Question 1 of 5: Nineteen short-answer items on error constants, stability, the root locus and frequency response

Question 1: Nineteen short-answer items on error constants, stability, the root locus and frequency response (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. Nineteen independent short-answer items, each supplying its own loop transfer function, sketch or design specification; the mark weight printed after each item is [1] or [2] and the nineteen weights total 25.

Find. The single best option for every item, together with the calculation or rule that forces it.

Approach. Six ideas cover all nineteen items: the static error constants and system type; the characteristic polynomial and the Routh test; the three admissibility rules of a root-locus sketch; the constant-damping ray as a design point; the angle and magnitude conditions of the locus; and the standard first- and second-order frequency-response asymptotes.

Items 1 and 2 — system type and the error constants. For a unity-feedback loop the error transform is $E(s)=R(s)/[1+G(s)]$, and the final-value theorem turns each input power of $t$ into one static error constant: $K_{p}=\lim_{s\to0}G$ for a step, $K_{v}=\lim_{s\to0}sG$ for a ramp, and $K_{a}=\lim_{s\to0}s^{2}G$ for a parabola. The number of free integrators in $G$ — the system type — decides which of these is finite, and the ramp and parabola components of a composite command are handled independently by superposition.

Items 3 and 4 — the characteristic polynomial does all the work. Both items close a given plant with unity feedback and ask a question about the resulting closed-loop poles, so both are answered from $1+G_{c}(s)G(s)=0$ written as a polynomial. For a second-order polynomial $s^{2}+2\zeta\omega_{n}s+\omega_{n}^{2}$ the critically damped case is exactly the double-root case, i.e. a vanishing discriminant; and for the same polynomial the Routh test collapses to the statement that every coefficient must be positive.

Item 5 — three tests reject three sketches. A drawing can be a root locus only if (i) the whole picture is symmetric about the real axis, because the closed-loop polynomial has real coefficients; (ii) a point on the real axis belongs to the locus only when the number of real poles and real zeros strictly to its right is odd; and (iii) exactly $n-m$ branches escape to infinity along asymptotes, where $n$ and $m$ count the finite poles and zeros. These three tests are enough here: no gain values, and no arithmetic, are required.

Items 6 to 9 — one design point, four questions. These four items share the plant $k/[(s+1)(s+2)(s+3)]$. A percent overshoot specification fixes the damping ratio through $\%OS=100\,e^{-\pi\zeta/\sqrt{1-\zeta^{2}}}$, the damping ratio fixes a radial line $\theta=\cos^{-1}\zeta$ in the $s$-plane, and the intersection of that line with the locus is the design point. Once the point is known the gain follows from the magnitude condition and the two time specifications follow from the point's real and imaginary parts alone.

Items 10 to 14 — the bookkeeping rules of the locus. Writing $G=K\,b(s)/a(s)$ gives the closed-loop transfer function $T=K\,b(s)/[a(s)+K\,b(s)]$: the numerator is the plant numerator times a scalar, so the closed-loop zeros never move, whereas the denominator changes with $K$ and the poles trace the locus. The remaining three items are the standard asymptote formulas — $n-m$ asymptotes at angles $(2q+1)180^\circ/(n-m)$ leaving the real axis at $\sigma_{a}=\left(\sum\text{poles}-\sum\text{zeros}\right)/(n-m)$ — plus the observation that a strictly proper transfer function has $n-m$ zeros at infinity.

Items 15 to 17 — specifications, and the angle condition run backwards. Item 15 turns two transient specifications into two inequalities on the single free gain and asks whether the resulting window is empty. Items 16 and 17 use the angle condition in reverse: the sum of the angles from all open-loop zeros minus the sum from all open-loop poles must be an odd multiple of $180^\circ$ at a point on the locus, so the contribution the compensator zero must supply is fixed by arithmetic, and its location then follows from simple trigonometry.

Items 18 and 19 — two standard frequency-response facts. A lightly damped second-order pole pair produces a magnitude peak $M_{r}=1/(2\zeta\sqrt{1-\zeta^{2}})$ near $\omega_{n}$, so the peak grows without bound as the damping falls; and a first-order pole $1/(1+j\omega\tau)$ has phase exactly $-45^\circ$ at its break frequency $\omega=1/\tau$, which is why the frequency axis of a normalised phase template is scaled by the break frequency.

1. Item 1 — identify the type, then use only the surviving error term. $G$ has a double pole at the origin, so the loop is Type 2 and both the constant and the ramp parts of $r(t)$ are tracked exactly. Writing $0.5t^{2}=t^{2}/2$ shows the third term is the unit parabola, whose transform is $1/s^{3}$, so $$K_{a}=\lim_{s\to0}s^{2}G(s)=\frac{10\times1}{5}=2 \qquad\Longrightarrow\qquad \boxed{e_{ss}=\frac{1}{K_{a}}=0.5}$$ Answer: option (a).
2. Item 2 — read the definition backwards. A finite non-zero error to a parabolic command requires $K_{a}=\lim_{s\to0}s^{2}G(s)$ to be finite and non-zero, which happens only when $G$ carries exactly two free integrators. That is a $\boxed{\text{Type 2 system}}$; Type 0 and Type 1 give infinite error and Type 3 gives zero error. Answer: option (c).
3. Item 3 — set the discriminant of the closed-loop polynomial to zero. Unity feedback around $k/[(s+1)(s+5)]$ gives $$1+\frac{k}{(s+1)(s+5)}=0\;\Longrightarrow\;s^{2}+6s+(5+k)=0 .$$ Critical damping is the double-root case, so $6^{2}=4(5+k)$ and $\boxed{k=4}$, placing both closed-loop poles at $s=-3$ ($\zeta=1$, $\omega_{n}=3$ rad/s). Answer: option (b).
4. Item 4 — positive coefficients are necessary and sufficient at second order. $$1+K(s+1)\frac{1}{(s+2)(s-1)}=0\;\Longrightarrow\;s^{2}+(1+K)s+(K-2)=0$$ Both coefficients must be positive, and $K-2>0$ is the binding condition, so the loop is stable exactly for $\boxed{K>2}$. Hence $K=1.10$ is unstable, because the constant term $K-2=-0.9$ is negative and puts one pole at $s=+0.365$; and $K=3$ is stable, the polynomial $s^{2}+4s+1$ having roots $-0.268$ and $-3.732$. Answer: option (c).
5. Item 5 — eliminate, do not construct. Sketch (d) shows a single complex pole with no conjugate partner, which no real-coefficient polynomial can produce. Sketches (b) and (c) both draw a heavy locus ray running left from the leftmost real critical point, where the number of real poles and zeros to the right is even — four in each case — so that ray cannot be part of a locus. Sketch (a) passes all three tests: the picture is conjugate-symmetric, the only real segment drawn is the one between the two real zeros (one critical point to its right, an odd count), and with $n=m=2$ there are no asymptotes, so both branches must terminate on the finite zeros exactly as drawn. Reading the sketch at its drawn scale (poles near $\pm j0.9$, zeros near $-2$ and $-0.7$) the two branches break into the real axis at $s=-1.145$ with $K=5.575$, which is the break-in the figure shows. Answer: option (a).
6. Item 6 — the overshoot specification becomes a ray. Inverting the overshoot relation, $$\zeta=\frac{-\ln(0.20)}{\sqrt{\pi^{2}+\ln^{2}(0.20)}}=0.4560 ,$$ so the design ray leaves the origin at $\cos^{-1}0.4560=62.87^\circ$. Marching along that ray until the three pole vectors sum to $180^\circ$ gives $\boxed{s=-0.866\pm1.690j}$ (the three pole vectors subtend $85.47^\circ+56.15^\circ+38.39^\circ=180.00^\circ$). Answer: option (c).
7. Item 7 — magnitude condition at the design point. $$K=\bigl|(s+1)(s+2)(s+3)\bigr|_{s=-0.866+1.690j} =1.6958\times2.0356\times2.7224=\boxed{9.397}$$ (Using the rounded option coordinates $-0.86\pm1.69j$ instead gives 9.427, so the offered value 9.398 is the intended one.) Answer: option (b).
8. Item 8 — peak time depends only on the imaginary part. $$T_{p}=\frac{\pi}{\omega_{d}}=\frac{\pi}{1.690}=\boxed{1.86\ \text{s}}$$ Answer: option (c).
9. Item 9 — settling time depends only on the real part. Using the usual 2% criterion, $$T_{s}=\frac{4}{|\sigma|}=\frac{4}{0.866}=\boxed{4.62\ \text{s}}$$ Answer: option (c).
10. Item 10 — separate numerator from denominator. With $G=K\,b(s)/a(s)$, $$T(s)=\frac{K\,b(s)}{a(s)+K\,b(s)} .$$ Multiplying the numerator by $K$ cannot move its roots, so the closed-loop zeros are the open-loop zeros for every gain; the denominator, however, changes with $K$ and its roots sweep the root locus. $\boxed{\text{Only statement 2 is true.}}$ Answer: option (c).
11. Item 11 — count the degrees. The numerator $(s^{2}-1)(s+3)$ is cubic and the denominator $s^{3}+2s^{2}+3s$ is cubic, so $n=m=3$ and $$n-m=\boxed{0\ \text{asymptotes}} .$$ Every branch terminates on one of the finite zeros $+1$, $-1$, $-3$; nothing escapes to infinity. Answer: option (a).
12. Item 12 — asymptote angles. For $k/[s(s+1)(s+2)]$, $n=3$ and $m=0$, so three asymptotes leave at $$\theta_{q}=\frac{(2q+1)180^\circ}{n-m}=60^\circ,\ 180^\circ,\ 300^\circ \equiv\boxed{60^\circ,\ -60^\circ,\ 180^\circ} .$$ Answer: option (c).
13. Item 13 — asymptote intercept for the same plant. $$\sigma_{a}=\frac{\sum\text{poles}-\sum\text{zeros}}{n-m} =\frac{(0-1-2)-0}{3}=\boxed{-1}$$ Answer: option (b).
14. Item 14 — zeros at infinity. A constant numerator means $m=0$ and a second-order denominator means $n=2$, so the transfer function magnitude falls like $1/\omega^{2}$ and there are $n-m=\boxed{2\ \text{zeros at infinity}}$. There is no finite zero, and none at the origin. Answer: option (d).
15. Item 15 — turn both specifications into bounds on the same gain. Closing $K/[s(s+5)]$ gives $s^{2}+5s+K$, so $\omega_{n}=\sqrt{K}$, $\zeta=2.5/\sqrt{K}$ and $\omega_{d}=\sqrt{K-6.25}$. The overshoot limit needs $\zeta\ge0.5912$, hence $K\le(2.5/0.5912)^{2}=17.89$; the peak-time limit needs $\omega_{d}\ge\pi$, hence $K\ge6.25+\pi^{2}=16.12$. The window $$\boxed{16.12\le K\le17.89}$$ is non-empty, so both specifications hold simultaneously — at $K=17$, for instance, $T_{p}=0.958$ s and $PO=9.12\%$. Answer: option (a).
16. Item 16 — the angle condition fixes the required contribution. At a point on the locus the zero angles minus the pole angles must equal an odd multiple of $180^\circ$: $$(60^\circ+\theta_{c})-(22.5^\circ+45^\circ)=\pm180^\circ .$$ The $+180^\circ$ branch demands $\theta_{c}=187.5^\circ$, which no real zero can supply because a real zero seen from an upper-half-plane point subtends an angle strictly between $0^\circ$ and $180^\circ$. The $-180^\circ$ branch gives $\theta_{c}=-172.5^\circ$, i.e. a magnitude of $\boxed{172.5^\circ}$. Answer: option (d).
17. Item 17 — convert the angle into a location. A vector from $z_{c}$ to $s=-1+j$ subtending $172.5^\circ$ points almost due left, so the zero lies to the right of the design point and the supplementary angle is $180^\circ-172.5^\circ=7.5^\circ$: $$\tan 7.5^\circ=\frac{1}{z_{c}+1} \;\Longrightarrow\; z_{c}=\frac{1}{\tan7.5^\circ}-1=\boxed{+6.60}$$ and a back-check gives $\arg(-1+j-6.596)=172.5^\circ$ exactly. Answer: option (d). Each offered location pairs with exactly one offered angle ($-1.52\leftrightarrow62.5^\circ$, $-0.96\leftrightarrow92.4^\circ$, $-0.08\leftrightarrow132.6^\circ$, $+6.59\leftrightarrow172.5^\circ$), which confirms the pairing of items 16 and 17.
18. Item 18 — the resonant peak against damping. For a standard second-order pair the peak magnitude is $$M_{r}=\frac{1}{2\zeta\sqrt{1-\zeta^{2}}},\qquad \zeta<\tfrac{1}{\sqrt2},$$ which is a decreasing function of $\zeta$ — 8.14 dB at $\zeta=0.2$ against 2.70 dB at $\zeta=0.4$. So the peak $\boxed{\text{increases as the damping ratio decreases}}$. Answer: option (a).
19. Item 19 — why the frequency axis is normalised. A first-order pole contributes $\arg[1/(1+j\omega\tau)]=-\tan^{-1}(\omega\tau)$, which is $$-\tan^{-1}(1)=\boxed{-45^\circ}\quad\text{at }\omega=1/\tau .$$ Dividing the frequency axis by the break frequency therefore places the $-45^\circ$ point of every first-order pole at $\omega=1$ rad/s, which is what makes one phase template reusable. Answer: option (a).

Two items deserve a comment beyond their option letters. Item 4's stability boundary is set by the constant term, not by the $s$ coefficient, because the plant already contributes a right-half-plane pole at $s=+1$: proportional-plus-zero control has to overcome that pole before it can do anything else, which is why the smaller of the two offered gains fails. Item 17 lands on a right-half-plane compensator zero, an unusual answer for a PD design, and it does so because the $+180^\circ$ branch of the angle condition is unreachable for the pole and zero angles the item stipulates.