Steady-State Errors of a Motor Position Servo: 22-Mec-A3 System Analysis and Control, December 2014, Q6

Question 6: Steady-State Errors of a Motor Position Servo (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. The unity-feedback position servo of Figure 4, with

$$G_{c}(s) = K, \qquad G(s) = \frac{1}{s(s+1)}, \qquad H(s) = 1,$$

and a load disturbance $D$ entering additively at the input of $G$. The reference inputs are a unit step $r(t)=u_{s}(t)$ and a unit ramp $r(t)=t$, applied in turn to $R$.

Find. The steady-state error $e_{ss}=\lim_{t\to\infty}[r(t)-c(t)]$ for each reference input, expressed in terms of $K$.

[Figure not reproduced: Figure 4 — redrawn from page 5. The error $E=R-C$ drives $G_{c}$; the disturbance $D$ is summed with the controller output ahead of the motor transfer function $G(s)$, and the loop is closed by unity feedback. See the official exam paper.]

Check: the disturbance input. The question asks only for errors due to $R$, so $D=0$ is taken throughout by superposition. The printed figure shows that the second summing junction receives only $G_{c}E$ and $D$ — the output $C$ is not returned to it. With the loop as drawn, the open-loop transfer function is simply $G_{c}G$.

Approach. Identify the system type from the number of open-loop poles at the origin, compute the relevant static error constant as a limit of the open-loop transfer function, and apply the standard error formulas. The final-value theorem is legitimate here only after confirming that the closed loop is stable.

Form the open-loop transfer function and confirm stability. With $D=0$, $$L(s) = G_{c}(s)G(s)H(s) = \frac{K}{s(s+1)} .$$ The closed-loop characteristic equation is $s(s+1)+K=0$, i.e. $s^{2}+s+K=0$. Both coefficients and the constant term are positive for every $K>0$, so the closed loop is stable for all positive gain and the final-value theorem may be applied. (Comparing with $s^{2}+2\zeta\omega_{n}s+\omega_{n}^{2}$ gives $\omega_{n}=\sqrt{K}$ and $\zeta=1/(2\sqrt{K})$: raising $K$ speeds the servo up but reduces its damping.)
Establish the system type. $L(s)$ has exactly one pole at the origin, so this is a type 1 system. Type-1 loops track a step with zero error and a ramp with a finite constant error — the classic position-servo behaviour.
Write the error transfer function. For unity feedback, $$E(s) = \frac{R(s)}{1+L(s)} = \frac{s(s+1)}{s^{2}+s+K}\,R(s),$$ and by the final-value theorem $e_{ss}=\lim_{s\to0}sE(s)$, valid because all closed-loop poles are in the left half-plane.
Unit step input. The position error constant is $$K_{p} = \lim_{s\to0}L(s) = \lim_{s\to0}\frac{K}{s(s+1)} = \infty ,$$ because of the free integrator. Hence $$e_{ss}^{\text{step}} = \frac{1}{1+K_{p}} = \frac{1}{1+\infty} \;\Longrightarrow\; \boxed{\;e_{ss}^{\text{step}} = 0\ \text{for every } K>0. \;}$$ Equivalently, substituting $R=1/s$ directly: $e_{ss}=\lim_{s\to0}s\cdot\dfrac{s(s+1)}{s^{2}+s+K}\cdot\dfrac{1}{s} =\lim_{s\to0}\dfrac{s(s+1)}{s^{2}+s+K}=0$. The servo eventually reaches any commanded position exactly, no matter how small the gain.
Unit ramp input. The velocity error constant is $$K_{v} = \lim_{s\to0}sL(s) = \lim_{s\to0}\frac{sK}{s(s+1)} = \lim_{s\to0}\frac{K}{s+1} = K .$$ Therefore $$\boxed{\;e_{ss}^{\text{ramp}} = \frac{1}{K_{v}} = \frac{1}{K}. \;}$$ The direct route confirms it: $e_{ss}=\lim_{s\to0}s\cdot\dfrac{s(s+1)}{s^{2}+s+K}\cdot\dfrac{1}{s^{2}} =\lim_{s\to0}\dfrac{s+1}{s^{2}+s+K}=\dfrac{1}{K}$.
Interpret the design trade-off. The ramp error is a steady lag of $1/K$ units behind a reference moving at unit velocity, so tracking a constant-speed command requires large $K$. But $\zeta=1/(2\sqrt{K})$ falls as $K$ rises: demanding $e_{ss}^{\text{ramp}}\le0.05$ forces $K\ge20$ and hence $\zeta\le0.112$, which corresponds to roughly 70 % overshoot. Proportional gain alone therefore cannot satisfy a tight accuracy specification and a reasonable damping specification simultaneously; the standard remedies are a lag or PI controller (which raises the type and drives the ramp error to zero) or rate feedback / lead compensation to restore damping.

Quantity	Result
Open-loop transfer function	$L(s)=K/[s(s+1)]$
System type	Type 1 (one free integrator)
Characteristic equation	$s^{2}+s+K=0$ — stable for all $K>0$
$\omega_{n}$, $\zeta$	$\sqrt{K}$, $1/(2\sqrt{K})$
Position error constant $K_{p}$	$\infty$
Unit-step steady-state error	$0$
Velocity error constant $K_{v}$	$K$
Unit-ramp steady-state error	$1/K$

Topic. System type, static error constants, and the accuracy-versus-damping trade-off in a proportional position servo.

Key relations. $e_{ss}=\lim_{s\to0}sR(s)/[1+L(s)]$; $K_{p}=\lim_{s\to0}L(s)$ with $e_{ss}^{\text{step}}=1/(1+K_{p})$; $K_{v}=\lim_{s\to0}sL(s)$ with $e_{ss}^{\text{ramp}}=1/K_{v}$; $K_{a}=\lim_{s\to0}s^{2}L(s)$ with $e_{ss}^{\text{parabola}}=1/K_{a}$; for a type-$N$ loop the error is zero for inputs of order below $N$, finite at order $N$, infinite above.

Why this works. Each free integrator in the open loop contributes an $s$ in the denominator of $L$, so $1+L$ behaves like $L$ near the origin and the error transfer function $1/(1+L)$ acquires a zero at $s=0$ of the same multiplicity. That zero cancels the pole of the reference transform one order at a time, which is exactly why a type-1 loop annihilates a step error but leaves a finite ramp error. The velocity constant $K_{v}$ measures how much loop gain survives after that cancellation, and here it equals the controller gain outright because the remaining plant factor has unity DC gain.

Common pitfall. Applying the final-value theorem before checking stability is the standard trap — it returns a finite, entirely fictitious "error" for an unstable loop. A second frequent error is treating the disturbance path as part of the reference loop; superposition requires $D=0$ when computing errors due to $R$. A third is forgetting to normalise the plant: had $G$ been $1/[s(s+4)]$, $K_{v}$ would be $K/4$ and the ramp error $4/K$, not $1/K$.

Source. Ogata, Modern Control Engineering 5th ed., §5-9 (steady-state errors and static error constants); Nise, Control Systems Engineering 8th ed., §7.2–7.3; Dorf & Bishop, Modern Control Systems 13th ed., Ch. 5.

Question 6 of 6: Steady-State Errors of a Motor Position Servo

Question 6: Steady-State Errors of a Motor Position Servo (25 marks)