Question 4 of 6: Transfer functions from step and frequency-response data

Question 4: Transfer functions from step and frequency-response data (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. Two separate identification experiments.

Find. The transfer functions $P(s)$ and $G(s)$.

[Figure not reproduced: Question 4(a): the measured unit-step response, redrawn from the exam chart. Three features fix the model - a 2 s dead time, an instantaneous step DOWN to -2 (direct feedthrough with a right-half-plane zero) and a single exponential rise to 6 with time constant 2 s. See the official exam paper.]

Approach. For (a) read the three independent features of the step record — dead time, initial jump and final value with its time constant — and reconstruct the transfer function as the Laplace transform of the response divided by the transform of a unit step. For (b) exploit the fact that a second-order system with no zeros has a phase of exactly $-90^\circ$ at its natural frequency, which decouples the two unknowns.

1. Part (a) — itemise what the chart actually says. Three features carry information. The response is identically zero for the first two seconds; at $t = 2$ it moves instantaneously and downwards, to $-2$; and thereafter it rises monotonically along a single exponential to $6$. Digitising the printed curve and fitting $y = 6 + (y_0 - 6)e^{-(t-2)/\tau}$ returns $y_0 = -2.0$ and $\tau = 1.99$ s, so the intended readings are $y_0 = -2$ and $\tau = 2$ s.
2. Translate each feature into structure. A flat segment before any response at all can only be transport delay, contributing a factor $e^{-2s}$. An instantaneous jump at the moment the response starts means the transfer function has a direct feedthrough term, i.e. numerator and denominator degrees are equal (bi-proper), since by the initial- value theorem the jump equals $\lim_{s\to\infty}P(s)$. A single exponential thereafter means one real pole. So the model has the form $e^{-2s}(\beta_1 s + \beta_0)/(s + 1/\tau)$.
3. Write the measured response and transform it. Shifting time so that the delay is stripped off, the response to a unit step is $$y(t) = 6 - 8e^{-t/2}, \qquad t \ge 0,$$ (final value 6, initial value $6 - 8 = -2$, total swing 8). Its transform is $$Y(s) = \frac{6}{s} - \frac{8}{s + 0.5}.$$
4. Divide by the transform of the input. Since the input is a unit step, $U(s) = 1/s$, the delay-free part of the plant is $Y(s)/U(s) = sY(s)$: $$P_0(s) = 6 - \frac{8s}{s+0.5} = \frac{6(s+0.5) - 8s}{s+0.5} = \frac{-2s + 3}{s + 0.5}.$$ Restoring the transport delay, $$\boxed{\;P(s) = e^{-2s}\,\frac{3 - 2s}{s + 0.5} = 6\,\frac{1 - \tfrac{2}{3}s}{1 + 2s}\,e^{-2s}\;}$$
5. Check the model against every feature. $P_0(0) = 3/0.5 = 6$ reproduces the final value; $\lim_{s\to\infty}P_0(s) = -2$ reproduces the initial jump; the pole at $s = -0.5$ gives $\tau = 2$ s; and the delay factor reproduces the two dead seconds. The numerator vanishes at $s = +1.5$: the plant has a right-half-plane zero, and that is precisely what the initial excursion in the wrong direction is telling us. Undershoot of this kind is the time-domain signature of an odd number of RHP zeros and is not a measurement artefact.
6. Part (b) — use the step test for the DC gain. A step of magnitude 2 produces a steady-state output of 10, so $$G(0) = \frac{10}{2} = 5 .$$ Writing the assumed structure with no finite zeros, $$G(s) = \frac{G(0)\,\omega_n^{2}}{s^{2} + 2\zeta\omega_n s + \omega_n^{2}},$$ which leaves two unknowns, $\omega_n$ and $\zeta$.
7. Use the phase reading to pin $\omega_n$. For this structure $$\angle G(j\omega) = -\arctan\!\frac{2\zeta\omega_n\omega}{\omega_n^{2} - \omega^{2}},$$ which passes through exactly $-90^\circ$ when the denominator's real part vanishes, i.e. when $\omega = \omega_n$ — independently of the damping. The measured $90^\circ$ lag at 8 rad/s therefore gives $$\boxed{\;\omega_n = 8\ \text{rad/s}\;}$$ This decoupling is what makes the problem solvable with only two measurements; attacking the magnitude equation first would leave two unknowns entangled.
8. Use the magnitude reading to pin $\zeta$. At $\omega = \omega_n$ the quadratic collapses to its imaginary part alone, so $$|G(j\omega_n)| = \frac{G(0)\,\omega_n^{2}}{2\zeta\omega_n^{2}} = \frac{G(0)}{2\zeta}.$$ The sinusoidal test gives $|G(j8)| = 15/2 = 7.5$, hence $$\boxed{\;\zeta = \frac{G(0)}{2\,|G(j\omega_n)|} = \frac{5}{15} = \frac{1}{3} = 0.3333\;}$$
9. Assemble and check. Substituting $G(0) = 5$, $\omega_n = 8$ and $\zeta = 1/3$, $$\boxed{\;G(s) = \frac{320}{s^{2} + \tfrac{16}{3}s + 64} = \frac{320}{s^{2} + 5.333s + 64}\;}$$ with poles at $s = -8/3 \pm j7.542$ ($\zeta = 1/3$, $\omega_d = 7.54$ rad/s). Evaluating directly confirms all three measurements: $G(0) = 5$, $|G(j8)| = 7.500$ and $\angle G(j8) = -90.00^\circ$. Because $\zeta \lt 1/\sqrt{2}$ the response is genuinely resonant, and $|G(j\omega_n)| = 7.5$ exceeding the DC gain of 5 is the expected amplification, not an inconsistency in the data.