Question 7 of 8: Rate-Feedback vs. PD Control — Same $K_p,T_d$, Very Different Outcomes

Question 7: Rate-Feedback vs. PD Control — Same $K_p,T_d$, Very Different Outcomes (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. Process $30/(s^2+10s+5)$ in both configurations; Q7.1: rate-feedback path $T_ds$ taps $Y(s)$ and is ADDED (per the source figure, flagged there as atypical) at the second summing node; Q7.2: standard unity feedback with a PD controller $K_p(T_ds+1)$ acting on the error.

Find. 1) $K_p,T_d$ for $PO=5\%,e_{ss(step)}=5\%$ in the Q7.1 configuration. 2) $G_{cl1}(s)$, $T_{settle}$. 3) $G_{cl2}(s)$ with the SAME $K_p,T_d$; compare.

Approach. Reduce Q7.1's block diagram to a standard 2nd-order closed-loop form (proportional gain sets $\omega_n^2$ via the DC gain / step-error spec; rate feedback sets $2\zeta\omega_n$), match to the $PO,e_{ss}$ specification, then reuse the SAME numeric $K_p,T_d$ in Q7.2's differently-structured loop and compare the resulting characteristic equations directly.

1. Part 1) — solving for $K_p,T_d$. With $e=R-Y$ and $u=K_pe+T_dsY$ (both added, as drawn), $Y=P(s)u\Rightarrow Y=P(K_p(R-Y)+T_dsY)\Rightarrow Y[1+K_pP-T_dsP]=K_pPR$, giving $$G_{cl1}(s)=\frac{30K_p}{s^2+(10-30T_d)s+(5+30K_p)}.$$ $PO=5\%\Rightarrow\zeta=-\ln(0.05)/\sqrt{\pi^2+\ln^2(0.05)}=\boxed{0.6901}$. $e_{ss(step)}=1-G_{cl1}(0)=5/(5+30K_p)=0.05\Rightarrow 5+30K_p=100\Rightarrow \boxed{K_p=95/30=19/6\approx3.167}$, which fixes $\omega_n^2=5+30K_p=100\Rightarrow\omega_n=10$. Matching $2\zeta\omega_n=10-30T_d$: $\ 2(0.6901)(10)=13.80=10-30T_d\Rightarrow$ $$\boxed{T_d=\frac{10-13.80}{30}=-0.1267\ \text{s}}$$ (negative — a direct consequence of the source diagram's atypical ADDITION at the rate-feedback node: reaching MORE damping than the $\zeta=0.5$ baseline the proportional-only loop already gives ($2\zeta\omega_n=10$ at $T_d=0$) requires REDUCING the $s^1$ coefficient below 10, which this diagram's sign convention needs a negative $T_d$ to do — physically equivalent to a CONVENTIONAL (subtracting) rate-feedback gain of $+0.1267$.)
2. Part 2) — $G_{cl1}(s)$ and settling time. Substituting, $$\boxed{G_{cl1}(s)=\frac{95}{s^2+13.80s+100}}$$ (poles at $-6.90\pm j7.24$, both stable), and $$T_{settle(\pm2\%)}=\frac{4}{\zeta\omega_n}=\frac{4}{6.901}=\boxed{0.580\ \text{s}}.$$
3. Part 3) — SAME $K_p,T_d$ in the PD configuration. For Q7.2, the derivative acts on the error itself: forward path $K_p(T_ds+1)\times30/(s^2+10s+5)$, giving $$G_{cl2}(s)=\frac{30K_p(T_ds+1)}{s^2+(10+30K_pT_d)s+(5+30K_p)} =\frac{30K_pT_d\,s+30K_p}{s^2+(10+30K_pT_d)s+100}.$$ With $30K_p=95$ and $T_d=-0.1267$: numerator $=-12.04s+95$; denominator $s^1$-coefficient $=10+30K_pT_d=10+95(-0.1267)=10-12.04=\boxed{-2.04}$ — NEGATIVE, so $$\boxed{G_{cl2}(s)=\frac{-12.04s+95}{s^2-2.04s+100}}$$ has poles at $+1.02\pm j9.95$ (verified directly): the closed loop is UNSTABLE, unlike the well-behaved, stable $G_{cl1}(s)$ above. Why the same numbers behave so differently: rate feedback (Q7.1) only ever enters the CHARACTERISTIC EQUATION (no closed-loop zero is created, since $T_d$ acts on $Y$, not on the forward-path signal reaching the output directly) — so a negative $T_d$ there simply reduces the $s^1$ coefficient, and the design compensates by choosing it negative enough to reach the target $\zeta$. PD control (Q7.2) instead applies $T_d$ to the ERROR: it both changes the SAME-LOOKING $s^1$ coefficient (with the OPPOSITE sign convention, $+30K_pT_d$ instead of $-30T_d$) AND adds a closed-loop ZERO at $s=-1/T_d=+7.89$ (a right-half-plane zero, since $T_d<0$). Reusing "the same $K_p,T_d$" therefore flips a well-damped, stable rate-feedback design into an outright unstable PD loop: $PO$ and settling time become meaningless (the step response diverges) rather than merely worse, and the rise-time comparison is moot. This starkly illustrates that rate (tachometer) feedback and PD control are NOT interchangeable just because they share the same nominal "$K_p,T_d$" labels — the physical signal each gain multiplies (the output vs. the error) fixes both the sign convention AND whether a closed-loop zero appears.