R-H root distribution of s⁶+3s⁵+4s⁴+6s³+5s²+3s+2

Result: the system is unstable. It has 2 roots in the LHP, a repeated pair on the $j\omega$ axis at $\pm j1$, and no roots in the RHP.

The Routh–Hurwitz criterion states that the number of roots in the right half of the s-plane equals the number of sign changes in the first column of the Routh array. A row of all zeros means there are roots placed symmetrically about the origin; these are found from the auxiliary equation.

$$q(s) = s^6+3s^5+4s^4+6s^3+5s^2+3s+2 = 0$$

All coefficients are present and positive, so the necessary condition is satisfied.

Routh array

The $s^3$ row is all zeros. Auxiliary equation from the $s^4$ row:

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

Replace the $s^3$ row by $8,\ 8$ and continue:

The $s^1$ row is zero again. New auxiliary equation $A_1(s) = 2s^2+2$, $\frac{dA_1}{ds} = 4s$:

First column: $1,\ 3,\ 2,\ 8,\ 2,\ 4,\ 2$. No sign change, so no root is in the right half plane.

Roots of the auxiliary equation

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

So $\pm j1$ is a repeated pair on the imaginary axis. Dividing $q(s)$ by $(s^2+1)^2$ leaves $s^2+3s+2 = (s+1)(s+2)$.

Location of roots

Comment on stability

There is no RHP root, but the imaginary-axis roots are repeated. Repeated poles on the $j\omega$ axis give terms like $t\sin t$ in the response, which grow without limit. Therefore the system is unstable. If the $j\omega$ roots were non-repeated, it would only be marginally stable.