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

Question 5 of 6: Lead-Compensator Design for Specified Damping and Natural Frequency

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

Notes on this paper

Paper format. National Exams, December 2017 — 16-Mec-A3, System Analysis and Control. Three hours, closed book; a Casio or Sharp approved calculator and semi-log graph paper are the only permitted aids. Six questions are printed; any four constitute a complete paper and all questions are of equal value (25 marks each). A table of Laplace transforms is supplied as page 5. All six questions are solved here.

Reference texts. K. Ogata, Modern Control Engineering, 5th ed. (Prentice Hall) — the source of this paper's problem style; N. S. Nise, Control Systems Engineering, 7th ed. (Wiley); R. C. Dorf and R. H. Bishop, Modern Control Systems, 13th ed. (Pearson); G. F. Franklin, J. D. Powell and A. Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed. (Pearson).

Check: conventions used throughout. All transfer functions are written in the Laplace variable s with zero initial conditions unless a question states otherwise. “Unity feedback” means $H(s)=1$, so the closed-loop transfer function is $T(s)=G(s)/[1+G(s)]$ and the error signal is $E(s)=R(s)-C(s)$. Root-locus gains are the gain $K$ exactly as it appears in the printed $G(s)$.

Question 5: Lead-Compensator Design for Specified Damping and Natural Frequency (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. A unity-feedback loop with a first-order (lead or lag) compensator $G_c(s)=K\dfrac{T_1s+1}{T_2s+1}$ in cascade with the plant $G_p(s)=\dfrac{10}{s(s+1)}$. Three parameters $K$, $T_1$, $T_2$ are free.

Find. Values of $K$, $T_1$ and $T_2$ such that the dominant closed-loop poles have $\zeta=0.5$ and $\omega_n=3$ rad/s.

Figure 5.1 — The compensated unity-feedback system of Question 5, with the designed parameter values annotated.

Approach. Three unknowns against only two specifications leaves one degree of freedom; use it in the standard way — let the compensator zero cancel the plant pole at $s=-1$, which reduces the loop to a pure second-order system whose two remaining parameters map directly onto $\zeta$ and $\omega_n$.

Write the target characteristic polynomial. Dominant poles with $\zeta=0.5$, $\omega_n=3$ require $$s^2+2\zeta\omega_n s+\omega_n^2=s^2+2(0.5)(3)s+3^2=s^2+3s+9=0$$ whose roots are $$s=-\zeta\omega_n\pm j\omega_n\sqrt{1-\zeta^2}=-1.5\pm j\,3\sqrt{0.75}=-1.5\pm j2.5981$$
Use the spare degree of freedom to cancel the plant pole. Setting the compensator zero at $s=-1/T_1$ on top of the plant pole at $s=-1$ means $$\frac{1}{T_1}=1\;\Longrightarrow\;\boxed{\;T_1=1\ \text{s}\;}$$ The open-loop transfer function then collapses to a second-order form: $$G(s)=K\,\frac{s+1}{T_2s+1}\cdot\frac{10}{s(s+1)}\cdot\frac{1}{1}=\frac{10K}{s\left(T_2s+1\right)}=\frac{10K/T_2}{s\left(s+1/T_2\right)}$$
Close the loop and match coefficients. With unity feedback the characteristic equation is $s(s+1/T_2)+10K/T_2=0$, i.e. $$s^2+\frac{1}{T_2}\,s+\frac{10K}{T_2}=0$$ Comparing term by term with $s^2+3s+9=0$: $$\frac{1}{T_2}=2\zeta\omega_n=3\;\Longrightarrow\;\boxed{\;T_2=\frac{1}{3}=0.3333\ \text{s}\;}$$ $$\frac{10K}{T_2}=\omega_n^2=9\;\Longrightarrow\;10K(3)=9\;\Longrightarrow\;\boxed{\;K=0.3\;}$$
Verify the design. Substituting back, the compensated open loop is $G(s)=3/[s(s+3)]$ and the closed-loop characteristic polynomial is $s^2+3s+9$, whose roots are $-1.5\pm j2.5981$. These give $$\omega_n=\sqrt{1.5^2+2.5981^2}=3.000\ \text{rad/s},\qquad \zeta=\frac{1.5}{3.000}=0.500\ \checkmark$$ Because $T_1=1$ cancels the plant pole exactly, these are the only two closed-loop poles — so “dominant” is satisfied trivially rather than approximately.
Report the resulting performance. The compensated loop remains type 1, with velocity error constant $$K_v=\lim_{s\to0}sG(s)=\lim_{s\to0}s\cdot\frac{10K}{s\left(T_2s+1\right)}=10K=3\ \text{s}^{-1}\;\Longrightarrow\;e_{ss,\text{ramp}}=\frac{1}{K_v}=0.333$$ and the standard second-order formulae give $$M_p=e^{-\pi\zeta/\sqrt{1-\zeta^2}}=e^{-\pi(0.5)/0.8660}=16.3\%,\qquad t_s\approx\frac{4}{\zeta\omega_n}=\frac{4}{1.5}=2.67\ \text{s}$$

Check: the pole-cancellation choice. The problem supplies three unknowns for two specifications, so the solution is not unique — any $(K,T_1,T_2)$ placing a dominant pair at $-1.5\pm j2.5981$ satisfies the stated requirement, with the third closed-loop pole then sitting somewhere on the negative real axis. Cancelling the plant pole with the compensator zero ($T_1=1$) is the standard textbook resolution because it makes the dominant pair exact rather than approximate and yields the clean values above. In hardware, exact cancellation is never achieved; a small residual pole–zero dipole near $s=-1$ would add a slow, low-amplitude tail to the step response without materially changing $\zeta$ or $\omega_n$. This assumption is stated here in accordance with Note 1 of the examination paper.

Quantity	Result
Target characteristic polynomial	$s^2+3s+9=0$
Dominant closed-loop poles	$s=-1.5\pm j2.5981$
Compensator zero constant	$T_1=1$ s (cancels the plant pole at $s=-1$)
Compensator pole constant	$T_2=1/3=0.3333$ s
Compensator gain	$K=0.3$
Compensated open loop	$G(s)=3/[s(s+3)]$
Velocity error constant	$K_v=3\ \text{s}^{-1}$, $e_{ss}=0.333$ for a unit ramp
Overshoot / settling time	$M_p=16.3\%$, $t_s\approx2.67$ s

Topic. Cascade compensator design by pole cancellation and coefficient matching to meet damping-ratio and natural-frequency specifications.

Key relations. Standard second-order form $s^2+2\zeta\omega_n s+\omega_n^2$; dominant poles $s=-\zeta\omega_n\pm j\omega_n\sqrt{1-\zeta^2}$; first-order compensator $G_c=K(T_1s+1)/(T_2s+1)$, which is a lead network when $T_1>T_2$; $K_v=\lim_{s\to0}sG(s)$; $M_p=e^{-\pi\zeta/\sqrt{1-\zeta^2}}$; $t_s\approx4/(\zeta\omega_n)$.

Why this works. Two performance specifications fix exactly two coefficients of a second-order characteristic polynomial, so a design problem with three free parameters is under-determined until one extra constraint is imposed. Pole–zero cancellation is the natural constraint here because it removes the plant's slowest dynamics from the closed loop entirely, leaving a polynomial whose order matches the number of specifications. Once that is done the design becomes pure coefficient matching — no root-locus search or iteration is needed. The result $T_1=1>T_2=1/3$ confirms the compensator is a lead network, which is the physically sensible outcome: the uncompensated plant $10/[s(s+1)]$ has $\omega_n=\sqrt{10}=3.16$ with $\zeta=0.158$, far too oscillatory, and phase lead is precisely what adds the missing damping.

Common pitfall. The most frequent error is failing to normalise the compensator into time-constant form before matching: $K(T_1s+1)/(T_2s+1)$ has unity gain at DC, whereas the pole–zero form $K'(s+z)/(s+p)$ does not, and mixing the two makes the computed $K$ wrong by a factor of $T_1/T_2$ (here a factor of 3). A second pitfall is forgetting to check that the cancelled pole is in the left half plane — cancelling an unstable pole is never legitimate, since the mode remains in the physical system and any modelling error leaves it uncontrolled. Finally, candidates sometimes match $1/T_2$ to $\omega_n$ rather than to $2\zeta\omega_n$; comparing the polynomial coefficient by coefficient avoids this entirely.

Source. Ogata, Modern Control Engineering, 5th ed., §6–6 (lead compensation) and Example A–6–9; Franklin, Powell and Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed., Ch. 5.

← Question 4 Next: Question 6 →

Quantity	Result
Target characteristic polynomial	\(s^2+3s+9=0\)
Dominant closed-loop poles	\(s=-1.5\pm j2.5981\)
Compensator zero constant	\(T_1=1\) s (cancels the plant pole at \(s=-1\))
Compensator pole constant	\(T_2=1/3=0.3333\) s
Compensator gain	\(K=0.3\)
Compensated open loop	\(G(s)=3/[s(s+3)]\)
Velocity error constant	\(K_v=3\ \text{s}^{-1}\), \(e_{ss}=0.333\) for a unit ramp
Overshoot / settling time	\(M_p=16.3\%\), \(t_s\approx2.67\) s