Root Locus and Stability Range for $GsHs=\frac{k}{ss^2+2s+5}$ | AI Research

Root Locus and Stability Range for $G(s)H(s)=\frac{k}{s(s^2+2s+5)}$

Verified Sources

Nov 4, 2026

We are given an open-loop transfer function
$G(s)H(s)=\frac{k}{s(s^2+2s+5)}=\frac{k}{s(s^2+2s+5)}.$

The root locus is the set of closed-loop pole locations as the gain $k$ varies. The closed-loop poles satisfy the characteristic equation $1+G(s)H(s)=0 \;\Rightarrow\; 1+\frac{k}{s(s^2+2s+5)}=0.$

Multiplying through by $s(s^2+2s+5)$ gives the characteristic polynomial $s(s^2+2s+5)+k=0 \;\Rightarrow\; s^3+2s^2+5s+k=0.$

Key terms: Root locus Characteristic equation Routh-Hurwitz Asymptotes.

Root Locus Fundamentals (rules + sketching)

1) Open-loop poles/zeros and basic root-locus facts

Write the loop transfer as $L(s)=G(s)H(s)=\frac{k}{s(s^2+2s+5)}.$

Open-loop poles come from the denominator:

$s=0$
$s^2+2s+5=0 \Rightarrow s=-1\pm j2$

There are no finite zeros (since the numerator is just $k$ ).

So:

Open-loop poles: $0,\,-1\pm j2$
Open-loop zeros: none

Number of poles is $n=3$ , number of zeros is $m=0$ , so the number of asymptotes is $n-m=3.$

The centroid (average location) of asymptotes is

=\frac{(0+(-1)+(-1))-0}{3}=\frac{-2}{3}.$$ Key root-locus quantities: - Number of asymptotes: $3$ - Centroid: $-\frac{2}{3}$

Step-by-step sketch of the root locus for $$s^3+2s^2+5s+k=0$$

1
Step 1
From $1+L(s)=0$ , obtain $s^3+2s^2+5s+k=0$ , so the closed-loop poles are the roots of this cubic for each $k\ge 0$ .
2
Step 2
At $k=0$ , the poles are exactly the open-loop poles: $s=0$ and $s=-1\pm j2$ .
3
Step 3
Real-axis points on the locus occur where the number of real-axis open-loop poles to the right is odd (and there are no real-axis zeros). Here the only real open-loop pole is at $s=0$ , so the real-axis locus can only lie on $(-\infty,0)$ .
4
Step 4
With $n-m=3$ , there are 3 asymptotes at angles $(2q+1)180^\circ/(n-m)$ for $q=0,1,2$ , i.e. $60^\circ, 180^\circ, 300^\circ$ measured from the positive real axis, all intersecting at the centroid $\sigma_a=-2/3$ .
5
Step 5
Use $k=-\,(s^3+2s^2+5s)$ from $s^3+2s^2+5s+k=0$ . Break points on the real axis satisfy $\frac{dk}{ds}=0$ i.e. $\frac{d}{ds}\left[-(s^3+2s^2+5s)\right]=0$ so $-\,(3s^2+4s+5)=0$ ⇒ $3s^2+4s+5=0$ .
6
Step 6
The discriminant is $\Delta=4^2-4\cdot 3\cdot 5=16-60=-44<0$ , so there are no real breakaway points; the complex-conjugate pair does not split on the real axis.
7
Step 7
Since there are no real-axis break points, one root moves along the real-axis segment $(-\infty,0)$ while the other two remain complex conjugates and move out along the other asymptote directions (from the $-1\pm j2$ starting locations).
8
Step 8
Sketch: start at $0$ (on real axis), and $-1\pm j2$ (complex pair). The real branch stays on $(-\infty,0)$ and heads toward the $180^\circ$ asymptote. The complex pair heads toward the $60^\circ$ and $300^\circ$ asymptotes emanating from centroid $-2/3$ .

Fast way to avoid heavy locus computations

Here, the polynomial is $s^3+2s^2+5s+k=0.$ Re-expressing $k=-(s^3+2s^2+5s)$ turns breakaway detection into solving $\frac{dk}{ds}=0$ , which is just a quadratic. If the quadratic has no real roots, there are no break points on the real axis.

2) Stability range of $k$ using Routh-Hurwitz

For the closed-loop characteristic polynomial $p(s)=s^3+2s^2+5s+k,$ the Routh array for a cubic $a_3 s^3+a_2 s^2+a_1 s+a_0$ gives the necessary and sufficient stability conditions:

$a_3>0$
$a_2>0$
$a_1>0$
$a_2 a_1 > a_3 a_0$

Here:

$a_3=1>0$
$a_2=2>0$
$a_1=5>0$
$a_0=k$

So the only nontrivial condition is $a_2 a_1 > a_3 a_0 \;\Rightarrow\; (2)(5) > (1)k \;\Rightarrow\; 10>k.$

Thus, for (strict) asymptotic stability: $k<10.$

What about $k=10$ ? Then $a_2 a_1=a_3 a_0$ so the Routh row yields a zero in the auxiliary condition, indicating poles on the imaginary axis (marginal stability). So:

Stable for $0\le k<10$ (and more generally for $k<10$ with $k$ not restricted to be nonnegative)
Marginal at $k=10$
Unstable for $k>10$

Key term: Asymptotic stability.

Stability regions vs gain k

Using cubic stability condition $2\cdot 5 > k$ .

How the poles move as k increases

Starting poles

k = 0

Poles at $s=0$ and $s=-1\pm j2$ ."

Stable motion

0 < k < 10

One pole moves along the real-axis locus on $(-\infty,0)$ ; the complex pair stays in the left-half plane."

Imaginary-axis crossing

k = 10

Marginal stability occurs (Routh condition becomes equality)."

Unstable motion

k > 10

At least one closed-loop pole moves into the right-half plane."

Common checks and edge cases

Root locus + stability quick self-test

1 / 4

Question · Term

Characteristic equation for this system

Click to reveal

Answer · Definition

$s^3+2s^2+5s+k=0$

Knowledge Check

Question 1 of 4

Q1Single choice

The closed-loop characteristic polynomial for $G(s)H(s)=\frac{k}{s(s^2+2s+5)}$ is:

$s^3+2s^2+5s+k=0$

$s^3+2s^2+5s-k=0$

$s^2+2s+5+k=0$

$s^4+2s^3+5s^2+k=0$

Explore Related Topics

LQR Optimal Control Input for $Q=I$, $R=1$

Simple Servo System Block Diagram: Finding $a$, $b$ from Overshoot and Time Constant, then $t_r$, $t_p$, $\omega_d$, $t_{ss}$

Transfer Function $C/R$ from a Signal Flow Graph (SFG) Using Mason’s Gain Formula

Browse all research articles

Root Locus and Stability Range for G(s)H(s)=ks(s2+2s+5)G(s)H(s)=\frac{k}{s(s^2+2s+5)}G(s)H(s)=s(s2+2s+5)k​