Kalman Decomposition & State-Feedback Control

Recap of Previous Material

The instructor briefly reminded us that the last session dealt with minimal realizations and how lack of controllability or observability inevitably leads to pole–zero cancellations in the transfer function. A fully controllable‐and‐observable 4-state system will exhibit a 4th-degree denominator, whereas an uncontrollable and/or unobservable one will show a reduced order because some poles are cancelled by zeros.

Why Kalman Decomposition?

A single LTI state–space model can always be rearranged—by a similarity transformation—into four independent subsystems that capture every possible combination of controllability and observability. This comprehensive rearrangement is called Kalman decomposition. Formally, every state vector x can be partitioned as
x \;\rightarrow\; \begin{bmatrix}z{co} & z{c\bar o} & z{\bar c o} & z{\bar c\bar o}\end{bmatrix}^T
where each symbolic sub-script indicates whether the corresponding block is controllable ((c)), observable ((o)), or their negations ((\bar c,\,\bar o)).

Structural Properties of the Decomposition

  • Input connections: only the controllable blocks ((z{co}, z{c\bar o})) feel the input u.

  • Output connections: only the observable blocks ((z{co}, z{\bar c o})) appear in y.

  • Decoupling zeros: carefully placed zero sub-matrices ensure that unobservable blocks do not leak into the output and uncontrollable blocks cannot be actuated—neither directly nor through internal couplings.

In canonical Kalman form the state, input and output equations become

\dot z = \begin{bmatrix}
A{co} & 0 & 0 & 0\ A{21} & A{c\bar o}&A{23}&A{24}\ 0 & 0 &A{\bar c o}&0\
0 & 0 &A{43}&A{\bar c\bar o}
\end{bmatrix}z +
\begin{bmatrix}B{co}\B{c\bar o}\0\0\end{bmatrix}u, \qquad
y = \begin{bmatrix}C{co}&0&C{\bar c o}&0\end{bmatrix}z.
Every zero block is indispensable: they sever the unwanted pathways from inputs to uncontrollable states and from unobservable states to outputs.

Obtaining the Form

  1. Compute the controllability matrix P=[B\; AB\;A^2B\;\ldots] and the observability matrix Q=\begin{bmatrix}C^T & A^T C^T & (A^T)^2C^T & \ldots\end{bmatrix}^T.

  2. Replace missing column vectors (for rank-deficient P) and missing row vectors (for rank-deficient Q) by independent basis vectors to build a transformation T that simultaneously exposes the four subspaces.

  3. Combine the two earlier “half-decompositions” (controllable/uncontrollable and observable/unobservable) to reach the full Kalman structure.

MATLAB automates the entire routine via

[Ahat,Bhat,Chat,That] = minreal(ss(A,B,C,D));   % or specialised kalman(A,B,C)

Worked Example – The Instrumentation Bridge Circuit

A classic bridge network (four resistive/capacitive legs, excited by a current source I) serves as a realistic example of a system that is both uncontrollable and unobservable for a specific choice of input/output:

Capacitive values: C1=C3=1\,\text{F},\;C2=2\,\text{F}; states are the capacitor voltages v1,v2,v3.

After applying KCL/KVL the state model is
\dot x = \begin{bmatrix}
-1 & 1/2 & 0\
1 & -1/2& 0\
0 & 0 & 0\end{bmatrix}x + \begin{bmatrix}1\-1/2\0\end{bmatrix}I,\qquad y = \begin{bmatrix}1 & -1 & 0\end{bmatrix}x
where x=\begin{bmatrix}v1 & v2 & v_3\end{bmatrix}^T.

Controllability check
P=[B\;AB\;A^2B]=\begin{bmatrix}1 & -1/2 & 1/4\-1/2 & 1/4 & -1/8\0 & 0 & 0\end{bmatrix},\qquad\text{rank}(P)=1
→ uncontrollable.

Observability check
Q=\begin{bmatrix}C\CA\CA^2\end{bmatrix}=\begin{bmatrix}1 & -1 & 0\-1/2 & 1/2 & 0\1/4 & -1/4 & 0\end{bmatrix},\qquad\text{rank}(Q)=1
→ unobservable.

Automatic Kalman transformation (MATLAB) yields
\hat A=\begin{bmatrix}
-1/2 & 0 & 0\0 & 0 & 0\0 & 0 & 0\end{bmatrix},\;\hat B=\begin{bmatrix}1/2\0\0\end{bmatrix},\;\hat C=\begin{bmatrix}3 & 0 & 0\end{bmatrix}
showing that only (z2) is simultaneously controllable and observable; (z1,z_3) are neither. The transfer function therefore collapses after two pole–zero cancellations:
H(s)=\frac{s(s+2)}{s(s+0.5)(s+2)}.

From Decomposition to Control – State Feedback

Once a plant is accessible via all its states we can stabilise or reshape dynamics with the elementary law
u = -Kx + r
where K (size m\times n) is the state-feedback gain and r is an external reference. Injecting the law into \dot x = Ax+Bu gives
\dot x = (A-BK)x + Br,
so the closed-loop matrix is A_c=A-BK.

Stability Refresher

• All eigenvalues of A in the left‐half complex plane ⇒ asymptotically stable.
• Any eigenvalue in the right-half plane ⇒ unstable.
• Simple eigenvalues on the imaginary axis ⇒ marginally stable; repeated ones again cause instability.

The inverted pendulum was cited: its linearised A has two zeros, one right-half plane pole (the unstable pendulum mode) and one left-half plane pole (stable downward mode).

Observer–Based Control (preview)

If some states cannot be measured economically we build an observer:
\dot{\hat x}=A\hat x+Bu+L(y-C\hat x)
and close the loop with u=-K\hat x+r. That concept is postponed; focus remains on full-state feedback for now.

Example – 2-State Pole Placement

Plant:
A=\begin{bmatrix}0&1\0&0\end{bmatrix},\;B=\begin{bmatrix}-1\1\end{bmatrix},\;u=-Kx=-\begin{bmatrix}k1&k2\end{bmatrix}x.
Closed-loop matrix:
Ac=A-BK=\begin{bmatrix}k1 & k2+1\-k1 & -k2\end{bmatrix}. Characteristic polynomial: \det(sI-Ac)=s^2+(k2-k1)s+k1=0. Because both k1,k_2 appear, and because the original controllability matrix has full rank, any desired quadratic polynomial can be imposed—hence poles (eigenvalues) can be placed arbitrarily. The bidirectional theorem follows:

• If ((A,B)) is controllable then state feedback allows arbitrary pole placement.
• Conversely, if arbitrary pole placement is feasible, the pair must be controllable.

Choosing the Desired Poles

Design must balance three practical themes:

  1. Stability – all closed-loop eigenvalues strictly left of the imaginary axis.

  2. Transient quality – rise time, settling time, overshoot, peak time.

  3. Steady-state accuracy – small or zero final error (often handled by integral action, not covered yet).

First-Order Guideline

A canonical first-order transfer function
G(s)=\frac{1}{1+\tau s}
has pole s=-1/\tau and settles in roughly Ts\approx4\tau. Therefore to achieve a chosen Ts pick
s{desired}\approx-\frac{4}{Ts}.

Second-Order Guideline

Standard form
G(s)=\frac{\omegan^2}{s^2+2\zeta\,\omegan s+\omega_n^2}
provides closed-form links between pole locations and time-domain indices:

Tr\,\approx\,\frac{2.16(\zeta+0.6)}{\omegan},\qquad Tp=\frac{\pi}{\omegan\sqrt{1-\zeta^2}},
Ts\,\approx\,\frac{4}{\zeta\,\omegan},\qquad \%OS=100\,e^{-\pi\zeta/\sqrt{1-\zeta^2}}.

Poles themselves are roots of
s^2+2\zeta\omegan s+\omegan^2=0 \;\Rightarrow\; s{1,2}=-\zeta\omegan\pm j\,\omegan\sqrt{1-\zeta^2}. Geometrically, writing \cos\theta=\zeta shows that the real axis makes angle \theta (the damping angle) with the pole vector of length \omegan. Small \theta (large (\zeta)) → overdamped/slow but monotone; \theta\to90^\circ ((\zeta\to0)) → lightly-damped oscillations.

Design Workflow (Summary)

  1. Specify performance ((Ts,\,Tr,\,\%OS)).

  2. Translate specs into desired poles (first or second-order rules, or numerical optimisation for higher order).

  3. Verify controllability of ((A,B)).

  4. Solve the pole-placement problem (e.g.
    K=\text{place}(A,B,\text{poles}) in MATLAB).

  5. Simulate & iterate; add observer if necessary.

Closing Remarks

Kalman decomposition isolates which states are even eligible for control or observation; once full controllability is confirmed, state feedback offers a direct, systematic path to relocate eigenvalues and thereby enforce stability and desired transient behaviour. The next lecture will complete the discussion for (\zeta>1) (over-damped design) and continue toward full observer-based controllers.