Stability range, and the stabilising effect of a feedback zero: 22-Mec-A3 System Analysis and Control, May 2013, Q4

Question 4: Stability range, and the stabilising effect of a feedback zero (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. Open-loop transfer function of a unity-feedback system

$$G(s)=\frac{K}{s\,(0.5s+1)\,(2s+1)}.$$

Find. (i) the range of $K$ for stability with unity feedback; (ii) show that replacing the feedback by $H(s)=s+3$ enlarges the stable range.

Approach. Form the characteristic polynomial for each feedback and apply Routh–Hurwitz; compare the resulting upper limits on $K$.

Unity feedback — characteristic polynomial. Since $s(0.5s+1)(2s+1)=s^{3}+2.5s^{2}+s$, $$1+G=0\;\Longrightarrow\;\boxed{\,s^{3}+2.5s^{2}+s+K=0\,}.$$
Unity feedback — Routh test.

$s^{3}$ $1$ $1$

$s^{2}$ $2.5$ $K$

$s^{1}$ $\dfrac{2.5-K}{2.5}$ $0$

$s^{0}$ $K$

Positivity needs $K>0$ and $2.5-K>0$, so $$\boxed{\,0<K<2.5\,}.$$ The imaginary-axis crossing is at $K=2.5$, $\omega=\sqrt{1/(0.5\cdot2)}=1$ rad/s.
Feedback $H(s)=s+3$ — characteristic polynomial. Now $1+G(s)H(s)=0$ gives $s(0.5s+1)(2s+1)+K(s+3)=0$, i.e. $$\boxed{\,s^{3}+2.5s^{2}+(1+K)s+3K=0\,}.$$ The feedback zero at $s=-3$ raises the $s^{1}$ coefficient from $1$ to $(1+K)$, adding phase lead.
Feedback $H(s)=s+3$ — Routh test.

$s^{3}$ $1$ $1+K$

$s^{2}$ $2.5$ $3K$

$s^{1}$ $\dfrac{2.5(1+K)-3K}{2.5}=\dfrac{2.5-0.5K}{2.5}$ $0$

$s^{0}$ $3K$

Positivity needs $K>0$ and $2.5-0.5K>0$, so $$\boxed{\,0<K<5\,}.$$
Comparison. The derivative (lead) action of the feedback zero doubles the upper gain limit from $2.5$ to $5$: $$\text{unity: }0<K<2.5\quad\longrightarrow\quad H=s+3:\ 0<K<5.$$ The added zero contributes positive phase, so the loop tolerates twice the gain before its high-frequency poles drive it unstable — the stability range is increased, as required.

$s^{3}$	$1$	$1$
$s^{2}$	$2.5$	$K$
$s^{1}$	$\dfrac{2.5-K}{2.5}$	$0$
$s^{0}$	$K$

$s^{3}$	$1$	$1+K$
$s^{2}$	$2.5$	$3K$
$s^{1}$	$\dfrac{2.5(1+K)-3K}{2.5}=\dfrac{2.5-0.5K}{2.5}$	$0$
$s^{0}$	$3K$

Feedback	Characteristic polynomial	Stable range
$H=1$	$s^{3}+2.5s^{2}+s+K$	$0<K<2.5$
$H=s+3$	$s^{3}+2.5s^{2}+(1+K)s+3K$	$0<K<5$

Topic: Routh–Hurwitz range and the stabilising influence of derivative (rate / lead) feedback.

Key relations: $1+G(s)H(s)=0$; for a cubic $s^{3}+a_2s^{2}+a_1s+a_0$ stability requires $a_2a_1>a_0$ (and all $a_i>0$).

Why this works: A feedback zero $s+3$ is proportional-plus-derivative sensing; the derivative term injects phase lead that offsets the lag of the two real plant poles, pushing the $j\omega$-axis crossing to a higher gain. Algebraically it lifts the $s^{1}$ coefficient, so the Routh product $a_2a_1$ outruns $a_0$ over a wider $K$ band.

Common pitfall: forgetting that the feedback also multiplies the numerator $K(s+3)$, so the constant term becomes $3K$ (not $K$); using $K$ there gives the wrong crossing gain. Practical rate feedback also amplifies noise — a pure $s+3$ is an idealisation.

Question 4 of 6: Stability range, and the stabilising effect of a feedback zero

Question 4: Stability range, and the stabilising effect of a feedback zero (equal value)