22-Mec-A3 System Analysis and Control · December 2017

Question 6 of 6: Bode Diagram of a Resonant Plant; Gain Design for a Specified Phase Margin

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

Question 6: Bode Diagram of a Resonant Plant; Gain Design for a Specified Phase Margin (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. Part (a): \(G(s)=10\left(s^2+0.4s+1\right)/\left[s\left(s^2+0.8s+9\right)\right]\) — an integrator, a lightly damped quadratic numerator and a lightly damped quadratic denominator. Part (b): the unity-feedback loop \(G(s)=K/\left[s\left(s^2+s+4\right)\right]\), with \(K\) to be chosen.

Find. (a) the asymptotic and exact Bode magnitude and phase plots, with all corner frequencies, damping ratios and the low-frequency asymptote identified; (b) the gain \(K\) giving a \(50^\circ\) phase margin, and the resulting gain margin.

Approach. For part (a), first normalise both quadratics into standard time-constant form so that the correct Bode gain is used, then identify each factor's corner frequency and damping ratio and superpose the asymptotes. For part (b), solve the phase condition analytically for the gain-crossover frequency, impose \(|G|=1\) there to obtain \(K\), then evaluate the gain at the \(-180^\circ\) phase-crossover frequency and cross-check against the Routh gain limit.

1. Normalise G(s) into Bode (time-constant) form. This is the step that fixes the low-frequency asymptote, and it is the one most often skipped. The numerator quadratic already has unit constant term, but the denominator quadratic has constant term 9, so $$G(s)=\frac{10\left(s^2+0.4s+1\right)}{9\,s\left(\dfrac{s^2+0.8s+9}{9}\right)}=\frac{10}{9}\cdot\frac{s^2+0.4s+1}{s\left[\left(\frac{s}{3}\right)^2+\frac{0.8}{9}s+1\right]}$$ The Bode gain is therefore \(K_B=10/9=1.1111\), i.e. \(20\log_{10}(1.1111)=0.915\ \text{dB}\) — not \(20\log_{10}10=20\) dB, which is what reading the printed numerator alone would suggest.
2. Extract the corner frequencies and damping ratios. Comparing each quadratic with \(s^2+2\zeta\omega_n s+\omega_n^2\): $$\text{numerator: }\ \omega_1=\sqrt{1}=1\ \text{rad/s},\quad 2\zeta_1\omega_1=0.4\;\Rightarrow\;\zeta_1=0.20$$ $$\text{denominator: }\ \omega_2=\sqrt{9}=3\ \text{rad/s},\quad 2\zeta_2\omega_2=0.8\;\Rightarrow\;\zeta_2=\frac{0.8}{6}=0.1333$$ Both are lightly damped, so both will produce pronounced departures from the asymptotes — a notch at the numerator corner and a peak at the denominator corner.
3. Assemble the asymptotic magnitude plot. At low frequency only the integrator and the Bode gain act, so \(|G|\approx K_B/\omega\): a \(-20\ \text{dB/decade}\) line passing through \(0.915\) dB at \(\omega=1\) rad/s (equivalently crossing 0 dB at \(\omega=K_B=1.111\) rad/s). The numerator quadratic adds \(+40\) dB/decade above \(\omega_1=1\), and the denominator quadratic subtracts \(40\) dB/decade above \(\omega_2=3\). The asymptote slopes are therefore $$\boxed{\;-20\ \text{dB/dec}\ (\omega<1)\;\to\;+20\ \text{dB/dec}\ (1<\omega<3)\;\to\;-20\ \text{dB/dec}\ (\omega>3)\;}$$ and at high frequency \(|G|\to 10/\omega\), confirming the final \(-20\) dB/decade slope.
4. Correct the asymptotes at the two resonances. A lightly damped quadratic departs from its asymptote by \(\mp20\log_{10}(2\zeta)\) at the corner. Evaluating the exact magnitude: $$\left|G(j0.991)\right|=-6.07\ \text{dB}\ \text{(anti-resonant notch, numerator)},\qquad \left|G(j3.012)\right|=+21.02\ \text{dB}\ \text{(resonant peak, denominator)}$$ The peak is over 20 dB above the asymptote intersection because \(\zeta_2=0.133\) is very low; a designer would regard this as a structural resonance that must be gain-stabilised or notched.
5. Assemble the phase plot. The integrator contributes a constant \(-90^\circ\). The numerator quadratic sweeps \(+180^\circ\) (rapidly, because \(\zeta_1\) is small) as \(\omega\) passes 1, and the denominator quadratic sweeps \(-180^\circ\) as \(\omega\) passes 3. The phase therefore starts at \(-90^\circ\), rises to a maximum of about \(+52^\circ\) between the two corners, and returns to \(-90^\circ\) at high frequency: $$\angle G(j0.01)=-89.8^\circ,\qquad \angle G(j100)=-89.8^\circ$$ Crucially the phase never reaches \(-180^\circ\), so this open loop has an infinite gain margin; the magnitude curve crosses 0 dB three times (at \(\omega=0.692,\ 1.348\) and \(10.73\) rad/s) because the notch and the peak drag it back and forth across the axis.

Part (b) is a frequency-domain design rather than a plotting exercise: the phase-margin specification pins the gain-crossover frequency first, and the magnitude condition then delivers the gain.

1. Write the phase of the open loop. With \(G(j\omega)=K/\left[j\omega\left(4-\omega^2+j\omega\right)\right]\), $$\angle G(j\omega)=-90^\circ-\arctan\!\left(\frac{\omega}{4-\omega^2}\right)$$ using the two-argument arctangent so that the branch stays correct past \(\omega=2\).
2. Impose the phase-margin specification. A phase margin of \(50^\circ\) means \(\angle G(j\omega_{gc})=-180^\circ+50^\circ=-130^\circ\), hence $$-90^\circ-\arctan\!\left(\frac{\omega}{4-\omega^2}\right)=-130^\circ\;\Longrightarrow\;\arctan\!\left(\frac{\omega}{4-\omega^2}\right)=40^\circ$$ Writing \(t=\tan40^\circ=0.83910\) and assuming \(4-\omega^2>0\) (verified below), this is \(t\left(4-\omega^2\right)=\omega\), i.e. the quadratic $$t\,\omega^2+\omega-4t=0\;\Longrightarrow\;\omega_{gc}=\frac{-1+\sqrt{1+16t^2}}{2t}=1.4910\ \text{rad/s}$$ Here \(4-\omega_{gc}^2=1.7771>0\), confirming the branch assumption.
3. Impose the magnitude condition to obtain K. At the gain-crossover frequency \(|G(j\omega_{gc})|=1\), and $$\left|G(j\omega)\right|=\frac{K}{\omega\left|\left(4-\omega^2\right)+j\omega\right|}=\frac{K}{\omega\sqrt{\left(4-\omega^2\right)^2+\omega^2}}$$ Substituting \(\omega_{gc}=1.4910\): $$K=\omega_{gc}\sqrt{\left(4-\omega_{gc}^2\right)^2+\omega_{gc}^2}=1.4910\times\sqrt{1.7771^2+1.4910^2}=1.4910\times2.3197$$ $$\boxed{\;K=3.459\;}$$
4. Locate the phase crossover and evaluate the gain margin. The phase reaches \(-180^\circ\) when \(\arctan\left[\omega/(4-\omega^2)\right]=90^\circ\), i.e. when the real part vanishes: $$4-\omega^2=0\;\Longrightarrow\;\omega_{pc}=2\ \text{rad/s}$$ There \(\left|G(j2)\right|=K/\left(2\times2\right)=K/4=0.8646\), so $$\text{GM}=\frac{1}{\left|G(j\omega_{pc})\right|}=\frac{4}{K}=\frac{4}{3.459}=1.1566\;\Longrightarrow\;\boxed{\;\text{GM}=20\log_{10}(1.1566)=1.26\ \text{dB}\;}$$
5. Cross-check the gain margin against Routh. The closed-loop characteristic equation is \(s^3+s^2+4s+K=0\), whose Routh \(s^1\) entry is \((1\times4-K)/1=4-K\), giving the stability limit \(K_{\max}=4\). The gain-margin ratio must therefore equal \(K_{\max}/K\): $$\frac{K_{\max}}{K}=\frac{4}{3.459}=1.1566\;\checkmark$$ This identity is an excellent free check on frequency-domain work, because it catches the very common reciprocal slip in the gain-margin definition.
6. Interpret the answer. A \(50^\circ\) phase margin normally signals a well-damped design, yet the gain margin here is only 1.26 dB — the loop tolerates barely a 16 % increase in gain before it oscillates. The two margins disagree because the lightly damped plant pole pair (\(\omega_n=2\), \(\zeta=0.25\)) makes the phase plunge through \(-180^\circ\) almost immediately after the gain crossover. Never quote one margin alone: a design meeting a phase-margin specification can still be fragile to gain variation, and this question is constructed precisely to make that point.