Stability Analysis Practice Questions

$\frac{Y(s)}{R(s)}=\frac{\frac{K}{s}}{1-\left(\frac{3}{s}+\frac{K}{s}\right)}=\frac{K}{s-(3+K)}$ For system to be stable,

$$\begin{aligned} 3+K& \lt 0 \\ \Rightarrow \quad K& \lt -3 \end{aligned}$$

Routh table is

$$\begin{array}{c|ccc} s^{5} & 2(1) & 4(2) & 2(1) \\ s^{4} & 1 & 2 & 1 \\ s^{3} & 0 & 0 & 0 \\ s^{2} & & & \\ s^{1} & & & \\ s^{0} & & & \end{array}$$ $$\begin{aligned} \therefore \frac{d}{d s}\left(s^{4}+2 s^{2}+1\right) &=0 \\ 4 s^{3}+4 s &=0 \\ s &=\pm j, s=\pm j \\ \frac{d}{d s}\left(s^{2}+1\right) &=0 \\ 2 s &=0 \\ \Rightarrow s &=0 \end{aligned}$$

Double roots on imaginary axis so system is unstable

$$\begin{aligned} T . F . &=\frac{G_{1} G_{2}}{1+G_{1} G_{2} H} \\ &=\frac{\frac{K(s-2)}{(s+2)^{2}}}{1+\frac{K(s-2)(s-2)}{(s+2)^{2}}} \\ &=\frac{K(s-2)}{(s+2)^{2}+K(s-2)^{2}} \end{aligned}$$

$\therefore$ Characteristic equation $=s^{2}+4+4 s+K s^{2}-4 K s+4 K$ $(1+K) s^{2}+K(4-4 K) s+4 K+4=0$ Routh table is

$$\begin{array}{l|ll} s^{2} & (1+K) & (4 K+4) \\ s^{1} & K(4-4 K) & 0 \\ s^{0} & 4(K+1) & \end{array}$$

For system to be stable,

$$\begin{aligned} 1-K& \gt 0\\ 1& \gt K \text{ and } K \geq 0 \quad \text{ (given \in question)} \end{aligned}$$

From third row,

$$\begin{aligned} K& \gt -1 \\ \therefore \quad 0 &\leq K \lt 1 \end{aligned}$$