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

Question 5 of 6: Steady-State Error Specifications for a Position Control System

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

Notes on this paper

Paper format. 07-Mec-A3 System Analysis and Control, National Exams May 2016 — 3 hours, closed book (Casio or Sharp approved calculator; semi-log graph paper is the only permitted aid). The rubric states that any four (4) questions constitute a complete paper and that all questions are of equal value, so each question is worth 25 marks of the 100 marked. Because this document is a study resource rather than an exam script, all six questions are solved in full.

Reference texts. K. Ogata, Modern Control Engineering, 5th ed. (Pearson) — Ch. 3–7; N. S. Nise, Control Systems Engineering, 8th ed. (Wiley) — Ch. 4, 6, 8, 10; R. C. Dorf and R. H. Bishop, Modern Control Systems, 13th ed. (Pearson) — Ch. 5–8; G. F. Franklin, J. D. Powell and A. Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed. (Pearson) — Ch. 3–6. A table of Laplace transform pairs is appended to the examination paper (pages 3–4 of the original), together with two sheets of 3-cycle semi-logarithmic graph paper for the Bode question.

Question 5: Steady-State Error Specifications for a Position Control System (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.

Check: the printed stem gives the closed-loop transfer function and then lists the two conditions (a) and (b) without repeating the instruction. Read against the standard form of this problem, the ask is: determine the constraints the coefficients $b_{0},b_{1},a_{1},a_{2}$ must satisfy so that (a) and (b) hold. That is the interpretation solved here, and a candidate should state the same assumption on the answer paper as the paper's own Note 1 invites. Both parts are answered as coefficient constraints, and a worked numerical design satisfying them is supplied.

Given. A closed-loop position servo, output and reference both in metres:

Item	Value / expression
Closed-loop transfer function	$T(s)=Y/R=(b_{0}s+b_{1})/(s^{2}+a_{1}s+a_{2})$
Error definition	$e(t)=r(t)-y(t)$ (unity reference comparison, metre/metre)
Step specification	$e_{ss}=0$ for $r(t)=R_{0}\cdot1(t)$
Ramp specification	$\|e_{ss}\|\le1\text{ mm}=10^{-3}\text{ m}$ for $\dot r=0.1\text{ m/s}$
Implied stability requirement	$a_{1}>0$ and $a_{2}>0$ (both roots in the left half-plane)

Find. The conditions on $b_{0},b_{1},a_{1},a_{2}$ that make the step error identically zero and hold the ramp error to 1 mm at a reference speed of 0.1 m/s.

Approach. Form the error transfer function $E/R=1-T$ directly from the given closed-loop transfer function — no loop decomposition is needed — then apply the Final Value Theorem to a step and to a ramp in turn, checking first that $T$ is stable so that the theorem is valid.

Form the error transfer function. Since $E=R-Y$ and $Y=T R$,$$\frac{E(s)}{R(s)}=1-T(s)=\frac{\left(s^{2}+a_{1}s+a_{2}\right)-\left(b_{0}s+b_{1}\right)}{s^{2}+a_{1}s+a_{2}}=\frac{s^{2}+(a_{1}-b_{0})s+(a_{2}-b_{1})}{s^{2}+a_{1}s+a_{2}}$$Everything that follows is read off this one expression. Note that both specifications are conditions on the numerator of $E/R$: the constant term controls step accuracy and the $s$-coefficient controls ramp accuracy.
Impose zero steady-state error to a step. For $r(t)=R_{0}\cdot1(t)$, $R(s)=R_{0}/s$ and$$e_{ss}=\lim_{s\to0}s\cdot\frac{R_{0}}{s}\cdot\frac{E(s)}{R(s)}=R_{0}\,\frac{a_{2}-b_{1}}{a_{2}}$$Setting this to zero for every step size requires$$\boxed{\;b_{1}=a_{2}\;}$$Equivalently $T(0)=b_{1}/a_{2}=1$: the closed-loop DC gain must be exactly unity, which is the physical statement that one metre commanded must produce one metre delivered. In loop terms this is what makes the system type 1.
Reduce the error function using that result. With $b_{1}=a_{2}$ the constant term of the numerator vanishes and an $s$ factors out:$$\frac{E(s)}{R(s)}=\frac{s\left[s+(a_{1}-b_{0})\right]}{s^{2}+a_{1}s+a_{2}}$$That single factor of $s$ is what guarantees the step error is zero and what makes the ramp error finite rather than unbounded.
Impose the ramp specification. For a ramp of slope $A=0.1\text{ m/s}$, $R(s)=A/s^{2}$ and$$e_{ss}=\lim_{s\to0}s\cdot\frac{A}{s^{2}}\cdot\frac{s\left[s+(a_{1}-b_{0})\right]}{s^{2}+a_{1}s+a_{2}}=A\,\frac{a_{1}-b_{0}}{a_{2}}$$Requiring $|e_{ss}|\le10^{-3}\text{ m}$ with $A=0.1\text{ m/s}$ gives$$\frac{a_{1}-b_{0}}{a_{2}}\le\frac{10^{-3}}{0.1}=10^{-2}\ \text{s}$$that is,$$\boxed{\;a_{2}\ge100\,(a_{1}-b_{0})\quad\Longleftrightarrow\quad K_{v}\equiv\frac{a_{2}}{a_{1}-b_{0}}\ge100\ \text{s}^{-1}\;}$$The quantity $a_{2}/(a_{1}-b_{0})$ is precisely the velocity error constant of the equivalent unity-feedback loop, so the specification is the familiar $K_{v}\ge A/e_{\max}$ in disguise.
Interpret the role of the numerator zero. The term $b_{0}s$ is a feed-forward (or derivative) path from the reference. It does not affect the step error at all, but it reduces $(a_{1}-b_{0})$ and therefore raises $K_{v}$ for a fixed denominator. With $b_{0}=0$ the requirement collapses to $a_{2}\ge100a_{1}$, which for a well-damped pair ($a_{1}=2\zeta\omega_{n}$, $a_{2}=\omega_{n}^{2}$) forces $\omega_{n}\ge200\zeta$ — an uncomfortably fast servo. Allowing $b_{0}\to a_{1}$ makes the ramp error vanish entirely, at the cost of a differentiating path that amplifies reference noise.
Produce a concrete design that meets both specifications. Choose a comfortable pair, $\zeta=0.707$ and $\omega_{n}=14.14\text{ rad/s}$, i.e.$$a_{1}=2\zeta\omega_{n}=20\ \text{s}^{-1},\qquad a_{2}=\omega_{n}^{2}=200\ \text{s}^{-2}$$Then $b_{1}=a_{2}=200$ from part (a), and setting the ramp error exactly on its 1 mm limit,$$b_{0}=a_{1}-\frac{a_{2}}{K_{v}}=20-\frac{200}{100}=18\ \text{s}^{-1}$$The resulting servo$$\frac{Y(s)}{R(s)}=\frac{18s+200}{s^{2}+20s+200}$$has $T(0)=1$ (zero step error) and $K_{v}=200/2=100\text{ s}^{-1}$, giving exactly $e_{ss}=0.1/100=1\text{ mm}$ at the specified reference speed. Both denominator coefficients are positive, so the design is stable and the Final Value Theorem used above was legitimate.

Figure 5.1 — Tracking-error transients of the worked design $T(s)=(18s+200)/(s^{2}+20s+200)$. The unit-step error starts at 1 and decays to zero (part a); the ramp error, plotted magnified 100 times, rises from zero and settles at 0.001 m = 1 mm (part b), exactly on specification.

Quantity	Result
Error transfer function	$E/R=\dfrac{s^{2}+(a_{1}-b_{0})s+(a_{2}-b_{1})}{s^{2}+a_{1}s+a_{2}}$
(a) Zero step error requires	$b_{1}=a_{2}$, i.e. closed-loop DC gain $T(0)=1$
Ramp error (with $b_{1}=a_{2}$)	$e_{ss}=A(a_{1}-b_{0})/a_{2}$
(b) Ramp specification requires	$a_{2}\ge100(a_{1}-b_{0})$, i.e. $K_{v}\ge100\text{ s}^{-1}$
Stability side-condition	$a_{1}>0$, $a_{2}>0$
Worked design	$a_{1}=20$, $a_{2}=200$, $b_{1}=200$, $b_{0}=18$
Design performance	$\zeta=0.707$, $\omega_{n}=14.14\text{ rad/s}$, $K_{v}=100\text{ s}^{-1}$, $e_{ss,\text{ramp}}=1.00\text{ mm}$

Topic. Steady-state error specifications expressed directly on closed-loop coefficients, and the static velocity error constant.

Key relations. $E/R=1-T$; Final Value Theorem $e_{ss}=\lim_{s\to0}sE(s)$; step error $=1-T(0)$; ramp error $=A\lim_{s\to0}\left[(1-T)/s\right]$; velocity error constant $K_{v}=A/e_{ss,\text{ramp}}$; and $K_{v}=a_{2}/(a_{1}-b_{0})$ for this second-order form.

Why this works. The usual textbook route to steady-state error goes through the open-loop transfer function and the system type, which requires knowing how the loop is decomposed into forward path and feedback. Here only the closed-loop transfer function is given, and it is not necessary to guess a decomposition: the error signal is defined by $e=r-y$, so $E/R=1-T$ is exact regardless of internal structure. Expanding $1-T$ shows the specifications map onto the low-order coefficients of its numerator — the constant term must vanish for zero step error, and the surviving $s$-coefficient sets the ramp error. This coefficient view also makes visible something the loop view hides: the reference-path zero $b_{0}$ improves ramp tracking without touching stability, because it lives in the numerator only.

Common pitfall. Applying the Final Value Theorem to a ramp before enforcing part (a). If $b_{1}\ne a_{2}$ the numerator of $E/R$ has a non-zero constant term, $sE(s)$ retains a pole at the origin, and the ramp error is infinite — so parts (a) and (b) must be solved in that order. A second pitfall is unit sloppiness: the specification mixes m/s with mm, and forgetting the $10^{-3}$ conversion inflates the permitted $K_{v}$ by a thousand.

Source. Ogata, Modern Control Engineering 5th ed., §5-8 (steady-state errors and static error constants); Nise, Control Systems Engineering 8th ed., Ch. 7; Franklin, Powell & Emami-Naeini, 8th ed., §4.2.

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