Cho hệ có phương trình đặc trưng \({s^3} + 20{s^2} + 10s + 100 = 0\)

Hệ thống không ổn định, có 3 nghiệm bên phải mặt phẳng phức

Hệ thống ổn định, không có nghiệm có phần thực dương

Hệ thống không ổn định, có 2 nghiệm bên phải mặt phẳng phức, 1 nghiệm bên trái mặt phẳng phức

Hệ thống không ổn định, có 1 nghiệm bên phải mặt phẳng phức, 2 nghiệm bên trái mặt phẳng phức

Đáp án đúng: B

The Routh-Hurwitz criterion is used to determine the stability of the system. The characteristic equation is \(s^3 + 20s^2 + 10s + 100 = 0\). We construct the Routh table as follows: | s^3 | 1 | 10 | |-----|-----|-----| | s^2 | 20 | 100 | | s^1 | 10 - (100/20) = 5 | 0 | | s^0 | 100 | | There are no sign changes in the first column (1, 20, 5, 100). However, note that for a 3rd order polynomial, we need all positive coefficients for stability. In this case, we have all positive coefficients in the first column of the Routh array. A more accurate conclusion would be that the system is not stable, and this is because the auxiliary equation is formed from the s^2 row: 20s^2 + 100 = 0, whose roots are +/- j*sqrt(5). To explain further, consider the auxiliary polynomial formed from the s^2 row: \(20s^2 + 100 = 0\). This implies \(s^2 = -5\), thus \(s = \pm j\sqrt{5}\). The roots are imaginary. The presence of roots on the imaginary axis (jω axis) indicates marginal stability, meaning oscillations will neither decay nor grow in amplitude. Since the question indicates an "unstable" system, the existence of imaginary roots does imply instability (or at least, not strictly stable). Therefore, the system is marginally stable (oscillatory) but can also be considered unstable for design purposes.