22-Elec-A2 Systems and Control · May 2018

Question 1 of 8: Servo Transfer Functions, Mason’s Rule and Routh–Hurwitz Stability

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, 16-Elec-A2 Systems and Control, May 2018 — 3 hours, closed book (approved Casio or Sharp calculator plus one double-sided handwritten 8.5 × 11″ formula sheet). A short Laplace-transform table is printed on page 2, and the $\zeta$–overshoot, $\zeta$–resonant-peak and phase-margin design charts plus the second-order model, controllability/observability and state-space transfer-function definitions are on page 3. Questions 1 and 2 are compulsory; five questions constitute a complete paper, so a candidate chooses three of Q3–Q8. Each question carries 20 marks, for 100 in total. All eight questions are worked below so the set is a complete study resource.

Reference texts. N. S. Nise, Control Systems Engineering (7th ed., Wiley) — block-diagram and signal-flow reduction with Mason’s rule (Ch. 5), steady-state error and the static error constants (Ch. 7), Routh–Hurwitz (Ch. 6), root locus (Ch. 8), cascade compensator design (Ch. 9), frequency response and gain/phase margins (Ch. 10), frequency-domain lead design (Ch. 11), state space and controllability/observability (Ch. 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — second-order correlations, dominant-poles modelling and pole placement by state feedback. Every block diagram, root locus, pole–zero map, Bode plot, closed-loop magnitude curve and step response below is redrawn as an inline figure.

Reading the supplied design charts. Page 3 gives $PO = 100\,e^{-\zeta\pi/\sqrt{1-\zeta^{2}}}$ for percent overshoot, $M_r/K_{dc} = 1/\left(2\zeta\sqrt{1-\zeta^{2}}\right)$ for the closed-loop resonant peak, and the phase-margin correlation $\Phi_m \approx 100\zeta$ (equivalently $\zeta \approx 0.01\,\Phi_m$). Those three relations, together with $T_{settle(\pm 2\%)} \approx 4/(\zeta\omega_n)$ and $T_{rise(0-100\%)} = (\pi - \cos^{-1}\zeta)/\omega_d$, carry most of the numerical work in Questions 2, 5, 7 and 8.

Question 1: Servo Transfer Functions, Mason’s Rule 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 loop of Figure Q1.1, with the following blocks and branch gains read directly from the diagram.

Element	Transfer function / gain
PI controller $G_c(s)$, $\tau_i = 0.1\ \text{s}$	$K_p\left(1 + \dfrac{1}{0.1s}\right) = \dfrac{K_p(s+10)}{s}$
Amplifier / calibration gain	$0.5$
Armature	$\dfrac{4}{s+1}$
Load	$\dfrac{10}{s+1}$
Tachometer minor loop (load speed back to the inner summer)	$-0.05$
Gear box	$\dfrac{1}{100}$
Torque disturbance $T_{dist}(s)$ injection	gain $-1$, entering between the armature and the load
Outer speed feedback	$-1$ (unity)

Find. (1) the reference-to-load-speed transfer function $G_{cl}(s)$ as a polynomial ratio in $K_p$; (2) the disturbance-to-load-speed transfer function $G_{dist}(s)$; and (3) the range of $K_p$ giving a stable closed loop, with the critical gain and the resulting oscillation frequency.

[Figure not reproduced: Figure Q1.1 (redrawn) — servomotor positioning system. The PI controller $G_c(s)=K_p(1+1/0.1s)$ feeds a 0.5 calibration gain; the armature and load lag blocks sit inside a tachometer minor loop of gain 0.05, the gear box scales by 1/100, and the torque disturbance enters with gain $-1$ between. See the official exam paper.]

Approach. Collapse the tachometer minor loop first to get the plant seen by the amplifier, multiply through the forward path to form the open-loop function, close the outer unity loop for $G_{cl}$, then repeat the reduction with $\Omega_{ref}=0$ and the disturbance as input for $G_{dist}$; finally apply Routh–Hurwitz to the common denominator.

Reduce the tachometer minor loop. The armature and load in cascade give $A(s)L(s) = \dfrac{4}{s+1}\cdot\dfrac{10}{s+1} = \dfrac{40}{(s+1)^{2}}$. Closing the $0.05$ tachometer loop around that cascade and following it with the gear box gives the plant from the amplifier output $u$ to the load speed: $$\frac{\Omega_{load}(s)}{U(s)} = \frac{1}{100}\cdot\frac{A L}{1 + 0.05\,A L} = \frac{1}{100}\cdot\frac{40/(s+1)^{2}}{1 + 2/(s+1)^{2}} = \frac{0.4}{(s+1)^{2}+2}$$ Expanding the denominator, $(s+1)^{2}+2 = s^{2}+2s+3$, so the plant is $0.4/(s^{2}+2s+3)$.
Form the open-loop transfer function. Multiplying the PI controller, the calibration gain and the plant along the forward path: $$G_{open}(s) = \underbrace{\frac{K_p(s+10)}{s}}_{G_c}\cdot 0.5 \cdot \frac{0.4}{s^{2}+2s+3} = \boxed{\;\frac{0.2\,K_p\,(s+10)}{s\left(s^{2}+2s+3\right)}\;}$$ This is a Type 1 loop (one free integrator, supplied by the PI action). As a check, setting $K_p = 3$ gives $0.6(s+10)/[s(s^{2}+2s+3)]$, which is exactly the $G_{open}(s)$ quoted in Question 2 — confirming both the reduction and the reading of the figure.
Close the outer unity loop. With $G_{cl} = G_{open}/(1+G_{open})$ and clearing the fraction, $$G_{cl}(s) = \frac{\Omega_{load}(s)}{\Omega_{ref}(s)} = \boxed{\;\frac{0.2\,K_p\,s + 2K_p}{s^{3} + 2s^{2} + \left(3 + 0.2K_p\right)s + 2K_p}\;}$$ The numerator is $0.2K_p(s+10)$ and the characteristic polynomial is $\Delta(s) = s^{3}+2s^{2}+(3+0.2K_p)s+2K_p$. Substituting $K_p=3$ gives $\Delta = s^{3}+2s^{2}+3.6s+6$, which factors as $(s+1.827)(s^{2}+0.173s+3.284)$ — again the factorisation printed in Question 2.
Disturbance transfer function. Set $\Omega_{ref}=0$ and drive the loop with $T_{dist}$, which enters after the armature with gain $-1$. Only the load block and the gear box lie between the injection point and the output, while both feedback paths still act, so the same characteristic polynomial appears: $$G_{dist}(s) = \frac{\Omega_{load}(s)}{T_d(s)} = \frac{-\,L/100}{1 + 0.05\,AL + AL\,G_c\,(0.5)/100} = \boxed{\;\frac{-0.1\,s\,(s+1)}{s^{3} + 2s^{2} + \left(3 + 0.2K_p\right)s + 2K_p}\;}$$ Two features are worth naming. The sign is negative because a positive load torque opposes the motion and slows the arm. More importantly the numerator carries a zero at the origin, so $G_{dist}(0)=0$: the integral action rejects any constant torque disturbance completely, leaving zero steady-state speed droop. That is the whole point of the I term in this loop.
Routh array for stability. Build the array on $\Delta(s) = s^{3}+2s^{2}+(3+0.2K_p)s+2K_p$: $$\begin{array}{c|cc} s^{3} & 1 & 3+0.2K_p \\ s^{2} & 2 & 2K_p \\ s^{1} & \dfrac{2(3+0.2K_p) - 2K_p}{2} & 0 \\ s^{0} & 2K_p & \end{array}$$ The $s^{0}$ row requires $K_p \gt 0$. The $s^{1}$ element simplifies to $3 + 0.1K_p - K_p = 3 - 0.9K_p$, which is positive while $K_p \lt 3/0.9 = 3.75$.
Stable range, critical gain and oscillation frequency. Combining the two conditions, $$\boxed{\;0 \lt K_p \lt 3.75\;}\qquad\Longrightarrow\qquad \boxed{\;K_{crit} = 3.75\;}$$ At $K_p = K_{crit}$ the $s^{1}$ row vanishes and the auxiliary equation formed from the $s^{2}$ row gives the imaginary-axis crossing: $$2s^{2} + 2K_{crit} = 0 \;\Longrightarrow\; s^{2} = -3.75 \;\Longrightarrow\; \boxed{\;\omega_{osc} = \sqrt{3.75} = 1.9365\ \text{rad/s}\;}$$ The same number follows independently from the $s^{1}$ coefficient, $\omega_{osc} = \sqrt{3+0.2K_{crit}} = \sqrt{3.75}$, and a numerical root solve at $K_p=3.75$ returns roots $-2$ and $\pm j1.9365$ — a purely imaginary pair, as marginal stability requires.

Quantity	Result
Closed-loop transfer function $G_{cl}(s)$	$\dfrac{0.2K_p(s+10)}{s^{3}+2s^{2}+(3+0.2K_p)s+2K_p}$
Disturbance transfer function $G_{dist}(s)$	$\dfrac{-0.1\,s(s+1)}{s^{3}+2s^{2}+(3+0.2K_p)s+2K_p}$
Stable gain range	$0 \lt K_p \lt 3.75$
Critical gain $K_{crit}$	$3.75$
Oscillation frequency $\omega_{osc}$	$1.9365\ \text{rad/s}$ ($0.3082\ \text{Hz}$)

Topic. Block-diagram reduction of a nested servo loop, and the Routh–Hurwitz test for the absolute stability of a third-order characteristic polynomial.

Key relations. Minor-loop closure $G/(1+GH)$; open loop $G_{open}=G_cG_p$; closed loop $G_{cl}=G_{open}/(1+G_{open})$; the characteristic equation $1+G_{open}(s)=0$; Routh’s criterion — the number of closed-loop poles in the right half plane equals the number of sign changes in the first column. For the cubic $s^{3}+a_2s^{2}+a_1s+a_0$ the array collapses to the compact condition $a_2a_1 \gt a_0$ with all $a_i \gt 0$.

Why this works. Reduction proceeds from the innermost loop outward because each closure replaces a sub-network with a single equivalent block, and every path from the input to the output must pass through it. The tachometer loop here converts the double lag $40/(s+1)^{2}$ into $0.4/(s^{2}+2s+3)$: it does not move the DC gain much but it stiffens the dynamics, which is exactly what rate feedback is for. Once the loop is a single ratio, stability is a property of the denominator alone, so the Routh array — which counts right-half-plane roots without factoring — answers the question in closed form as a function of the unknown gain. The disturbance path shares that denominator because feedback acts on the loop, not on the particular input, which is why one Routh test settles stability for both transfer functions at once.

Common pitfall. Two traps recur here. First, students often read the $-1$ near the disturbance arrow as a minor feedback loop around the armature; it is the injection gain of the disturbance itself, and treating it as a loop produces the denominator $s^{2}+6s+7$ instead of $s^{2}+2s+3$ and poisons every later part. The value of $G_{open}$ quoted in Question 2 is the built-in check against that error — always test a reduction against any transfer function the paper hands you. Second, the $s^{1}$ Routh element is often written as $2(3+0.2K_p)-2K_p$ without dividing by the pivot; the sign condition is unchanged here, but the habit fails on higher-order arrays.

Source. Nise, Control Systems Engineering 7th ed., §5.2–5.3 (block-diagram reduction) and Ch. 6 (Routh–Hurwitz, including the auxiliary-polynomial treatment of a vanishing row); Ogata, Modern Control Engineering 5th ed., §5.4.

← Paper overview Next: Question 2 →