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Self control principle learning notes - system stability analysis (1) - BIBO stability and Routh criterion

2022-07-27 19:03:00 【Miracle Fan】

Definition ： If a system is subjected to a bounded input or disturbance, its response is finite .
necessary and sufficient condition ：
$y(t)=\int_0^tg(\tau)u(t-\tau)d\tau\Rightarrow |y(t)|\le \int _0^t|g(\tau)|\cdot |u(t-\tau)|d\tau \le M\int_0^t|g(\tau)|d\tau$
If you want to make y（t） bounded , Then the necessary and sufficient condition is $|g(\tau)|$ Absolutely integrable
Non integrable examples ： $g(t)=\frac{1}{t-1}$
about CLTIS, Satisfy BIBO Then there is only the pole of the left half plane ：
Assume that the transfer function is impulse response Laplace Transformation , therefore ：
$G(s)=\int_0^\infty g(t)e^{-st}dt\Rightarrow|G(s)|\le\int_0^\infty |g(t)|\cdot|e^{-st}|dt=\int_0^\infty|g(t)|\cdot|e^{-\sigma t}|dt\\ \Longrightarrow if \quad\sigma \ge 0,\quad |G(s)||_{s=\sigma+iw}\rightarrow\infty\le\int^\infty_0|g(t)|\cdot|e^{-\sigma t}|dt\le\int_0^\infty|g(t)|dt$

$∣ g (t) ∣$ unbounded , And BIBO Stabilize contradictions , All only when $\sigma<0$ , Meet the conditions .

Definition ： When t It goes to infinity , The response produced by the initial conditions tends to 0.
Sufficient and necessary conditions for stability ：
- about $\forall s_i,Re(s_i)<0 when ,CLTIS Asymptotic stability$
- $Re(s_i)>0||\text{ There are multiple virtual roots },CLTIS unstable$
- There are only single virtual roots , other $Re(s_j)<0$ ,LTIS Critical delimitation .
Related examples ：

about LTIS,BIBO、 Zero input stability requires eigenvalues to be located on the left of the complex plane
about LTIS, Stability only depends on the inherent properties of the system （ The eigenvalue ）, It has nothing to do with external conditions .
Stability has a local characteristic （ Multiple stable points ）, But only in time-varying systems and nonlinear systems , Time invariant systems are global .

All coefficients of the characteristic equation of the system are greater than 0

List some Routh surface
$s^n\quad a_n \quad a_{n-2}\quad a_{n-4} \quad \dots$
$s^{n-1}\quad a_{n-1} \quad a_{n-3}\quad a_{n-5} \quad \dots$
$s^{n-2}\quad b_1 \quad b_2\quad b_3 \dots$
$s^{n-3}\quad c_1 \quad c_2\quad \dots$
…………
$s^0 \quad h_1$
Conventional calculation
$b_1=-\frac{ \begin{bmatrix} a_{n}& a_{n-2} \\ a_{n-1}& a_{n-3} \end{bmatrix} } {a_{n-1}} \quad b_2=-\frac{ \begin{bmatrix} a_{n}& a_{n-4} \\ a_{n-1}& a_{n-5} \end{bmatrix} } {a_{n-1}}\\ c_1=-\frac{ \begin{bmatrix} a_{n-1}& a_{n-3} \\ b_1&b_2 \end{bmatrix} } {b_1} \quad c_2=-\frac{ \begin{bmatrix} a_{n-1}& a_{n-5} \\ b_1&b_3 \end{bmatrix} } {b_1}$
Judge the stability method ：
The number of changes of the coefficient symbol in the first column , That is, the characteristic root is in the right half s Number of planes
Necessary and sufficient conditions for stability ： The coefficients in the first column of the table are all greater than 0

Use small $\epsilon$ Method instead of zero value term , Continue to solve according to the conventional method .
If the first column is all positive , The system is not asymptotically stable , But there are pure virtual roots , Critical stability

indicate s The plane has a real root symmetrical to the origin , Or conjugate virtual root

Take all as 0 Previous line of , Take its coefficient as the auxiliary equation （ Only take even times ）
Derivation of auxiliary equation , The coefficient substitution is 0 That's ok
Continue with the normal steps
Solve the auxiliary equation to get the symmetric root