22-Mec-A3 System Analysis and Control · May 2016

Question 4 of 6: Bode Magnitude and Phase Asymptotes

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

Notes on this paper

Paper format. 07-Mec-A3 System Analysis and Control, National Exams May 2016 — 3 hours, closed book (Casio or Sharp approved calculator; semi-log graph paper is the only permitted aid). The rubric states that any four (4) questions constitute a complete paper and that all questions are of equal value, so each question is worth 25 marks of the 100 marked. Because this document is a study resource rather than an exam script, all six questions are solved in full.

Reference texts. K. Ogata, Modern Control Engineering, 5th ed. (Pearson) — Ch. 3–7; N. S. Nise, Control Systems Engineering, 8th ed. (Wiley) — Ch. 4, 6, 8, 10; R. C. Dorf and R. H. Bishop, Modern Control Systems, 13th ed. (Pearson) — Ch. 5–8; G. F. Franklin, J. D. Powell and A. Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed. (Pearson) — Ch. 3–6. A table of Laplace transform pairs is appended to the examination paper (pages 3–4 of the original), together with two sheets of 3-cycle semi-logarithmic graph paper for the Bode question.

Question 4: Bode Magnitude and Phase Asymptotes (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. An open-loop transfer function already written in time-constant (Bode) form, so the leading constant is the Bode gain:

Factor	Type	Corner $\omega$ (rad/s)	Asymptotic effect
$100$	Bode gain $K_{B}$	—	$+40\text{ dB}$ level shift
$1/s$	Integrator (type 1)	—	$-20\text{ dB/dec}$ from DC; $-90^{\circ}$ at all $\omega$
$1/(0.5s+1)$	Real pole, $\tau=0.5\text{ s}$	$2$	Adds $-20\text{ dB/dec}$; $-90^{\circ}$ over $0.2\to20$
$1/(0.1s+1)$	Real pole, $\tau=0.1\text{ s}$	$10$	Adds $-20\text{ dB/dec}$; $-90^{\circ}$ over $1\to100$

Find. The straight-line (asymptotic) magnitude and phase plots, and — as the natural engineering payoff of the sketch — the gain and phase crossover frequencies and the resulting stability margins.

Approach. Take the low-frequency integrator asymptote as the starting line, break its slope by $-20\text{ dB/dec}$ at each corner in increasing order, build the phase from $-90^{\circ}$ plus a $-45^{\circ}$/decade ramp centred on each corner, then read off the crossovers and refine them with the exact expressions.

Confirm the form and fix the low-frequency asymptote. Both first-order factors are already normalised to $(\tau s+1)$, so no rescaling is needed and $K_{B}=100$. Well below the first corner,$$|L(j\omega)|\approx\frac{100}{\omega}\quad\Longrightarrow\quad 20\log_{10}|L|\approx40-20\log_{10}\omega\ \text{dB}$$This is a $-20\text{ dB/dec}$ line passing through $+40\text{ dB}$ at $\omega=1$ and through $0\text{ dB}$ at $\omega=100$. At the left edge of the plot, $\omega=0.1$, it sits at $+60\text{ dB}$.
Break the magnitude at each corner. The slopes accumulate as$$\begin{array}{lll}0.1<\omega<2:&-20\text{ dB/dec}\\ 2<\omega<10:&-40\text{ dB/dec}\\ \omega>10:&-60\text{ dB/dec}\end{array}$$The asymptote levels follow directly: at $\omega=2$ it has fallen to $40-20\log_{10}2=+34.0\text{ dB}$; at $\omega=10$ to $34.0-40\log_{10}5=+6.0\text{ dB}$; thereafter it descends at $60\text{ dB}$ per decade.
Construct the phase asymptotes. The integrator contributes a constant $-90^{\circ}$. Each real pole contributes $0^{\circ}$ below one-tenth of its corner, $-90^{\circ}$ above ten times it, and a straight $-45^{\circ}$/decade ramp between:$$\angle L(j\omega)=-90^{\circ}-\arctan(0.5\omega)-\arctan(0.1\omega)$$so the phase runs from $-90^{\circ}$ at very low frequency, through $-180^{\circ}$, to an asymptote of $-270^{\circ}$. The ramps begin at $\omega=0.2$ and $\omega=1$ and end at $\omega=20$ and $\omega=100$ respectively; they overlap between $1$ and $20$, where the composite slope is $-90^{\circ}$/decade.
Locate the gain crossover. Reading the asymptotes, $0\text{ dB}$ is crossed on the $-60\text{ dB/dec}$ segment at about $12.6\text{ rad/s}$. Solving $|L(j\omega)|=1$ exactly,$$\frac{100}{\omega\sqrt{1+0.25\omega^{2}}\;\sqrt{1+0.01\omega^{2}}}=1\;\Longrightarrow\;\boxed{\;\omega_{gc}=11.40\ \text{rad/s}\;}$$The asymptotic reading overestimates $\omega_{gc}$ by about 10%, which is the expected penalty when two corners lie less than a decade apart — sketch with asymptotes, but evaluate with the exact magnitude.
Locate the phase crossover in closed form. The $-180^{\circ}$ point satisfies $\arctan(0.5\omega)+\arctan(0.1\omega)=90^{\circ}$, which for two positive arctangents happens exactly when the product of their arguments is unity:$$(0.5\omega)(0.1\omega)=1\;\Longrightarrow\;\omega_{pc}=\sqrt{20}=4.472\ \text{rad/s}$$No numerical search is needed — a useful closed form for any type-1 plant with two real poles.
Evaluate the margins and pass judgement. At $\omega_{pc}$,$$|L(j\omega_{pc})|=\frac{100}{4.472\sqrt{1+5}\sqrt{1+0.2}}=8.333\;\Longrightarrow\;\text{GM}=-20\log_{10}(8.333)=-18.4\ \text{dB}$$and at $\omega_{gc}$ the phase is $-90^{\circ}-\arctan5.699-\arctan1.140=-218.8^{\circ}$, so$$\boxed{\;\text{GM}=-18.4\ \text{dB},\qquad \text{PM}=-38.8^{\circ}\ \Rightarrow\ \text{closed loop UNSTABLE}\;}$$Both margins are negative because $\omega_{gc}>\omega_{pc}$. This is a legitimate answer, not an arithmetic slip, and it is confirmed independently by Routh: the unity-feedback characteristic polynomial is $0.05s^{3}+0.6s^{2}+s+K$, whose stability limit is $K_{\max}=(0.6)(1)/0.05=12$. The gain margin as a ratio is $12/100=0.12$, i.e. $-18.4\text{ dB}$ — identical to the frequency-domain figure, which is the cleanest possible cross-check. The loop gain would have to be reduced by a factor of 8.33 to reach the stability boundary.

Figure 4.1 — Asymptotic (dashed) and exact (solid) Bode plots of $L(s)=100/[s(0.1s+1)(0.5s+1)]$. Corners at $\omega=2$ and $\omega=10\text{ rad/s}$ break the slope from $-20$ to $-40$ to $-60\text{ dB/dec}$; the phase falls from $-90^{\circ}$ to $-270^{\circ}$. Gain crossover ($11.40\text{ rad/s}$) lies above phase crossover ($4.472\text{ rad/s}$), so both margins are negative.

Quantity	Result
Bode gain	$K_{B}=100$ ($+40\text{ dB}$); type 1, so $K_{v}=100\text{ s}^{-1}$
Corner frequencies	$\omega=2\text{ rad/s}$ and $\omega=10\text{ rad/s}$
Magnitude slopes	$-20$, then $-40$ (above 2), then $-60\text{ dB/dec}$ (above 10)
Asymptote levels	$+60\text{ dB}$ at $\omega=0.1$; $+34.0\text{ dB}$ at $\omega=2$; $+6.0\text{ dB}$ at $\omega=10$
Phase range	$-90^{\circ}$ (low $\omega$) to $-270^{\circ}$ (high $\omega$)
Gain crossover $\omega_{gc}$	$11.40\text{ rad/s}$ (asymptotic estimate $12.6$)
Phase crossover $\omega_{pc}$	$\sqrt{20}=4.472\text{ rad/s}$
Gain margin	$-18.4\text{ dB}$ (ratio $0.12$)
Phase margin	$-38.8^{\circ}$
Verdict	Unity-feedback closed loop is unstable; $K_{\max}=12$

Topic. Asymptotic (straight-line) Bode construction and the reading of stability margins from it.

Key relations. Time-constant form $L=K_{B}\prod(\tau_{z}s+1)/[s^{N}\prod(\tau_{p}s+1)]$; magnitude in dB $=20\log_{10}|L|$; each real pole adds $-20\text{ dB/dec}$ above $\omega=1/\tau$ and $-45^{\circ}$/decade centred there; type-1 integrator gives $-20\text{ dB/dec}$ and $-90^{\circ}$ throughout; $\text{GM}=-20\log_{10}|L(j\omega_{pc})|$, $\text{PM}=180^{\circ}+\angle L(j\omega_{gc})$.

Why this works. Taking logarithms converts the product of factors into a sum, so each pole and zero contributes an independent, universal shape that can be superposed by eye. The corner frequency is where the two asymptotes of a single factor intersect, and the maximum error of the straight-line magnitude for one real pole is only $3\text{ dB}$ (at the corner itself) — small enough that the sketch is quantitatively useful, which is why the examination supplies semi-log paper rather than a computer. The phase, however, is the weaker approximation: its ramp spans two decades, so when corners are closer together than that the ramps overlap and the straight-line phase can be several degrees off, which is exactly why $\omega_{gc}$ read from the asymptotes ($12.6$) differs noticeably from the exact value ($11.40$).

Common pitfall. Failing to normalise a factor written in pole-zero form. Had the plant been given as $100/[s(s+10)(s+2)]$ the Bode gain would be $100/(10\times2)=5$, i.e. $+14\text{ dB}$, not $+40\text{ dB}$ — a 26 dB error that shifts the whole magnitude plot. Here the factors are already in $(\tau s+1)$ form, so the printed 100 is $K_{B}$. A second pitfall is reporting a positive gain margin out of habit: when $\omega_{gc}$ exceeds $\omega_{pc}$ both margins are negative, and the correct engineering statement is that the loop must be de-tuned (or compensated) before it can be closed.

Source. Ogata, Modern Control Engineering 5th ed., §7-2 (Bode diagrams) and §7-6 (stability margins); Nise, Control Systems Engineering 8th ed., §10.2 and §10.7; Franklin, Powell & Emami-Naeini, 8th ed., §6.1 and §6.4.

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