Question 4 of 6: Motor Position Control — Velocity Constant, Damping Ratio and Ramp Error

Question 4: Motor Position Control — Velocity Constant, Damping Ratio and Ramp Error (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. Part (a): a motor position plant $G(s)=A/[s(s+a)]$ closed with negative unity feedback, with design targets $K_{v}=20\ \text{s}^{-1}$ and $\zeta=0.707$; both $A$ and $a$ are free. Part (b): the block diagram printed on page 3 of the paper — a negative unity-feedback loop whose forward path is $A/[s(\tau s+1)]$, driven by the ramp $r(t)=r_{0}t\,\mathbf{1}(t)$.

Find. (a) the numerical values of $A$ and $a$ that meet both specifications simultaneously; (b) the type number of the system in the figure and its steady-state tracking error to the ramp.

Approach. Close the loop to expose the standard second-order form $\omega_{n}^{2}/(s^{2}+2\zeta\omega_{n}s+\omega_{n}^{2})$, read $\omega_{n}$ and $\zeta$ off the coefficients, write $K_{v}$ from the open-loop transfer function, and solve the two resulting equations; for part (b) count the integrators to fix the type and apply the matching error constant.

Part (a) — choosing $A$ and $a$

1. Write the velocity constant. The loop is type 1, so the relevant error constant is$$K_{v}=\lim_{s\to0}s\,G(s)=\lim_{s\to0}\frac{sA}{s(s+a)}=\frac{A}{a}$$Setting this equal to the specification gives the first design equation$$\frac{A}{a}=20\quad\Longrightarrow\quad A=20a$$
2. Close the loop and identify the second-order parameters. With negative unity feedback,$$\frac{Y(s)}{R(s)}=\frac{G(s)}{1+G(s)}=\frac{A}{s^{2}+as+A}$$Comparing with the canonical denominator $s^{2}+2\zeta\omega_{n}s+\omega_{n}^{2}$ gives$$\omega_{n}=\sqrt{A},\qquad 2\zeta\omega_{n}=a\quad\Longrightarrow\quad \zeta=\frac{a}{2\sqrt{A}}$$
3. Solve the two equations together. Substituting the damping specification $\zeta=0.707$ and then $A=20a$,$$\frac{a}{2\sqrt{A}}=0.707\;\Longrightarrow\;a=1.414\sqrt{A}=1.414\sqrt{20a}\;\Longrightarrow\;a^{2}=2(20a)=40a$$Discarding the trivial root $a=0$ leaves$$\boxed{\;a=40\ \text{s}^{-1},\qquad A=20a=800\ \text{s}^{-2}\;}$$
4. Check both specifications and report the resulting dynamics. With these values $K_{v}=800/40=20\ \text{s}^{-1}$ as required, $\omega_{n}=\sqrt{800}=28.28\ \text{rad/s}$ and $\zeta=40/(2\times28.28)=0.7071$ — both targets met exactly. The resulting closed-loop behaviour is the textbook $\zeta=1/\sqrt{2}$ response: overshoot $M_{p}=\exp\!\left(-\pi\zeta/\sqrt{1-\zeta^{2}}\right)=e^{-\pi}=4.32\,\%$, and a $2\,\%$ settling time $T_{s}\approx4/(\zeta\omega_{n})=4/20=0.20\ \text{s}$.

Part (b) — type number and ramp error of the figured system

1. Read the loop transfer function off the block diagram. The figure shows a single forward-path block $A/[s(\tau s+1)]$ between the summing junction and the output $Y$, with the output fed back unmodified to the negative input. The loop transfer function is therefore$$L(s)=\frac{A}{s(\tau s+1)}$$
2. Determine the system type. $L(s)$ has exactly one pole at the origin, so$$\boxed{\;\text{the system is TYPE 1}\;}$$Type 1 implies zero steady-state error to a step reference and a finite, non-zero error to a ramp — which is what part (b) asks for.
3. Form the error transfer function. For negative unity feedback the error is $E(s)=R(s)/[1+L(s)]$. With the ramp $r(t)=r_{0}t\,\mathbf{1}(t)$, whose transform is $R(s)=r_{0}/s^{2}$,$$E(s)=\frac{r_{0}}{s^{2}}\cdot\frac{1}{1+\dfrac{A}{s(\tau s+1)}}=\frac{r_{0}(\tau s+1)}{s\left[s(\tau s+1)+A\right]}$$
4. Apply the Final Value Theorem. Provided the closed loop is stable — which for $\tau>0$ and $A>0$ it is, since $\tau s^{2}+s+A$ has all-positive coefficients — every pole of $sE(s)$ lies in the open left half-plane and the theorem is legitimate:$$e_{ss}=\lim_{s\to0}sE(s)=\lim_{s\to0}\frac{r_{0}(\tau s+1)}{s(\tau s+1)+A}=\frac{r_{0}}{A}$$
5. Present the result through the velocity constant. The same answer follows in one line from the error-constant table, since $K_{v}=\lim_{s\to0}sL(s)=A$:$$\boxed{\;e_{ss}=\frac{r_{0}}{K_{v}}=\frac{r_{0}}{A}\;}$$The time constant $\tau$ does not appear: it shapes the transient and sets how quickly the error settles, but it has no influence whatever on the steady-state ramp error, which is governed by the DC behaviour of the loop alone.