Old spec 3.5 examples, stability short cutsTutorialOld spec ENGR2022:210 min

ENGR202 (old spec) Control 3.5 Hurwitz Stability Criterion, stability short cut examples (slides 9-10)

Using the stability short cuts where possible (and the Hurwitz criterion or the poles where they are not enough), decide whether each characteristic equation is stable, marginally stable or unstable.

Formulas you may need
  • Necessary conditions for stability: all coefficients exist (non-zero) and have the same sign (learn this)
  • If these hold, the Hurwitz criterion (or the poles) is still needed (learn this)
  1. (a)
    s4+3s3+5s2+10=0s^4 + 3s^3 + 5s^2 + 10 = 0
  2. (b)
    s5+4s4+2s3+9s2+3s=0s^5 + 4s^4 + 2s^3 + 9s^2 + 3s = 0
  3. (c)
    5s2−2s+2=05s^2 - 2s + 2 = 0
  4. (d)
    −3s3−2s2−4s−10=0-3s^3 - 2s^2 - 4s - 10 = 0 (careful!)
  5. (e)
    (s+2)(s+3)=0(s + 2)(s + 3) = 0