Old spec 3.5 example, fourth order Hurwitz determinantTutorialOld spec ENGR2022:210 min

ENGR202 (old spec) Control 3.5 Hurwitz Stability Criterion, fourth order example (slides 4-6)

A system has the characteristic equation 2.5s4+1.3s3+0.02s2−s+5=02.5s^4 + 1.3s^3 + 0.02s^2 - s + 5 = 0 Write down the 4th order Hurwitz determinant and use its principal minors to determine the stability.

Formulas you may need
  • Hurwitz determinant for a0sn+a1sn−1+⋯+an=0a_0s^n + a_1s^{n-1} + \dots + a_n = 0: rows (a1,a3,a5,… )(a_1, a_3, a_5, \dots), (a0,a2,a4,… )(a_0, a_2, a_4, \dots), then the same pair shifted one column right, and so on; stable if all principal minors Δ1,…,Δn>0\Delta_1, \dots, \Delta_n > 0 (learn this)
  • Necessary condition: all coefficients present and of the same sign (learn this)