Question 6 of 8: Routh–Hurwitz Stability of a PID Loop

Question 6: Routh–Hurwitz Stability of a PID Loop (20 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. Unit-feedback loop, $G_{PID}(s) = K_p(1 + 5/s + 0.2s)$ (i.e. $\tau_i = 0.2$ s, $\tau_d = 0.2$ s) in cascade with $G(s) = 5/(s+1)^2$.

Find. The closed-loop transfer function as a polynomial ratio in $K_p$; every critical gain at which the loop is marginally stable together with the corresponding oscillation frequency; and the complete range of $K_p$ giving stable operation.

Approach. Form the open-loop product over a common denominator, build the closed-loop characteristic polynomial, construct the Routh array, and solve the $s^1$ element for zero — which here is a quadratic in $K_p$, so two critical gains and a banded stability region are to be expected. Each oscillation frequency comes from the auxiliary polynomial formed on the $s^2$ row.

1. Form the open-loop transfer function. Placing the controller over a common denominator, $$G_{PID}(s) = K_p\,\frac{0.2s^2 + s + 5}{s}$$ so that $$G_{PID}(s)G(s) = \frac{5K_p\,(0.2s^2+s+5)}{s(s+1)^2}$$
2. Close the loop. For unit feedback $G_{cl} = G_{PID}G/(1+G_{PID}G)$, so $$G_{cl}(s) = \frac{5K_p(0.2s^2+s+5)}{s(s+1)^2 + 5K_p(0.2s^2+s+5)}$$ Expanding $s(s+1)^2 = s^3+2s^2+s$ and $5K_p(0.2s^2+s+5) = K_p s^2 + 5K_ps + 25K_p$, $$\boxed{\,G_{cl}(s) = \frac{K_p\,(s^2+5s+25)}{s^3 + (2+K_p)s^2 + (1+5K_p)s + 25K_p}\,}$$
3. Build the Routh array. For $Q(s) = s^3 + (2+K_p)s^2 + (1+5K_p)s + 25K_p$:

4. Reduce the $s^1$ element. Expanding the numerator, $$(2+K_p)(1+5K_p) - 25K_p = 2 + 10K_p + K_p + 5K_p^2 - 25K_p = 5K_p^2 - 14K_p + 2$$ Since $2+K_p \gt 0$ and $25K_p \gt 0$ for any positive gain, the only way the array can lose a sign is through this quadratic.
5. Solve for the critical gains. Setting $5K_p^2 - 14K_p + 2 = 0$, $$K_p = \frac{14 \pm \sqrt{196-40}}{10} = \frac{14 \pm \sqrt{156}}{10} \;\Rightarrow\; \boxed{\,K_{crit,1} = 0.151, \qquad K_{crit,2} = 2.649\,}$$ Two positive roots means the loop is conditionally stable: it becomes marginally stable twice as the gain is raised.
6. Find the oscillation frequencies. At a critical gain the $s^1$ row vanishes and the auxiliary polynomial is formed from the $s^2$ row — retaining its pivot: $$(2+K_p)s^2 + 25K_p = 0 \;\Longrightarrow\; \omega_{osc} = \sqrt{\frac{25K_p}{2+K_p}}$$ Evaluating at each root, $$\omega_{osc,1} = \sqrt{\frac{25(0.151)}{2.151}} = \boxed{1.325\ \text{rad/s}}, \qquad \omega_{osc,2} = \sqrt{\frac{25(2.649)}{4.649}} = \boxed{3.774\ \text{rad/s}}$$ Factoring the characteristic polynomial numerically at each gain confirms a purely imaginary root pair at exactly these frequencies (together with a real root at $-2.151$ and $-4.649$ respectively).
Figure 6.1 — The Routh s^1 row numerator 5Kp^2 - 14Kp + 2 against gain. It is negative only between its two roots, so the stable set is the union of 0 < Kp < 0.151 and Kp > 2.649.
7. Determine the stable range. The quadratic $5K_p^2 - 14K_p + 2$ opens upwards, so it is positive outside its roots and negative between them. The $s^1$ element must be positive for stability, hence $$\boxed{\,0 \lt K_p \lt 0.151 \quad \text{or} \quad K_p \gt 2.649\,}$$ The safe operating region is a union of two intervals, not a single band: raising the gain from zero the loop is stable, goes unstable at 0.151, and recovers stability at 2.649. Spot checks confirm this — at $K_p = 0.1$ and $K_p = 5$ all three roots lie in the left half plane, whereas at $K_p = 1$ two roots have positive real parts. In practice one would operate well inside the upper band, since the lower band gives very sluggish response and both boundaries are uncomfortably close together.