Error Constants and Steady-State Errors: 22-Mec-A3 System Analysis and Control, December 2018, Q7

Question 7: Error Constants and Steady-State Errors (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 whose forward path is the cascade of $2/(s+1)$ and $1/(s+4)$, read from the block diagram printed with the question, so that

$$G(s)=\frac{2}{s+1}\cdot\frac{1}{s+4}=\frac{2}{(s+1)(s+4)}=\frac{2}{s^{2}+5s+4}$$

Find. The position, velocity and acceleration error constants $K_{p}$, $K_{v}$, $K_{a}$, and the steady-state error to a unit step, a unit ramp and a unit parabolic input.

The system of Question 7: two cascaded first-order blocks in the forward path, closed with unity negative feedback.

Approach. Confirm the closed loop is stable (the error constants are Final-Value limits and mean nothing otherwise), identify the system type from the number of integrators in $G(s)$, evaluate the three constants from their defining limits, and convert each into a steady-state error.

Check stability first. The characteristic equation is $1+G(s)=0$, i.e.

$$(s+1)(s+4)+2=s^{2}+5s+6=(s+2)(s+3)=0$$

Both closed-loop poles, $s=-2$ and $s=-3$, are in the left half plane, so the loop is stable (and in fact overdamped, so the response contains no oscillation). Every error result below rests on the Final Value Theorem, which would be invalid without this step.

Identify the system type. The type number is the count of poles of $G(s)$ at the origin. Here $G(s)=2/[(s+1)(s+4)]$ has poles at $s=-1$ and $s=-4$ only, so there is no integrator in the forward path and this is a type 0 system. That single fact already dictates the pattern of the answers: finite $K_{p}$, zero $K_{v}$ and $K_{a}$.
Evaluate the three error constants from their definitions.

$$K_{p}=\lim_{s\to0}G(s)=\frac{2}{(1)(4)}=\boxed{0.5}$$$$K_{v}=\lim_{s\to0}sG(s)=\lim_{s\to0}\frac{2s}{(s+1)(s+4)}=\boxed{0}$$$$K_{a}=\lim_{s\to0}s^{2}G(s)=\lim_{s\to0}\frac{2s^{2}}{(s+1)(s+4)}=\boxed{0}$$

The velocity and acceleration constants vanish because each extra factor of $s$ goes to zero while $G(0)$ stays finite — the plant has no integrator to keep the limit alive.

Convert each constant into a steady-state error. For unity feedback, $E(s)=R(s)/[1+G(s)]$ and $e_{ss}=\lim_{s\to0}sE(s)$. Applying this to the three standard inputs:

$$e_{\text{step}}=\frac{1}{1+K_{p}}=\frac{1}{1+0.5}=\boxed{\frac{2}{3}\approx0.667}$$$$e_{\text{ramp}}=\frac{1}{K_{v}}=\frac{1}{0}=\boxed{\infty},\qquad e_{\text{parabolic}}=\frac{1}{K_{a}}=\frac{1}{0}=\boxed{\infty}$$

Interpret the numbers as an engineer would.

The step result is severe: with $K_{p}=0.5$ the output settles at only $K_{p}/(1+K_{p})=1/3$ of the commanded value, leaving two-thirds of the command as permanent error. This is not a subtle offset but a gross one, and it is a direct consequence of the low DC loop gain of $0.5$ — a proportional loop can only reduce error by the factor $1+K_{p}$, and here that factor is a mere $1.5$. The ramp and parabolic errors grow without bound, meaning the output falls progressively further behind a moving command and never catches up.

The remedy follows straight from the type number. Raising the forward gain shrinks the step error but can never eliminate it, since $1/(1+K_{p})$ is zero only in the limit. Inserting an integrator — a PI controller, $G_{c}(s)=K_{c}(1+1/T_{i}s)$ — raises the loop to type 1, which drives the step error to exactly zero and makes the ramp error finite at $1/K_{v}$. That is the standard reason integral action appears in position-control loops of this kind.

Question 7 — final results
Quantity	Value
Open-loop transfer function	$G(s)=2/[(s+1)(s+4)]$
Closed-loop poles	$s=-2$, $s=-3$ (stable, overdamped)
System type	Type 0 (no integrator)
Position error constant $K_{p}$	$0.5$
Velocity error constant $K_{v}$	$0$
Acceleration error constant $K_{a}$	$0$
Error to a unit step	$2/3\approx0.667$
Error to a unit ramp	$\infty$
Error to a unit parabola	$\infty$

Concept & Theory

Topic. Steady-state error analysis: system type and the static error constants.

Key relations. $E(s)=\dfrac{R(s)}{1+G(s)}$ for unity feedback; $K_{p}=\lim_{s\to0}G(s)$, $K_{v}=\lim_{s\to0}sG(s)$, $K_{a}=\lim_{s\to0}s^{2}G(s)$; $e_{\text{step}}=1/(1+K_{p})$, $e_{\text{ramp}}=1/K_{v}$, $e_{\text{parab}}=1/K_{a}$; type $N$ = number of poles of $G(s)$ at the origin.

Why this works. The three test inputs $1/s$, $1/s^{2}$, $1/s^{3}$ probe the loop gain at successively deeper orders near $s=0$, so the error to each is governed by how many powers of $s$ the loop gain can supply there. A type $N$ system tracks the first $N$ input classes with zero error, the $(N+1)$th with a finite error, and everything beyond with unbounded error. That is the whole content of the error-constant table, and it explains why adding an integrator — raising the type by one — shifts the entire pattern one column to the right.

Common pitfall. Quoting error constants without first confirming closed-loop stability: they are Final-Value limits and are meaningless for an unstable loop, no matter how cleanly they evaluate. A second common slip is using $e_{\text{step}}=1/K_{p}$ instead of $1/(1+K_{p})$ — the $+1$ belongs only to the step case, since $1+G(s)\to1+K_{p}$ while $sG(s)$ and $s^{2}G(s)$ tend to $K_{v}$ and $K_{a}$ alone.

Source. Ogata, Modern Control Engineering, 5th ed., §5.8 (steady-state errors in unity-feedback control systems); Nise, Control Systems Engineering, 8th ed., §7.2–7.4; Franklin, Powell & Emami-Naeini, 8th ed., §4.2.

Question 7 of 7: Error Constants and Steady-State Errors

Question 7: Error Constants and Steady-State Errors (25 marks)