Question 5 of 8: Robot-joint servo: Mason reduction & step response

Question 5 — Robot-joint servo: Mason reduction & step response [10 + 5 + 5]

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. $G_c=15+2.5s=2.5(s+6)$; calibration $30$; motor/arm $\dfrac{18}{(s+5)(s+15)}$ with local feedback $-0.5$; gearbox $\dfrac{1}{30s}$.

Find. $G_{cl}(s)$; Type and $K_{pos},K_v,K_a$; $\theta(t)$.

[Figure not reproduced: Figure Q5.1 redrawn — two nested loops; the calibration $\times30$ and gearbox $\times\tfrac{1}{30}$ cancel exactly. See the official exam paper.]

Part 1 — Closed-loop transfer function [10]

Approach. Close the inner $-0.5$ loop, cascade the forward elements (the $30$ and $1/30$ cancel), then close the unity outer loop.

1. Inner loop. $\dfrac{18/[(s+5)(s+15)]}{1+0.5\cdot18/[(s+5)(s+15)]}=\dfrac{18}{s^2+20s+84}$.
2. Forward path. $G_c\cdot30\cdot\dfrac{18}{s^2+20s+84}\cdot\dfrac{1}{30s} =\dfrac{45(s+6)}{s(s^2+20s+84)}=L(s)$.
3. Close the unity loop. $$G_{cl}=\frac{L}{1+L}=\frac{45(s+6)}{s^3+20s^2+129s+270}.$$ The hint $s=-9$ is a root, and the denominator factors as $(s+9)(s+5)(s+6)$.
4. Pole–zero cancellation. The numerator zero at $-6$ cancels the denominator pole at $-6$: $$\boxed{G_{cl}(s)=\frac{45}{(s+5)(s+9)}=\frac{45}{s^2+14s+45}.}$$ DC gain $=1$; two real poles $\Rightarrow$ overdamped.

Part 2 — Type & error constants [5]

1. Type. The open loop $L(s)=\dfrac{45(s+6)}{s(s^2+20s+84)}$ has one pole at the origin $\Rightarrow$ $\boxed{\text{Type 1}}$.
2. Constants. $K_{pos}=\lim_{s\to0}L=\infty$; $K_v=\lim_{s\to0}sL=\dfrac{45(6)}{84}=\boxed{3.21}$; $K_a=\lim_{s\to0}s^2L=0$. Hence $e_{ss}(\text{ramp})=1/K_v=0.311$.

Part 3 — Unit-step response [5]

1. Partial fractions. $\Theta(s)=\dfrac{45}{s(s+5)(s+9)}=\dfrac{1}{s}-\dfrac{2.25}{s+5}+\dfrac{1.25}{s+9}$.
2. Inverse transform. $$\boxed{\theta(t)=1-2.25\,e^{-5t}+1.25\,e^{-9t},\quad t\ge0.}$$ Check $\theta(0)=0$, $\theta(\infty)=1$ — a clean overdamped rise, no overshoot.