22-Elec-B2 Advanced Control Systems · December 2019

Question 5 of 5: Bode diagram and stability margins

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

Notes on this paper

Paper format. 16-Elec-B2 Advanced Control Systems, December 2019 — a three-hour open-book examination; any non-communicating calculator is permitted. The cover page states “Any four questions constitute a complete paper. Only the first four questions as they appear in your answer book will be marked” and “All questions are of equal value (25%)”. The sitting prints five questions, so each carries 25 marks and a candidate answers four. Question 1 is a seventeen-item multiple-choice block whose per-item weights are printed in the margin as [1], [2] or [3] and total exactly 25. A table of inverse Laplace transforms and a table of Laplace/z transforms are appended. All five questions are worked below, because this set is a study resource rather than a timed sitting.

Reference texts. N. S. Nise, Control Systems Engineering, 8th ed. (Ch. 4 time response, percent overshoot, peak time and settling time; Ch. 6 Routh–Hurwitz stability; Ch. 7 steady-state error and system type; Ch. 8 root-locus sketching rules including break-away/break-in points and angles of departure; Ch. 10 frequency response and stability margins); R. C. Dorf and R. H. Bishop, Modern Control Systems, 13th ed. (Ch. 5, 6, 7, 9); K. Ogata, Modern Control Engineering, 5th ed. (Ch. 5 transient response, Ch. 6 root-locus analysis, Ch. 7 frequency-response analysis); G. F. Franklin, J. D. Powell and A. Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed. (Ch. 3, 5, 6). These are the references listed for exam code 16-Elec-B2 in the Engineers Canada syllabus, and this paper’s vocabulary (“break-in point”, “intercept of the asymptotes”, “angle of departure”) follows Nise closely.

Check — three source readings checked against the printed paper.

(1) Question 1, item 2. The printed sentence reads “where $\theta(s)$, $T(s)$, $J$ and $K$ represent torque, angular displacement, moment of inertia and spring constant, respectively”. The paper really does print the first two descriptions in the wrong order. The drawing beside it settles the physics — $T(t)$ is the applied torque acting on the inertia $J$, and $\theta$ is the resulting angular displacement, which is also the only reading under which $\theta(s)/T(s)$ has the units of a compliance. The answer does not depend on the naming at all, because only the denominator is used.

(2) Question 1, item 16. The margins are read off a printed Bode pair, so they are inherently approximate. The printed Bode pair gives a gain crossover near $\omega \approx 3$ rad/s where the phase curve sits near $-116^\circ$, and a phase crossover near $\omega \approx 10$ rad/s where the magnitude curve is near $-21$ dB. Both readings point at the same option. The graph is coarse enough that the honest statement is “approximately”, exactly as the question words it.

(3) Question 2. The pole-zero map prints its two crosses to the right of the imaginary axis, at $s = 1 \pm j1$, and its two circles on the negative real axis at $-2$ and $-3$. This is the whole point of the question, because an open-loop plant with two right-half-plane poles is unstable until enough gain is applied, which is the opposite of the usual root-locus habit.

Question 5: Bode diagram and stability margins (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. A unity-feedback loop whose forward path is a type-1 plant with two finite poles and the gain set to 5.

Given data
Quantity	Value
Forward path	$G(s) = K/[\,s(s+1)(s+5)\,]$
Gain	$K = 5$
Feedback	unity, negative
Corner frequencies	$1$ and $5$ rad/s
System type	1 (one pole at the origin)

Question 5: the printed unity-feedback loop with $G(s) = K/[s(s+1)(s+5)]$ and $K = 5$.

Find. (a) the asymptotic and exact Bode magnitude and phase plots, and (b) the gain margin, the phase margin and the frequency at which each is measured.

Approach. Normalise each factor to the Bode standard form so the corner frequencies and the low-frequency gain can be read off directly; add the four component asymptotes; then locate the two crossover frequencies and measure the two margins there.

Part (a) — normalise the transfer function as the hint asks. Factor each pole so that its constant term is unity: $$G(s) = \frac{5}{s(s+1)(s+5)} = \frac{5}{s\,(s+1)\cdot 5\left(\frac{s}{5}+1\right)} = \boxed{\frac{1}{s\,(1+s)\left(1+0.2s\right)}}$$ The normalised gain is $K_{\text{norm}} = 5/(1 \times 5) = 1$, i.e. $0$ dB, so the constant term contributes a horizontal line on the axis and the whole low-frequency behaviour comes from the integrator.
Sketch the four components separately. The gain contributes $0$ dB and $0^\circ$ everywhere. The integrator $1/s$ contributes a straight $-20$ dB/decade line passing through $0$ dB at $\omega = 1$ rad/s, and a constant $-90^\circ$. The pole at $-1$ contributes $0$ dB below its corner and $-20$ dB/decade above it, with phase running from $0^\circ$ through $-45^\circ$ at $\omega = 1$ to $-90^\circ$. The pole at $-5$ does the same about $\omega = 5$ rad/s.
Add the asymptotes to get the overall magnitude. Slopes accumulate: $$\text{slope} = \begin{cases} -20\ \text{dB/dec}, & \omega \lt 1 \\ -40\ \text{dB/dec}, & 1 \lt \omega \lt 5 \\ -60\ \text{dB/dec}, & \omega \gt 5 \end{cases}$$ Exact values at a few checkpoints, from $|G| = 5/\left[\omega\sqrt{1+\omega^{2}}\sqrt{25+\omega^{2}}\right]$, are $+19.96$ dB at $\omega = 0.1$, $+5.01$ dB at $\omega = 0.5$, $-3.18$ dB at $\omega = 1$ and $-13.65$ dB at $\omega = 2$. The phase, from $\angle G = -90^\circ - \arctan\omega - \arctan(\omega/5)$, passes $-96.9^\circ$ at $\omega = 0.1$ and $-146.3^\circ$ at $\omega = 1$.
Part (b) — locate the phase crossover in closed form. The phase reaches $-180^\circ$ when the two arctangents together make up $90^\circ$, which happens exactly when the product of their arguments is one: $$\arctan\frac{\omega}{1} + \arctan\frac{\omega}{5} = 90^\circ \iff \frac{\omega}{1}\cdot\frac{\omega}{5} = 1 \iff \omega_{pc} = \sqrt{(1)(5)}$$ $$\boxed{\omega_{pc} = \sqrt{5} = 2.236\ \text{rad/s}}$$ This closed form — $\omega_{pc} = \sqrt{ab}$ for any plant $K/[s(s+a)(s+b)]$ — is worth carrying, because it does not depend on $K$ at all.
Evaluate the gain margin there. Substituting $\omega_{pc}$, $$|G(j\omega_{pc})| = \frac{K}{ab\,(a+b)} = \frac{5}{(1)(5)(6)} = \frac{1}{6}$$ $$GM = \frac{1}{|G(j\omega_{pc})|} = 6 \quad\Longrightarrow\quad \boxed{GM = 6 = 15.56\ \text{dB at } \omega_{pc} = 2.236\ \text{rad/s}}$$ Equivalently the loop is stable up to $K_{\max} = ab(a+b) = 30$, which is precisely the Routh–Hurwitz limit of $s^{3}+6s^{2}+5s+K$; at that gain the closed-loop poles sit at $\pm j\sqrt{5}$, matching $\omega_{pc}$.
Locate the gain crossover. The magnitude condition $|G(j\omega)| = 1$ gives $$\omega^{2}\left(1+\omega^{2}\right)\left(25+\omega^{2}\right) = 25$$ which has no tidy closed form; solving numerically, $$\boxed{\omega_{gc} = 0.7793\ \text{rad/s}}$$ The magnitude there is $1.0000$ to four figures, confirming the root.
Measure the phase margin at the gain crossover. $$\angle G(j\omega_{gc}) = -90^\circ - \arctan(0.7793) - \arctan\!\left(\frac{0.7793}{5}\right) = -90^\circ - 37.93^\circ - 8.86^\circ = -136.79^\circ$$ $$\boxed{PM = 180^\circ - 136.79^\circ = 43.21^\circ \text{ at } \omega_{gc} = 0.779\ \text{rad/s}}$$
Interpret the two margins together. Both are positive, so the closed loop is stable, and comfortably so: the gain may rise by a factor of six, or an extra $43^\circ$ of phase lag may be tolerated, before the loop reaches the edge. The rule of thumb $\zeta \approx PM/100 = 0.43$ predicts about 22 % overshoot from the dominant pair, and the closed-loop bandwidth will sit a little above $\omega_{gc}$, near 1 rad/s — a modest, well-damped design. Note that the two margins are read at different frequencies (0.779 and 2.236 rad/s); quoting both at one frequency is the classic error.

Question 5(a) and (b): the Bode diagram of $5/[s(s+1)(s+5)]$ with the straight-line asymptotes dashed. The phase margin is measured at $\omega_{gc} = 0.779$ rad/s and the gain margin at $\omega_{pc} = 2.236$ rad/s.

Question 5 — results
Part	Quantity	Symbol	Result
(a)	Normalised form	—	$1/[s(1+s)(1+0.2s)]$, $K_{\text{norm}} = 0$ dB
(a)	Corner frequencies	—	$1$ and $5$ rad/s
(a)	Asymptotic slopes	—	$-20 \to -40 \to -60$ dB/dec
(b)	Gain crossover	$\omega_{gc}$	$0.7793\ \text{rad/s}$
(b)	Phase margin	$PM$	$43.21^\circ$
(b)	Phase crossover	$\omega_{pc}$	$2.236\ \text{rad/s}$
(b)	Gain margin	$GM$	$6 = 15.56\ \text{dB}$
check	Limiting gain	$K_{\max}$	$30$ (Routh limit)

Topic. Asymptotic Bode construction from normalised factors, and the measurement of gain and phase margins at their two distinct crossover frequencies.

Key relations. Normalised form $G = K_{\text{norm}}/[s\prod(1+s/p_i)]$ with $K_{\text{norm}} = K/\prod p_i$; each simple pole adds $-20$ dB/decade above its corner and $-45^\circ$ at it; for $K/[s(s+a)(s+b)]$, $\omega_{pc} = \sqrt{ab}$, $|L(j\omega_{pc})| = K/[ab(a+b)]$, $GM = ab(a+b)/K$ and $K_{\max} = ab(a+b)$; $PM = 180^\circ + \angle L(j\omega_{gc})$.

Why this works. Taking logarithms turns a product of factors into a sum, so a Bode plot of a factored transfer function is the superposition of a handful of standard shapes — which is exactly what the hint is asking the candidate to exploit. The phase-crossover closed form follows from the arctangent addition identity, and the fact that it agrees with the Routh limit is not a coincidence: both describe the gain at which a pair of closed-loop poles sits on the imaginary axis.

Common pitfall. Forgetting to normalise, which puts the low-frequency asymptote at the wrong height (here $20\log_{10}5 = 14$ dB too high) and shifts $\omega_{gc}$ by a factor of five; and reading both margins at the same frequency. A third is quoting the gain margin as a negative number: $GM = -|G(j\omega_{pc})|_{\text{dB}}$, so a magnitude of $-15.56$ dB is a gain margin of $+15.56$ dB.

Source. Nise, Control Systems Engineering, 8th ed., §10.2–10.4 (asymptotic sketching of Bode plots) and §10.7 (gain and phase margins from Bode plots); Ogata, Modern Control Engineering, 5th ed., §7.2 and §7.6; Franklin, Powell and Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed., §6.4.

← Question 4 Back to the paper →

Quantity	Value
Forward path	\(G(s) = K/[\,s(s+1)(s+5)\,]\)
Gain	\(K = 5\)
Feedback	unity, negative
Corner frequencies	\(1\) and \(5\) rad/s
System type	1 (one pole at the origin)

Part	Quantity	Symbol	Result
(a)	Normalised form	—	\(1/[s(1+s)(1+0.2s)]\), \(K_{\text{norm}} = 0\) dB
(a)	Corner frequencies	—	\(1\) and \(5\) rad/s
(a)	Asymptotic slopes	—	\(-20 \to -40 \to -60\) dB/dec
(b)	Gain crossover	\(\omega_{gc}\)	\(0.7793\ \text{rad/s}\)
(b)	Phase margin	\(PM\)	\(43.21^\circ\)
(b)	Phase crossover	\(\omega_{pc}\)	\(2.236\ \text{rad/s}\)
(b)	Gain margin	\(GM\)	\(6 = 15.56\ \text{dB}\)
check	Limiting gain	\(K_{\max}\)	\(30\) (Routh limit)