R-H stability of s⁵+2s⁴+2s³+4s²+s+2

Result: RHP = 0, $j\omega$ axis = 4 ($\pm j1$ repeated), LHP = 1 ($s=-2$). The system is unstable.

The Routh array is formed. A row of zeros is handled with the auxiliary polynomial, whose roots are the roots symmetric about the origin.

$$q(s) = s^5 + 2s^4 + 2s^3 + 4s^2 + s + 2 = 0$$

Routh array

Zero row. Auxiliary polynomial from the $s^4$ row:

$$A(s) = 2s^4 + 4s^2 + 2, \qquad \frac{dA}{ds} = 8s^3 + 8s$$

New auxiliary polynomial $A_1(s) = 2s^2+2$, $\frac{dA_1}{ds} = 4s$:

First column: $1,\ 2,\ 8,\ 2,\ 4,\ 2$. There is no sign change, so no pole is in the RHP.

Poles on the imaginary axis

$$2s^4 + 4s^2 + 2 = 2(s^2+1)^2 = 0 \Rightarrow s = \pm j1,\ \pm j1$$

The remaining factor is $\frac{q(s)}{(s^2+1)^2} = s + 2$, giving $s = -2$.

Stability

The poles on the $j\omega$ axis are repeated, so the response contains $t\sin t$ terms that grow with time. The system is unstable, even though no pole is in the RHP.