22-Elec-B2 Advanced Control Systems · May 2017

Question 4 of 5: Bode plots and stability margins for $K/[s(s+2)(s+10)]$

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

Notes on this paper

Paper format. 16-Elec-B2 Advanced Control Systems, May 2017 — a three-hour open-book examination. 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%)”. This 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] or [2] and total exactly 25. Tables of inverse Laplace transforms and of Laplace/z transforms are appended to the paper. 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 and the rise-time/peak-time/settling-time relations, Ch. 6 Routh–Hurwitz stability, Ch. 7 steady-state error and system type, Ch. 8 root-locus sketching rules, Ch. 9 root-locus design of cascade compensators, 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 design, Ch. 7 frequency-response design); 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 (percent overshoot, settling time, asymptote intercept, “summation of the angles”) follows Nise closely.

Check — two source readings that were checked against the printed paper.

(1) Question 1, item 5. The paper prints the $s^2$ first-column entry of the Routh table as $(1-4\varepsilon)/\varepsilon$, which is exactly what the Routh–Hurwitz recursion gives, and which is confirmed independently by the printed $s^1$ entry. The table on the paper is correct and internally consistent. The answer is unaffected by the alternative form $\varepsilon/(1-4\varepsilon)$, because both are positive for small $\varepsilon>0$.

(2) Question 4. The block diagram prints the plant as $10/[s(s+2)(s+10)]$ with no $K$ shown, while part (a) says “the open-loop transfer function when $K = 20$”. This solution takes the direct reading, $G(s)=K/[s(s+2)(s+10)]$ with $K = 20$, so the printed 10 is one sample value of the same numerator gain. The alternative reading — a fixed plant gain of 10 multiplied by $K$, i.e. $G(s)=10K/[s(s+2)(s+10)]$ — is carried through in a callout under part (b) so that both mark schemes are covered.

Question 4: Bode plots and stability margins for $K/[s(s+2)(s+10)]$ (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.

Quantity	Value
Loop transfer function	$G(s) = K/[s(s+2)(s+10)]$, unity feedback
Gain for parts (a) and (b)	$K = 20$
Poles	$s = 0$ (integrator), $s = -2$, $s = -10$
Corner frequencies	$\omega = 2$ rad/s and $\omega = 10$ rad/s
System type	Type 1 (one pole at the origin)

Find. (a) the normalised form and the component-by-component Bode asymptotes, then the composite plot; (b) the gain margin and phase margin with the frequencies at which each is measured, for $K = 20$; (c) the largest $K$ for which the closed loop remains stable.

Approach. Normalise each factor to the form $(1 + s/p)$ so that every component has unity low-frequency gain, add the asymptotes ($-20$ dB/dec per pole above its corner, $-20$ dB/dec everywhere for the integrator), then locate the phase crossover in closed form using the fact that $\tan^{-1}(\omega/a) + \tan^{-1}(\omega/b) = 90^\circ$ exactly when $\omega^2 = ab$, and the gain crossover numerically.

Part (a) — normalise the transfer function. Factor the DC gain out of each pole term so every factor is unity at $\omega = 0$: $$G(s) = \frac{20}{s(s+2)(s+10)} = \frac{20}{s\cdot 2\left(1+\dfrac{s}{2}\right)\cdot 10\left(1+\dfrac{s}{10}\right)} = \frac{20/20}{s\left(1+\dfrac{s}{2}\right)\left(1+\dfrac{s}{10}\right)} ,$$ $$\boxed{G(s) = \frac{1}{s\left(1+\dfrac{s}{2}\right)\left(1+\dfrac{s}{10}\right)}}$$ The normalised gain is exactly $1$, i.e. $0$ dB, so the “overall gain after normalization” component is a flat line on the 0 dB axis with $0^\circ$ phase. This also identifies the velocity error constant: $K_v = \lim_{s\to0}sG(s) = 1\ \text{s}^{-1}$, so the steady-state error to a unit ramp is $1/K_v = 1$.

Part (a) continued — the four components. Sketched separately, as the question asks:

Component	Magnitude asymptote	Phase
Normalised gain $1$	flat at $0$ dB	$0^\circ$ at all $\omega$
Integrator $1/s$	$-20$ dB/dec through $0$ dB at $\omega = 1$	$-90^\circ$ at all $\omega$
Pole $1/(1+s/2)$	$0$ dB below $\omega=2$, then $-20$ dB/dec	$0^\circ \to -45^\circ$ at $2 \to -90^\circ$
Pole $1/(1+s/10)$	$0$ dB below $\omega=10$, then $-20$ dB/dec	$0^\circ \to -45^\circ$ at $10 \to -90^\circ$

Adding them gives a composite magnitude that falls at $-20$ dB/dec up to $\omega = 2$, at $-40$ dB/dec from 2 to 10, and at $-60$ dB/dec above 10; and a composite phase that starts at $-90^\circ$ and ends at $-270^\circ$.

Part (a) continued — tabulate the composite curve. Straight-line asymptotes and exact values at the frequencies worth marking on the grid:

$\omega$ (rad/s)	Asymptote (dB)	Exact $\|G\|$ (dB)	Exact $\angle G$
0.1	$+20.00$	$+19.99$	$-93.44^\circ$
1	$0.00$	$-1.01$	$-122.28^\circ$
2	$-6.02$	$-9.20$	$-146.31^\circ$
10	$-33.98$	$-37.16$	$-213.69^\circ$
100	$-93.98$	$-94.02$	$-263.14^\circ$

The asymptote and the exact curve differ by the expected $3$ dB at each corner, growing to about $3.2$ dB where the two corners interact.

Part (b) — phase crossover in closed form. The phase is $$\angle G(j\omega) = -90^\circ - \tan^{-1}\!\frac{\omega}{2} - \tan^{-1}\!\frac{\omega}{10},$$ so $\angle G = -180^\circ$ requires $\tan^{-1}(\omega/2) + \tan^{-1}(\omega/10) = 90^\circ$. Two angles sum to $90^\circ$ exactly when the product of their tangents is 1, so $$\frac{\omega}{2}\cdot\frac{\omega}{10} = 1 \quad\Longrightarrow\quad \omega_{pc}^2 = 2\times 10 = 20,$$ $$\boxed{\omega_{pc} = \sqrt{20} = 4.472\ \text{rad/s}}$$ This is independent of $K$ — gain scaling moves the magnitude curve, never the phase curve.
Part (b) continued — gain margin. At $\omega_{pc}$ the loop transfer function is real and negative, and its magnitude has a tidy closed form. Writing $a = 2$, $b = 10$ and using $\omega_{pc}^2 = ab$, $$\left|G(j\omega_{pc})\right| = \frac{K}{ab(a+b)} = \frac{20}{2\times 10\times 12} = \frac{20}{240} = 0.08333 ,$$ so $$\boxed{GM = \frac{1}{0.08333} = 12.0 = 21.58\ \text{dB at }\omega_{pc} = 4.472\ \text{rad/s}}$$ The loop gain can be raised by a factor of 12 before instability.
Part (b) continued — gain crossover and phase margin. The gain crossover has no closed form (it is a cubic in $\omega^2$), so solve $|G(j\omega)| = 1$ numerically: $$\frac{20}{\omega\sqrt{\omega^2+4}\,\sqrt{\omega^2+100}} = 1 \quad\Longrightarrow\quad \boxed{\omega_{gc} = 0.907\ \text{rad/s}}$$ (At $\omega = 0.9$ the magnitude is 1.009 and at $\omega = 0.92$ it is 0.977, bracketing the root.) The phase there is $$\angle G(j0.907) = -90^\circ - \tan^{-1}(0.4535) - \tan^{-1}(0.0907) = -90^\circ - 24.42^\circ - 5.18^\circ = -119.58^\circ ,$$ $$\boxed{PM = 180^\circ - 119.58^\circ = 60.4^\circ \text{ at } \omega_{gc} = 0.907\ \text{rad/s}}$$ A phase margin near $60^\circ$ corresponds to a closed-loop damping of roughly $\zeta \approx PM/100 = 0.60$ and about 10% overshoot, so this loop is comfortably damped — consistent with the large gain margin.
Part (c) — maximum gain for stability. Instability begins when the magnitude at the (gain-independent) phase crossover reaches unity, i.e. when the gain margin is exhausted: $$\left|G(j\omega_{pc})\right| = \frac{K}{ab(a+b)} = 1 \quad\Longrightarrow\quad \boxed{K_{\max} = ab(a+b) = 2\times 10\times 12 = 240}$$ Equivalently $K_{\max} = K \times GM = 20 \times 12 = 240$. The Routh test gives the same number without any frequency response: the characteristic polynomial is $s^3 + 12s^2 + 20s + K$, and the cubic condition $a_2a_1 > a_0$ reads $12 \times 20 > K$, i.e. $K < 240$. At $K = 240$ the closed-loop poles are $-12$ and $\pm j4.472$ — sustained oscillation at exactly $\omega_{pc}$, which is the cross-check that ties the two methods together.

Question 4: Bode plot of $G(s) = 20/[s(s+2)(s+10)]$ with the straight-line asymptotes dashed. Gain crossover at $\omega_{gc} = 0.907$ rad/s gives $PM = 60.4^\circ$; phase crossover at $\omega_{pc} = 4.472$ rad/s gives $GM = 21.58$ dB. Corner frequencies are 2 and 10 rad/s, where the slope steepens from $-20$ to $-40$ and then to $-60$ dB/dec.

Check — the alternative reading of the printed numerator. If the printed $10$ is taken as a fixed plant gain that $K$ then multiplies, i.e. $G(s) = 10K/[s(s+2)(s+10)]$, then $K = 20$ means a loop gain of 200 and the answers become: $\omega_{pc} = 4.472$ rad/s unchanged (the phase curve never depends on gain); $GM = 240/200 = 1.2 = 1.58$ dB; $\omega_{gc} = 4.078$ rad/s; $PM = 3.9^\circ$; and $K_{\max} = 240/10 = 24$. That reading makes part (c) pedagogically pointed — the paper would be asking about a loop deliberately parked just under its stability limit — so it is worth carrying. The method, the corner frequencies and $\omega_{pc}$ are identical under either reading; only the two margins and the numerical value of $K_{\max}$ differ.

Quantity	Result
(a) Normalised loop	$G(s) = 1/\bigl[s(1+s/2)(1+s/10)\bigr]$; normalised gain $0$ dB
(a) Asymptote slopes	$-20$ dB/dec ($\omega<2$), $-40$ ($2<\omega<10$), $-60$ ($\omega>10$)
(a) Phase span	$-90^\circ$ at low $\omega$ to $-270^\circ$ at high $\omega$
(b) Phase-crossover frequency	$\omega_{pc} = \sqrt{20} = 4.472$ rad/s
(b) Gain margin	$12.0$ (ratio) $= 21.58$ dB
(b) Gain-crossover frequency	$\omega_{gc} = 0.907$ rad/s
(b) Phase margin	$60.4^\circ$
(c) Maximum stable gain	$K_{\max} = 240$ (Routh: $12\times20 > K$)
Oscillation at $K_{\max}$	$s = \pm j4.472$, third pole at $-12$
Alternative reading $10K/[\cdots]$	$GM = 1.58$ dB, $PM = 3.9^\circ$, $\omega_{gc}=4.078$, $K_{\max}=24$

Topic. Frequency-response analysis of a Type-1 third-order loop: normalised Bode construction from components, and gain/phase margins as the measure of distance from instability.

Key relations. For $L = K/[s(s+a)(s+b)]$: $\omega_{pc} = \sqrt{ab}$ exactly (because $\tan^{-1}(\omega/a)+\tan^{-1}(\omega/b) = 90^\circ \Leftrightarrow \omega^2 = ab$); $|L(j\omega_{pc})| = K/[ab(a+b)]$; hence $GM = ab(a+b)/K$ and $K_{\max} = ab(a+b)$, which is also the Routh limit of $s^3+(a+b)s^2+ab\,s+K$. Asymptote rules: $-20$ dB/dec per pole above its corner, $3$ dB error at the corner, $-45^\circ$ phase at the corner. Margin–damping rule of thumb $\zeta \approx PM/100$ for $PM \lesssim 70^\circ$.

Why this works. The Nyquist criterion for this open-loop-stable plant reduces to a distance measurement along the $-1$ direction, and the Bode plot separates that measurement into two independent one-dimensional questions: at the frequency where the phase is $-180^\circ$, how far below unity is the gain (that is the gain margin); and at the frequency where the gain is unity, how far above $-180^\circ$ is the phase (that is the phase margin). Because gain scaling translates the magnitude curve vertically and leaves the phase curve untouched, $\omega_{pc}$ is a property of the plant alone, and the maximum gain follows by inspection — which is why parts (b) and (c) are really one calculation. The tangent identity makes $\omega_{pc}$ exact for this pole pattern, so only the gain crossover needs iteration.

Common pitfall. Reading $\omega_{gc}$ and $\omega_{pc}$ off the straight-line asymptotes rather than the exact curve: the asymptote is $3$ dB optimistic at each corner, and here it puts the gain crossover at $\omega = 1$ instead of $0.907$, which shifts the phase margin by about $3^\circ$. The related error is quoting the gain margin as a ratio where decibels were wanted, or vice versa — $12$ and $21.58$ are the same margin. Finally, do not expect a phase crossover to exist at all for every plant: a two-pole Type-1 loop never reaches $-180^\circ$, so its gain margin is infinite, and reporting a finite number there is a common loss of marks in this series.

Source. Nise, Control Systems Engineering, 8th ed., §10.2 (asymptotic Bode construction, component by component), §10.7 (gain and phase margins from Bode plots), §10.10 (relation between phase margin and damping); Ogata, Modern Control Engineering, 5th ed., §7.2 and §7.4; Franklin, Powell and Emami-Naeini, 8th ed., §6.4–6.5.

← Question 3 Next: Question 5 →

Component	Magnitude asymptote	Phase
Normalised gain \(1\)	flat at \(0\) dB	\(0^\circ\) at all \(\omega\)
Integrator \(1/s\)	\(-20\) dB/dec through \(0\) dB at \(\omega = 1\)	\(-90^\circ\) at all \(\omega\)
Pole \(1/(1+s/2)\)	\(0\) dB below \(\omega=2\), then \(-20\) dB/dec	\(0^\circ \to -45^\circ\) at \(2 \to -90^\circ\)
Pole \(1/(1+s/10)\)	\(0\) dB below \(\omega=10\), then \(-20\) dB/dec	\(0^\circ \to -45^\circ\) at \(10 \to -90^\circ\)

\(\omega\) (rad/s)	Asymptote (dB)	Exact \(\|G\|\) (dB)	Exact \(\angle G\)
0.1	\(+20.00\)	\(+19.99\)	\(-93.44^\circ\)
1	\(0.00\)	\(-1.01\)	\(-122.28^\circ\)
2	\(-6.02\)	\(-9.20\)	\(-146.31^\circ\)
10	\(-33.98\)	\(-37.16\)	\(-213.69^\circ\)
100	\(-93.98\)	\(-94.02\)	\(-263.14^\circ\)

Quantity	Result
(a) Normalised loop	\(G(s) = 1/\bigl[s(1+s/2)(1+s/10)\bigr]\); normalised gain \(0\) dB
(a) Asymptote slopes	\(-20\) dB/dec (\(\omega<2\)), \(-40\) (\(2<\omega<10\)), \(-60\) (\(\omega>10\))
(a) Phase span	\(-90^\circ\) at low \(\omega\) to \(-270^\circ\) at high \(\omega\)
(b) Phase-crossover frequency	\(\omega_{pc} = \sqrt{20} = 4.472\) rad/s
(b) Gain margin	\(12.0\) (ratio) \(= 21.58\) dB
(b) Gain-crossover frequency	\(\omega_{gc} = 0.907\) rad/s
(b) Phase margin	\(60.4^\circ\)
(c) Maximum stable gain	\(K_{\max} = 240\) (Routh: \(12\times20 > K\))
Oscillation at \(K_{\max}\)	\(s = \pm j4.472\), third pole at \(-12\)
Alternative reading \(10K/[\cdots]\)	\(GM = 1.58\) dB, \(PM = 3.9^\circ\), \(\omega_{gc}=4.078\), \(K_{\max}=24\)

Question 4 of 5: Bode plots and stability margins for \(K/[s(s+2)(s+10)]\)

Question 4: Bode plots and stability margins for \(K/[s(s+2)(s+10)]\) (25 marks)

Quantity	Value
Loop transfer function	\(G(s) = K/[s(s+2)(s+10)]\), unity feedback
Gain for parts (a) and (b)	\(K = 20\)
Poles	\(s = 0\) (integrator), \(s = -2\), \(s = -10\)
Corner frequencies	\(\omega = 2\) rad/s and \(\omega = 10\) rad/s
System type	Type 1 (one pole at the origin)