22-Mec-A3 System Analysis and Control · May 2016

Question 3 of 6: Routh Stability Range and Root Locus for System 1.0

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 07-Mec-A3 System Analysis and Control, National Exams May 2016 — 3 hours, closed book (Casio or Sharp approved calculator; semi-log graph paper is the only permitted aid). The rubric states that any four (4) questions constitute a complete paper and that all questions are of equal value, so each question is worth 25 marks of the 100 marked. Because this document is a study resource rather than an exam script, all six questions are solved in full.

Reference texts. K. Ogata, Modern Control Engineering, 5th ed. (Pearson) — Ch. 3–7; N. S. Nise, Control Systems Engineering, 8th ed. (Wiley) — Ch. 4, 6, 8, 10; R. C. Dorf and R. H. Bishop, Modern Control Systems, 13th ed. (Pearson) — Ch. 5–8; G. F. Franklin, J. D. Powell and A. Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed. (Pearson) — Ch. 3–6. A table of Laplace transform pairs is appended to the examination paper (pages 3–4 of the original), together with two sheets of 3-cycle semi-logarithmic graph paper for the Bode question.

Question 3: Routh Stability Range and Root Locus for System 1.0 (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 block diagram of System 1.0 (reproduced below from page 2 of the paper):

[Figure not reproduced: Figure 3.1 — System 1.0 as printed on the examination paper: forward path $K\,(s+3)/[s(s^{2}+4s+5)]$, negative feedback through $1/(s+1)$. See the official exam paper.]

Element	Transfer function
Forward gain	$K$ (real, adjustable)
Plant	$G_{p}(s)=\dfrac{s+3}{s\,(s^{2}+4s+5)}$
Feedback	$H(s)=\dfrac{1}{s+1}$ (non-unity, first-order lag)
Loop gain	$L(s)=K\,G_{p}(s)H(s)=\dfrac{K(s+3)}{s\,(s+1)(s^{2}+4s+5)}$

Find. (a) the complete range of $K$ giving a stable closed loop, by the Routh–Hurwitz criterion; (b) the root locus versus $K$, annotated with departure and arrival angles, the breakaway and break-in points with their gains, and the imaginary-axis crossings with their gain and frequency.

Approach. Form the characteristic polynomial $1+L(s)=0$, build the Routh array symbolically in $K$ and require every first-column entry to be positive; then construct the locus from the standard rules (real-axis segments, asymptote centroid and angles, $dK/ds=0$ for breakaway/break-in, the angle criterion for departure, and the Routh auxiliary polynomial for the axis crossing).

Part (a) — Routh–Hurwitz

Form the characteristic polynomial. Setting $1+L(s)=0$ and clearing denominators,$$s(s+1)(s^{2}+4s+5)+K(s+3)=0$$Expanding the product $s(s+1)(s^{2}+4s+5)=(s^{2}+s)(s^{2}+4s+5)=s^{4}+5s^{3}+9s^{2}+5s$ gives$$\Delta(s)=s^{4}+5s^{3}+9s^{2}+(5+K)s+3K=0$$A necessary condition is immediate: every coefficient must be positive, so $K>0$ at the very least.
Build the Routh array in terms of $K$.$$\begin{array}{c|ccc}s^{4}&1&9&3K\\ s^{3}&5&5+K&0\\ s^{2}&b_{1}&3K&\\ s^{1}&c_{1}&&\\ s^{0}&3K&&\end{array}$$with$$b_{1}=\frac{5(9)-1(5+K)}{5}=\frac{40-K}{5},\qquad c_{1}=\frac{b_{1}(5+K)-5(3K)}{b_{1}}=(5+K)-\frac{75K}{40-K}$$
Impose positivity of the whole first column. The $s^{0}$ row requires $3K>0$, i.e. $K>0$. The $s^{2}$ row requires $40-K>0$. The binding condition is the $s^{1}$ row; multiplying through by the positive quantity $(40-K)$,$$(5+K)(40-K)-75K>0\;\Longrightarrow\;200+35K-K^{2}-75K>0\;\Longrightarrow\;K^{2}+40K-200<0$$The positive root of $K^{2}+40K-200=0$ is$$K_{\max}=\frac{-40+\sqrt{1600+800}}{2}=\frac{-40+20\sqrt{6}}{2}=-20+10\sqrt6$$
Collect the range. Since $-20+10\sqrt6=4.4949<40$, the $s^{2}$ condition is not binding and the answer is$$\boxed{\;0<K<-20+10\sqrt6=4.495\;}$$A numerical spot check confirms it: at $K=4$ the closed-loop roots are all in the left half-plane, while at $K=5$ a conjugate pair has crossed into the right half-plane.

Part (b) — Root locus

Identify the open-loop poles and zeros. From $L(s)=K(s+3)/[s(s+1)(s^{2}+4s+5)]$ the four poles are $s=0,\;-1,\;-2\pm j1$ (since $s^{2}+4s+5=(s+2)^{2}+1$) and the single finite zero is $s=-3$. Hence $n=4$, $m=1$ and $n-m=3$ branches go to infinity.
Real-axis segments. A point on the real axis belongs to the $K>0$ locus when the number of real poles and zeros to its right is odd. Counting from the right: the interval $(-1,\,0)$ has one (the pole at the origin) and is on the locus; $(-3,\,-1)$ has two and is off; $(-\infty,\,-3)$ has three and is on.
Asymptote centroid and angles.$$\sigma_{a}=\frac{\sum p_{i}-\sum z_{j}}{n-m}=\frac{\bigl(0-1-2-2\bigr)-(-3)}{3}=\frac{-5+3}{3}=-\frac{2}{3}=-0.667$$$$\theta_{a}=\frac{(2k+1)180^{\circ}}{3}=60^{\circ},\;180^{\circ},\;300^{\circ}$$
Angle of departure from the complex poles. The angle criterion at $s_{0}=-2+j1$ gives$$\theta_{d}=180^{\circ}+\sum\angle(s_{0}-z_{j})-\sum_{i\neq0}\angle(s_{0}-p_{i})$$With $\angle(s_{0}+3)=\angle(1+j1)=45^{\circ}$ and the pole contributions $\angle(s_{0}-0)=153.43^{\circ}$, $\angle(s_{0}+1)=135^{\circ}$, $\angle(s_{0}+2+j1)=\angle(j2)=90^{\circ}$,$$\theta_{d}=180^{\circ}+45^{\circ}-\left(153.43^{\circ}+135^{\circ}+90^{\circ}\right)=-153.43^{\circ}$$so$$\boxed{\;\theta_{d}=-153.4^{\circ}\text{ at }s=-2+j1,\qquad +153.4^{\circ}\text{ at }s=-2-j1\;}$$The branch therefore leaves the complex pole heading down and to the left, consistent with the sketch below.
Angle of arrival at the finite zero. Applying the mirror-image criterion at $s=-3$, the four pole vectors contribute $180^{\circ}+180^{\circ}+(-135^{\circ})+135^{\circ}=360^{\circ}$, so$$\theta_{a,\,z}=180^{\circ}+360^{\circ}\equiv180^{\circ}$$i.e. the branch arrives at the zero along the real axis from the left, as it must, since $-3$ is the right-hand end of the real-axis segment $(-\infty,-3)$.
Breakaway and break-in points. Writing $K(s)=-\dfrac{s(s+1)(s^{2}+4s+5)}{s+3}$ and solving $dK/ds=0$ yields four roots; only the two real ones lying on $K>0$ segments are physical:$$s_{b}=-0.4363\;\text{(in }(-1,0)\text{)}\quad\text{and}\quad s_{r}=-3.6503\;\text{(in }(-\infty,-3)\text{)}$$Substituting each back into $K(s)$,$$\boxed{\;\text{breakaway: }s=-0.436,\;K=0.331\qquad\text{break-in: }s=-3.650,\;K=55.39\;}$$The remaining pair of $dK/ds=0$ roots, $s=-1.623\pm j0.710$, returns a complex gain and therefore belongs to no real-$K$ locus at all — it is discarded.
Imaginary-axis crossings. At $K=K_{\max}$ the $s^{1}$ row vanishes and the auxiliary polynomial from the $s^{2}$ row gives the crossing frequency:$$\frac{40-K}{5}\,s^{2}+3K=0\;\Longrightarrow\;s^{2}=-\frac{15K}{40-K}$$At $K=4.4949$ this is $s^{2}=-1.8990$, so$$\boxed{\;s=\pm j1.378\ \text{rad/s at }K=4.495\;}$$Solving the quartic at that gain confirms the picture: the roots are $\pm j1.378$ together with a well-damped pair at $-2.500\pm j0.923$.

Figure 3.2 — Root locus of $1+K(s+3)/[s(s+1)(s^{2}+4s+5)]=0$ for $K>0$. Crosses are open-loop poles, the circle is the zero at $s=-3$. Two branches leave the poles at $0$ and $-1$, break away at $s=-0.436$ ($K=0.331$) and cross the imaginary axis at $\pm j1.378$ when $K=4.495$; the branches from $-2\pm j1$ depart at $\mp153.4^{\circ}$, meet the real axis at the break-in point $s=-3.650$ ($K=55.39$), and thereafter one runs to the zero at $-3$ while the other escapes along the $180^{\circ}$ asymptote.

The locus and the Routh result agree exactly, which is the intended cross-check of the question: the gain at which the branches pierce the imaginary axis is precisely the $K_{\max}$ delivered by the $s^{1}$ row, and the frequency of that crossing is precisely the root of the auxiliary polynomial. Note also the qualitative message — the loop is stable only over a rather narrow band of gain, and the limit is imposed not by the plant alone but by the extra lag $1/(s+1)$ in the measurement path, which contributes an additional $90^{\circ}$ of phase at high frequency.

Quantity	Result
Characteristic equation	$s^{4}+5s^{3}+9s^{2}+(5+K)s+3K=0$
(a) Stable gain range	$0<K<-20+10\sqrt6=4.495$
Open-loop poles / zero	$0,\;-1,\;-2\pm j1$ / $-3$
Asymptotes	$\sigma_{a}=-0.667$; $60^{\circ},180^{\circ},300^{\circ}$
Departure angle at $-2+j1$	$-153.4^{\circ}$ (and $+153.4^{\circ}$ at $-2-j1$)
Arrival angle at zero $-3$	$180^{\circ}$
Breakaway point	$s=-0.436$, $K=0.331$
Break-in point	$s=-3.650$, $K=55.39$
Imaginary-axis crossing	$s=\pm j1.378$ at $K=4.495$
Roots at $K=K_{\max}$	$\pm j1.378$ and $-2.500\pm j0.923$

Topic. Routh–Hurwitz stability testing with a symbolic gain, and root-locus construction for a loop with non-unity feedback.

Key relations. Characteristic equation $1+KG_{p}(s)H(s)=0$; Routh first-column positivity; auxiliary polynomial from the row above a vanishing row gives the imaginary-axis roots; centroid $\sigma_{a}=(\sum p-\sum z)/(n-m)$; asymptote angles $(2k+1)180^{\circ}/(n-m)$; breakaway/break-in from $dK/ds=0$; departure angle $\theta_{d}=180^{\circ}+\sum\angle(s_{0}-z)-\sum\angle(s_{0}-p)$.

Why this works. The root locus is the set of points where the loop gain has magnitude one and phase an odd multiple of $180^{\circ}$; every construction rule is a consequence of that single angle condition evaluated in a limiting position. The Routh test, by contrast, counts right-half-plane roots without finding them, using the sign pattern of the first column. The two methods must and do agree at the boundary, and using them together is the cheapest available error check under exam conditions: if the $K$ from your auxiliary polynomial differs from the $K$ from your $s^{1}$ row, one of the two is arithmetically wrong. A practical point specific to this problem: because $H(s)\ne1$, the loop transfer function used for the locus is $KG_{p}H$, but the closed-loop transfer function is $KG_{p}/(1+KG_{p}H)$ — the zero at $-3$ appears in both, whereas the feedback pole at $-1$ appears as a closed-loop zero and does not.

Common pitfall. Sketching the locus on the three plant poles only and forgetting that the feedback lag contributes a fourth pole at $s=-1$; this drops the asymptote count from three to two and yields a stability limit roughly an order of magnitude too high. The second classic error is accepting all four roots of $dK/ds=0$ as breakaway points — only roots that lie on a $K>0$ segment (equivalently, that return a positive real $K$) are genuine.

Source. Ogata, Modern Control Engineering 5th ed., §5-6 (Routh) and §6-2/6-3 (root-locus rules); Nise, Control Systems Engineering 8th ed., Ch. 6 and Ch. 8; Dorf & Bishop, 13th ed., §7.2–7.4.

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