Root locus of k(s−1)/[s(s−1)]

$$G(s)H(s) = \frac{k(s-1)}{s(s-1)}$$

The factor $(s-1)$ appears in both numerator and denominator. It must not be cancelled before forming the characteristic equation, because the cancelled pole at $s = +1$ is still a real mode of the system.

Characteristic equation

$$\begin{aligned} 1 + G(s)H(s) &= 0 \\ s(s-1) + k(s-1) &= 0 \\ (s-1)(s+k) &= 0 \end{aligned}$$

Closed-loop poles: $s = +1$ (for every $k$) and $s = -k$.

Root locus construction

• Open-loop poles: $0$ and $+1$. Open-loop zero: $+1$.
• At $s = +1$ the pole and zero coincide, so a branch "starts and ends" at the same point. One closed-loop pole stays fixed at $s = +1$ for all $k$.
• The reduced part $k/s$ has a pole at $0$ and no finite zero. Its real-axis locus is the whole negative real axis, with asymptote angle $180^\circ$.
• That branch starts at $s = 0$ ($k = 0$) and moves left along the negative real axis, with the pole at $s = -k$ (e.g. $-5$ for $k = 5$). There is no breakaway point and no $j\omega$ crossing.

               jw
               |
 <=============x-----⊗------> σ
  s = -k       0    +1
 (moves left)       (fixed pole,
                    pole-zero overlap)

Comments

• If the pair is cancelled first, one gets $G = k/s$, which seems stable for all $k > 0$. That conclusion is wrong.
• The closed-loop pole at $s = +1$ lies in the right half-plane for every $k$. The system is therefore unstable for all values of $k$. Its output contains a growing mode $e^{t}$ that the gain cannot move.
• Pole-zero cancellation in the right half-plane is never used to stabilise a system. Exact cancellation is impossible in practice, and the hidden unstable mode still exists internally.

Answer: The root locus is a fixed point at $s = +1$ plus a branch from $0$ to $-\infty$ on the negative real axis. The system is unstable for all $k$.