Question 1 of 5: Multiple choice — seventeen items

Question 1: Multiple choice — seventeen items (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. Seventeen independent items drawn from time response, root-locus construction, steady-state error and frequency response, with the printed weights shown below.

Find. The best answer to each of the seventeen items, with the reasoning that earns the mark rather than the letter alone.

Approach. Each item is settled from one standard relation — the overshoot/peak-time pair, the static error constants, the root-locus construction rules, or the definition of the two stability margins — and every numerical item is recomputed rather than recognised.

Item 1 — reading %OS and \(T_p\) off a step response. The plotted response settles on a final value of 1.0 and reaches its first peak at 1.4. Percent overshoot is measured from the final value, not from the axis: \(\%OS = \dfrac{c_{\max}-c_{\text{final}}}{c_{\text{final}}}\times 100 = \dfrac{1.4-1.0}{1.0}\times 100 = 40\,\%\). The peak occurs at \(t = 4\) s, so \(T_p = 4\) s. The distractor “140 %” is the peak value expressed as a percentage, and “20 seconds” is roughly where the response has finished settling, not where it peaks. Answer (c). As a cross-check, \(40\,\%\) overshoot corresponds to \(\zeta = -\ln(0.40)/\sqrt{\pi^{2}+\ln^{2}(0.40)} = 0.280\), and \(T_p = \pi/(\omega_n\sqrt{1-\zeta^{2}}) = 4\) s then gives \(\omega_n = 0.818\) rad/s — a slow, lightly damped system, which is exactly what the plot shows.

[Figure not reproduced: Item 1: the printed unit-step response, redrawn from \(\zeta = 0.280\), \(\omega_n = 0.818\) rad/s. The overshoot is measured from the final value of 1.0, and the peak time is the abscissa of the first maximum. See the official exam paper.]

Item 2 — sizing \(J\) for a settling time. Comparing the given transfer function with the standard second-order denominator \(s^{2}+2\zeta\omega_n s+\omega_n^{2}\) gives \(2\zeta\omega_n = 1/J\), so the exponential envelope decays as \(e^{-\zeta\omega_n t}\) with \(\zeta\omega_n = 1/(2J)\). The 2 % settling time is \(T_s = 4/(\zeta\omega_n) = 8J\); setting \(8J = 3\) s gives $$\boxed{J = \tfrac{3}{8} = 0.375\ \text{kg}\cdot\text{m}^{2}}$$ Answer (c). Note that the spring constant \(K\) never enters: it fixes \(\omega_n\) and hence the oscillation frequency, but the settling time depends only on the real part of the poles, which the damper alone controls.

Item 3 — do the closed-loop zeros move with gain? For a unity-feedback loop with forward path \(KG_0(s) = K\,n(s)/d(s)\), \(T(s) = \dfrac{K n(s)}{d(s) + K n(s)}\). The numerator is \(K n(s)\), so the closed-loop zeros are the roots of \(n(s)\) — the open-loop zeros — whatever \(K\) is; only the overall scale factor changes. The poles are what move, and tracing them is precisely what a root locus does. Answer (c) No.

Item 4 — phase of four poles at the origin. A factor \(1/s\) contributes \(\angle (j\omega)^{-1} = -90^\circ\) at every frequency. Four such factors contribute \(4 \times (-90^\circ) = -360^\circ\), constant with frequency. Answer (b). The magnitude contribution would be \(-80\) dB/decade, and it is worth remembering that the phase of a pole at the origin is flat, unlike the phase of a pole at \(-a\), which swings from \(0^\circ\) to \(-90^\circ\) about the corner.

Item 5 — steady-state error to a parabolic command. The open-loop function \(G(s) = 10(s+1)/[s^{2}(s+5)]\) has two poles at the origin, so the loop is type 2: it tracks a step and a ramp with zero error, and only the \(0.5t^{2}\) term produces a finite error. The acceleration error constant is $$K_a = \lim_{s \to 0} s^{2}G(s) = \lim_{s \to 0}\frac{10(s+1)}{s+5} = \frac{10}{5} = 2$$ For a command \(r(t) = \tfrac{1}{2}a t^{2}\) the error is \(e_{ss} = a/K_a\); here \(a = 1\) because \(0.5t^{2} = \tfrac{1}{2}(1)t^{2}\), so $$\boxed{e_{ss} = \frac{1}{K_a} = \frac{1}{2} = 0.5}$$ Answer (a). The constant “5” and the ramp term “\(-t\)” in the command contribute nothing, which is the trap: a type-2 loop annihilates both of them.

Item 6 — how many asymptotes? The forward path is \(k(s^{2}-1)(s+3)/(s^{3}+2s^{2}+3s)\), so there are \(n = 3\) finite open-loop poles (\(s = 0\) and \(s = -1 \pm j\sqrt{2}\)) and \(m = 3\) finite open-loop zeros (\(s = +1,\,-1,\,-3\)). The number of asymptotes is \(n - m = 0\): every branch of the locus ends on a finite zero, so no branch runs off to infinity and there is nothing for an asymptote to describe. Answer (a).

Item 7 — asymptote angles. For \(k/[s(s+1)(s+2)]\) there are \(n = 3\) finite poles and \(m = 0\) finite zeros, so three branches go to infinity along asymptotes at $$\theta_a = \frac{(2k+1)180^\circ}{n-m} = \frac{(2k+1)180^\circ}{3}, \qquad k = 0, 1, 2$$ which gives \(60^\circ\), \(180^\circ\) and \(300^\circ \equiv -60^\circ\). Answer (c). Their common intercept is \(\sigma_a = \dfrac{\sum \text{poles} - \sum \text{zeros}}{n-m} = \dfrac{0-1-2}{3} = -1\).

Item 8 — zeros of a constant-over-quadratic transfer function. A transfer function \(G(s) = c/(s^{2}+as+b)\) has no finite zero, but the notion of a zero is properly counted including the point at infinity: a system with \(n\) poles and \(m\) finite zeros has \(n-m\) zeros at infinity, because \(G(s) \to 0\) as \(s \to \infty\) along \(1/s^{2}\). Here \(n-m = 2-0 = 2\). Answer (d). Option (b) is the naive reading and is exactly what the item is testing against; the count of zeros at infinity is what fixes the number of root-locus asymptotes, so it is not a pedantic distinction.

Item 9 — when may a pole be neglected? Two circumstances let a pole be dropped from a reduced-order model. First, if its real part is much larger in magnitude than that of the dominant poles — the usual rule of thumb is a factor of five or more — its exponential dies out long before the dominant response develops. Second, if the pole sits close to a zero, the pole-zero pair very nearly cancels and the residue of that mode is small even though its time constant may be slow. Answer (b), which is the only option naming both. Option (a) offers “on the real axis”, which is irrelevant, and option (c) has the geometry backwards: a pole near the imaginary axis is slow and therefore dominant.

Item 10 — what does the imaginary part generate? A complex pair \(s = -\sigma \pm j\omega_d\) produces the time function \(e^{-\sigma t}\cos(\omega_d t + \phi)\). The real part \(\sigma\) sets the exponential envelope, i.e. the time constant \(1/\sigma\); the imaginary part \(\omega_d\) is the radian frequency of the damped oscillation. Answer (c).

Item 11 — the break-in point. The open-loop function is \(G(s) = K(s+1)/(s^{2}+5s+10)\), so the two open-loop poles are \(s = -2.5 \pm j1.936\) (complex, and stable) and the single finite zero is at \(s = -1\). On the real axis the locus exists where an odd number of real poles and zeros lie to the right, which here is the whole ray \(\sigma \le -1\). The branches leave the complex poles, curve into that ray and meet it at the break-in point, found from \(dK/ds = 0\) with \(K = -\dfrac{s^{2}+5s+10}{s+1}\): $$\frac{dK}{ds} = 0 \;\Longrightarrow\; (2s+5)(s+1) - (s^{2}+5s+10) = 0 \;\Longrightarrow\; s^{2}+2s-5 = 0$$ $$s = -1 \pm \sqrt{6} \;\Longrightarrow\; s = +1.449 \ \text{or}\ s = -3.449$$ Only the root that lies on the locus is admissible, so $$\boxed{\sigma_{\text{break-in}} = -1-\sqrt{6} = -3.449}$$ Answer (c). The rejected root \(+1.449\) is not on the locus (no real-axis locus exists to the right of the zero), which is the routine discipline this item is checking.

Item 12 — intercept of the asymptotes. For \(G(s) = K/[(s+1)(s+2)(s+3)]\) there are three finite poles and no finite zero, so $$\sigma_a = \frac{\sum \text{finite poles} - \sum \text{finite zeros}}{n-m} = \frac{(-1)+(-2)+(-3)-0}{3-0} = \frac{-6}{3} = -2$$ Answer (b). Option (d), \(-6\), is the un-divided numerator and is the standard slip on this formula.

Item 13 — range of \(K\) for stability. The closed-loop characteristic polynomial is $$(s+1)(s+2)(s+3) + K = s^{3} + 6s^{2} + 11s + (6+K)$$ The Routh array is $$\begin{array}{c|cc} s^{3} & 1 & 11 \\ s^{2} & 6 & 6+K \\ s^{1} & \dfrac{66-(6+K)}{6} = \dfrac{60-K}{6} & 0 \\ s^{0} & 6+K & \end{array}$$ A first column with no sign change requires \(60-K \gt 0\) and \(6+K \gt 0\), so $$\boxed{-6 \lt K \lt 60}$$ Answer (c). At the upper limit the \(s^{1}\) row vanishes and the auxiliary polynomial \(6s^{2}+66 = 0\) gives the imaginary-axis crossing at \(s = \pm j\sqrt{11} = \pm j3.317\) — the frequency at which the loop would oscillate. The lower limit is easy to lose: negative \(K\) is admissible down to \(-6\), where the constant term changes sign, which is why option (a) and option (d) are both incomplete rather than simply wrong.

Item 14 — the resonant peak of an under-damped system. For a standard second-order system with \(\zeta \lt 1/\sqrt{2}\) the closed-loop magnitude has a peak \(M_r = \dfrac{1}{2\zeta\sqrt{1-\zeta^{2}}}\) at \(\omega_r = \omega_n\sqrt{1-2\zeta^{2}}\). The peak height grows without bound as \(\zeta \to 0\): \(M_r = 1.36\) at \(\zeta = 0.4\), \(2.55\) at \(\zeta = 0.2\), \(5.03\) at \(\zeta = 0.1\). So the peak increases as the damping ratio decreases. Answer (a).

Item 15 — the high-frequency slope of a complex pole pair. Well above the corner, a quadratic factor \(1/[(s/\omega_n)^{2}+2\zeta(s/\omega_n)+1]\) behaves as \((\omega_n/\omega)^{2}\), i.e. the magnitude falls by a factor of 100 per decade, which is \(-40\) dB/decade. An octave is a doubling of frequency, so the same asymptote is \(-40 \times \log_{10}2 = -12.04 \approx -12\) dB/octave. Answer (c). Option (b), \(-6\) dB/octave, is the same slope written for a single real pole, and option (d), \(-20\) dB/decade, is that single pole again — both correct statements about the wrong system.

Item 16 — margins from a printed Bode diagram. The two margins are read at two different frequencies and must not be confused. The gain crossover is where the magnitude curve passes 0 dB; on the printed plot this is near \(\omega \approx 3\) rad/s, and the phase curve there is near \(-116^\circ\), so $$PM = 180^\circ + \angle G(j\omega_{gc}) = 180^\circ - 116^\circ = 64^\circ$$ The phase crossover is where the phase curve passes \(-180^\circ\), near \(\omega \approx 10\) rad/s, and the magnitude there is about \(-21\) dB, so $$GM = -\,|G(j\omega_{pc})|_{\text{dB}} = +21\ \text{dB}$$ $$\boxed{PM \approx 64^\circ, \qquad GM \approx 21\ \text{dB}}$$ Answer (d). The distractor \(116^\circ\) is the raw phase reading quoted as if it were the margin, and \(3\) dB is what the gain margin would be if the magnitude curve were still near 0 dB at the phase crossover; both are the errors this item exists to catch. A gain margin of 21 dB is a factor of \(10^{21/20} = 11.2\) in gain, so the loop tolerates an elevenfold gain increase before it oscillates — a comfortably stable design.

Item 17 — what a vanishing margin means. A gain margin close to unity (0 dB) or a phase margin close to zero puts the Nyquist plot within a whisker of the \(-1\) point, so the closed-loop poles sit close to the imaginary axis and the damping ratio is close to zero. The response is not unstable, but it rings for a long time: it is highly oscillatory. Answer (c). The useful rule of thumb is \(\zeta \approx PM/100\) for phase margins up to about \(60^\circ\), so a phase margin of a few degrees corresponds to a damping ratio of a few hundredths.