Question 1 of 5: Multiple choice — fifteen items

Question 1: Multiple choice — fifteen 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 fifteen items and the data printed with each, together with the mark weight shown in brackets on the paper:

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

Approach. Each item is decided by one standard result — impedance reflection through a gear train for item 1; block-diagram and signal-flow reduction for items 2 and 3; Routh–Hurwitz sign conditions for items 4 and 11; static error constants and system type for items 5 and 6; the second-order damping relations for item 7; the root-locus construction rules for items 8, 9, 10, 12 and 13; and the standard controller and Bode-asymptote definitions for items 14 and 15 — so the work below applies the governing relation item by item and boxes the letter.

[Figure not reproduced: Item 1: the rotational system as printed. The input shaft carries the applied torque \(T(t)\), an inertia of 3 kg·m² and a spring of 3 N·m/rad to ground; its gear \(N_1=5\) meshes with \(N_3=25\) (upper branch: 200 kg·m² and a 1000 N·m·s/rad damper to the wall) and. See the official exam paper.]

1. Item (1) — reflect every impedance onto the \(\theta_2\) shaft. A rotational impedance seen across a gear pair is multiplied by the square of the tooth ratio of the destination gear to the source gear, and an applied torque is multiplied by the ratio itself. Taking the \(N_2=50\) shaft (the one that carries \(\theta_2\)) as the destination: $$\left(\frac{N_2}{N_1}\right)^{\!2} = \left(\frac{50}{5}\right)^{\!2} = 100, \qquad \left(\frac{N_2}{N_3}\right)^{\!2} = \left(\frac{50}{25}\right)^{\!2} = 4 .$$ The input shaft contributes \(J_1s^2+K_1 = 3s^2+3\), which reflects to \(100(3s^2+3)=300s^2+300\); the damper branch contributes \(200s^2+1000s\), which reflects to \(800s^2+4000s\); and the \(\theta_2\) shaft contributes \(200s^2+250\) unchanged.
2. Item (1) (continued) — sum the reflected impedances and the reflected torque. Adding the three contributions, $$\big[(300+200+800)s^2 + 4000s + (300+250)\big]\theta_2(s) = \frac{N_2}{N_1}\,T(s) = 10\,T(s),$$ so $$\boxed{\frac{\theta_2(s)}{T(s)} = \frac{10}{1300s^2+4000s+550} \quad\text{— option (a)}}$$ The three coefficients each check independently: \(3(100)+200+200(4)=1300\), \(1000(4)=4000\) and \(3(100)+250=550\). Note that option (b) has an \(s\) in its numerator, which is dimensionally impossible for a displacement-per-torque transfer function, and options (c) and (d) reflect the damper branch with the wrong ratio.
3. Item (2) — write the closed-loop transfer function of the printed structure. Let \(p\) be the plant output and \(P(s)=(s-1)/(s+1)\). The first summer forms \(r - K_2 p\), so \(p = P(r-K_2p)\) and \(p = Pr/(1+K_2P)\); the second summer adds the feed-forward path, \(y = p + K_1 r\). Hence $$\frac{y}{r} = K_1 + \frac{P}{1+K_2P} = \frac{s(1+K_1+K_1K_2) + (-1+K_1-K_1K_2)}{s(1+K_2)+(1-K_2)} .$$

4. Item (2) (continued) — force the numerator to vanish at \(s=-1\). Substituting \(s=-1\) into the numerator above, $$-(1+K_1+K_1K_2) + (-1+K_1-K_1K_2) = -2 - 2K_1K_2 = 0 \quad\Longrightarrow\quad K_1K_2 = -1 .$$ The condition is on the product alone. Option (a) offers \(K_1=1,K_2=-1\) and \(K_1=-1,K_2=1\), both of which give \(K_1K_2=-1\); options (b) and (c) give a product of zero and (d) gives \(+1\). With \(K_1K_2=-1\) the numerator collapses to \(K_1(s+1)\), so the zero sits exactly at \(s=-1\) as required: $$\boxed{\text{(a) } K_1=1,\;K_2=-1 \text{ or } K_1=-1,\;K_2=1}$$
5. Item (3) — read the cascade off the graph. The graph is a plain cascade: three input branches with gains \(a\), \(b\), \(c\) converge on \(x_4\), then a single branch of gain \(d\) reaches \(x_5\) and a single branch of gain \(e\) reaches \(x_6\). There is no loop and no second forward path, so no Mason determinant is needed — multiply along the one path: $$x_4 = ax_1+bx_2+cx_3, \qquad x_5 = d\,x_4, \qquad \boxed{x_6 = e\,x_5 = de\,(ax_1+bx_2+cx_3) \text{ — option (a)}}$$ Options (b) and (c) add the cascade gains instead of multiplying them, which would be right for parallel branches, not for a series chain.

6. Item (4) — close the loop and apply Routh–Hurwitz. The loop gain is \(G(s)H(s)=K(s^2-2s+2)/(s^2+2s+4)\), so $$1+GH = 0 \;\Longrightarrow\; (s^2+2s+4) + K(s^2-2s+2) = (1+K)s^2 + (2-2K)s + (4+2K) = 0 .$$ For a second-order polynomial the Routh array reduces to the statement that all three coefficients share a sign, so the three conditions are \(1+K>0\), \(2-2K>0\) and \(4+2K>0\), i.e. \(K>-1\), \(K<1\) and \(K>-2\). Their intersection is $$\boxed{-1 < K < 1}$$ and under the paper’s own preamble (\(K>0\)) this narrows to \(0<K<1\).

7. Item (5) — static position error constant of a Type 1 loop. For a unity-feedback loop the step error is \(e_{ss}=1/(1+K_p)\) with \(K_p=\lim_{s\to0}G(s)\). A Type 1 loop carries one pole at the origin, so \(K_p=\infty\) and $$\boxed{e_{ss} = \frac{1}{1+\infty} = 0 \text{ — option (a)}}$$ Option (c) is the general formula rather than its value here, and the “second-order” qualifier is a distractor: the order does not enter, only the number of poles at the origin does.
8. Item (6) — velocity error constant, then solve for \(K\). A 20% error to a unit ramp means \(e_{ss}=0.20=1/K_v\), so \(K_v=5\). For \(G(s)=K(s+5)/[s(s+6)(s+7)(s+8)]\), $$K_v = \lim_{s\to0} sG(s) = \frac{5K}{6\cdot 7\cdot 8} = \frac{5K}{336} .$$ Setting this to 5 gives $$\boxed{K = \frac{5 \times 336}{5} = 336 \text{ — option (d)}}$$ Checking back: \(K_v = 5(336)/336 = 5\) and \(e_{ss}=1/5=0.20\) exactly.

9. Item (7) — normalise the denominator before reading \(\zeta\). The damping ratio belongs to the characteristic polynomial, so divide by the leading coefficient: $$Ks^2+s+6 = 0 \;\Longrightarrow\; s^2 + \frac{s}{K} + \frac{6}{K} = 0 \;\Longrightarrow\; \omega_n = \sqrt{6/K}, \qquad 2\zeta\omega_n = \frac{1}{K} .$$ Therefore $$\zeta = \frac{1/K}{2\sqrt{6/K}} = \frac{1}{2\sqrt{6K}} ,$$ and demanding \(\zeta = 0.5\) gives \(\sqrt{6K}=1\), i.e. $$\boxed{K = \tfrac16 \text{ — option (c)}}$$ The trap is to read \(2\zeta\omega_n = 1\) straight off the un-normalised polynomial; that gives \(\zeta = 1/(2\sqrt{6/K}) \cdot K\) and lands on option (d). Substituting back: \(\omega_n = \sqrt{36}=6\) rad/s and \(2(0.5)(6)=6 = 1/K\) with \(K=1/6\). ✓
10. Item (8) — asymptote centroid. The centroid is $$\sigma_a = \frac{\sum \text{finite poles} - \sum \text{finite zeros}}{n_p - n_z} .$$ Here the poles are \(-3\) and \(-5\) and the single zero is \(-6\), so \(n_p-n_z = 2-1 = 1\) and $$\boxed{\sigma_a = \frac{(-3-5)-(-6)}{1} = \frac{-2}{1} = -2 \text{ — option (a)}}$$ With one excess pole there is a single asymptote at \(180^\circ\), which is simply the negative real axis to the left of \(-6\); the centroid quoted above is the formula’s value and is what the item asks for. Option (b), \(-1\), is what one gets by wrongly including a pole at the origin — the plant here has none.
11. Item (9) — branch bookkeeping. A root locus has \(n_p\) branches, one starting at each open-loop pole; \(n_z\) of them terminate on the finite open-loop zeros and the remaining \(n_p-n_z\) run off to infinity along the asymptotes. Hence $$\boxed{n_p - n_z \text{ segments do not terminate on zeros — option (c)}}$$ For the plant of item 8 that number is \(2-1=1\), which is exactly the single asymptote counted there — the two items cross-check one another.
12. Item (10) — where an added zero pulls the locus. Adding a left-half-plane open-loop zero at \(-z\) subtracts an angle \(\arg(s+z)\) from the angle balance, which the locus must recover by moving the branches toward that zero, i.e. to the left. Since a branch sitting further left has a more negative real part, the closed-loop poles become better damped and the range of \(K\) over which the loop stays stable widens (this is exactly the mechanism of a PD or lead compensator). Hence $$\boxed{\text{(a) the left, and the system becomes more stable}}$$ Option (d) has the correct direction with the wrong consequence, which is the distractor this item is built around.
13. Item (11) — a single zero in the Routh first column. When one entry of the first column is zero but the rest of the row is not, the next row’s formation divides by zero. The standard remedy is to replace the zero by a small positive quantity \(\varepsilon\), complete the array, and take the limit \(\varepsilon\to0^+\), counting sign changes as they then appear. Hence $$\boxed{\text{(b) substitute a small positive number for the zero and complete the array}}$$ Option (a) is wrong — the criterion still works. (An entire row of zeros is a different case, handled with the auxiliary polynomial formed from the row above, and it signals symmetric roots about the origin.)

14. Item (12) — solve \(dK/ds = 0\) on the real axis. On the locus \(K = -s(s+5)(s^2+6s+17.76)\). Expanding, $$K(s) = -\left(s^4 + 11s^3 + 47.76s^2 + 88.8s\right),$$ so the breakaway condition \(dK/ds=0\) is $$4s^3 + 33s^2 + 95.52s + 88.8 = 0 .$$ This cubic has one real root and one complex pair; the real root is the breakaway point: $$\boxed{\sigma = -1.830, \qquad K = 58.76 \text{ — option (a)}}$$ Substituting \(\sigma=-1.8299\) back gives \(K = -(-1.8299)(3.1701)(17.76-10.9794+3.3485) = 58.759\). ✓ The open-loop poles are \(0\), \(-5\) and \(-3\pm j2.960\), and the breakaway lies on the real-axis segment between \(0\) and \(-5\), as it must.

[Figure not reproduced: Item 13: the locus as printed on the paper — two poles at \(-2\) and \(-4\) with a circular complex portion, all of it in the left half-plane. See the official exam paper.]

15. Item (13) — test the three statements against the sketch. Statement 1 is true: exactly two × markers are printed, so the open-loop system is second order and the locus has two branches. Statement 3 is true: the entire sketch lies strictly to the left of the imaginary axis, so both closed-loop roots have negative real parts for every gain shown and the loop is stable for all \(K>0\). Statement 2 is false: the circular portion off the real axis is precisely where the two closed-loop roots form a complex conjugate pair, i.e. where the response is underdamped; the loop is overdamped only on the real-axis portion, which corresponds to small gain, and nothing on the sketch calibrates the gain scale so that “\(K>1\)” could be checked at all. Hence $$\boxed{\text{statements 1 and 3 — option (b)}}$$

16. Item (14) — the standard PID form. A PID controller sums a proportional term, an integral term whose strength is set by the integral time \(T_i\), and a derivative term whose strength is set by the derivative time \(T_d\): $$\boxed{G(s) = K\!\left[1 + \frac{1}{T_i s} + T_d s\right] \text{ — option (a)}}$$ The integral action must appear as \(1/(T_is)\) (a pole at the origin, so that a constant error integrates without bound) and the derivative action as \(T_ds\) (a zero at the origin). Options (b) and (d) invert one of the two, and (c) inverts both, which would give two poles at the origin and no derivative action at all.
17. Item (15) — asymptotic slope of a complex pole pair. Well above the corner frequency a quadratic factor \(1/(s^2+2\zeta\omega_ns+\omega_n^2)\) behaves as \(1/s^2\), whose magnitude falls by a factor of 100 — that is 40 dB — per decade. Per octave the same slope is $$-40\log_{10} 2 = -12.04 \approx -12 \text{ dB/octave},$$ so $$\boxed{-12 \text{ dB/octave — option (d)}}$$ Options (b) and (c) are the same slope written two ways for a single real pole (\(-20\) dB/decade \(= -6\) dB/octave), which is the intended trap: the item asks for the pair, so the slope doubles.