22-Elec-A2 Systems and Control · May 2018

Question 5 of 8: Lead Controller Design in the Frequency Domain

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

Notes on this paper

Paper format. National Exams, 16-Elec-A2 Systems and Control, May 2018 — 3 hours, closed book (approved Casio or Sharp calculator plus one double-sided handwritten 8.5 × 11″ formula sheet). A short Laplace-transform table is printed on page 2, and the $\zeta$–overshoot, $\zeta$–resonant-peak and phase-margin design charts plus the second-order model, controllability/observability and state-space transfer-function definitions are on page 3. Questions 1 and 2 are compulsory; five questions constitute a complete paper, so a candidate chooses three of Q3–Q8. Each question carries 20 marks, for 100 in total. All eight questions are worked below so the set is a complete study resource.

Reference texts. N. S. Nise, Control Systems Engineering (7th ed., Wiley) — block-diagram and signal-flow reduction with Mason’s rule (Ch. 5), steady-state error and the static error constants (Ch. 7), Routh–Hurwitz (Ch. 6), root locus (Ch. 8), cascade compensator design (Ch. 9), frequency response and gain/phase margins (Ch. 10), frequency-domain lead design (Ch. 11), state space and controllability/observability (Ch. 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — second-order correlations, dominant-poles modelling and pole placement by state feedback. Every block diagram, root locus, pole–zero map, Bode plot, closed-loop magnitude curve and step response below is redrawn as an inline figure.

Reading the supplied design charts. Page 3 gives $PO = 100\,e^{-\zeta\pi/\sqrt{1-\zeta^{2}}}$ for percent overshoot, $M_r/K_{dc} = 1/\left(2\zeta\sqrt{1-\zeta^{2}}\right)$ for the closed-loop resonant peak, and the phase-margin correlation $\Phi_m \approx 100\zeta$ (equivalently $\zeta \approx 0.01\,\Phi_m$). Those three relations, together with $T_{settle(\pm 2\%)} \approx 4/(\zeta\omega_n)$ and $T_{rise(0-100\%)} = (\pi - \cos^{-1}\zeta)/\omega_d$, carry most of the numerical work in Questions 2, 5, 7 and 8.

Question 5: Lead Controller Design in the Frequency Domain (20 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, plant $G(s) = 1/[s(s+1)(s+2.7)]$, lead controller $G_c(s) = K_c(\tau s+1)/(\alpha\tau s+1) = (a_1s+a_0)/(b_1s+1)$ with $\alpha \lt 1$. Design targets:

Specification	Required value
Steady-state error to a unit ramp	$e_{ss(ramp)} = 0.27\ \text{V/V}$
Percent overshoot	$PO \le 25\%$
Settling time	$T_{settle(\pm 2\%)} \le 2.5\ \text{s}$

Find. The uncompensated and required error constants; the uncompensated phase margin and crossover frequency together with the targets for the compensated loop; the lead parameters $\alpha$, $\tau$ (hence $a_1$, $a_0$, $b_1$) and $G_c(s)$; and the resulting closed-loop step-response specifications.

[Figure not reproduced: Question 5 loop structure (redrawn): a lead controller $G_c(s) = (a_1s+a_0)/(b_1s+1)$ in cascade with the Type 1 plant $1/[s(s+1)(s+2.7)]$, closed by unity negative feedback. See the official exam paper.]

Check: Figure Q5.1 does not match the stated plant. The printed chart shows a magnitude of 40 dB at $\omega = 0.01$ rad/s, a 0 dB crossing near 0.7 rad/s, and a phase that bottoms out near $-185^{\circ}$ before returning towards $-90^{\circ}$ at high frequency. The given $G(s) = 1/[s(s+1)(s+2.7)]$ has 31.4 dB at 0.01 rad/s, crosses 0 dB at 0.347 rad/s, and its phase falls monotonically to $-270^{\circ}$. The printed chart is therefore an idealised or borrowed plot. Every number below is computed analytically from the exact transfer function given in the question, and the redrawn Bode plot is the true response of that transfer function. A candidate reading the chart in the exam room would obtain the same design method with slightly different graphical values, which is acceptable — lead designs are not unique.

Approach. The ramp specification fixes the controller DC gain; the overshoot specification fixes the target phase margin through $\Phi_m \approx 100\zeta$, and the settling-time specification fixes the target crossover. Apply the standard lead formulas, then verify the achieved margin on the actual loop and iterate, because the plant phase rolls off steeply near the new crossover.

Error constants and the controller DC gain. The plant has one free integrator, so the loop is Type 1 and the position constant is infinite both before and after compensation: $$K_{pos,u} = \lim_{s\to0}G(s) = \infty = K_{pos,c} \qquad\Longrightarrow\qquad e_{ss(step)} = 0\% \ \text{in both cases}$$ The constant that actually governs the stated ramp requirement is the velocity constant: $$K_{v,u} = \lim_{s\to0}sG(s) = \frac{1}{1 \times 2.7} = 0.3704 \;\Longrightarrow\; e_{ss(ramp),u} = 2.70\ \text{V/V}$$ $$K_{v,c} = \frac{1}{0.27} = 3.7037 \;\Longrightarrow\; \boxed{\;K_c = a_0 = \frac{K_{v,c}}{K_{v,u}} = \frac{3.7037}{0.3704} = 10\;}$$ The required tenfold lift is exactly the DC gain of the lead controller, since $G_c(0) = a_0 = K_c$.
Uncompensated margins. Solving $|G(j\omega)| = 1$ numerically on the exact transfer function gives the gain-crossover frequency, and the phase there follows from $\angle G = -90^{\circ} - \tan^{-1}\omega - \tan^{-1}(\omega/2.7)$: $$\omega_{cp,u} = 0.347\ \text{rad/s}, \qquad \Phi_{m,u} = 180^{\circ} + \angle G(j\omega_{cp,u}) = \boxed{\;63.5^{\circ}\;}$$ The uncompensated loop is comfortably stable but far too slow, and it misses the ramp specification by a factor of ten.
Targets for the compensated loop. The overshoot limit converts to a damping ratio, and thence to a phase margin through the page-3 correlation: $$\zeta = \frac{-\ln(PO/100)}{\sqrt{\pi^{2}+\ln^{2}(PO/100)}} = \frac{1.3863}{3.4340} = 0.4037 \;\Longrightarrow\; \Phi_m \approx 100\zeta = 40.4^{\circ}$$ Allowing a little margin, take $\Phi_{m,c} = 45^{\circ}$. The settling-time limit fixes the required bandwidth: $$\zeta\omega_n \ge \frac{4}{2.5} = 1.6 \;\Longrightarrow\; \omega_n \ge \frac{1.6}{0.4037} = 3.96\ \text{rad/s}$$ and converting natural frequency to crossover through $\omega_{cp} = \omega_n\sqrt{\sqrt{1+4\zeta^{4}}-2\zeta^{2}}$ gives $$\boxed{\;\Phi_{m,c} = 45^{\circ}, \qquad \omega_{cp,c} \approx 3.38\ \text{rad/s}\;}$$
Phase deficiency after the gain lift. Raising the gain to $K_c = 10$ pushes the crossover out to where the plant phase has already collapsed. Solving $|10\,G(j\omega)| = 1$: $$\omega = 1.644\ \text{rad/s}, \qquad \Phi_m = 180^{\circ} + \angle G(j1.644) = -0.03^{\circ}$$ The gain-scaled loop is therefore on the verge of instability, and the lead must supply essentially the whole 45°. Adding the customary correction $\varepsilon \approx 8^{\circ}$ for the crossover shift, $$\phi_{max} = \Phi_{m,c} - \Phi_{m,\text{gain-scaled}} + \varepsilon = 45 - (-0.03) + 8 = 53.0^{\circ}$$
Standard lead parameters. From $\alpha = (1-\sin\phi_{max})/(1+\sin\phi_{max})$, $$\alpha = \frac{1-\sin 53.0^{\circ}}{1+\sin 53.0^{\circ}} = 0.1118$$ The peak phase must occur at the new crossover, which is where the gain-scaled magnitude equals $10\log_{10}\alpha = -9.52$ dB. Solving that condition gives $\omega_m = 2.709$ rad/s, and $$\tau = \frac{1}{\omega_m\sqrt{\alpha}} = 1.1042 \;\Longrightarrow\; a_1 = K_c\tau = 11.04, \quad a_0 = 10, \quad b_1 = \alpha\tau = 0.1234$$
Check the achieved margin — and iterate. Evaluating the actual compensated loop $G_cG$ at its true crossover gives $$\omega_{cp} = 2.709\ \text{rad/s}, \qquad \Phi_m = 28.2^{\circ}$$ which falls well short of the 45° target. This is not an arithmetic slip: the one-shot formula assumes the plant phase is roughly constant across the crossover shift, and for this triple-pole plant it drops by tens of degrees. The fix is a more aggressive lead. Taking $\alpha = 0.05$ (a 20:1 lead ratio, peak phase $64.8^{\circ}$) with $\tau = 0.8415$: $$\boxed{\;G_c(s) = \frac{8.415\,s + 10}{0.04207\,s + 1}\;} \qquad (\alpha = 0.05,\ \tau = 0.8415,\ K_c = 10)$$ which achieves $$\omega_{cp,c} = 2.391\ \text{rad/s}, \qquad \boxed{\;\Phi_{m,c} = 39.0^{\circ}\;}$$ This is the recommended design, and the numbers in part 4 use it.
Compensated closed-loop specifications. The step error is zero because the loop is still Type 1, and the ramp error is the design value by construction: $$e_{ss(step)} = \boxed{\;0\%\;}, \qquad e_{ss(ramp)} = \frac{1}{K_{v,c}} = \boxed{\;0.27\ \text{V/V}\;}$$ Converting the achieved margin through the page-3 correlations, $\zeta \approx 0.01\Phi_m = 0.390$ and $\omega_n = \omega_{cp}/\sqrt{\sqrt{1+4\zeta^{4}}-2\zeta^{2}} = 2.778$, so $$PO = 100\,e^{-\zeta\pi/\sqrt{1-\zeta^{2}}} = 26.4\%, \qquad T_{settle(\pm2\%)} = \frac{4}{\zeta\omega_n} = 3.69\ \text{s}$$ Simulating the exact fourth-order compensated closed loop confirms the model is in the right region and slightly optimistic: $$\boxed{\;PO_{exact} = 31.0\%, \qquad T_{settle,exact} = 3.88\ \text{s}\;}$$

Open-loop frequency response for Question 5, computed from the exact $G(s)$. Dashed blue: uncompensated, crossing 0 dB at 0.347 rad/s with $63.5^{\circ}$ of margin. Amber: after the tenfold gain lift demanded by the ramp specification — crossover moves to 1.644 rad/s and the margin collapses to nearly zero. Green: with the recommended lead, crossover 2.391 rad/s and $39.0^{\circ}$ of margin restored.

Quantity	Result
$K_{pos,u}$ and $K_{pos,c}$	both $\infty$ (Type 1)
$K_{v,u}$ / $e_{ss(ramp),u}$	$0.3704$ / $2.70\ \text{V/V}$
$K_{v,c}$ required / controller DC gain $K_c=a_0$	$3.7037$ / $10$
$\omega_{cp,u}$ / $\Phi_{m,u}$	$0.347\ \text{rad/s}$ / $63.5^{\circ}$
Targets $\omega_{cp,c}$ / $\Phi_{m,c}$	$3.38\ \text{rad/s}$ / $45^{\circ}$
Lead parameters $\alpha$, $\tau$	$0.05$, $0.8415\ \text{s}$
Controller $G_c(s)$	$\dfrac{8.415s+10}{0.04207s+1}$
Achieved $\omega_{cp}$ / $\Phi_m$	$2.391\ \text{rad/s}$ / $39.0^{\circ}$
$e_{ss(step\%)}$ / $e_{ss(ramp)}$	$0\%$ / $0.27\ \text{V/V}$
$PO$ (model / exact)	$26.4\%$ / $31.0\%$
$T_{settle(\pm2\%)}$ (model / exact)	$3.69\ \text{s}$ / $3.88\ \text{s}$

Check: the three specifications cannot all be met by a single-stage lead at this gain. The ramp specification forces $K_c = 10$, which by itself drives the phase margin of this triple-pole plant to zero. Sweeping the lead parameters shows that the best achievable margin with the peak placed at crossover is $33.8^{\circ}$, and even an aggressive 20:1 lead reaches only $39.0^{\circ}$ — short of the $40.4^{\circ}$ that $PO \le 25\%$ requires, with an exact overshoot of 31.0 % against the 25 % limit and a settling time of 3.88 s against the 2.5 s limit. The design above is the best single-stage lead available and is what the question asks for; meeting the full specification set would require either a two-stage (lead–lead) compensator, a lead–lag, or relaxation of the ramp accuracy. Stating this trade-off explicitly is the professionally correct answer, and the exam rubric invites exactly that under its instruction to record any assumptions.

Topic. Frequency-domain design of a cascade lead compensator to meet simultaneous steady-state accuracy and transient-response specifications.

Key relations. $K_v = \lim_{s\to0}sG_cG$ and $e_{ss(ramp)}=1/K_v$; $\Phi_m \approx 100\zeta$; $\zeta = -\ln(PO/100)/\sqrt{\pi^{2}+\ln^{2}(PO/100)}$; $T_{settle}\approx4/(\zeta\omega_n)$; $\omega_{cp} = \omega_n\sqrt{\sqrt{1+4\zeta^{4}}-2\zeta^{2}}$; and the lead pair $\alpha = (1-\sin\phi_{max})/(1+\sin\phi_{max})$ with $\omega_m = 1/(\tau\sqrt{\alpha})$, the peak phase being placed at the intended new crossover where the uncompensated magnitude equals $10\log_{10}\alpha$.

Why this works. A lead network contributes positive phase in a band around $\omega_m$ while lifting the high-frequency gain, so placing that band at the crossover buys back the margin that the accuracy-driven gain increase destroyed. The design splits cleanly because the two requirements act at opposite ends of the spectrum: $K_c$ is set purely by the low-frequency behaviour (the error constant), while $\alpha$ and $\tau$ shape the response near crossover and leave the DC gain untouched. The correlations $\Phi_m\approx100\zeta$ and $T_{settle}\approx4/(\zeta\omega_n)$ are second-order results, so they are estimates for a fourth-order loop — which is precisely why the achieved margin must be checked on the real loop rather than assumed, and why the exact simulation here returns 31 % overshoot where the model predicted 26 %.

Common pitfall. Trusting the one-shot formula without verifying the achieved phase margin. The standard procedure assumes the plant phase is nearly flat over the crossover shift; with three poles clustered below 3 rad/s the phase falls steeply, the true crossover lands where the plant has lost far more phase than the correction $\varepsilon$ anticipated, and the design silently misses its target by 17°. The second pitfall is answering part 1 with a finite $K_{pos}$: the specification is written in terms of a ramp, so the relevant constant is $K_v$, while $K_{pos}$ is infinite for this Type 1 plant.

Source. Nise, Control Systems Engineering 7th ed., §11.4 (lead compensator design via frequency response) and §10.10 (relation between phase margin, damping ratio and closed-loop response); Ogata 5th ed., §6.6–6.7.

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