Question 2 of 5: Multiple choice — eighteen items

Question 2: Multiple choice — eighteen 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. The eighteen items and the data printed with each:

Find. The single best option for each of the eighteen items, with the working that settles it.

Approach. Items 1, 4 and 10 are reduction problems — count paths or collapse the diagram algebraically before testing anything. Items 2, 3, 5, 6 and 7 are computations on a characteristic polynomial or an error transfer function. Items 8–9 and 11–18 are decided by a definition, but three of them (6, 8 and 18) hide a subtlety worth stating explicitly, so those get a sentence of justification beyond the letter.

1. Item (1) — enumerate the forward paths. A forward path visits each node at most once and travels only in the direction of the arrows. Numbering the nodes \(1\) (input) to \(5\) (output) from left to right, the forward-directed branches are the chain \(1\!\to\!2\!\to\!3\!\to\!4\!\to\!5\) plus the single skip branch \(3\!\to\!5\). Every remaining branch points leftward (\(4\!\to\!2\), \(3\!\to\!2\), \(5\!\to\!2\)) or is the self-loop at node 4, and none of those can appear in a forward path. That leaves $$P_1: 1\!\to\!2\!\to\!3\!\to\!4\!\to\!5, \qquad P_2: 1\!\to\!2\!\to\!3\!\to\!5,$$ so $$\boxed{2 \text{ forward paths — option (b)}}$$ The three feedback branches and the self-loop are what make the Mason determinant of this graph non-trivial, but they contribute nothing to the path count that this item asks for.
2. Item (2) — critical damping means a vanishing discriminant. Closing unity feedback around \(2K/(s^2+7s+2)\), $$s^2+7s+2+2K = 0 .$$

3. Item (2) (continued) — solve for \(K\). A repeated real root requires \(b^2 = 4ac\): $$7^2 = 4(2+2K) \quad\Longrightarrow\quad 49 = 8+8K \quad\Longrightarrow\quad \boxed{K = \frac{41}{8} = 5.125 \text{ — option (b)}}$$ The repeated root is then \(s = -7/2 = -3.5\) twice, which lies in the open left half-plane, so the critically damped loop is also stable — worth checking, because a discriminant can vanish on the unstable side as well. Equivalently \(\omega_n = \sqrt{2+2(5.125)} = 3.5\) rad/s and \(2\zeta\omega_n = 7\) gives \(\zeta = 1\) exactly. ✓
4. Item (3) — a sign change in the coefficients settles it without a Routh array. The denominator is $$s^6+4s^5+11s^4-32s^3+40s^2+64s-48 ,$$ whose leading coefficient is \(+1\) and whose constant term is \(-48\). A necessary condition for all roots to lie in the left half-plane is that every coefficient of the characteristic polynomial share the same sign; the \(-32s^3\) and \(-48\) terms violate it, so at least one root lies in the right half-plane and $$\boxed{\text{the system is unstable — option (c)}}$$ Factoring numerically confirms three right-half-plane roots (\(0.604\) and \(1.253\pm j1.251\)) against three in the left half-plane, so the instability is not marginal. Note also that the numerator zero at \(s=+2\) is irrelevant to stability: a right-half-plane zero slows a system down but never destabilises it.
5. Item (4) — collapse the nested loops before testing \(p\). Reading the printed diagram, let \(x\) be the output of \(1/(s+1)\), \(u\) the output of the middle summer and \(y = C\) the output of \(1/(s+p)\). Then \(u = x - y\) and \(y = u/(s+p)\), so \(y(s+p+1) = x\). The lower summer forms \(f = u + y = (x-y)+y = x\), and that signal is what feeds back negatively to the input summer, so \(x = (R-x)/(s+1)\), i.e. \(x = R/(s+2)\). Hence $$\frac{C(s)}{R(s)} = \frac{1}{(s+2)\,(s+p+1)} .$$ The closed-loop poles are \(-2\) and \(-(p+1)\), so stability requires \(p+1>0\): $$\boxed{p > -1 \text{ — option (b)}}$$ The pleasant surprise is that the inner cross-coupling cancels exactly (\(f=x\)), which is why the answer contains no cross term between the two blocks.

6. Item (5) — read \(\zeta\omega_n\) and apply the 2% settling formula. Comparing \(s^2+1.6s+4\) with \(s^2+2\zeta\omega_ns+\omega_n^2\), $$\omega_n = \sqrt4 = 2 \text{ rad/s}, \qquad 2\zeta\omega_n = 1.6 \;\Longrightarrow\; \zeta = 0.4, \qquad \zeta\omega_n = 0.8 \text{ s}^{-1}.$$ The envelope of the underdamped step response decays as \(e^{-\zeta\omega_n t}\), and reaching a 2% band takes about four time constants: $$\boxed{T_s = \frac{4}{\zeta\omega_n} = \frac{4}{0.8} = 5 \text{ s — option (d)}}$$ Option (b), 2.5 s, is the 2%-band answer for the 5% constant \(3/\zeta\omega_n\) misapplied, and option (a) simply echoes the \(1.6\) from the polynomial.
7. Item (6) — the loop is Type 1, so the actuating error vanishes. The loop gain is $$L(s) = K\,\frac{s+40}{s(s+10)}\cdot\frac{1}{s+20} = \frac{K(s+40)}{s(s+10)(s+20)} ,$$ which carries exactly one pole at the origin. The error measured at the summer is \(E = R - HC\), so \(E/R = 1/(1+L)\) and for \(R = 1/s\), $$e_{ss} = \lim_{s\to0} s\cdot\frac1s\cdot\frac{1}{1+L(s)} = \frac{1}{1+\infty} \;\Longrightarrow\; \boxed{e_{ss} = 0 \text{ — option (c)}}$$

8. Item (7) — build the error from \(1-T\). With the closed-loop transfer function given directly, \(E(s) = R(s)\,[1-T(s)]\). Here $$1-T(s) = \frac{(s^3+2s^2+4s+4)-4(s+1)}{s^3+2s^2+4s+4} = \frac{s^3+2s^2}{s^3+2s^2+4s+4} = \frac{s^2(s+2)}{s^3+2s^2+4s+4} .$$ The double zero at the origin is the structural fact that decides the item: \(T(0)=4/4=1\), so the loop tracks a step and a ramp exactly, and only the \(t^2\) term can leave an error.
9. Item (7) (continued) — apply the final-value theorem. Transforming \(r(t) = 3 - t + \tfrac14 t^2\) term by term gives \(R(s) = 3/s - 1/s^2 + 1/(2s^3)\), so $$sR(s) = 3 - \frac1s + \frac{1}{2s^2}, \qquad e_{ss} = \lim_{s\to0} sR(s)\,[1-T(s)] = \lim_{s\to0} \frac{\left(3s^2 - s + \tfrac12\right)(s+2)}{s^3+2s^2+4s+4} .$$ Every negative power of \(s\) has been absorbed by the \(s^2\) in \(1-T\), so the limit is just a substitution: $$\boxed{e_{ss} = \frac{(1/2)(2)}{4} = \frac14 \text{ — option (c)}}$$ Evaluating the same expression at \(s = 10^{-3}\) and \(s = 10^{-5}\) gives \(0.2503\) and \(0.2500\), confirming the limit numerically.
10. Item (8) — find the false statement. Options (a), (c) and (d) are all standard root-locus properties. Option (b) is false: the locus is symmetric about the real axis, because the characteristic polynomial has real coefficients and its complex roots therefore occur in conjugate pairs. There is no mechanism that would make it symmetric about the imaginary axis — indeed such symmetry would force a right-half-plane root for every left-half-plane one, so no stable loop could exist. Hence $$\boxed{\text{(b) — the locus is symmetric about the REAL axis}}$$
11. Item (9) — the \(K\to0\) endpoint. The characteristic equation \(1+KG(s)H(s)=0\) can be written \(\mathrm{den}(s) + K\,\mathrm{num}(s) = 0\). Setting \(K=0\) leaves \(\mathrm{den}(s)=0\), whose roots are precisely the open-loop poles. Hence $$\boxed{\text{(d) the roots coincide with the open-loop poles}}$$ The mirror fact — that \(K\to\infty\) drives \(n_z\) of the roots onto the open-loop zeros and the rest to infinity — is what makes poles the \(K=0\) markers and zeros the \(K=\infty\) markers on every locus plot.

[Figure not reproduced: Item 10: the printed locus redrawn. Five × markers (poles at \(0\), \(-1\), \(-2\) and \(-3\pm j2\)) and two ○ markers (zeros near \(-0.7\pm j1\)). See the official exam paper.]

12. Item (10) — count the × markers, not the branches you can see. The number of closed-loop poles equals the degree of the characteristic polynomial, which equals the number of open-loop poles (the denominator degree), because \(\mathrm{den}+K\,\mathrm{num}\) has the degree of \(\mathrm{den}\) whenever \(n_p>n_z\). The plot prints three poles on the real axis (\(0\), \(-1\), \(-2\)) and a complex pair at \(-3\pm j2\), so $$\boxed{5 \text{ closed-loop poles — option (a)}}$$ The two finite zeros do not change the count; they absorb two of the five branches, leaving \(n_p - n_z = 3\) to escape along asymptotes, which is what the plot shows.
13. Item (11) — assess the assertion and the reason separately. The assertion is true: a PID controller places a pole at the origin and two zeros, and the resulting closed-loop polynomial can have a right-half-plane root at high gain. The reason is also true: closing a PID loop around a second-order plant gives a cubic characteristic polynomial. And the reason explains the assertion, because every second-order polynomial with positive coefficients is stable, whereas a cubic \(s^3+a_2s^2+a_1s+a_0\) can have positive coefficients and still be unstable (it needs \(a_2a_1>a_0\)); raising the order from 2 to 3 is precisely what creates the possibility of instability. Hence $$\boxed{\text{(a) both correct, and R is the correct explanation of A}}$$
14. Item (12) — what integration measures. \(\int_0^t e(\tau)\,d\tau\) is the accumulated area between the error curve and the time axis, which is why integral action keeps building output while any error persists and therefore drives the steady-state error of a step to zero. Hence $$\boxed{\text{(b) the area under the curve}}$$
15. Item (13) — test all three statements. (1) is true: negative feedback divides the error by \(1+L\), which exceeds unity wherever the loop gain is large. (2) is true: \(|1/(1+L)|\) is below unity at low frequency and can exceed unity near the crossover region, so feedback reduces gain in the band where it acts and can amplify it elsewhere — the “water-bed” effect. (3) is true: adding the loop changes the characteristic polynomial, and enough gain around a plant with three or more poles produces right-half-plane roots. Hence $$\boxed{\text{all three — option (a)}}$$
16. Item (14) — phase margin from the crossover phase. By definition the phase margin is the additional phase lag that would put the loop on the verge of instability at the gain crossover frequency: $$\boxed{\mathrm{PM} = 180^\circ + \angle L(j\omega_{gc}) = 180^\circ - 105^\circ = 75^\circ \text{ — option (d)}}$$ Option (a), \(23^\circ\), and option (c), \(60^\circ\), are unrelated; the trap in items of this shape is to quote \(105^\circ\) itself, which is the phase, not the margin.
17. Item (15) — resonant peak versus damping. For a second-order system with \(\zeta<1/\sqrt2\) the closed-loop magnitude peaks at \(\omega_r = \omega_n\sqrt{1-2\zeta^2}\) with height $$M_r = \frac{1}{2\zeta\sqrt{1-\zeta^2}} .$$ \(M_r\) grows without bound as \(\zeta\to0\) and falls to unity at \(\zeta = 1/\sqrt2\), so the peak $$\boxed{\text{increases as the damping ratio decreases — option (a)}}$$ (For \(\zeta \ge 1/\sqrt2\) there is no peak at all, which is the boundary worth remembering.)
18. Item (16) — what the resonance frequency indicates. \(\omega_r\) sits just below the natural frequency \(\omega_n\) and scales with it, and \(\omega_n\) sets the time scale of the response (\(T_p = \pi/\omega_n\sqrt{1-\zeta^2}\), \(T_s \approx 4/\zeta\omega_n\)). A higher resonance frequency therefore means a faster system: $$\boxed{\text{(a) speed of response}}$$
19. Item (17) — the high-frequency asymptote crossing 0 dB. For \(G(s) = K/[s(Js+F)]\), the \(-40\) dB/decade asymptote is the one that holds well above the corner \(\omega = F/J\), where \(Js\) dominates \(F\) and $$|G(j\omega)| \approx \left|\frac{K}{J(j\omega)^2}\right| = \frac{K}{J\omega^2} .$$ Setting that to unity (0 dB) gives $$\boxed{\omega^2 = \frac{K}{J} \text{ — option (b)}}$$ A companion result worth knowing is that the low-frequency \(-20\) dB/decade asymptote crosses 0 dB at \(\omega = K/F\).
20. Item (18) — which plots display frequency response. A Bode pair plots \(|G(j\omega)|\) and \(\angle G(j\omega)\) against \(\log\omega\); a Nyquist diagram plots the same complex number \(G(j\omega)\) parametrically in the complex plane. Both are frequency-response plots, so $$\boxed{\text{(d) both (a) and (b)}}$$ Strictly, option (c) is also a frequency-response plot — a Nyquist diagram simply is the polar plot of the loop transfer function, extended to negative frequencies — so (b) and (c) name the same object and (d) is the most complete of the printed choices.