Question 3 of 5: Mason’s rule and PD compensator design

Question 3: Mason’s rule and PD compensator design (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.

Find. Part 1: the loop gains, forward-path gain, non-touching-loop product and the Mason transfer function \(Y/R\). Part 2: the pole-angle sum at the design point, the angle the compensator zero must contribute, and the zero’s location on the real axis.

Approach. Part 1 is a direct application of Mason’s gain formula: list the one forward path, the three loops, decide which loops share no node, assemble \(\Delta\) and divide. Part 2 is the angle condition of the root locus: the design point lies on the compensated locus only if the angles of all open-loop vectors sum to an odd multiple of \(180^\circ\), which fixes the zero’s angle and hence, by simple trigonometry, its position.

1. Part 1 — the single forward path. Only one route runs from \(R(s)\) to \(Y(s)\) in the direction of the arrows, so there is exactly one forward-path gain: $$P_1 = 1\cdot K_1\cdot\frac{1}{s+1}\cdot 3\cdot\frac{1}{2s-4}\cdot K_2\cdot\frac{10}{s-10} = \frac{30K_1K_2}{(s+1)(2s-4)(s-10)} .$$
2. Part 1 (continued) — the three loop gains. Each feedback branch closes one loop with the forward gains it spans: $$L_1 = -\frac{K_1K_3}{s+1}, \qquad L_2 = -\frac{K_2K_5}{2s-4}, \qquad L_3 = -\frac{30K_1K_2K_4}{(s+1)(2s-4)(s-10)} = -K_4P_1 .$$ \(L_1\) runs through the \(K_1\) and \(1/(s+1)\) blocks, \(L_2\) through \(1/(2s-4)\) and \(K_2\), and \(L_3\) is the outer loop, which encircles the whole chain.
3. Part 1 (continued) — which loops are non-touching. \(L_1\) occupies the nodes at the input end of the chain and \(L_2\) those in the middle; they share no node, so they are non-touching and contribute the product $$L_1L_2 = \frac{K_1K_2K_3K_5}{(s+1)(2s-4)} .$$ \(L_3\) touches every node of the chain, so it touches both \(L_1\) and \(L_2\); there is no non-touching triple. The graph determinant is therefore $$\Delta = 1 - (L_1+L_2+L_3) + L_1L_2 ,$$ and because the forward path touches all three loops, its cofactor is \(\Delta_1 = 1\).
4. Part 1 (continued) — assemble Mason’s formula. With \(Y/R = P_1\Delta_1/\Delta\), multiplying numerator and denominator by \((s+1)(2s-4)(s-10)\) clears every fraction and gives $$\boxed{\frac{Y(s)}{R(s)} = \frac{30K_1K_2} {\begin{array}{l}(s+1)(2s-4)(s-10) + K_1K_3(2s-4)(s-10)\\ \qquad + K_2K_5(s+1)(s-10) + 30K_1K_2K_4 + K_1K_2K_3K_5(s-10)\end{array}}}$$ Expanded, the denominator is \(2s^3 + (2K_1K_3 + K_2K_5 - 22)s^2 + (16 - 24K_1K_3 - 9K_2K_5 + K_1K_2K_3K_5)s + (40 + 40K_1K_3 - 10K_2K_5 + 30K_1K_2K_4 - 10K_1K_2K_3K_5)\). Two structural checks: setting all five gains to zero returns the open-loop chain denominator \(2s^3-22s^2+16s+40 = (s+1)(2s-4)(s-10)\), and the numerator contains no \(K_3\), \(K_4\) or \(K_5\), as it must — feedback gains can move poles but never create a forward-path zero.

Part 2 turns from analysis to design. The plant is Type 1 with poles at \(0\), \(-10\) and \(-20\); the desired closed-loop pair sits at \(s_d = -12.78 + j24.94\), which has \(\zeta = 12.78/|s_d| = 0.456\) and \(|s_d| = 28.02\) rad/s — well outside the uncompensated locus, so a compensator zero is needed to bend the locus onto that point.

5. Part 2(a) — sum the angles from the three open-loop poles to the design point. Each vector runs from a pole to \(s_d\); its angle is measured from the positive real axis. $$\angle(s_d - 0) = \angle(-12.78 + j24.94) = 180^\circ - \arctan\frac{24.94}{12.78} = 180^\circ - 62.87^\circ = 117.13^\circ$$ $$\angle(s_d + 10) = \angle(-2.78 + j24.94) = 180^\circ - \arctan\frac{24.94}{2.78} = 180^\circ - 83.64^\circ = 96.36^\circ$$ $$\angle(s_d + 20) = \angle(7.22 + j24.94) = \arctan\frac{24.94}{7.22} = 73.85^\circ$$ Adding, $$\boxed{\sum \theta_{\text{poles}} = 117.13^\circ + 96.36^\circ + 73.85^\circ = 287.35^\circ}$$ The two vectors from poles to the left of the design point’s real part have obtuse angles and the one from the pole to its right is acute — a useful sanity check on the signs.
6. Part 2(b) — the angle the compensator zero must supply. A point lies on the root locus of a positive-gain loop when the angles of the open-loop vectors satisfy $$\sum \theta_{\text{zeros}} - \sum \theta_{\text{poles}} = (2k+1)180^\circ .$$ With one compensator zero and no compensator pole, and taking \(k=-1\) so that the required angle is in range, $$\theta_z = \sum\theta_{\text{poles}} - 180^\circ = 287.35^\circ - 180^\circ = \boxed{107.35^\circ}$$ Because \(\theta_z\) exceeds \(90^\circ\), the vector from the zero to \(s_d\) points up and to the left: the zero must lie to the right of \(-12.78\), i.e. between the origin and the design point’s real part.
7. Part 2(c) — convert the angle into a position. Put the zero at \(s = -z_c\) with \(z_c>0\). The vector to the design point is \((z_c - 12.78) + j24.94\), so $$\tan\theta_z = \frac{24.94}{z_c - 12.78} \quad\Longrightarrow\quad z_c = 12.78 + \frac{24.94}{\tan 107.35^\circ} = 12.78 + \frac{24.94}{-3.2119} = 12.78 - 7.79 .$$ Hence $$\boxed{z_c = 4.99 \quad\text{— the PD zero sits at } s = -4.99}$$ so \(G_c(s) = K(s+4.99)\), i.e. \(K_p = 4.99K\) and \(K_d = K\). The angle condition closes exactly: \(\angle(s_d + 4.99) = 107.35^\circ\) and \(107.35^\circ - 287.35^\circ = -180^\circ\). ✓ The question does not ask for the gain, but for completeness the magnitude condition \(|G_cG|=1\) gives $$K = \frac{|s_d||s_d+10||s_d+20|}{10\,|s_d+4.99|} = \frac{28.02 \times 25.09 \times 25.96}{10 \times 26.16} = 69.88 ,$$ and substituting \(K=69.88\), \(z_c=4.99\) into \(s(s+10)(s+20) + 10K(s+z_c)\) does return \(s_d\) as a root, which confirms the design.