Question 4 of 6: Motor position servo with and without rate feedback

Question 4: Motor position servo with and without rate feedback

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 servo of Fig. 3: forward path $K/[s(s+1)]$; an inner (minor) loop that feeds the output back through $K_g s$ into a second summing junction; and an outer unity position feedback loop. Data are collected below.

Find. (a) $K$ for $\zeta=0.5$ with $K_g=0$, and the resulting unit-ramp error; (b) $K$ for a unit-ramp error of 0.1 with $K_g=0$, and the resulting $\zeta$; (c) with that $K$, the value of $K_g$ giving $\zeta=0.5$, and how the ramp error then compares with part (b).

Approach. Reduce the inner loop algebraically to obtain a single equivalent forward path, compare the resulting closed-loop denominator with the standard form $s^2+2\zeta\omega_n s+\omega_n^2$ to relate $\zeta$ to the gains, and read the velocity error constant off the equivalent open loop.

1. Reduce the inner loop. Let $E'$ be the output of the second summing junction, so $E'=E-K_g s\,C$ and $C=\dfrac{K}{s(s+1)}E'$. Eliminating $E'$, $$\frac{C}{E}=\frac{K/[s(s+1)]}{1+K K_g s/[s(s+1)]} =\frac{K}{s(s+1)+KK_g s}=\frac{K}{s\,[\,s+(1+KK_g)\,]}$$ The rate loop leaves the integrator untouched and simply moves the plant pole from $-1$ to $-(1+KK_g)$.
2. Write the closed-loop denominator and the error constant once, generally. Closing the outer unity loop, $$\frac{C}{R}=\frac{K}{s^2+(1+KK_g)s+K} \;\Longrightarrow\; \omega_n=\sqrt{K},\qquad 2\zeta\omega_n=1+KK_g$$ and from the equivalent open loop, $K_v=\lim_{s\to0}sC/E$, so $$K_v=\frac{K}{1+KK_g},\qquad e_{ss}(\text{ramp})=\frac{1}{K_v}=\frac{1+KK_g}{K}$$ Every part of the question is now a substitution into these two relations.
3. Part (a): choose $K$ for $\zeta=0.5$ with $K_g=0$. With $K_g=0$ the damping relation reads $2\zeta\sqrt{K}=1$, hence $$K=\frac{1}{4\zeta^2}=\frac{1}{4(0.5)^2}=1$$ so $\omega_n=1\ \text{rad/s}$ and the closed-loop poles are $-0.5\pm j0.866$. The ramp error follows from $K_v=K=1$: $$\boxed{\;K=1,\qquad e_{ss}(\text{ramp})=1.0\;}$$
4. Part (b): choose $K$ for a ramp error of 0.1. Still with $K_g=0$, $e_{ss}=1/K$, so a specification of 0.1 demands $K_v=10$ and therefore $$K=10,\qquad \omega_n=\sqrt{10}=3.162\ \text{rad/s}$$ The damping that comes with it is fixed — it is not a free choice: $$\zeta=\frac{1}{2\sqrt{K}}=\frac{1}{2\sqrt{10}}=0.158$$ $$\boxed{\;K=10,\qquad \zeta=0.158\;}$$ This is a badly under-damped servo: the poles are at $-0.5\pm j3.12$ and the step response would overshoot by roughly 60 %.
5. Part (c): recover the damping with rate feedback. Keeping $K=10$ so that $\omega_n=3.162\ \text{rad/s}$ is unchanged, the damping relation $2\zeta\omega_n=1+KK_g$ with $\zeta=0.5$ gives $$1+10K_g=2(0.5)(3.162)=3.162 \quad\Longrightarrow\quad \boxed{\;K_g=0.2162\ \text{s}\;}$$ The closed-loop poles move from $-0.5\pm j3.12$ to $-1.581\pm j2.739$, i.e. onto the $\zeta=0.5$ radial line.
6. Compare the steady-state errors. Rate feedback is inside the error path, so it degrades the velocity error constant: $$K_v=\frac{K}{1+KK_g}=\frac{10}{3.162}=3.162\ \text{s}^{-1} \quad\Longrightarrow\quad e_{ss}(\text{ramp})=\frac{1}{3.162}=0.3162$$ The error has grown from 0.1 to 0.316, a factor of $\sqrt{10}=3.16$ worse. In other words the damping bought in part (c) is paid for in tracking accuracy: to hold both $\zeta=0.5$ and $e_{ss}=0.1$ the designer would have to change the plant or add a proportional-plus-integral or lead compensator, not merely a tachometer.