Question 2 of 6: Routh’s Criterion Applied to Two Unity-Feedback Loops

Question 2: Routh’s Criterion Applied to Two Unity-Feedback Loops (25 marks)

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. Two open-loop transfer functions, each to be closed with negative unity feedback so that the characteristic equation is $1+L(s)=0$:

In part (b) the gain $K$ is the free parameter and time is measured in milliseconds, so every root and every frequency obtained below carries the unit $\text{ms}^{-1}$ (equivalently $\text{rad/ms}$).

Find. (a) whether the closed loop is stable, and if not how many closed-loop poles lie in the right half-plane; (b) the complete range of $K$ over which the tape-drive loop is stable.

Approach. Clear each characteristic equation to a polynomial in $s$, build the Routh array, and read stability from the first column — the number of sign changes equals the number of right-half-plane roots, and in part (b) the requirement that every first-column entry stay positive converts directly into inequalities on $K$.

Part (a)

1. Form the closed-loop characteristic polynomial. With negative unity feedback the characteristic equation is $1+L(s)=0$; multiplying through by the open-loop denominator clears the fraction:$$s^{2}(s+1)+2(s+4)=0\quad\Longrightarrow\quad s^{3}+s^{2}+2s+8=0$$Every coefficient is positive, so the elementary necessary condition is satisfied — but for a third-order or higher polynomial positivity is necessary, not sufficient, and the Routh test must still be run.
2. Build the Routh array. The first two rows are the alternate coefficients of the polynomial, and each later row is formed from the two rows above it:$$\begin{array}{c|cc} s^{3} & 1 & 2\\ s^{2} & 1 & 8\\ s^{1} & b_{1} & 0\\ s^{0} & 8 & \end{array}$$with the $s^{1}$ entry given by the usual determinant rule$$b_{1}=\frac{(1)(2)-(1)(8)}{1}=\frac{2-8}{1}=-6$$
3. Count the sign changes in the first column. Reading down the first column gives the sequence $1,\;1,\;-6,\;8$. The sign changes twice, from $+$ to $-$ and back from $-$ to $+$, so$$\boxed{\;\text{two closed-loop poles lie in the right half-plane}\;\Rightarrow\;\text{the loop is UNSTABLE}\;}$$
4. Confirm the count independently. Factoring the cubic numerically gives the closed-loop poles $s=-2$ and $s=+0.500\pm j1.936$. The complex pair indeed lies in the right half-plane, so the Routh count of two is exact, and the closed-loop response contains a growing oscillation at roughly $1.94\ \text{rad/s}$.

The physical reading is that a double integrator in the forward path is too much free integration for this compensator: the single zero at $s=-4$ supplies at most $90^{\circ}$ of phase lead, which cannot recover the $180^{\circ}$ lag that $1/s^{2}$ imposes at the crossover frequency.

Part (b)

1. Expand the characteristic polynomial in $K$. Clearing $1+G(s)=0$ gives $s(s+0.5)(s+1)(s^{2}+0.4s+4)+K(s+4)=0$. Multiplying the quartet of factors out,$$(s+0.5)(s+1)=s^{2}+1.5s+0.5$$and then$$(s^{2}+1.5s+0.5)(s^{2}+0.4s+4)=s^{4}+1.9s^{3}+5.1s^{2}+6.2s+2$$Multiplying by the remaining $s$ and adding $K(s+4)$ yields the fifth-order characteristic polynomial$$s^{5}+1.9s^{4}+5.1s^{3}+6.2s^{2}+(2+K)s+4K=0$$
2. Lay out the first two Routh rows. Alternate coefficients fill the $s^{5}$ and $s^{4}$ rows:$$\begin{array}{c|ccc} s^{5} & 1 & 5.1 & 2+K\\ s^{4} & 1.9 & 6.2 & 4K \end{array}$$
3. Compute the $s^{3}$ row. Applying the determinant rule column by column,$$b_{1}=\frac{(1.9)(5.1)-(1)(6.2)}{1.9}=\frac{3.49}{1.9}=1.8368,\qquad b_{2}=\frac{(1.9)(2+K)-(1)(4K)}{1.9}=\frac{3.8-2.1K}{1.9}=2-1.1053K$$
4. Compute the $s^{2}$ row. Using the $s^{4}$ and $s^{3}$ rows,$$c_{1}=\frac{b_{1}(6.2)-1.9\,b_{2}}{b_{1}}=6.2-\frac{3.8-2.1K}{1.8368}=4.1313+1.1433K,\qquad c_{2}=4K$$
5. Compute the $s^{1}$ row and impose positivity. The last non-trivial entry is$$d_{1}=\frac{c_{1}b_{2}-b_{1}c_{2}}{c_{1}}$$Multiplying the requirement $d_{1}>0$ through by the positive quantity $c_{1}$ gives a quadratic in $K$:$$(2-1.1053K)(4.1313+1.1433K)-4(1.8368)K>0\;\Longrightarrow\;-1.2636K^{2}-9.6270K+8.2625>0$$
6. Solve for the gain margin and collect every condition. The positive root of $1.2636K^{2}+9.6270K-8.2625=0$ is$$K^{*}=\frac{-9.6270+\sqrt{9.6270^{2}+4(1.2636)(8.2625)}}{2(1.2636)}=0.7787$$The remaining first-column entries impose only $4K>0$ (from the $s^{0}$ row), while $b_{1}>0$ and $c_{1}>0$ hold automatically for positive $K$. The complete stability range is therefore$$\boxed{\;0<K<0.779\;}$$
7. Locate the marginal condition. At $K=K^{*}$ the $s^{1}$ row vanishes and the auxiliary polynomial formed from the $s^{2}$ row, $c_{1}s^{2}+c_{2}=0$, gives the crossing frequency:$$\omega=\sqrt{\frac{4K^{*}}{4.1313+1.1433K^{*}}}=\sqrt{\frac{3.1147}{5.0215}}=0.7876$$Because time is in milliseconds this is $0.788\ \text{rad/ms}=788\ \text{rad/s}$, i.e. about $125\ \text{Hz}$ — a fast structural resonance of the tape transport, which is exactly what the lightly damped factor $s^{2}+0.4s+4$ represents.

The pole map confirms the algebra: at the critical gain the lightly damped quadratic factor is not the pair that reaches the imaginary axis. It is instead a slower pair, born of the interaction between the integrator and the two real poles, that crosses first at $0.788\ \text{rad/ms}$.