Question 4 of 8: Steady-State Error Analysis with Reference and Disturbance Ramps

Question 4: Steady-State Error Analysis with Reference and Disturbance Ramps (20 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 Question 2 servo, with

Find. The system Type and all three error constants with their errors at $K = 1$ and again at $K = K_{op}$; the operating gain that meets the ramp specification; the gain margin there; and the total steady-state error when both a reference ramp and a disturbance ramp act together.

Approach. Convert the two-loop system into its equivalent unity-feedback open loop, read the Type and the error constants from that, then superpose the reference-driven error and the disturbance-driven output using the final value theorem.

1. Build the equivalent unity-feedback open loop. Any closed loop can be written as $G_{cl} = G_{eq}/(1 + G_{eq})$, so $G_{eq} = G_{cl}/(1 - G_{cl})$. With the Question 2 result, the denominator minus the numerator is $s^4 + 15s^3 + 50s^2 + 5s = s(s^3 + 15s^2 + 50s + 5)$, giving $$G_{eq}(s) = \frac{5K(s + 0.5)}{s\,(s^3 + 15 s^2 + 50 s + 5)}.$$ This is the loop the error constants must be computed from — not the raw forward path, which would ignore the inner feedback.
2. Read the System Type. $G_{eq}$ has exactly one pole at the origin, so $$\boxed{\text{the system is Type 1.}}$$ That single integrator comes from the PI controller; the plant integrator $1/s$ is inside the inner loop and so does not add a second free integration to the equivalent open loop.
3. Error constants at $K = 1$. For a Type 1 loop $K_{pos} = \lim_{s \to 0} G_{eq}(s) = \infty$ and hence $e_{ss(step)} = 1/(1 + K_{pos}) = 0$. The velocity constant is $$K_v = \lim_{s\to 0} s\,G_{eq}(s) = \frac{5K \times 0.5}{5} = \frac{K}{2} = 0.5 \ \text{at } K = 1,$$ so $e_{ss(ramp)} = 1/K_v = 2.00$ V/V. The acceleration constant is $K_a = \lim_{s\to 0}s^2 G_{eq}(s) = 0$, so $e_{ss(parab)} = \infty$. $$\boxed{K_{pos} = \infty,\ e_{ss(step)} = 0\%;\quad K_v = 0.5,\ e_{ss(ramp)} = 2.00;\quad K_a = 0,\ e_{ss(parab)} = \infty.}$$
4. Part 2a — the operating gain. Zero step error is automatic for a Type 1 loop, so only the ramp specification binds. Requiring $e_{ss(ramp)} = 1/K_v = 2/K = 0.04$ gives $$\boxed{K_{op} = \frac{2}{0.04} = 50.}$$ This is a satisfying independent confirmation of the value inferred in Question 2 by matching the transfer function that Question 3 supplies.
5. Part 2b — stability and gain margin at $K_{op}$. Question 2 established the stable range $0 \lt K \lt 126.68$, and $K_{op} = 50$ lies inside it, so the system remains stable. The gain margin is the factor by which the gain may still be raised before instability: $$G_m = \frac{K_{crit}}{K_{op}} = \frac{126.68}{50} = 2.53\ \text{V/V} \;\;(= 20\log_{10} 2.53 = 8.07\ \text{dB}).$$ A margin of about 2.5 is modest but workable for a servo drive.
6. Part 3 — error constants at $K_{op} = 50$. The Type is unchanged, so $K_{pos}$ is still infinite and $K_a$ still zero; only $K_v$ scales with the gain: $$K_v = \frac{K_{op}}{2} = 25, \qquad e_{ss(ramp)} = \frac{1}{25} = 0.04\ \text{V/V},$$ as designed. So $K_{pos} = \infty$ with $e_{ss(step)} = 0\%$, $K_v = 25$ with $e_{ss(ramp)} = 0.04$, and $K_a = 0$ with $e_{ss(parab)} = \infty$.
7. Part 4, first contribution — the reference ramp. The error is linear in the inputs, so treat the two sources separately and add. For $r(t) = 2t\cdot 1(t)$, i.e. $R(s) = 2/s^2$, the error transfer function is $1 - G_{cl}$ and the final value theorem gives $$e_{ss,r} = \lim_{s \to 0} s\,\frac{2}{s^2}\,\big[1 - G_{cl}(s)\big] = \frac{2}{K_v} = \frac{2}{25} = 0.080\ \text{V/V}.$$ A ramp twice as steep produces twice the lag error, exactly as expected.
8. Part 4, second contribution — the disturbance ramp. For $d(t) = 10t \cdot 1(t)$, i.e. $D(s) = 10/s^2$, the output caused by the disturbance is $Y_d = G_{dist}D$, and $$y_{ss,d} = \lim_{s \to 0} s\,\frac{10}{s^2}\,\frac{s(s+5)}{s^4 + 15s^3 + 50s^2 + 255s + 125} = \frac{10 \times 5}{125} = 0.400\ \text{V}.$$ The single $s$ in the disturbance numerator cancels one power of $s$ from the ramp, which is why a constant torque would give zero but a torque ramp gives a finite offset.
9. Combine the two contributions. Error is reference minus output, so the disturbance-induced output subtracts from the error: $$e_{ss,total} = e_{ss,r} - y_{ss,d} = 0.080 - 0.400 = -0.320\ \text{V/V}.$$ $$\boxed{e_{ss,total} = -0.320\ \text{V/V} \;\; (\text{magnitude } 0.32).}$$ The negative sign is physically meaningful: the growing torque disturbance drives the load faster than the reference asks, so the arm ends up ahead of its commanded velocity by 0.32 V/V, and the disturbance term dominates the reference lag by a factor of five.