Routh–Hurwitz stability range of a fourth-order loop: 22-Mec-A3 System Analysis and Control, May 2013, Q3

Question 3: Routh–Hurwitz stability range of a fourth-order loop (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Unity-feedback loop with

$$G(s)=\frac{K(s+2)}{s\,(s+1)\,(s^{2}+2s+2)}.$$

The plant has poles at $0,\,-1,\,-1\pm j1$ and a zero at $-2$.

Find. the range of $K$ for closed-loop stability.

Approach. Build the closed-loop characteristic polynomial $1+G=0$, then require every first-column entry of the Routh array to be positive.

Characteristic polynomial. Expanding $s(s+1)(s^{2}+2s+2)=s^{4}+3s^{3}+4s^{2}+2s$ and adding $K(s+2)$, $$\boxed{\,s^{4}+3s^{3}+4s^{2}+(2+K)s+2K=0\,}.$$
Routh array. Tabulating the coefficients:

$s^{4}$ $1$ $4$ $2K$

$s^{3}$ $3$ $2+K$ $0$

$s^{2}$ $b_1=\dfrac{12-(2+K)}{3}=\dfrac{10-K}{3}$ $2K$

$s^{1}$ $c_1=\dfrac{20-10K-K^{2}}{3\,b_1}$ $0$

$s^{0}$ $2K$
Positivity conditions. Requiring the first column $>0$:
- $s^{0}$: $2K>0\Rightarrow K>0$;
- $s^{2}$: $b_1>0\Rightarrow 10-K>0\Rightarrow K<10$;
- $s^{1}$: $20-10K-K^{2}>0\Rightarrow K^{2}+10K-20<0\Rightarrow K<-5+3\sqrt5=1.708.$
The $s^{1}$ row is the binding constraint (it fails well before $K=10$).
Stability range. Intersecting the conditions, $$\boxed{\,0<K<3\sqrt5-5\approx1.708\,}.$$ At $K=1.708$ the $s^{1}$ entry vanishes and a conjugate pole pair crosses onto the imaginary axis (marginal stability); above it the loop is unstable.

$s^{4}$	$1$	$4$	$2K$
$s^{3}$	$3$	$2+K$	$0$
$s^{2}$	$b_1=\dfrac{12-(2+K)}{3}=\dfrac{10-K}{3}$	$2K$
$s^{1}$	$c_1=\dfrac{20-10K-K^{2}}{3\,b_1}$	$0$
$s^{0}$	$2K$

Quantity	Value
Characteristic polynomial	$s^{4}+3s^{3}+4s^{2}+(2+K)s+2K$
Binding Routh condition	$K^{2}+10K-20<0$
Stable range	$0<K<3\sqrt5-5\approx1.708$

Topic: Routh–Hurwitz stability criterion for a parameter range.

Key relations: closed-loop poles are the roots of $1+G(s)=0$; the system is stable iff all first-column Routh entries share the same (positive) sign; a first-column zero signals an imaginary-axis crossing.

Why this works: Routh–Hurwitz counts right-half-plane roots from sign changes in the first column without factoring the quartic, so a symbolic array yields the exact gain window in one pass. The lower limit $K>0$ comes from the constant term (product of poles); the upper limit from the $s^{1}$ row, where an oscillatory mode reaches the $j\omega$ axis.

Common pitfall: forgetting the low-side limit — students often report only $K<1.708$ and miss that $K\le0$ is also unstable (the $s^{0}$ entry $2K$ turns negative). Also, arithmetic slips when forming $b_1$; always keep it symbolic before substituting.

Question 3 of 6: Routh–Hurwitz stability range of a fourth-order loop

Question 3: Routh–Hurwitz stability range of a fourth-order loop (equal value)