Question 7 of 8: Controller Design by Pole Placement, Response Specifications (20 marks)

Question 7: Controller Design by Pole Placement, Response Specifications (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. A positioning loop under PI control with rate (tachometer) feedback, per Figure Q7.1.

Find. (1) the closed-loop transfer function and characteristic equation in terms of $K_p$, $\tau_d$, $\tau_i$; (2) the required $\zeta$, $\omega_n$, $K_{dc}$; (3) the controller parameters that place the poles with a pole-zero cancellation.

Approach. Reduce the loop, then match the resulting cubic characteristic polynomial term by term against the desired product of the dominant quadratic and a third pole placed exactly on the closed-loop zero.

1. Reduce the loop (part 1). With $G_c$ in the forward path and $H$ in the feedback path, $G_{cl}=\dfrac{G_cG_p}{1+G_cG_pH}$. Writing $G_c=K_p\dfrac{\tau_is+1}{\tau_is}$,$$G_{cl}(s)=\frac{K_p(\tau_is+1)}{\tau_is(s^2+8s+5)+K_p(\tau_is+1)(1+\tau_ds)}.$$Note the closed-loop zero at $s=-1/\tau_i$, contributed by the PI controller. Expanding and dividing through by $\tau_i$,$$\boxed{Q(s)=s^3+\left(8+K_p\tau_d\right)s^2+\left(5+K_p\frac{\tau_i+\tau_d}{\tau_i}\right)s+\frac{K_p}{\tau_i}=0}$$
2. Convert the transient specs (part 2). From $PO=5\%$,$$\zeta=\frac{-\ln(0.05)}{\sqrt{\pi^2+\ln^2(0.05)}}=0.6901,$$and from $T_{s(\pm2\%)}=4/(\zeta\omega_n)=1$ s,$$\zeta\omega_n=4\;\Rightarrow\;\omega_n=\frac{4}{0.6901}=5.796\ \text{rad/s}.$$The integral action makes the loop Type 1, so the step error is zero automatically and$$\boxed{\zeta=0.690,\quad\omega_n=5.796\ \text{rad/s},\quad K_{dc}=1}$$Setting $s=0$ in $G_{cl}$ confirms $K_{dc}=K_p/K_p=1$ for any parameter choice — the error specification costs no design freedom.
3. Write the desired characteristic polynomial (part 3). Placing the third pole at $s=-a$ so that it cancels the closed-loop zero requires $a=1/\tau_i$. The target is$$Q_{des}(s)=(s^2+2\zeta\omega_ns+\omega_n^2)(s+a)=(s^2+8s+33.596)(s+a),$$since $2\zeta\omega_n=8$ and $\omega_n^2=33.596$.
4. Match the constant term. Comparing $s^0$ coefficients, $K_p/\tau_i=K_pa=\omega_n^2a$, so$$\boxed{K_p=\omega_n^2=33.60}$$
5. Match the $s^2$ term. $8+K_p\tau_d=2\zeta\omega_n+a=8+a$, hence $K_p\tau_d=a$ — note this simple form arises only because the plant’s $s$ coefficient (8) happens to equal $2\zeta\omega_n$ (8). In general $K_p\tau_d=2\zeta\omega_n+a-c$ for a plant $s^2+cs+d$.
6. Match the $s^1$ term to find $a$. Using $K_p(\tau_i+\tau_d)/\tau_i=K_p+K_p\tau_da=K_p+a^2$, the $s^1$ comparison gives$$5+K_p+a^2=\omega_n^2+2\zeta\omega_na\;\Longrightarrow\;5+\omega_n^2+a^2=\omega_n^2+8a,$$so the $\omega_n^2$ terms cancel and the condition collapses onto the plant’s own coefficients:$$\boxed{a^2-8a+5=0\;\Rightarrow\;a=\frac{8\pm\sqrt{44}}{2}=7.317\ \text{or}\ 0.683}$$
7. Choose the root and finish. Both roots place the dominant pair exactly and cancel exactly, so either is mathematically valid. Choose the larger, $a=7.317$: the cancelled mode is then fast, so if the cancellation is imperfect in practice (component tolerance, plant drift) the residual tail decays quickly instead of leaving a slow $0.68$ rad/s creep. Then$$\tau_i=\frac{1}{a}=0.1367\ \text{s},\qquad \tau_d=\frac{a}{K_p}=\frac{7.317}{33.596}=0.2178\ \text{s}.$$Substituting back reproduces $Q(s)=s^3+15.317s^2+92.13s+245.8$, whose roots are $-4\pm j4.195$ and $-7.317$ — the pair at exactly $\zeta=0.690$, $\omega_n=5.796$, and the third pole exactly on the zero. Simulation confirms $PO=5.00\%$ and $T_{s(\pm2\%)}=1.03$ s.