$)	drift already helps	Ex 1
B	Unstable scalar plant ($A>0$)	control must stabilize	Ex 2
C	Cheap control limit ($R\to 0$)	gain blows up, fast poles	Ex 3
D	Expensive control limit ($R\to \infty$)	gain $\to$ minimal, mirror trick	Ex 3
E	Zero state penalty ($Q=0$) on stable plant	degenerate: do nothing	Ex 4
F	Kalman: clean vs noisy sensor ($V$ small/large)	trust sensor vs model	Ex 5
G	2×2 plant, full separation, poles = union	matrix case, no coupling	Ex 6
H	Real-world word problem (spacecraft rate damping)	translate physics $\to$ matrices	Ex 7
I	Exam twist: uncontrollable / degenerate $B$	Riccati has no stabilizing $P$	Ex 8

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Ex 1 — Cell A: stable scalar plant

Forecast: the plant already decays on its own ($A=-1$). Will the optimal controller push hard or barely nudge? Guess the sign of the pole before reading.

1. Write CARE for scalars. With scalars everything commutes, so $A^{\top }P+PA=2AP=-2P$ and $PB{R}^{-1}{B}^{\top }P=P^2$. $-2P-P^2+2=0\Rightarrow P^2+2P-2=0.$ Why this step? CARE is the optimality condition from the HJB equation; solving it is the whole LQR problem.
2. Take the positive root (the value function $x^{\top }Px$ must be a bowl, so $P>0$): $P=\frac{-2+\sqrt{4+8}}{2}=-1+\sqrt{3}\approx \mathrm{0.732.}$ Why this step? Of the two algebraic roots only $P⪰0$ gives a real cost-to-go and a stabilizing gain.
3. Gain and closed loop: $K=R^{-1}{B}^{\top }P=P\approx 0.732$, so $A-BK=-1-0.732=-\mathrm{1.732.}$ Why this step? $u=-Kx$ is the state-feedback law; $A-BK$ is the pole of the controlled system.

Verify: the plant was already stable ($-1$) and the optimum only made it more stable ($-1.732<0$) — modest push because the plant helps itself. ✅ Units: $P,K$ dimensionless here (all matrices scalar).

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Ex 2 — Cell B: unstable scalar plant

Forecast: an unstable plant with $A=2$ must be pulled below zero. Will $K$ exceed $2$? Guess.

1. CARE: $2AP-P^2+Q=4P-P^2+1=0\Rightarrow P^2-4P-1=0$. Why this step? Same scalar CARE template; only $A=2$ changed.
2. Positive root: $P=2+\sqrt{5}\approx 4.236$. Why this step? Pick $P>0$ for the stabilizing solution.
3. Gain: $K=P=4.236$. Closed loop: $A-BK=2-4.236=-2.236$. Why this step? The controller must overcome the $+2$ drift and then some to reach a negative pole.

Verify: indeed $K=4.236>2$, exactly as an unstable plant demands — the controller "cancels" the $+2$ and adds $-2.236$ of decay. ✅

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Ex 3 — Cells C & D: cheap vs expensive control limits

Forecast: with free control we expect an infinitely aggressive pole; with priceless control we expect the mildest stabilizing pole. What is that mildest pole for an unstable $A=1$?

1. General scalar CARE with $R$ kept: $2P-P^2/R+Q=0\Rightarrow P^2-2RP-RQ=0$, so $P=R+\sqrt{R^2+RQ},K=\frac{P}{R}=1+\sqrt{1+\frac{Q}{R}}.$ Why this step? Keeping $R$ symbolic lets us take clean limits instead of guessing.
2. Cheap control $R\to 0$: $K=1+\sqrt{1+Q/R}\to \infty$; pole $A-BK\to -\infty$. Why this step? Free control means no reason to hold back — infinite gain, instantaneous correction. (Physically impossible, hence we never set $R=0$.)
3. Expensive control $R\to \infty$: $K\to 1+\sqrt{1}=2$; pole $A-BK\to 1-2=-1$. Why this step? This is the famous mirror-image result: the cheapest-effort stabilizer just reflects the unstable pole $+1$ across the imaginary axis to $-1$.

Verify: at $Q=1$, take $R=1$ → $K=1+\sqrt{2}\approx 2.414$ (matches parent Example 1); take $R=100$ → $K=1+\sqrt{1.01}\approx 2.005\to 2$. Limits bracket the parent's answer. ✅

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Ex 4 — Cell E: degenerate $Q=0$ on a stable plant

Forecast: $Q=0$ says "state error is free." On an already-stable plant, is there any reason to spend control effort?

1. CARE with $Q=0$: $-2P-P^2=0\Rightarrow P(P+2)=0$. Why this step? Same template with $Q=0$; factor it.
2. Choose $P⪰0$: the roots are $P=0$ and $P=-2$. Only $P=0$ is admissible. Why this step? Cost-to-go cannot be negative; $P=-2$ is rejected.
3. Gain: $K=P=0$, so $u^{\star }=0$. Closed loop $=A=-1$. Why this step? With no state penalty and a stable plant, the optimal action is do nothing and let the plant decay for free.

Verify: $P=0\Rightarrow J=x^{\top }Px=0$: with a decaying plant and no penalties, the total cost really is zero. ✅

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Ex 5 — Cell F: Kalman clean vs noisy sensor

Forecast: which sensor makes us lean on the measurement (big $L$) and which makes us trust the model (small $L$)?

1. FARE for scalars: $2A\Sigma -{\Sigma }^2{C}^2/V+W=-2\Sigma -{\Sigma }^2/V+1=0$, i.e. ${\Sigma }^2+2V\Sigma -V=0\Rightarrow \Sigma =-V+\sqrt{V^2+V}.$ Why this step? FARE is the dual of CARE (swap $A\to A^{\top }$, $B\to C^{\top }$, $Q\to W$, $R\to V$).
2. Clean sensor $V=0.01$: $\Sigma =-0.01+\sqrt{0.0001+0.01}=-0.01+\sqrt{0.0101}\approx 0.0905$. Then $L=\Sigma C{V}^{-1}=0.0905/0.01\approx 9.05$. Why this step? Tiny $V$ means the sensor is trustworthy — large $L$ pulls the estimate hard toward $y$.
3. Noisy sensor $V=100$: $\Sigma =-100+\sqrt{10000+100}=-100+\sqrt{10100}\approx 0.4988$. Then $L=0.4988/100\approx 0.004988$. Why this step? Huge $V$ means the sensor lies — small $L$ tells the filter to trust its model.

Verify: $L(V=0.01)\approx 9.05$ is roughly $1800\times$ larger than $L(V=100)\approx 0.005$ — monotone with sensor quality, exactly as "large $V$ → small $L$" predicts. ✅

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Ex 6 — Cell G: 2×2 plant, separation gives union of poles

Forecast: because everything is diagonal, each axis is an independent scalar problem. Predict that the 4 poles are just two copies of a scalar answer, once per axis.

1. Diagonal ⇒ two scalar CAREs. Axis 1: $-2P-P^2+1=0\Rightarrow P=-1+\sqrt{2}$, $K_1=P\approx 0.414$, pole $-1-0.414=-1.414$. Why this step? A diagonal system decouples the state-feedback problem into independent channels.
2. Axis 2: $-4P-P^2+1=0\Rightarrow P=-2+\sqrt{5}\approx 0.236$, $K_2\approx 0.236$, pole $-2-0.236=-2.236$. Why this step? Same, with $A=-2$.
3. Estimator poles by duality: $C=I,W=V=I$ mirrors the CAREs, giving $L_1\approx 0.414,L_2\approx 0.236$ and estimator poles $-1.414,-2.236$. Why this step? FARE = CARE here (symmetric numbers), so estimator poles equal controller poles.
4. Union of poles. By the block-triangular structure the LQG poles are $\{-1.414,-2.236\}\cup \{-1.414,-2.236\}.$ Why this step? The $0$ block in $\left[\begin{array}{cc}A-BK& BK\\ 0& A-LC\end{array}\right]$ makes eigenvalues the union — the separation principle, applied.

Verify: all four poles are strictly negative → closed loop stable, and the set is exactly (controller poles) ∪ (estimator poles). ✅

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Ex 7 — Cell H: real-world word problem (spacecraft rate damping)

Forecast: the axis already de-spins slowly ($A=-0.5$). Will the controller add strong or gentle damping given the heavy $Q=4$?

1. Translate physics to matrices: $A=-0.5,B=1,C=1$. Rate is the state; torque is the input. Why this step? GNC problems are just state-space in disguise — identify $A,B,C$ first.
2. CARE: $2AP-P^2+Q=-P-P^2+4=0\Rightarrow P^2+P-4=0\Rightarrow P=\frac{-1+\sqrt{17}}{2}\approx 1.5616$. Gain $K=P\approx 1.5616$. Closed loop $=-0.5-1.5616=-2.0616$. Why this step? Heavy $Q=4$ demands tight rate control, so we expect a fast pole — confirmed.
3. FARE: $2A\Sigma -{\Sigma }^2/V+W=-\Sigma -{\Sigma }^2+0.25=0\Rightarrow {\Sigma }^2+\Sigma -0.25=0\Rightarrow \Sigma =\frac{-1+\sqrt{2}}{2}\approx 0.2071$. Kalman gain $L=\Sigma /V\approx 0.2071$. Estimator pole $-0.5-0.2071=-0.7071$. Why this step? Gyro noise ($V=1$) is comparable to process noise, so a moderate $L$ blends model and sensor.
4. Assemble the controller: run $\dot{\hat{\omega }}=-0.5\hat{\omega }+u+L(y-\hat{\omega })$, then command $u=-K\hat{\omega }$. Why this step? Separation lets us bolt the two pieces together directly.

Verify: controller pole $-2.06$ (fast damping, as heavy $Q$ demands) and estimator pole $-0.707$ are both stable; units of $K$ are $[\text{torque/inertia}]/[\text{rate}]={\text{s}}^{-1}$, matching a rate-damping gain. ✅

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Ex 8 — Cell I: exam twist, degenerate/uncontrollable $B$

Forecast: with $B=0$ the control does nothing. Can LQR possibly stabilize an unstable, uncontrollable plant?

1. Plug $B=0$ into CARE: $2AP-PB{R}^{-1}{B}^{\top }P+Q=2P-0+1=0\Rightarrow P=-\frac{1}{2}$. Why this step? The quadratic term vanishes because $B=0$; we're left with a linear equation.
2. Reject the root. $P=-\frac{1}{2}<0$ is not a valid cost-to-go — there is no $P⪰0$. Why this step? No admissible $P$ means the infinite-horizon cost is $+\infty$: the unstable mode is uncontrollable and cannot be regulated.
3. Interpret: LQR requires $(A,B)$ stabilizable. With $B=0$ and $A=1>0$, the diverging mode is untouchable, so $J={\int }_0^{\infty }{x}^{2}dt=\infty$ for any input. Why this step? This is the standard exam gotcha: check stabilizability before trusting a Riccati solution.

Verify: the only algebraic root $P=-1/2$ violates $P⪰0$; consistent with $J=\infty$ for an unstable uncontrollable mode. ✅

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Recall Quick self-test

Scalar $A=2,B=1,Q=1,R=1$: what is $K$? ::: $K=2+\sqrt{5}\approx 4.236$ (Ex 2) Mirror-pole limit $R\to \infty$ for $A=1$: closed-loop pole? ::: $-1$ (reflect $+1$) (Ex 3) $Q=0$, $A=-1$: optimal control? ::: $u=0$, do nothing (Ex 4) Huge sensor noise $V$: is $L$ big or small? ::: small — trust the model (Ex 5) $B=0,A>0$: does LQR have a solution? ::: no — uncontrollable unstable mode, $J=\infty$ (Ex 8)