Routh–Hurwitz gain limit and the poles at that limit: 22-Mec-A3 System Analysis and Control, December 2013, Q2

Question 2: Routh–Hurwitz gain limit and the poles at that limit

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. The single-loop system of Fig. 1: reference $R$, error $E$, a gain block $K$ in cascade with $G(s)=1/[s(s+1)(s+4)]$, output $C$, and unity negative feedback.

Find. The range of $K$ for closed-loop stability, and the locations of the closed-loop poles on the imaginary axis when $K$ sits exactly at that limit.

Figure 2.1 — The single-loop configuration of Fig. 1: forward path $KG(s)$ with unity negative feedback, so the characteristic equation is $1+KG(s)=0$.

Approach. Form the characteristic polynomial from $1+KG(s)=0$, build the Routh array with $K$ carried symbolically, require every entry in the first column to be positive, and then use the auxiliary polynomial formed from the row above the vanishing row to locate the imaginary-axis roots.

Form the characteristic equation. For unity feedback the closed-loop poles are the roots of $1+KG(s)=0$, i.e. $$s(s+1)(s+4)+K=0 \quad\Longrightarrow\quad s^3+5s^2+4s+K=0$$ Expanding is essential here: $s(s+1)(s+4)=s(s^2+5s+4)=s^3+5s^2+4s$.
Build the Routh array. With $a_3=1,\;a_2=5,\;a_1=4,\;a_0=K$: $$\begin{array}{c|cc} s^3 & 1 & 4\\ s^2 & 5 & K\\ s^1 & \dfrac{5(4)-K}{5}=\dfrac{20-K}{5} & 0\\ s^0 & K & \end{array}$$
Apply the sign condition. The system is stable if and only if every entry of the first column is strictly positive. The $s^0$ row requires $K \gt 0$; the $s^1$ row requires $20-K \gt 0$. Together, $$\boxed{\;0 \lt K \lt 20\;}$$ so the stability limit is $K_{\max}=20$.
Locate the imaginary-axis poles at the limit. At $K=20$ the $s^1$ row vanishes, which is exactly the signature of a pair of roots on the imaginary axis. The auxiliary polynomial is formed from the row above: $$A(s)=5s^2+K=5s^2+20=0 \quad\Longrightarrow\quad s^2=-4 \quad\Longrightarrow\quad s=\pm j2$$ The system therefore sustains an undamped oscillation at $\omega=2\ \text{rad/s}$, i.e. $f = \omega/2\pi = 0.318\ \text{Hz}$.
Find the third pole and confirm. Dividing the characteristic polynomial at $K=20$ by the auxiliary factor, $$s^3+5s^2+4s+20=(s^2+4)(s+5)$$ so the three closed-loop poles at the stability limit are $$\boxed{\;s=+j2,\qquad s=-j2,\qquad s=-5\;}$$ The single remaining pole is well damped, so the marginal behaviour is a clean 2 rad/s sinusoid superimposed on a fast-decaying transient.

Testing the result numerically confirms the boundary: at $K=19.5$ the dominant pair lies at $-0.0087\pm j1.978$ (stable), while at $K=20.5$ it has moved to $+0.0086\pm j2.021$ (unstable). The crossing is exactly at $K=20$, as the Routh array predicts.

Quantity	Result
Characteristic equation	$s^3+5s^2+4s+K=0$
Stability range	$0 \lt K \lt 20$
Limiting gain	$K_{\max}=20$
Imaginary-axis poles at $K_{\max}$	$s=\pm j2$ ($\omega=2$ rad/s, 0.318 Hz)
Third pole at $K_{\max}$	$s=-5$

Topic. The Routh–Hurwitz criterion applied to a gain-parameterised characteristic polynomial, and the auxiliary polynomial as a locator of imaginary-axis roots.

Key relations. Closed-loop characteristic equation $1+KG(s)H(s)=0$; for a cubic $s^3+a_2s^2+a_1s+a_0$ the Routh conditions reduce to $a_0 \gt 0$ and $a_2a_1 \gt a_0$; the auxiliary polynomial from the row above a vanishing row has the imaginary-axis roots as its own roots.

Why this works. The Routh test counts sign changes in the first column and returns the number of closed-loop poles in the right half plane without ever factoring the polynomial. When an entire row goes to zero the array has detected a symmetric root pattern about the origin — for a system with otherwise damped roots that pattern is a conjugate pair sitting precisely on the imaginary axis, which is the boundary between stability and instability. That is why the same calculation that gives the gain limit also gives the frequency of the resulting oscillation.

Common pitfall. Forgetting that $K$ appears in the constant term, so that the lower bound $K \gt 0$ is a genuine part of the answer, and quoting "$K \lt 20$" alone. A second trap is reading the oscillation frequency off the $s^1$ row instead of forming the auxiliary polynomial from the $s^2$ row.

Source. Nise, Control Systems Engineering, 8th ed., §6.2–6.4 (Routh–Hurwitz, entire row of zeros); Ogata, Modern Control Engineering, 5th ed., §5.6.

Question 2 of 6: Routh–Hurwitz gain limit and the poles at that limit

Question 2: Routh–Hurwitz gain limit and the poles at that limit