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

Question 6 of 6: Error Constants and Steady-State Errors

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

Question 6: 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-negative-feedback loop whose forward path is two cascaded blocks written in operator notation, with $D\equiv d/dt$ (so $D\to s$ under the Laplace transform, for zero initial conditions):

Find. The error constants $K_{p}$, $K_{v}$, $K_{a}$ and the corresponding steady-state errors $e_{ss}$.

Approach. Combine the cascade into a single open-loop transfer function, identify the system type from the number of poles at the origin, confirm closed-loop stability so the limits are meaningful, then evaluate the three standard error-constant limits.

1. Form the open-loop transfer function. Cascaded blocks multiply, and the twos cancel:$$G(s)=\frac{3}{2s}\cdot\frac{2}{s+4}=\frac{3}{s(s+4)}$$There is exactly one pole at the origin, so this is a type-1 system. The type number, not the gain, decides which errors are finite.
2. Confirm the closed loop is stable. The error constants are limits of a steady-state expression, so they only mean anything if the transients die. Here$$1+G(s)=0\;\Longrightarrow\;s^{2}+4s+3=0\;\Longrightarrow\;s=-1,\,-3$$Both roots are in the left half plane, so the Final Value Theorem may be applied.
3. Position error constant. $$K_{p}=\lim_{s\to0}G(s)=\lim_{s\to0}\frac{3}{s(s+4)}=\boxed{\infty}$$The integrator makes the DC loop gain unbounded.
4. Velocity error constant. Multiplying by $s$ cancels the pole at the origin exactly, leaving a finite limit:$$K_{v}=\lim_{s\to0}sG(s)=\lim_{s\to0}\frac{3}{s+4}=\frac{3}{4}=\boxed{0.75\ \text{s}^{-1}}$$
5. Acceleration error constant. $$K_{a}=\lim_{s\to0}s^{2}G(s)=\lim_{s\to0}\frac{3s}{s+4}=\boxed{0}$$One integrator is not enough to track an accelerating command.
6. Convert to steady-state errors. With unity feedback, $E(s)=R(s)/[1+G(s)]$, and the standard results follow from the Final Value Theorem:$$e_{ss}^{\text{step}}=\frac{1}{1+K_{p}}=\frac{1}{1+\infty}=\boxed{0}$$$$e_{ss}^{\text{ramp}}=\frac{1}{K_{v}}=\frac{1}{0.75}=\frac{4}{3}=\boxed{1.333}$$$$e_{ss}^{\text{parabola}}=\frac{1}{K_{a}}=\frac{1}{0}=\boxed{\infty}$$
7. Cross-check the ramp error directly. Rather than trusting the formula, evaluate the limit from the error transfer function. With $$\frac{E(s)}{R(s)}=\frac{1}{1+G(s)}=\frac{s(s+4)}{s^{2}+4s+3}$$and a unit ramp $R(s)=1/s^{2}$,$$e_{ss}=\lim_{s\to0}s\cdot\frac{s(s+4)}{s^{2}+4s+3}\cdot\frac{1}{s^{2}}=\lim_{s\to0}\frac{s+4}{s^{2}+4s+3}=\frac{4}{3}=1.333$$in agreement with the error-constant route.
8. Interpret physically. The loop tracks a constant command exactly, follows a constant-velocity command with a fixed lag of 1.333 units, and cannot follow an accelerating command at all — the error grows without bound. Raising the forward gain would shrink the ramp error proportionally (it is $1/K_{v}$) but can never make it zero, and it would not change the parabolic result at all: only adding a second integrator, making the loop type 2, does that.