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

Question 3 of 6: Bode Diagrams — a Lag Network and a Three-Pole Loop

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

Question 3: Bode Diagrams — a Lag Network and a Three-Pole Loop (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. Two open-loop transfer functions, both already written in time-constant (Bode) form — every binomial reads $(\tau s+1)$ — so the leading constants 15 and 25 are the Bode gains and no renormalisation is needed.

Find. Asymptotic and exact magnitude ("attenuation") curves plus phase curves for each loop, with the corrections evaluated at the break points.

Approach. Convert the gain to decibels for the low-frequency asymptote, locate each corner frequency at $\omega=1/\tau$, add $-20\ \text{dB/decade}$ per pole and $+20\ \text{dB/decade}$ per zero beyond its corner, then compute the exact magnitude at each corner to place the correction. Phase is the sum of $\pm\tan^{-1}(\tau\omega)$ terms.

Part (a)

1. Identify the gain and the corner frequencies. The Bode gain is $K=15$, so the low-frequency asymptote sits at$$20\log_{10}15=23.52\ \text{dB}$$The pole $\tau_{p}=0.01\ \text{s}$ breaks at $\omega=1/0.01=100\ \text{rad/s}$ and the zero $\tau_{z}=0.001\ \text{s}$ breaks at $\omega=1/0.001=1000\ \text{rad/s}$.
2. Build the asymptotic magnitude. Below 100 rad/s the curve is flat at 23.52 dB. The pole comes first, so between 100 and 1000 rad/s the asymptote falls at $-20\ \text{dB/decade}$ — one full decade, hence a 20 dB drop. Above 1000 rad/s the zero contributes $+20\ \text{dB/decade}$, cancelling the pole, and the asymptote is flat again at$$20\log_{10}\!\left(15\times\frac{0.001}{0.01}\right)=20\log_{10}(1.5)=3.52\ \text{dB}$$This is a lag network: high frequencies are attenuated by a factor $\alpha=\tau_{z}/\tau_{p}=0.1$ relative to low frequencies.
3. Correct the curve at the break points. Evaluating the exact magnitude,$$|GH(j100)|=\frac{15\sqrt{1+(0.1)^{2}}}{\sqrt{1+1}}\;\Rightarrow\;20.55\ \text{dB}$$$$|GH(j1000)|=\frac{15\sqrt{1+1}}{\sqrt{1+100}}\;\Rightarrow\;6.49\ \text{dB}$$so the true curve lies 2.97 dB below the asymptote at the pole corner and 2.97 dB above it at the zero corner. These fall slightly short of the textbook 3.01 dB because the corners are only one decade apart and each feature already feels the other.
4. Sketch the phase. $\angle GH=\tan^{-1}(0.001\omega)-\tan^{-1}(0.01\omega)$, which starts at $0^{\circ}$, dips to a maximum lag and returns to $0^{\circ}$. The extremum sits at the geometric mean of the corners,$$\omega_{m}=\sqrt{(100)(1000)}=316.2\ \text{rad/s}$$where the standard lag-network result gives$$\phi_{m}=-\sin^{-1}\!\left(\frac{1-\alpha}{1+\alpha}\right)=-\sin^{-1}\!\left(\frac{0.9}{1.1}\right)=\boxed{-54.90^{\circ}}$$at which frequency the magnitude is 13.52 dB — exactly midway between the two flat asymptotes, as symmetry requires.

Part (b)

1. Identify the gain and corner frequencies. The Bode gain is $K=25$:$$20\log_{10}25=27.96\ \text{dB}$$The three poles break at $\omega=1/0.2=5$, $\omega=1/0.1=10$ and $\omega=1/0.01=100\ \text{rad/s}$. There are no zeros and the loop is type 0, so the curve starts flat.
2. Assemble the asymptotic slopes. Each pole steepens the slope by a further $-20\ \text{dB/decade}$ beyond its corner:

3. Correct the curve at each break point. Evaluating $|GH(j\omega)|=25/\prod\sqrt{1+(\tau_{i}\omega)^{2}}$ exactly:$$|GH(j5)|=23.97\ \text{dB},\qquad|GH(j10)|=17.92\ \text{dB},\qquad|GH(j100)|=-21.13\ \text{dB}$$The first two corners are only half a decade apart, so each correction exceeds the isolated-pole value of 3.01 dB (3.99 dB and 4.13 dB respectively); the well-separated third corner is corrected by very nearly the textbook amount.
4. Locate the gain crossover and the phase margin. Setting $|GH|=1$ (0 dB) and solving numerically gives$$\omega_{gc}=33.52\ \text{rad/s}$$At that frequency the phase, accumulated by hand as $-\left[\tan^{-1}(0.2\omega)+\tan^{-1}(0.1\omega)+\tan^{-1}(0.01\omega)\right]$, is $-173.43^{\circ}$, so$$\text{PM}=180^{\circ}-173.43^{\circ}=\boxed{+6.57^{\circ}}$$
5. Locate the phase crossover and the gain margin. The phase reaches $-180^{\circ}$ at $\omega_{pc}=39.37\ \text{rad/s}$, where $|GH|=-2.84\ \text{dB}$, so$$\text{GM}=\frac{1}{|GH(j\omega_{pc})|}=1.386\;\Rightarrow\;\boxed{+2.84\ \text{dB}}$$
6. Cross-check the gain margin against Routh–Hurwitz. This is the cheapest guard against a reciprocal slip. Closing the loop, $1+GH=0$ gives$$0.0002s^{3}+0.023s^{2}+0.31s+(1+K)=0$$whose Routh $s^{1}$ row vanishes at $K_{\max}=(0.023)(0.31)/0.0002-1=34.65$. The gain margin as a ratio must equal $K_{\max}/K=34.65/25=1.386$ — which it does exactly.