Question 8 of 8: Series-Configuration PID Design by Pole Placement

Question 8: Series-Configuration PID Design by Pole Placement (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. Series PID $C(s)=K_p(1+K_i/s)(K_ds+1)$, process $P(s)=1/(s^2+7s+8)$, unity negative feedback; design targets $PO=15\%$, $T_{settle(\pm2\%)}=2$ s; the third (real) closed-loop pole is to sit exactly at $-K_i$ (pole–zero cancellation with the controller's own $(s+K_i)$ zero).

Find. 1) $G_{cl}(s)$ and $Q(s)$. 2) $\zeta,\omega_n$. 3) $K_p,K_d,K_i$. 4) The correct root of the resulting quadratic, justified.

Approach. Write $C(s)P(s)$ as a single ratio and form $Q(s)=1+C(s)P(s)=0$ cleared of fractions. Solve $PO,T_{settle}$ for $\zeta,\omega_n$ by the standard formulas. Set the DESIRED characteristic polynomial to $(s+K_i)(s^2+2\zeta\omega_ns+\omega_n^2)$ (dominant pair plus the chosen real pole at $-K_i$) and match its coefficients term-by-term against $Q(s)$'s actual coefficients.

1. Part 1) — closed-loop TF and characteristic equation. $C(s)=K_p(s+K_i)(K_ds+1)/s$, so $$G_{cl}(s)=\frac{C(s)P(s)}{1+C(s)P(s)}=\frac{K_p(s+K_i)(K_ds+1)}{s(s^2+7s+8)+K_p(s+K_i)(K_ds+1)},$$ $$\boxed{Q(s)=s^3+(7+K_pK_d)s^2+(8+K_p+K_pK_iK_d)s+K_pK_i=0.}$$
2. Part 2) — $\zeta,\omega_n$ from the transient spec. $PO=15\%\Rightarrow\zeta=\dfrac{-\ln(0.15)}{\sqrt{\pi^2+\ln^2(0.15)}}\approx\boxed{0.517}$. $T_{settle(\pm2\%)}=4/(\zeta\omega_n)=2\Rightarrow\zeta\omega_n=2$, so $\boxed{\omega_n=2/0.517\approx3.87\ \text{rad/s}}$ (and $\zeta\omega_n=2$ exactly — a clean number that will re-appear as the dominant pair's real part).
3. Part 3) — match the desired cubic and solve for the gains. Desired: $(s+K_i)(s^2+4s+14.97)$ (using $2\zeta\omega_n=4$, $\omega_n^2=14.97$). Matching the CONSTANT term first is the cleanest entry point: $$K_pK_i=\omega_n^2K_i\ \Rightarrow\ \boxed{K_p=\omega_n^2\approx14.97}$$ (the $K_i$ cancels directly, so $K_p$ is pinned down without needing $K_i$ at all). Matching the $s^2$ and $s^1$ coefficients gives $K_pK_d=K_i+2\zeta\omega_n-7=K_i-3$ and $8+K_p+K_pK_iK_d=\omega_n^2+2\zeta\omega_nK_i=14.97+4K_i$; substituting $K_pK_d=K_i-3$ into the second equation and simplifying leaves a single quadratic in $K_i$: $$K_i^2-7K_i+8=0\ \Rightarrow\ \boxed{K_i=5.562\ \text{or}\ K_i=1.438.}$$ Correspondingly $K_d=(K_i-3)/K_p$: $K_d\approx0.171$ (for $K_i=5.562$) or $K_d\approx-0.104$ (for $K_i=1.438$).
4. Part 4) — choosing the physically sensible root. Substituting both $(K_i,K_d)$ pairs back into $Q(s)$ confirms both place the dominant complex pair exactly at $-2\pm j3.312$ as designed — the two solutions differ only in WHERE the third real pole (at $-K_i$) ends up. For $K_i=5.562$: third pole at $-5.56$, which is $5.56/2\approx2.8\times$ farther from the imaginary axis than the dominant pair's real part ($-2$) — a reasonable, if modest, separation, AND $K_d\approx0.171>0$ (a physically ordinary positive derivative gain). For $K_i=1.438$: the third pole sits at $-1.44$, which is CLOSER to the imaginary axis than the dominant pair itself, so it would actually dominate the response (defeating the entire premise of a 2nd-order dominant-pole design), and it forces $K_d\approx-0.104\lt0$, a negative derivative gain that has no ordinary physical justification in a series PID compensator. $\boxed{\text{Choose }K_i=5.562,\ K_d=0.171,\ K_p=14.97}$ — the actual (3rd-order) response will settle a little slower and overshoot a little more than the pure dominant-pole prediction, since the third pole, while reasonably separated, is not overwhelmingly far away.