Question 2 of 8: Servo Block Reduction, Disturbance Transfer Function and Routh–Hurwitz Stability

Question 2: Servo Block Reduction, Disturbance Transfer Function and Routh–Hurwitz Stability (20 marks, compulsory)

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 servo of Figure Q2.1, read directly off the exam drawing:

Find. $G_{cl}(s) = Y/R$ and $G_{dist}(s) = Y/D$ as polynomial ratios in $s$ with $K$ as a parameter, then the full stable range of $K$ together with the critical gain and the oscillation frequency at that gain.

[Figure not reproduced: Figure Q2.1 (redrawn) — robot-joint servo. Note the two unity negative feedback paths from Y(s): the inner one closes around the whole plant chain, the outer one around the PI controller as well. See the official exam paper.]

Approach. Collapse the inner loop first, then the outer loop, keeping the disturbance entry point separate so that both transfer functions come from the same denominator; then apply the Routh–Hurwitz array to that denominator and take the auxiliary polynomial at the critical gain.

1. Name the plant products. Let $P_1 = \dfrac{5}{s+5}$ and $P_2 = \dfrac{1}{s(s+10)}$, so the whole chain from the inner summer to the output is $P_1 P_2 = \dfrac{5}{\Delta}$ with $$\Delta(s) = s(s+5)(s+10) = s^3 + 15 s^2 + 50 s.$$ The disturbance enters between $P_1$ and $P_2$, so it sees only $P_2$ on its way to the output.
2. Write the two node equations. Call $u_1$ the PI output. The inner summer forms $u_1 - Y$, and the disturbance adds after $P_1$, so $$Y = P_1 P_2 (u_1 - Y) + P_2 D, \qquad u_1 = G_{PI}(R - Y).$$ Eliminating $u_1$ and collecting all $Y$ terms on the left, $$Y\,\big(1 + P_1 P_2 + G_{PI} P_1 P_2\big) = G_{PI} P_1 P_2\,R + P_2\,D.$$ Both transfer functions therefore share the denominator $1 + P_1 P_2 + G_{PI} P_1 P_2$ — the extra $P_1P_2$ term is the signature of the inner loop and is exactly what a single-loop formula would miss.
3. Form $G_{cl}(s)$. Substituting the blocks and multiplying numerator and denominator by $2s\Delta$ to clear fractions, $$G_{cl}(s) = \frac{5K(2s+1)}{2s\Delta + 10 s + 5K(2s+1)}.$$ Expanding $2s\Delta = 2s^4 + 30 s^3 + 100 s^2$ and dividing throughout by 2 to make the denominator monic, $$\boxed{G_{cl}(s) = \frac{2.5K(2s+1)}{s^4 + 15 s^3 + 50 s^2 + (5 + 5K)s + 2.5K} = \frac{5K\,(s + 0.5)}{s^4 + 15 s^3 + 50 s^2 + (5 + 5K)s + 2.5K}.}$$ The PI zero appears unchanged at $s = -1/\tau_i = -0.5$, as it must.
4. Cross-check the reduction against the transfer function the paper supplies. Question 3 quotes the closed loop at the operating gain as $G_{cl} = 250(s+0.5)/(s^4 + 15s^3 + 50s^2 + 255s + 125)$. Setting $K = 50$ in the boxed result gives $5K = 250$, $5 + 5K = 255$ and $2.5K = 125$ — an exact match, term for term. The block reduction is therefore correct, and the operating gain implied by the rest of the paper is $K_{op} = 50$ (Question 4 independently derives the same value from the ramp-error specification).
5. Form $G_{dist}(s)$. From the same node equation the disturbance numerator is $P_2$ rather than $G_{PI}P_1P_2$. Multiplying by the same $2s\Delta$, the numerator becomes $2s\Delta/[s(s+10)] = 2s(s+5)$, so $$\boxed{G_{dist}(s) = \frac{2s(s+5)}{2s^4 + 30 s^3 + 100 s^2 + (10 + 10K)s + 5K} = \frac{s(s+5)}{s^4 + 15 s^3 + 50 s^2 + (5+5K)s + 2.5K}.}$$ The factor $s$ in the numerator is the integral action doing its job: a constant torque disturbance is rejected completely in steady state, because $G_{dist}(0) = 0$. A ramp disturbance, however, is not — Question 4 part 4 exploits precisely that.
6. Build the Routh array. The characteristic polynomial is the common denominator, with coefficients $a_4 = 1$, $a_3 = 15$, $a_2 = 50$, $a_1 = 5 + 5K$, $a_0 = 2.5K$. The array is $$b_1 = \frac{15 \times 50 - (5 + 5K)}{15} = \frac{745 - 5K}{15}, \qquad c_1 = (5 + 5K) - \frac{15\,(2.5K)}{b_1}.$$ The first-column entries are $1$, $15$, $b_1$, $c_1$ and $a_0 = 2.5K$.
7. Impose positivity of the whole first column. $a_0 = 2.5K \gt 0$ requires $K \gt 0$. $b_1 \gt 0$ requires $K \lt 149$. The binding condition is $c_1 \gt 0$, which after clearing the fraction becomes the quadratic $$(5 + 5K)\,\frac{745 - 5K}{15} \gt 37.5\,K \;\Longleftrightarrow\; 25K^2 - 3140K + 3725 \lt 0 .$$ Solving, the positive root that matters is $K_{crit} = \tfrac{251}{4} + \tfrac{3}{4}\sqrt{7265} = 126.68$.
8. State the stable range and the critical gain. Because $c_1$ is positive for small $K$ and changes sign once, the loop is stable on a single interval: $$\boxed{0 \lt K \lt K_{crit} = 126.7, \qquad K_{crit} = 126.68.}$$ The operating gain $K_{op} = 50$ therefore sits comfortably inside the stable region, with a gain margin of $126.68/50 = 2.53$ V/V (8.07 dB).
9. Find the oscillation frequency from the auxiliary polynomial. At $K = K_{crit}$ the $c_1$ row vanishes and the row above it, the $s^2$ row, supplies the auxiliary polynomial $A(s) = b_1 s^2 + a_0$. Its roots are the purely imaginary closed-loop poles, so $$\omega_{osc} = \sqrt{\frac{a_0}{b_1}} = \sqrt{\frac{2.5 \times 126.68}{(745 - 5 \times 126.68)/15}} = \sqrt{\frac{316.7}{7.443}} = 6.52\ \text{rad/s}.$$ Factoring the quartic numerically at $K = K_{crit}$ returns roots $\pm j\,6.524$, $-14.49$ and $-0.514$, confirming both the frequency and the fact that the remaining two poles stay in the left half-plane: $$\boxed{\omega_{osc} = 6.52\ \text{rad/s at } K_{crit} = 126.7.}$$