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22-Mec-A3 System Analysis and Control · December 2018

Question 4 of 7: Bode Plots and the Gain Margin at 45° Phase Margin

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

Notes on this paper

Paper format. National Examination 16-Mec-A3 — System Analysis and Control, December 2018. Three hours; closed book; no aids other than semi-log graph paper and an approved Casio or Sharp calculator. Seven questions are printed and any four constitute a complete paper, all of equal value (25 marks each). Only the first four questions appearing in the answer book are marked. A table of Laplace transforms is supplied with the paper. All seven questions are solved here, so that the set works as a complete study resource rather than one candidate’s four choices.

Reference texts. K. Ogata, Modern Control Engineering, 5th ed. (Pearson) — Chs. 5 (transient response), 6 (root locus), 7 (frequency response); N. S. Nise, Control Systems Engineering, 8th ed. (Wiley) — Chs. 4, 6, 7, 8, 10; G. F. Franklin, J. D. Powell and A. Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed. (Pearson) — Chs. 3, 5, 6; R. C. Dorf and R. H. Bishop, Modern Control Systems, 13th ed. (Pearson). These are the standard references for the 16-Mec-A3 syllabus; the exam is closed book, so every result below is obtained with the supplied Laplace-transform table and hand methods only.

Question 4: Bode Plots and the Gain Margin at 45° 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. Unity-feedback loop with

$$G(s)=\frac{1}{s(s+1)\left[\dfrac{s^{2}}{25}+0.4\dfrac{s}{5}+1\right]}$$

The bracketed quadratic is already in time-constant (normalised) form: comparing it with $s^{2}/\omega_n^{2}+2\zeta s/\omega_n+1$ gives

Factor inventory for the Bode construction
FactorTypeCorner frequencySlope contribution
$1/s$Integrator (type 1)—$-20$ dB/dec from $\omega\to0$
$1/(s+1)$Real pole, $\tau=1$ s$\omega=1$ rad/sa further $-20$ dB/dec
$1/[(s/5)^{2}+0.08s+1]$Quadratic pole pair$\omega_n=5$ rad/sa further $-40$ dB/dec
Damping of the pair$2\zeta/\omega_n=0.08$$\zeta=0.2$resonant peak $\approx+8$ dB

Find. (a) the magnitude and phase Bode plots of $G(j\omega)$; (b) the gain margin that results once a gain $K$ is inserted to make the phase margin exactly $45^{\circ}$.

0.1110100-120-80-40040frequency omega (rad/s, log scale)dB0.1110100-360-270-180-90frequency omega (rad/s, log scale)deg-20 dB/dec-40-80 dB/decGM = 14.9 dBPM = 45 degsolid: K = 1dashed: K = 1.115omega_gc = 0.867omega_pc = 2.887
Bode plots of the loop. Solid: the given $G(j\omega)$ ($K=1$). Dashed: the magnitude after the gain is raised to $K=1.115$ to set a $45^{\circ}$ phase margin. The green markers locate the gain crossover, the magenta markers the phase crossover and the resulting gain margin.

Approach. Build the asymptotic magnitude from the factor inventory and add the phase contributions; then find the frequency where the phase is $-135^{\circ}$ (this fixes the gain crossover for a $45^{\circ}$ phase margin), set $K$ to force $|KG|=1$ there, and measure the surviving magnitude at the $-180^{\circ}$ phase crossover. Finally, cross-check the gain margin against Routh’s criterion applied to the same loop.

  1. (a) Sketch the magnitude asymptotes. With unity Bode gain, the low-frequency asymptote is $|G|\approx1/\omega$, which passes through 0 dB at $\omega=1$ rad/s and falls at $-20$ dB/dec. Each corner then steepens it:
$$\text{slope}=\begin{cases}-20\ \text{dB/dec}, & \omega<1\\-40\ \text{dB/dec}, & 1<\omega<5\\-80\ \text{dB/dec}, & \omega>5\end{cases}$$

Because $\zeta=0.2$ is light, the exact curve departs sharply from the asymptotes near $\omega_n=5$: the resonant peak rises about $-20\log_{10}(2\zeta)=+7.96$ dB above the corner. This peak is the dominant feature of the plot and the reason the answer to part (b) is not obvious by eye.

  1. (a) Assemble the phase. Adding the contributions of the integrator, the real pole and the quadratic pair,
$$\angle G(j\omega)=-90^{\circ}-\tan^{-1}\omega-\tan^{-1}\!\left(\frac{0.08\,\omega}{1-\omega^{2}/25}\right)$$

The phase starts at $-90^{\circ}$, passes $-135^{\circ}$ near $\omega=0.87$, and because the lightly damped pair swings a full $-180^{\circ}$ over a narrow band around $\omega=5$, the total runs rapidly down to $-360^{\circ}$. The arctangent of the quadratic term must be evaluated as a two-argument angle: once $\omega>5$ the denominator $1-\omega^{2}/25$ turns negative and the angle continues past $-90^{\circ}$ toward $-180^{\circ}$ instead of folding back.

  1. (b) Locate the gain crossover demanded by a 45° phase margin. Phase margin is $\text{PM}=180^{\circ}+\angle G(j\omega_{gc})$, so $\text{PM}=45^{\circ}$ requires the phase to be $-135^{\circ}$ at the crossover. Since a pure gain shifts magnitude but never phase, that frequency is fixed by $G$ alone. Solving $\angle G(j\omega)=-135^{\circ}$ gives
$$\boxed{\omega_{gc}=0.8666\ \text{rad/s}}$$

At that frequency the quadratic factor is still close to unity, so the condition is dominated by the integrator and the pole at $-1$: the requirement is essentially $\tan^{-1}\omega\approx45^{\circ}$, which is why the answer lands just below $1$ rad/s.

  1. (b) Set the gain so the crossover actually occurs there. Evaluating the magnitude,
$$|G(j\omega_{gc})|=\frac{1}{\omega\sqrt{1+\omega^{2}}\sqrt{(1-\omega^{2}/25)^{2}+(0.08\omega)^{2}}}\Bigg|_{\omega=0.8666}=0.8968$$

so the loop needs to be lifted slightly:

$$K=\frac{1}{0.8968}=\boxed{1.115}\qquad(+0.95\ \text{dB})$$
  1. (b) Find the phase crossover in closed form. Set $\angle G=-180^{\circ}$, i.e. $\tan^{-1}\omega+\tan^{-1}[0.08\omega/(1-\omega^{2}/25)]=90^{\circ}$. Two first-quadrant angles sum to $90^{\circ}$ exactly when the product of their tangents is $1$:
$$\omega\cdot\frac{0.08\,\omega}{1-\omega^{2}/25}=1\;\Longrightarrow\;0.08\omega^{2}=1-0.04\omega^{2}\;\Longrightarrow\;\omega_{pc}=\sqrt{\tfrac{1}{0.12}}=\boxed{2.887\ \text{rad/s}}$$

No numerical search is needed. Note that the phase reaches $-180^{\circ}$ well below the resonance at $\omega_n=5$, because the lightly damped pair begins contributing phase lag long before its corner.

  1. (b) Read off the gain margin. The magnitude of the compensated loop at the phase crossover is
$$|KG(j\omega_{pc})|=1.115\times0.16071=0.17924\;\Longrightarrow\;\text{GM}=\frac{1}{0.17924}=5.580$$$$\boxed{\text{GM}=5.58=14.9\ \text{dB at }\omega_{pc}=2.887\ \text{rad/s}}$$
  1. Cross-check the margin with Routh’s criterion. The gain margin is by definition the factor by which $K$ may rise before instability, so it must equal $K_{\max}/K$ from a Routh test on the same loop. The characteristic equation $s(s+1)(s^{2}/25+0.08s+1)+K=0$ expands to $0.04s^{4}+0.12s^{3}+1.08s^{2}+s+K=0$, or after dividing by $0.04$,
$$s^{4}+3s^{3}+27s^{2}+25s+25K=0$$

The Routh condition for a quartic $s^4+as^3+bs^2+cs+d$ is $d<c(ab-c)/a^{2}$, giving $25K_{\max}<155.56$, hence $K_{\max}=6.222$. Then $K_{\max}/K=6.222/1.115=5.580$, which reproduces the Bode gain margin exactly. This is the cheapest available guard against the classic reciprocal slip (quoting $1/\text{GM}$, or $-14.9$ dB).

Both margins are positive, so the loop is stable with a reasonable stability cushion at this gain: $45^{\circ}$ of phase margin and almost 15 dB of gain margin is a serviceable, if not generous, design point.

Question 4 — final results
QuantityValue
Corner frequencies$\omega=1$ rad/s (real pole); $\omega_n=5$ rad/s, $\zeta=0.2$
Asymptote slopes$-20$, then $-40$, then $-80$ dB/dec
Resonant peak$\approx+7.96$ dB near $\omega=5$ rad/s
Gain crossover for PM $=45^{\circ}$$\omega_{gc}=0.8666$ rad/s
Required gain$K=1.115$ ($+0.95$ dB)
Phase crossover$\omega_{pc}=\sqrt{1/0.12}=2.887$ rad/s
(b) Gain margin$\mathbf{5.58}$ (ratio) $=\mathbf{14.9}$ dB
Routh cross-check$K_{\max}=6.222$; $K_{\max}/K=5.58$ ✓