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

Question 3 of 6: Steady-State Error — Recovering G ( s ) and the Effect of Gain and Damping

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 3: Steady-State Error — Recovering G(s) and the Effect of Gain and Damping (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. Part (a): a unity-feedback loop specified by its closed-loop transfer function $T(s)=(Ks+b)/(s^2+as+b)$, with $a$, $b$ and $K$ positive constants. Part (b): the classic inertia-plus-viscous-damping servo $G(s)=K/[s(Js+B)]$, where $J$ is the load inertia, $B$ the viscous friction coefficient and $K$ the loop gain.

Find. (a) the open-loop $G(s)$ consistent with the given $T(s)$, and a proof that the unit-ramp steady-state error equals $(a-K)/b$; (b) a reasoned discussion of how $K$ and $B$ separately affect the ramp error, plus sketched ramp responses at three gain levels.

Approach. For part (a), invert the closure formula — if $T=G/(1+G)$ then $G=T/(1-T)$ — then read the system type off the resulting $G$ and apply the velocity-error-constant definition. For part (b), express $K_v$, $\omega_n$ and $\zeta$ in terms of the physical parameters and reason about the trade-off they encode.

Invert the closed-loop relation. For unity feedback $T=G/(1+G)$, which rearranges to $G=T/(1-T)$. Substituting the given $T$, $$G(s)=\frac{\dfrac{Ks+b}{s^2+as+b}}{1-\dfrac{Ks+b}{s^2+as+b}}=\frac{Ks+b}{\left(s^2+as+b\right)-\left(Ks+b\right)}$$ The constant $b$ cancels in the denominator and an $s$ factors out: $$G(s)=\frac{Ks+b}{s^2+(a-K)s}\;\Longrightarrow\;\boxed{\;G(s)=\frac{Ks+b}{s\left[s+(a-K)\right]}\;}$$
Identify the system type. The single free $s$ in the denominator means the loop contains exactly one integrator, so this is a type-1 system. Type 1 gives zero steady-state error to a step, a finite error to a ramp, and infinite error to a parabola — so a finite ramp error is precisely what we should expect. Note also that stability requires $a>K$, which is what keeps the error expression positive.
Evaluate the velocity error constant. By definition $K_v=\lim_{s\to 0} s\,G(s)$: $$K_v=\lim_{s\to 0}\;s\cdot\frac{Ks+b}{s\left[s+(a-K)\right]}=\lim_{s\to 0}\frac{Ks+b}{s+(a-K)}=\frac{b}{a-K}$$
Convert to the steady-state ramp error. For a unit ramp $R(s)=1/s^2$, the final-value theorem applied to $E(s)=R(s)/[1+G(s)]$ gives $$e_{ss}=\lim_{s\to 0}\,s\cdot\frac{1/s^2}{1+G(s)}=\lim_{s\to 0}\frac{1}{s+sG(s)}=\frac{1}{K_v}$$ and therefore $$\boxed{\;e_{ss}=\frac{1}{K_v}=\frac{a-K}{b}\;}$$ which is the required result. An independent route confirms it: since $E/R=1-T$, $$1-T(s)=\frac{s^2+as+b-Ks-b}{s^2+as+b}=\frac{s\left[s+(a-K)\right]}{s^2+as+b}$$ so $e_{ss}=\lim_{s\to0}s\cdot\frac{1}{s^2}\cdot\frac{s[s+(a-K)]}{s^2+as+b}=\frac{a-K}{b}$, in agreement.
Read the design message of part (a). The numerator gain $K$ appears in the error expression with a minus sign, so a larger closed-loop zero coefficient reduces the ramp error without moving any closed-loop pole (the denominator $s^2+as+b$ is untouched by $K$). This is feed-forward: the zero $Ks$ injects a derivative of the reference that anticipates the ramp. It is the cheapest way to buy tracking accuracy, because it costs nothing in stability margin — but it does amplify reference noise.

Part (b) applies the same error machinery to a physical servo, and then asks for the transient picture that accompanies each gain choice.

Compute the error constant for the servo. With $G(s)=K/[s(Js+B)]$, again type 1, $$K_v=\lim_{s\to0}s\cdot\frac{K}{s(Js+B)}=\frac{K}{B}\;\Longrightarrow\;\boxed{\;e_{ss}=\frac{1}{K_v}=\frac{B}{K}\;}$$ So the ramp error is directly proportional to the viscous friction and inversely proportional to the loop gain. The inertia $J$ does not appear at all: it governs the transient, not the steady state.
Express the transient parameters. The closed loop is $$T(s)=\frac{K}{Js^2+Bs+K}=\frac{K/J}{s^2+(B/J)s+K/J}$$ so, comparing with the standard form $s^2+2\zeta\omega_n s+\omega_n^2$, $$\omega_n=\sqrt{\frac{K}{J}},\qquad \zeta=\frac{B}{2\sqrt{KJ}},\qquad 2\zeta\omega_n=\frac{B}{J}$$ The product $2\zeta\omega_n$ is fixed by the hardware alone: raising $K$ cannot change the decay rate of the envelope, only the frequency of oscillation.
Discuss the effect of increasing K. Raising the gain reduces $e_{ss}=B/K$ proportionally — ten times the gain gives one tenth the ramp lag. But $\zeta=B/(2\sqrt{KJ})$ falls as $1/\sqrt{K}$, so the response becomes progressively more oscillatory and the overshoot grows. With $J=B=1$: $K=0.5$ gives $\zeta=0.707$ with a large lag $e_{ss}=2$; $K=2$ gives $\zeta=0.354$, $e_{ss}=0.5$; $K=10$ gives $\zeta=0.158$, $e_{ss}=0.1$ but a visibly ringing approach to the ramp. Accuracy and damping are in direct conflict through $K$.
Discuss the effect of increasing B. Raising the viscous friction increases $e_{ss}=B/K$ proportionally — it makes tracking worse — while simultaneously raising $\zeta$ and so improving damping. It leaves $\omega_n=\sqrt{K/J}$ untouched. Physically, friction dissipates the energy that would otherwise sustain oscillation, but it also demands a permanently non-zero error to generate the torque needed to keep the load moving at constant velocity. Increasing $B$ is therefore the opposite trade to increasing $K$, and neither alone can deliver good damping and small error; that requires reshaping the loop with derivative, lead or PI compensation.

Figure 3.1 — Unit-ramp responses of $T(s)=K/(Js^2+Bs+K)$ with $J=B=1$, for a small ($K=0.5$), medium ($K=2$) and large ($K=10$) gain. The steady-state lag behind the ramp is $e_{ss}=B/K$ — 2, 0.5 and 0.1 respectively — while the approach becomes increasingly oscillatory as $K$ grows.

Quantity	Result
(a) Open-loop transfer function	$G(s)=\dfrac{Ks+b}{s\left[s+(a-K)\right]}$ (type 1)
(a) Velocity error constant	$K_v=b/(a-K)$
(a) Unit-ramp steady-state error	$e_{ss}=1/K_v=(a-K)/b$
(b) Servo error constant	$K_v=K/B$, so $e_{ss}=B/K$
(b) Transient parameters	$\omega_n=\sqrt{K/J}$, $\zeta=B/(2\sqrt{KJ})$
(b) Effect of raising $K$	$e_{ss}$ falls as $1/K$; $\zeta$ falls as $1/\sqrt{K}$ — more accurate, less damped
(b) Effect of raising $B$	$e_{ss}$ rises $\propto B$; $\zeta$ rises $\propto B$ — less accurate, better damped
(b) Illustrative values ($J=B=1$)	$K=0.5\!:\,e_{ss}=2,\ \zeta=0.707$; $K=2\!:\,0.5,\ 0.354$; $K=10\!:\,0.1,\ 0.158$

Topic. Steady-state error analysis: system type, static error constants, and the gain–damping trade-off in a positional servomechanism.

Key relations. Inverse closure $G=T/(1-T)$; error transfer function $E/R=1/(1+G)=1-T$ for unity feedback; static constants $K_p=\lim_{s\to0}G$, $K_v=\lim_{s\to0}sG$, $K_a=\lim_{s\to0}s^2G$; errors $e_{step}=1/(1+K_p)$, $e_{ramp}=1/K_v$, $e_{parabola}=1/K_a$; second-order standard form $\omega_n^2/(s^2+2\zeta\omega_n s+\omega_n^2)$.

Why this works. The static error constants are nothing more than the final-value theorem evaluated on $E(s)=R(s)/[1+G(s)]$ for each standard input, and the system type — the number of poles of $G$ at the origin — decides which constant is finite. One integrator in the loop makes $K_p$ infinite (zero step error) and $K_v$ finite (constant ramp lag), which is exactly the behaviour a positional servo exhibits: it settles perfectly on a fixed target but always trails a moving one. The trade-off in part (b) is structural rather than accidental: $e_{ss}=B/K$ and $\zeta=B/(2\sqrt{KJ})$ both contain $B/K$-like dependence, so any pure-gain adjustment that improves one degrades the other. Recognising that is the whole motivation for dynamic compensation.

Common pitfall. In part (a) the frequent error is to assume $G(s)$ is simply the numerator over the denominator of $T(s)$, or to guess an open-loop split by inspection; the algebraic inversion $G=T/(1-T)$ is the only correct route and the cancellation of $b$ is what reveals the integrator. A second trap is quoting $K_v$ without first checking closed-loop stability — the static constants come from the final-value theorem, which is meaningless if any pole lies in the right half plane. In part (b), many candidates state that increasing $K$ “makes the system faster and more accurate” and stop; the marks are in recognising that $2\zeta\omega_n=B/J$ is gain-independent, so the envelope decay rate does not improve, only the oscillation frequency rises.

Source. Ogata, Modern Control Engineering, 5th ed., §5–8 to §5–10 (steady-state errors in unity-feedback control systems) and Example A–5–14; Dorf and Bishop, Modern Control Systems, 13th ed., Ch. 5.

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Quantity	Result
(a) Open-loop transfer function	\(G(s)=\dfrac{Ks+b}{s\left[s+(a-K)\right]}\) (type 1)
(a) Velocity error constant	\(K_v=b/(a-K)\)
(a) Unit-ramp steady-state error	\(e_{ss}=1/K_v=(a-K)/b\)
(b) Servo error constant	\(K_v=K/B\), so \(e_{ss}=B/K\)
(b) Transient parameters	\(\omega_n=\sqrt{K/J}\), \(\zeta=B/(2\sqrt{KJ})\)
(b) Effect of raising \(K\)	\(e_{ss}\) falls as \(1/K\); \(\zeta\) falls as \(1/\sqrt{K}\) — more accurate, less damped
(b) Effect of raising \(B\)	\(e_{ss}\) rises \(\propto B\); \(\zeta\) rises \(\propto B\) — less accurate, better damped
(b) Illustrative values (\(J=B=1\))	\(K=0.5\!:\,e_{ss}=2,\ \zeta=0.707\); \(K=2\!:\,0.5,\ 0.354\); \(K=10\!:\,0.1,\ 0.158\)