State-space introduction
Introducido| English | Significado |
|---|---|
| state space | espacio de estados |
| eigenvalues/ˈaɪɡənvæljuːz/ | eigenvalues |
| transition matrix/trænˈsɪʃn ˈmeɪtrɪks/ | matriz de transición |
| controllability | controlabilidad |
Four loops at once
- A balancing robot or a drone runs several loops at once. One transfer function, one input and one output, cannot describe it.
- State space 状态空间 models every internal variable together, and turns design into linear algebra.
The state equations
- Write the system as $n$ first-order equations: $\dot{x} = Ax + Bu$ and $y = Cx + Du$.
- The states are $n$ stores of memory. $A$ says how they push each other, $B$ where the input enters, and $C$ what you measure.
Match each matrix to its job.
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
D, usually zero, is a direct path from input to output.
Eigenvalues are poles
- $\det(sI - A) = 0$ is the characteristic equation, so the eigenvalues 特征值 of $A$ are the system's poles.
- With no input, $x(t) = \Phi(t)\,x(0)$. The transition matrix 状态转移矩阵 $\Phi(t) = e^{At} = I + At + \dfrac{(At)^{2}}{2!} + \cdots$.
In companion form, s² + 5s + 6 gives A =
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
The top row shifts the states; the bottom row holds the negated coefficients, last first. Its eigenvalues are −2 and −3.
Worked example: a companion model
- $G = \dfrac{8}{s^{2} + 6s + 8}$ gives $A = \begin{bmatrix} 0 & 1 \\ -8 & -6 \end{bmatrix}$, $B = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$, $C = [8 \;\; 0]$.
- Checks: the eigenvalues of $A$ are −2 and −4, the poles; and the DC gain $-CA^{-1}B = 8/8 = 1$ matches $G(0)$.
G = 6/(s² + 5s + 6) in companion form has C = [6 0]. What DC gain does −C·A^(−1)·B give?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
It must match $G(0) = 6/6 = 1$. The check catches a wrong C at once.
Worked example: the free response
- By the series, $\Phi(0.5) = \begin{bmatrix} 0.600 & 0.116 \\ -0.930 & -0.097 \end{bmatrix}$. Its eigenvalues are $e^{-1} = 0.368$ and $e^{-2} = 0.135$: each pole decays on its own.
- Every trajectory bends onto the slow direction $[1, -2]^{T}$, so the $-2$ pole owns the tail of the response.
![State trajectories of 8/[(s+2)(s+4)] in the phase plane, bending onto the slow eigenvector.](/handout-media/university_electrical_engineering/assets/10-state-portrait.png?v=1793069544)
A pole at −2 becomes which eigenvalue of Φ(1)? Give 3 decimal places.
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$e^{-2 \times 1} = 0.135$. Each pole becomes its own decay factor over the time step.
Pole placement
- Full-state feedback $u = -K_{fb}x$ gives the closed loop $A - BK_{fb}$, and you choose its characteristic polynomial directly.
- It works when the input can reach every state direction: controllability 能控性, checked by the rank of $[B \;\; AB \;\; \cdots]$.
With A = [[0, 1], [−8, −6]] and B = [0, 1]ᵀ, the system is controllable.
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
$AB = [1, -6]^{T}$, and $[B \;\; AB]$ has rank 2, so every pole can be placed.
Worked example: both poles at −10
- $A - BK_{fb}$ has characteristic polynomial $s^{2} + (6 + k_{2})s + (8 + k_{1})$. Match $s^{2} + 20s + 100$: $k_{1} = 92$, $k_{2} = 14$.
- $[B \;\; AB]$ has rank 2, so both poles can be placed. Root locus searched along a ray; this solves two equations.
Place both poles of the same model at −5 (target s² + 10s + 25). What is k₁?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$8 + k_{1} = 25$ gives $k_{1} = 17$; $6 + k_{2} = 10$ gives $k_{2} = 4$.
Three common slips
- Forgetting $C$. The transfer function is $C(sI - A)^{-1}B + D$; dropping $C$ changes the gain.
- $e^{At}$ entry by entry. The matrix exponential comes from the series, not from each entry.
- Placing poles without checking controllability. If the rank drops, no $K_{fb}$ exists.
The model's C changes from [8 0] to [1 0]. What changes?
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
C only chooses what is measured. The poles are set by A; the gain is $-CA^{-1}B$.
Recap
- $\dot{x} = Ax + Bu$, $y = Cx + Du$; the eigenvalues of $A$ are the poles.
- $e^{At}$ is the free response.
- $u = -K_{fb}x$ places the poles, when the system is controllable.