Root locus of k(s+4)/[s(s²+2s+2)] and stable k

$$G(s) = \frac{k(s+4)}{s(s^2+2s+2)}$$

Characteristic equation: $s^3 + 2s^2 + (2+k)s + 4k = 0$.

Root locus construction

1. Poles: $0,\ -1 \pm j1$. Zero: $-4$. With $n = 3$ and $m = 1$, two branches go to infinity.
2. Real-axis locus: $[-4, 0]$. The pole at $0$ travels to the zero at $-4$.
3. Asymptotes:

$$\sigma_A = \frac{(0 - 1 - 1) - (-4)}{2} = +1, \qquad \theta_A = \pm 90^\circ$$

The asymptotes lie in the RHP at $\sigma = +1$, so the complex branches will cross the $j\omega$-axis. 4. Breakaway: $dk/ds = 0 \Rightarrow 2s^3 + 14s^2 + 16s + 8 = 0$. Roots: $-5.72$ (gives $k < 0$) and $-0.64 \pm j0.54$ (complex). Hence no breakaway point for $k > 0$. 5. Angle of departure from $-1 + j1$: from zero $-4$: $18.43^\circ$; from pole $0$: $135^\circ$; from pole $-1-j1$: $90^\circ$.

$$\phi_d = 180^\circ + 18.43^\circ - (135^\circ + 90^\circ) = -26.57^\circ$$

6. $j\omega$ crossing (Routh):

$k_{mar} = 2$. Auxiliary equation: $2s^2 + 8 = 0 \Rightarrow s = \pm j2$.

                               | ..
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                               ..
                              .X
                             ..|
                        x..... |
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------o........................x------------------
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                        x..... |
                             ..|
                              .X
                               ..
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     -4                -1      0     1
x pole   o zero (-4)
X jw crossing (+-j2, k = 2)

The complex poles leave at $\mp 26.6^\circ$, curve right, cross the $j\omega$-axis at $\pm j2$, and approach the vertical asymptotes at $\sigma = +1$.

Range of k for stability

First column positive: $4k > 0$ and $2 - k > 0$.

Answer: The system is stable for 0 < k < 2. At k = 2 it oscillates at ω = 2 rad/s; for k > 2 it is unstable.