---
title: 'Advanced: state feedback and observers'
url: https://doc.liz6.com/en/theory/02-control-theory/10-state-feedback-and-observers
locale: en
area: theory
tags:
- Control theory
- Theory
date: 2026-09-10
modified: 2026-09-10
description: A thermal model may need only temperature, but motion needs position and velocity even when the sensor reports position alone. This chapter develops state feedback and estimation using dynamic models, poles and basic matrix operations.
---
# Advanced: state feedback and observers
A thermal model may need only temperature, but motion needs position and velocity even when the sensor reports position alone. This chapter develops state feedback and estimation using [dynamic models](02-dynamic-models.md), [poles](05-transfer-functions-and-poles.md) and basic matrix operations.
## Two first-order equations
A unit-mass, undamped motion plant satisfies $\ddot q=u$. Define $x=[q,v]^\mathsf T$, $v=\dot q$:
$$\dot x=Ax+Bu,\quad y=Cx,\qquad A=\begin{bmatrix}0&1\\0&0\end{bmatrix},\quad B=\begin{bmatrix}0\\1\end{bmatrix},\quad C=\begin{bmatrix}1&0\end{bmatrix}.$$
The first row advances position using velocity; the second advances velocity using acceleration. One measured output does not reduce the system to one state. Here $C$ is an output matrix, not the controller transfer function used earlier.
## Controllability and stabilizability
The controllability matrix is $\mathcal C=[B,AB,\ldots,A^{n-1}B]$. Here
$$\mathcal C=\begin{bmatrix}0&1\\1&0\end{bmatrix}$$
has rank 2. Inputs can influence both states. This finite-time reachability result concerns an unconstrained linear model; bounded actuators cannot achieve arbitrary transitions arbitrarily fast.
For $A=\operatorname{diag}(-1,1)$ and $B=[1,0]^\mathsf T$, the second unstable state cannot be influenced. No first-channel feedback can stabilize it. Stabilizing feedback only requires stabilizability: uncontrollable modes must already be stable, a weaker condition than full controllability.
## Place state-feedback poles
Set $u=-Kx=-k_1q-k_2v$. The characteristic polynomial of $A-BK$ is
$$s^2+k_2s+k_1.$$
Desired poles −1 and −2 give $s^2+3s+2$, hence $K=[2,3]$. Position feedback restores the origin; velocity feedback supplies damping. This is regulation to zero. Tracking a nonzero reference needs a feasible reference equilibrium and feedforward, or integral augmentation.
## Observability and an observer
The observability matrix is $\mathcal O=[C;CA;\ldots;CA^{n-1}]$. With position measurement,
$$\mathcal O=\begin{bmatrix}1&0\\0&1\end{bmatrix}$$
has rank 2. One position measurement cannot determine velocity, but its time history together with known input can. Measuring only velocity, $C=[0,1]$, gives rank 1: an initial absolute-position offset is invisible.
Differencing noisy positions can amplify noise. An observer predicts with the model and corrects using measurement innovation:
$$\dot{\hat x}=A\hat x+Bu+L(y-C\hat x).$$
For exact models without noise, error $\tilde x=x-\hat x$ obeys $\dot{\tilde x}=(A-LC)\tilde x$. Choosing $L=[9,20]^\mathsf T$ places error poles at −4 and −5.
**Advanced: state feedback and observers · Experiment**
Double integrator, K=[2,3], true initial state [1,0], estimate [0,0]. L=[2α,α²] gives repeated poles −α. Measurement noise is a fixed-amplitude sin(30t), not white noise; no action saturation.
The true state starts at position 1, velocity 0; the estimate starts at zero. Control uses $u=-K\hat x$. Vary observer speed and compare true and estimated states and action. Optional noise is a fixed sinusoid for reproducibility, not random white noise.
## What separation guarantees
In coordinates $[x,\tilde x]$,
$$\frac d{dt}\begin{bmatrix}x\\\tilde x\end{bmatrix}=\begin{bmatrix}A-BK&BK\\0&A-LC\end{bmatrix}\begin{bmatrix}x\\\tilde x\end{bmatrix}.$$
This block-triangular matrix has the union of controller and observer-error eigenvalues. The exact unconstrained LTI model therefore permits separate designs. Saturation, model errors, delays and noise need additional analysis.
Faster observer poles typically mean stronger measurement gains and more noise in estimates and action. Kalman filtering chooses gains from explicit process and measurement covariance models; its optimality depends on the stated assumptions. Detectability permits asymptotic estimation when any unobservable modes are already stable.
## Check your understanding
Can velocity alone reveal the absolute position of a stationary object?
Reasoning
No. Different position offsets with identical velocity produce identical outputs. Add a position reference or acknowledge the unobservable state.
Reasoning
The polynomial is $(s+\alpha)^2$, giving repeated poles at $-\alpha$. Gains grow with $\alpha$ and $\alpha^2$, so noise, discretization and model uncertainty limit useful speed.