Stability of 56/[s(s⁴+7s³+6s²+42s+8)] feedback system

Result: the system is marginally stable. It has roots at $\pm j\sqrt2$ and $\pm j2$, one root at $s=-7$, and none in the RHP.

For unity negative feedback, the characteristic equation is $1+G(s)=0$. Its Routh array is formed and the first column is examined.

Characteristic equation

$$\begin{aligned} s(s^4+7s^3+6s^2+42s+8) + 56 &= 0 \\ s^5 + 7s^4 + 6s^3 + 42s^2 + 8s + 56 &= 0 \end{aligned}$$

Routh array

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

$$A(s) = 7s^4 + 42s^2 + 56, \qquad \frac{dA}{ds} = 28s^3 + 84s$$

First column: $1,\ 7,\ 28,\ 21,\ 9.33,\ 56$. No sign change, so no root is in the RHP.

Roots on the imaginary axis

$$7s^4+42s^2+56 = 7(s^2+2)(s^2+4) = 0 \Rightarrow s = \pm j1.414,\ \pm j2$$

The fifth root: $\frac{q(s)}{(s^2+2)(s^2+4)} = s+7$, so $s=-7$.

Conclusion

The poles on the $j\omega$ axis are simple (non-repeated), and none are in the RHP. The system is marginally stable: its response contains sustained oscillations at 1.414 rad/s and 2 rad/s.