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. The seventeen items and their data, with the printed mark weights:

Find. The single best option for each of the seventeen items, with the calculation or the structural argument that settles it.

Approach. Each item is decided by one standard result — system type and static error constants for 1, 7 and 17; the second-order pole/damping relations for 2, 3, 4 and 13–16; Routh–Hurwitz sign counting for 5, 6 and 11; and the asymptotic Bode rules for 9 and 10 — so the work below applies the governing relation item by item and boxes the letter.

1. Item (a) — system type for a finite parabolic error. For a unity-feedback loop the steady-state error to \(r(t)=\tfrac12 A t^2\) is \(e_{ss}=A/K_a\) with the acceleration error constant \(K_a=\lim_{s\to 0} s^2 G(s)\). \(K_a\) is finite and non-zero only when \(G(s)\) carries exactly two poles at the origin, which is the definition of a Type 2 loop: a Type 0 or Type 1 loop gives \(K_a=0\) and an infinite error, and a Type 3 loop gives \(K_a=\infty\) and zero error. Hence \(\boxed{\text{(c) Type 2 system}}\).
2. Item (b) — gain for a critically damped second-order loop. The loop is unity feedback around \(k/[(s+1)(s+5)]\):

3. Item (b) (continued) — close the loop and set the discriminant to zero. The characteristic polynomial is $$(s+1)(s+5)+k = s^2 + 6s + (5+k).$$ Critical damping means a repeated real root, i.e. a vanishing discriminant: $$6^2 - 4(5+k) = 0 \quad\Longrightarrow\quad 36 = 20 + 4k \quad\Longrightarrow\quad \boxed{k = 4 \text{ — option (b)}}$$ The repeated root is then \(s = -3\) (twice), which is in the open left half-plane, so the loop is stable as the item requires. Equivalently \(\omega_n=\sqrt{5+k}=3\) and \(\zeta = 6/(2\omega_n) = 1\).
4. Item (c) — what the real part of a pole contributes. A conjugate pair \(s = -\sigma \pm j\omega_d\) produces the time function \(e^{-\sigma t}\!\left(C\cos\omega_d t + D\sin\omega_d t\right)\). The real part \(\sigma\) appears only in the exponential envelope, whose time constant is \(1/\sigma\); the imaginary part \(\omega_d\) sets the radian frequency and the residues set the amplitude. So the real part generates the \(\boxed{\text{(a) time constant of an exponential response}}\). This is the same decomposition used in the inverse-Laplace table appended to the paper.
5. Item (d) — rise time of \(T(s)=16/(s^2+3s+16)\). Matching the standard second-order form \(\omega_n^2/(s^2+2\zeta\omega_n s + \omega_n^2)\): $$\omega_n = \sqrt{16} = 4\ \text{rad/s}, \qquad \zeta = \frac{3}{2\omega_n} = \frac{3}{8} = 0.375 .$$ Rise time has no closed form, so use Nise’s normalised curve fit for the 0–100% rise time of an underdamped second-order step response, $$\omega_n T_r = 1.76\zeta^3 - 0.417\zeta^2 + 1.039\zeta + 1 .$$ Substituting \(\zeta = 0.375\) gives \(\omega_n T_r = 0.09281 - 0.05864 + 0.38963 + 1 = 1.4238\), so $$\boxed{T_r = \frac{1.4238}{4} = 0.356\ \text{s — option (a)}}$$ The distractor 2.33 s is roughly the settling time \(4/(\zeta\omega_n)=2.67\) s, which is what a candidate who confuses the two relations would report.
6. Item (e) — reading pole locations from the printed Routh array. The first two rows carry the coefficients of the original quintic, alternating: \(s^5:[1,\,4,\,3]\) and \(s^4:[-1,\,-4,\,-2]\), so the polynomial is $$P(s) = s^5 - s^4 + 4s^3 - 4s^2 + 3s - 2 .$$ The \(s^3\) first-column entry vanished and was replaced by the small quantity \(\varepsilon\) (the standard epsilon method for a zero in the first column with a non-zero row). Counting sign changes down the first column with \(\varepsilon\to 0^{+}\):

7. Item (e) (continued) — count the sign changes. The sequence \(+,-,+,+,+,-\) contains three sign changes, so three roots lie in the right half-plane and the remaining two lie in the left half-plane. Repeating the count with \(\varepsilon\to 0^{-}\) gives \(+,-,-,-,+,-\), again three changes, which is the consistency check the epsilon method requires. The \(s^0\) entry is non-zero and no row of zeros occurred, so there are no roots on the imaginary axis. Numerically the roots of \(P(s)\) are \(0.6180\pm j1.6180\), \(0.3820\), \(-0.5000\pm j1.3229\) — three in the right half-plane, two in the left — confirming $$\boxed{\text{(b) 3 rhp, 2 lhp}}$$
8. Item (f) — range of \(K\) for \(G(s)=K(s+6)/[s(s+1)(s+4)]\). Closing the unity-feedback loop, $$T(s) = \frac{K(s+6)}{s(s+1)(s+4) + K(s+6)} = \frac{K(s+6)}{s^3 + 5s^2 + (4+K)s + 6K}.$$ For a cubic \(s^3+a_2s^2+a_1s+a_0\) the Routh conditions collapse to the familiar pair \(a_0>0\) and \(a_2a_1>a_0\) — which is what “without forming the Routh table” is asking for. Here $$6K > 0 \;\Longrightarrow\; K > 0, \qquad 5(4+K) > 6K \;\Longrightarrow\; 20 + 5K > 6K \;\Longrightarrow\; K < 20,$$ so the closed loop is stable exactly for $$\boxed{0 < K < 20}$$ At \(K = 20\) the loop is marginally stable with an imaginary-axis pair at \(s = \pm j\sqrt{4+K} = \pm j4.899\). Spot checks confirm the interval: \(K = 19.5\) gives all roots in the left half-plane, \(K = 20.5\) gives a right-half-plane pair.
9. Item (f) (continued) — none of the printed options is this range. The four choices offered are \(K>-2\), \(0<K<1\), \(K>-1\) and \(1<K<2\); none of them is \(0<K<20\). The option set is the one that belongs to a second-order plant — for example \(K/[(s+1)(s+2)]\) has the characteristic polynomial \(s^2+3s+(2+K)\) and is stable precisely for \(K>-2\), which is choice (a), and \(K/(s+1)^2\) gives \(K>-1\), which is choice (c). That is also the only kind of system for which the instruction “without forming the Routh table” is literally true, since a quadratic needs only all-positive coefficients. The item as printed therefore appears to pair a third-order plant with a second-order answer key. If a letter must be entered, (b) \(0<K<1\) is the only choice every member of which is genuinely a stabilising gain, and its lower limit \(K=0\) is the true lower limit; it is nonetheless a proper subset of the real answer, not the answer.
10. Item (g) — gain for a specified parabolic error. The input is \(r(t)=10t^2u(t)\), so \(R(s) = 10\cdot 2/s^3 = 20/s^3\). Writing the parabolic input as \(A/s^3\) gives \(A = 20\). The plant \(G(s)=K(s+3)/[s^2(s+7)]\) is Type 2, so the acceleration error constant is $$K_a = \lim_{s\to 0} s^2 G(s) = \lim_{s\to 0}\frac{K(s+3)}{s+7} = \frac{3K}{7},$$ and the steady-state error is \(e_{ss} = A/K_a = 20\cdot 7/(3K) = 140/(3K)\). Setting this equal to the required 0.061, $$\boxed{K = \frac{140}{3(0.061)} = \frac{140}{0.183} = 765.0\ \text{— option (d)}}$$ Back-substituting, \(K_a = 3(765.03)/7 = 327.87\) and \(e_{ss}=20/327.87 = 0.0610\), as demanded. The trap is dropping the factor 2 from \(\mathcal{L}\{t^2\}=2/s^3\), which halves \(A\) and returns roughly 382 — not an offered option, which is itself a hint that the factor is needed. Because 765.03 is printed exactly as choice (d), the fifth option (e) “None of the above” is not required for this item; it is present only as a guard against the three arithmetic distractors (a), (b) and (c), none of which any consistent reading of the question produces.
11. Item (h) — is a pole at \(-3.001\) cancelled by a zero at \(-3.0\)? The residue of a closed-loop pole \(p\) in \(T(s)=N(s)/D(s)\) is proportional to \(N(p)\), and a zero distant \(\delta\) from the pole makes that residue proportional to \(\delta\). Here \(\delta = 0.001\), some three orders of magnitude smaller than the spacing to any other singularity, so the residue of the \(-3.001\) mode is negligible and the pair behaves as a cancellation for every practical purpose. Nise’s working rule is that a pole and zero within about a factor of 0.1 of the neighbouring pole spacing may be treated as cancelling; 0.001 is far inside that. Hence \(\boxed{\text{(a) Yes}}\).
12. Item (i) — total Bode magnitude slope of four poles. Each simple pole contributes an asymptotic slope of \(-20\) dB/decade above its corner frequency, because \(|1/(1+j\omega/p)| \to p/\omega\) and \(20\log_{10}(1/\omega)\) falls 20 dB per decade. Four poles therefore contribute $$4 \times (-20\ \text{dB/dec}) = \boxed{-80\ \text{dB/dec — option (d)}}$$ The sign matters: poles subtract magnitude, zeros add it, so the \(+80\) distractor is the four-zero answer.
13. Item (j) — effect of a pure delay on the Bode magnitude plot. A transport delay has the frequency response \(e^{-j\omega T}\), whose magnitude is $$\left|e^{-j\omega T}\right| = 1 \quad\text{for every }\omega ,$$ i.e. \(0\) dB at all frequencies: a delay is all-pass. Its entire effect is on phase, a lag of \(\omega T\) radians growing without bound. So the magnitude plot shows \(\boxed{\text{(d) No change at all}}\). Options (a) and (b) describe the phase curve, not the magnitude curve. This all-pass property is what makes the delay questions elsewhere in the B2 series tractable: the gain crossover keeps its delay-free value while only the phase crossover moves.
14. Item (k) — stability of \(G_c=K(s+1)\) around \(G=1/[(s+2)(s-1)]\). The plant already has a right-half-plane pole at \(s=+1\). Closing the loop, $$(s+2)(s-1) + K(s+1) = s^2 + s - 2 + Ks + K = s^2 + (1+K)s + (K-2).$$ For a quadratic all coefficients must be positive, so \(1+K>0\) and \(K-2>0\), i.e. \(K>2\). Therefore \(K = 1.10\) leaves the loop unstable (roots \(-3.766\) and \(+0.266\), one still in the right half-plane) while \(K = 3\) makes it stable (roots \(-3.303\) and \(-0.697\)): $$\boxed{\text{(c) Unstable for } K=1.10 \text{ and stable for } K=3}$$ The structural point is that gain alone can pull a single unstable pole into the left half-plane here, because the compensator zero at \(-1\) bends the locus leftward.
15. Item (l) — which sketch can be a root locus. Test each candidate against the two rules that any locus must satisfy: the number of branches running to infinity is \(n-m\) (poles minus zeros) and they leave along asymptotes at \((2k+1)180^\circ/(n-m)\); and a real-axis point lies on the locus only if the number of real poles and zeros strictly to its right is odd.
    • (b) shows three poles and one zero, so \(n-m=2\) and the two infinite branches must leave at \(\pm 90^\circ\) — yet the sketch runs a real-axis branch off to \(-\infty\), and it puts that branch to the left of the leftmost zero, where the odd-count rule fails. Rejected twice over.
    • (c) shows a double pole at the origin and two real zeros, so \(n=m=2\) and no branch may reach infinity; the sketch nevertheless draws the real-axis locus running to \(-\infty\), and it leaves the segment between the two zeros — the one stretch that does satisfy the odd-count rule — off the locus. Rejected.
    • (d) shows a single unpaired complex pole with a vertical locus segment. A real rational transfer function has conjugate-symmetric poles and a conjugate-symmetric locus, so no such diagram can arise. Rejected.
    • (a) shows a conjugate pole pair on the \(j\omega\) axis and two real zeros, so \(n=m=2\): no infinite branch, which the sketch respects. The two branches leave the imaginary axis, curve left, meet on the real axis at a break-in point between the two zeros, and then split along that segment into the zeros — and the segment between the zeros is exactly where the odd-count rule puts the real-axis locus. Every rule is satisfied.
    Hence \(\boxed{\text{(a)}}\). Redrawing (a) from an actual transfer function of that pole–zero pattern confirms the shape:

16. Items (m)–(p) — damping ratio demanded by 20% overshoot. Percent overshoot fixes \(\zeta\) through $$\%OS = 100\,e^{-\pi\zeta/\sqrt{1-\zeta^2}} \quad\Longleftrightarrow\quad \zeta = \frac{-\ln(\%OS/100)}{\sqrt{\pi^2 + \ln^2(\%OS/100)}} .$$ With \(\%OS = 20\), \(\ln 0.20 = -1.6094\) and $$\zeta = \frac{1.6094}{\sqrt{9.8696 + 2.5903}} = \frac{1.6094}{3.5299} = 0.4559 ,$$ so the design point lies on the ray at \(\theta = \cos^{-1}(0.4559) = 62.87^\circ\) from the negative real axis.
17. Item (m) — find the point where that ray meets the locus. Test the four offered coordinates against \(\zeta = -\sigma/|s|\): \(-1.25\pm 0.8j\) gives \(\zeta = 1.25/1.484 = 0.842\); \(-1\pm 1.25j\) gives \(0.625\); \(-0.5\pm 2.5j\) gives \(0.196\); and $$-0.86 \pm 1.69j:\quad |s| = \sqrt{0.86^2+1.69^2} = 1.8965,\qquad \zeta = \frac{0.86}{1.8965} = 0.4535 \approx 0.4559 .$$ Only the third matches. Confirming with the angle condition for \(K/[(s+1)(s+2)(s+3)]\) at \(s=-0.86+1.69j\), the pole angles are \(85.28^\circ\), \(54.03^\circ\) and \(40.70^\circ\), summing to \(180.01^\circ\) — the point is on the locus to within the rounding of the printed coordinates. Hence $$\boxed{s = -0.86 \pm 1.69j \text{ — option (c)}}$$
18. Item (n) — gain at that point. The magnitude condition \(|K G(s)| = 1\) gives \(K\) as the product of the distances from the poles to the design point: $$K = |s+1|\,|s+2|\,|s+3| = |0.14+1.69j|\;|1.14+1.69j|\;|2.14+1.69j| = 1.6958 \times 2.0383 \times 2.7268 ,$$ $$\boxed{K = 9.398 \text{ — option (b)}}$$ Cross-check with the coefficient relations for \(s^3+6s^2+11s+(6+K)\): the root sum must be \(-6\), so the third closed-loop pole is at \(-6-2(-0.86) = -4.28\), and the root product must be \(-(6+K)\), giving \((0.86^2+1.69^2)(4.28) = 3.5957 \times 4.28 = 15.39 = 6+K\), i.e. \(K = 9.39\). The two routes agree.
19. Item (o) — peak time. Peak time depends only on the damped natural frequency, i.e. on the imaginary part of the dominant pair: $$T_p = \frac{\pi}{\omega_d} = \frac{\pi}{1.69} = \boxed{1.86\ \text{s — option (c)}}$$
20. Item (p) — settling time. The 2% settling time depends only on the real part: $$T_s = \frac{4}{\zeta\omega_n} = \frac{4}{|\sigma|} = \frac{4}{0.86} = \boxed{4.65\ \text{s — option (c), 4.6}}$$ The third closed-loop pole at \(-4.28\) is five times faster than the dominant pair, which is what licenses treating the response as second order for items o and p. The locus and the design point are shown below.

21. Item (q) — what a forward-path gain moves. With unity feedback, \(T(s) = KG(s)/[1+KG(s)]\). Writing \(G = N/D\) gives \(T = KN/(D+KN)\): the numerator of \(T\) is \(KN\), so the closed-loop zeros are the open-loop zeros for every \(K\) — scaling the numerator by \(K\) does not move its roots. The denominator \(D+KN\) is precisely the root-locus polynomial, so the closed-loop poles move with \(K\); that is what a root locus is. Statement 1 is false and statement 2 is true: $$\boxed{\text{(c) Only 2 is true}}$$

Collecting the seventeen decisions: