Root locus for positive and negative gain: 22-Mec-A3 System Analysis and Control, May 2014, Q3

Question 3: Root locus for positive and negative gain (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. An open-loop function $KG(s)H(s)=50K/[(s+1)(s+2)(s+10)]$ with three real open-loop poles at $s=-1,\ -2,\ -10$, no finite zeros ($n = 3$, $m = 0$), and a gain $K$ that may take either sign.

Find. (a) the root-locus sketches for $K>0$ and $K<0$; (b) every $j\omega$-axis crossing with the gain at which it occurs; (c) the complete range of $K$ giving a stable closed loop.

Root locus of the closed loop. Solid blue: K > 0 (breakaway at −1.485, jω crossing at ±j5.657 when K = 7.92). Dashed red: K < 0 (crossing at the origin when K = −0.4).

Approach. Apply the standard construction rules (real-axis segments, centroid and asymptotes, breakaway points) to sketch both branches of the locus, then obtain the $j\omega$ crossings exactly from the Routh array of the closed-loop characteristic polynomial, which yields both the critical gain and the crossing frequency.

Form the closed-loop characteristic equation. With $1+KG(s)H(s)=0$, $$(s+1)(s+2)(s+10)+50K = s^{3}+13s^{2}+32s+20+50K = 0$$ so the gain enters only the constant term — the situation that makes the Routh analysis of part (b) particularly clean.
Real-axis segments. For $K>0$ a real point lies on the locus when the number of real poles and zeros to its right is odd: that gives the segment $-2 \le \sigma \le -1$ and the ray $\sigma \le -10$. For $K<0$ the rule inverts to an even count, giving the complementary sets $\sigma \ge -1$ and $-10 \le \sigma \le -2$. The negative-gain locus therefore runs into the right half-plane immediately from the pole at $s=-1$.
Centroid and asymptotes. With $n-m = 3$ branches going to infinity, the asymptotes meet on the real axis at $$\sigma_a=\frac{\sum p_i-\sum z_j}{n-m}=\frac{-1-2-10}{3}=-4.333$$ For $K>0$ the asymptote angles are $\theta=\pm 60^\circ,\ 180^\circ$; for $K<0$ they are $\theta=0^\circ,\ \pm120^\circ$, which is why the negative-gain branches head into the right half-plane along the real axis.
Breakaway point on the positive-gain locus. Writing $K = -(s+1)(s+2)(s+10)/50$ and setting $dK/ds=0$ gives $3s^{2}+26s+32=0$, whose roots are $s=-1.485$ and $s=-7.181$. Only the first lies on a $K>0$ segment, so the two branches leaving $-1$ and $-2$ meet and break away at $$\boxed{\;s_b = -1.485,\qquad K_b = 0.0425\;}$$ after which they turn into the complex plane and eventually cross the imaginary axis.
Positive-gain $j\omega$ crossing by Routh. The array of $s^{3}+13s^{2}+32s+(20+50K)$ has the first-column entry $$\frac{13\times 32-(20+50K)}{13}=\frac{396-50K}{13}$$ which vanishes at $$\boxed{\;K = \frac{396}{50}=7.92\;}$$ Substituting into the auxiliary polynomial from the $s^{2}$ row, $13s^{2}+(20+50K)=13s^{2}+416=0$, gives $s^{2}=-32$ and hence the crossing frequency $$\boxed{\;s=\pm j\sqrt{32}=\pm j5.657\ \text{rad/s}\;}$$
Negative-gain $j\omega$ crossing. The remaining first-column entry is the constant term $20+50K$, which vanishes at $K=-0.4$. The auxiliary condition here is simply $s = 0$: the branch travelling right from the pole at $-1$ reaches the origin at $$\boxed{\;K=-0.4,\quad s=0\;}$$ and enters the right half-plane for any more negative gain. A crossing at the origin rather than at a finite $\pm j\omega$ is the hallmark of the constant term changing sign.
Assemble the stable range. Between the two crossings all three Routh first-column entries are positive, and outside them one changes sign, so $$\boxed{\;-0.4 < K < 7.92\;}$$ Spot checks confirm the interval: at $K=3$ the closed-loop roots are $-11.503$ and $-0.749\pm j3.771$ (stable), while at $K=8$ they are $-13.020$ and $+0.0099\pm j5.680$ (unstable, though only just — $K=8$ sits barely above the $K=7.92$ boundary), and at $K=-0.5$ one root has moved to $s=+0.147$.

The sketch above shows both families on one axis set: the solid blue branches are the conventional $K>0$ locus (breakaway at $-1.485$, complex excursion, crossing at $\pm j5.657$), and the dashed red branches are the $K<0$ locus, whose right-most branch runs straight out along the positive real axis. Practically, this plant tolerates a useful positive gain but almost no negative gain — a sign inversion anywhere in the loop wiring would destabilise it at once.

Quantity	Value
Open-loop poles	−1, −2, −10 (n = 3, m = 0)
Asymptote centroid σ_a	−4.333
Asymptote angles (K > 0 / K < 0)	±60°, 180° / 0°, ±120°
Breakaway point (K > 0)	s = −1.485 at K = 0.0425
Upper jω crossing	s = ±j5.657 at K = 7.92
Lower jω crossing	s = 0 at K = −0.4
Stable gain range	−0.4 < K < 7.92

Topic. Root-locus construction for both gain signs, and imaginary-axis crossings via Routh–Hurwitz.

Key relations. Magnitude and angle criteria $|KGH| = 1$, $\angle GH = 180^\circ(2k+1)$ for $K>0$ and $\angle GH = 360^\circ k$ for $K<0$; centroid $\sigma_a=(\sum p-\sum z)/(n-m)$; asymptote angles $(2k+1)180^\circ/(n-m)$ and $360^\circ k/(n-m)$ respectively; breakaway from $dK/ds=0$; crossing gain from the vanishing Routh row with the frequency from the auxiliary polynomial.

Why this works. The locus is the set of $s$ satisfying the angle criterion, and the criterion differs by $180^\circ$ between positive and negative gain — which is why the two loci are exactly complementary on the real axis and why their asymptote fans are rotated by $180^\circ/(n-m)$. The Routh route to the crossings is exact rather than graphical: when a first-column entry vanishes, the corresponding auxiliary polynomial (formed from the row above) divides the characteristic polynomial and its roots are the imaginary-axis pair, so the gain and frequency drop out together with no sketching error.

Common pitfall. Sketching only the $K>0$ locus and asserting stability for all small $K$ — the negative-gain branch here goes unstable at only $K=-0.4$. A second error is discarding the $s=0$ crossing because it is not a complex pair; a real crossing at the origin is a genuine stability boundary. A third is taking both roots of $dK/ds=0$ as breakaway points: $s=-7.181$ lies on a negative-gain segment (it is a break-in for $K<0$).

Source. Nise, Control Systems Engineering, 8th ed., Ch. 8.4–8.7 (root-locus rules and the negative-gain locus); Ogata, Modern Control Engineering, 5th ed., §6.2–6.4.

Question 3 of 6: Root locus for positive and negative gain

Question 3: Root locus for positive and negative gain (25 marks)