Question 4 of 6: Identifying a second-order plant from three experiments

Question 4: Identifying a second-order plant from three experiments (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. Three measurements on a plant assumed to be second order with no finite zeros: one static gain test and two frequency-response points.

Find. The DC gain $K$, natural frequency $\omega_{n}$ and damping ratio $\zeta$; the percentage overshoot of the identified model to a unit step; and whether the unity-feedback loop formed with an integrating controller is stable.

Approach. Take each experiment in turn: the step test gives the DC gain, the 90-degree phase point identifies $\omega_{n}$ exactly, and the remaining phase measurement then yields $\zeta$. Apply the standard overshoot formula, then run a Routh test on the loop formed with the integrator.

1. Static gain from the step test. For a linear system the steady-state output of a step of magnitude $u_{0}$ is $P(0)u_{0}$, so$$K=P(0)=\frac{y_{ss}}{u_{0}}=\frac{10}{2}=5.$$The model is therefore $P(s)=\dfrac{5\omega_{n}^{2}}{s^{2}+2\zeta\omega_{n}s+\omega_{n}^{2}}$, with two unknowns left.
2. The 90-degree point identifies $\omega_{n}$ exactly. The phase of the standard second-order form is$$\angle P(j\omega)=-\arctan\!\left(\frac{2\zeta\omega_{n}\omega}{\omega_{n}^{2}-\omega^{2}}\right),$$which reaches exactly $-90^{\circ}$ when the real part of the denominator vanishes, that is when $\omega=\omega_{n}$ — independently of $\zeta$. The measured 90-degree lag at 8 rad/s therefore gives$$\boxed{\,\omega_{n}=8\ \text{rad/s}\,}$$This is the observation the question is really testing: one phase reading pins the natural frequency without any curve fitting.
3. The 15-degree point gives $\zeta$. Substituting $\omega=5$ rad/s and $\omega_{n}=8$ rad/s into the same phase relation,$$\tan 15^{\circ}=\frac{2\zeta(8)(5)}{8^{2}-5^{2}}=\frac{80\zeta}{39}\;\Longrightarrow\;\zeta=\frac{39\tan 15^{\circ}}{80}=\frac{39(0.26795)}{80}$$$$\boxed{\,\zeta=0.1306\,}$$so $2\zeta\omega_{n}=2.0901$ s$^{-1}$. The plant is very lightly damped, which is consistent with the phase running from 15 degrees to 90 degrees over a frequency change of only 3 rad/s.
4. Assemble the identified model. Putting the three parameters together,$$\boxed{\,P(s)=\frac{5(64)}{s^{2}+2.0901\,s+64}=\frac{320}{s^{2}+2.0901\,s+64}\,}$$As a check, this model reproduces both measured phases exactly: $\angle P(j8)=-90.00^{\circ}$ and $\angle P(j5)=-15.00^{\circ}$.
5. (b) Percentage overshoot. For a second-order system with no finite zeros the first peak of the unit-step response overshoots by$$\mathrm{PO}=\exp\!\left(\frac{-\pi\zeta}{\sqrt{1-\zeta^{2}}}\right)=\exp\!\left(\frac{-\pi(0.1306)}{\sqrt{1-0.1306^{2}}}\right)=\exp(-0.4139)=0.6610$$$$\boxed{\,\mathrm{PO}=66.10\ \%\,}$$Because the DC gain is 5, the response settles at 5 and peaks at $5(1+0.6610)=8.305$, at a peak time $t_{p}=\pi/\omega_{n}\sqrt{1-\zeta^{2}}=0.3961$ s, as plotted above.
6. (c) Close the loop with the integrator. Cascading $C(s)=1/s$ gives the loop gain $L(s)=\dfrac{320}{s(s^{2}+2.0901s+64)}$, so the closed-loop characteristic polynomial is$$s^{3}+2\zeta\omega_{n}s^{2}+\omega_{n}^{2}s+K\omega_{n}^{2}=s^{3}+2.0901\,s^{2}+64\,s+320=0.$$
7. Apply the Routh test. All coefficients are positive, which is necessary but not sufficient. The $s^{1}$ row is$$\frac{a_{2}a_{1}-a_{3}a_{0}}{a_{2}}=\frac{(2.0901)(64)-(1)(320)}{2.0901}=\frac{-186.23}{2.0901}=-89.10\lt 0,$$so the first column changes sign twice and the loop has two right-half-plane poles. Solving the cubic numerically confirms it: the roots are $1.1243\pm j8.5142$ and $-4.3386$.$$\boxed{\,\text{Unstable: two closed-loop poles at }1.1243\pm j8.5142\,}$$
8. Say why, in design terms. Writing the same Routh condition symbolically, the loop is stable if and only if $2\zeta\omega_{n}\omega_{n}^{2}\gt K\omega_{n}^{2}$, i.e. $2\zeta\omega_{n}\gt K$. Here $2\zeta\omega_{n}=2.0901$ while $K=5$, so the plant fails the test by a factor of about 2.4. The integrator adds a further 90 degrees of lag on top of the 180 degrees the plant itself contributes at high frequency, and with so little damping there is no frequency at which the gain has fallen below unity before the phase reaches $-180^{\circ}$. A practical remedy is to precede the integrator with a lead network, or to use a PI controller with a proportional term large enough to move the crossover below the lightly damped resonance.