Find K and b for K(s+1)/(s³+bs²+3s+1)

Result (assuming oscillation at 2 rad/s): $K = 1$, $b = 0.5$. In general, $K = \frac{3b-1}{1-b}$ and $\omega=\sqrt{\frac{2}{1-b}}$.

Assumption: the end of the question is missing. It is taken in its usual form: "...so that the system oscillates at a frequency of 2 rad/s". The general relations are also derived, so any other given frequency can be substituted.

Characteristic equation

$$\begin{aligned} s^3+bs^2+3s+1 + K(s+1) &= 0 \\ s^3 + bs^2 + (3+K)s + (1+K) &= 0 \end{aligned}$$

Routh array

Conditions for sustained oscillation

The $s^1$ row must be zero:

$$b(3+K) = 1+K \qquad \text{...(1)}$$

Auxiliary equation (from the $s^2$ row) with $s=j\omega$:

$$bs^2 + (1+K) = 0 \Rightarrow \omega^2 = \frac{1+K}{b} \qquad \text{...(2)}$$

From (1) and (2): $\omega^2 = 3 + K$.

Substitute $\omega = 2$ rad/s

$$\begin{aligned} 4 &= 3 + K \Rightarrow K = 1 \\ b &= \frac{1+K}{\omega^2} = \frac{2}{4} = 0.5 \end{aligned}$$

Check: $s^3+0.5s^2+4s+2 = (s^2+4)(s+0.5)$. The roots are $\pm j2$ and $-0.5$, so the system oscillates at 2 rad/s.

Answer: $K = 1$, $b = 0.5$. In general: $K = \frac{3b-1}{1-b}$ and $\omega = \sqrt{\frac{2}{1-b}}$. The system is stable when $b(3+K) > 1+K$.