22-Mec-A3 System Analysis and Control · Undated paper

Question 6 of 6: Root-locus plots for two open-loop transfer functions

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

Notes on this paper

Paper format. 16-Mec-A3 System Analysis and Control, National Exam, May 2019 sitting. Three hours, closed book, Casio or Sharp approved calculator only. Six questions; any four constitute a complete paper and all questions are of equal value (25 marks each). A table of Laplace transforms is supplied with the paper. Every one of the six questions is worked below.

Reference texts. Ogata, Modern Control Engineering, 5th ed. (Chs. 5–8); Nise, Control Systems Engineering, 8th ed. (Chs. 4, 6, 8, 10); Dorf & Bishop, Modern Control Systems, 13th ed.; Franklin, Powell & Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed.

Question 6: Root-locus plots for two open-loop transfer functions (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. Two open-loop transfer functions with adjustable gain $K^{\prime}\ge0$: (a) one pole at $s=-5$ and one zero at $s=-1$; (b) poles at $s=-1$ and $s=-6$ with a zero at $s=-3$. Unity feedback is understood, so the characteristic equation is $1+G(s)H(s)=0$ in both cases.

Find. The root loci, several scaled points on each, and confirmation of those points by solving the characteristic equations directly.

Approach. Apply the standard construction rules — number of branches, real-axis segments, asymptotes, breakaway points from $dK^{\prime}/ds=0$ — and then solve the characteristic equation in closed form, which for these low-order systems is exact and gives the check the question asks for.

Count branches and locate real-axis segments for (a). With $n=1$ pole and $m=1$ zero there is a single branch; $n-m=0$, so there are no asymptotes and the branch terminates on the finite zero. A point on the real axis belongs to the locus when the number of real poles and zeros to its right is odd, which here selects the segment $-5\le\sigma\le-1$.
Solve the characteristic equation for (a) exactly. From $s+5+K^{\prime}(s+1)=0$, $$\boxed{s=-\frac{5+K^{\prime}}{1+K^{\prime}}}$$ a single real root that moves monotonically from $s=-5$ at $K^{\prime}=0$ towards the zero at $s=-1$ as $K^{\prime}\to\infty$, never leaving the real axis and never leaving the left half-plane. The system is therefore stable for every positive gain.

Scaling several points on the branch, as the question requires, gives:

$K^{\prime}$	0	1	3	9	$\to\infty$
Closed-loop root $s$	−5.000	−3.000	−2.000	−1.400	−1.000

Part (a): root locus of K'(s+1)/(s+5). Crosses mark open-loop poles, circles mark zeros. The single branch travels along the real axis from the pole at −5 to the zero at −1.

Set up (b). Now $n=2$ and $m=1$, so there are two branches and $n-m=1$ asymptote at $180^\circ$: one branch ends on the finite zero at $s=-3$ and the other escapes to $-\infty$ along the negative real axis. The odd-count rule places the locus on the segments $-3\le\sigma\le-1$ and $\sigma\le-6$.
Test for breakaway points. Writing the gain as a function of $s$, $K^{\prime}=-(s^2+7s+6)/(s+3)$, and differentiating, $$\frac{dK^{\prime}}{ds}=-\frac{s^2+6s+15}{(s+3)^2}=0\quad\Longrightarrow\quad s=-3\pm j\sqrt{6}$$ Both roots are complex, and substituting either back gives a complex $K^{\prime}$. There is therefore no breakaway or break-in point at any real gain, and the two branches stay on the real axis for the whole locus. This is worth stating explicitly, because the usual expectation for a two-pole plant is a breakaway into a complex pair.
Confirm from the characteristic equation. Expanding $(s+1)(s+6)+K^{\prime}(s+3)=0$ gives $$s^2+\left(7+K^{\prime}\right)s+\left(6+3K^{\prime}\right)=0$$ whose discriminant is $$\Delta=\left(7+K^{\prime}\right)^2-4\left(6+3K^{\prime}\right)=K^{\prime\,2}+2K^{\prime}+25$$ The discriminant of that quadratic in $K^{\prime}$ is $4-100<0$, so $\Delta>0$ for every real $K^{\prime}$: the roots are always real, exactly as the breakaway test predicted.
Confirm stability for all gain. Both coefficients $7+K^{\prime}$ and $6+3K^{\prime}$ are positive for every $K^{\prime}>0$, so by the Routh criterion for a quadratic $$\boxed{\text{the loop of part (b) is stable for all }K^{\prime}>0}$$ with the dominant root migrating from $s=-1$ towards the zero at $s=-3$ — so increasing the gain speeds the response up to a limit of three times the open-loop dominant rate, and no further.

Scaling several points, again as the question requires:

$K^{\prime}$	0	2	10	$\to\infty$
Dominant root	−1.000	−1.628	−2.479	−3.000
Second root	−6.000	−7.372	−14.521	$\to-\infty$

Part (b): root locus of K'(s+3)/[(s+1)(s+6)]. Both branches remain on the real axis for every positive gain; one terminates on the zero at −3, the other runs off to minus infinity along the 180° asymptote.

Quantity	(a) $K^{\prime}(s+1)/(s+5)$	(b) $K^{\prime}(s+3)/[(s+1)(s+6)]$
Branches	1	2
Real-axis segments	$-5\le\sigma\le-1$	$-3\le\sigma\le-1$ and $\sigma\le-6$
Asymptotes	none ($n=m$)	one, at $180^\circ$
Breakaway / break-in	none	none ($dK^{\prime}/ds=0$ has only complex roots $-3\pm j2.449$)
Closed-loop roots	$s=-(5+K^{\prime})/(1+K^{\prime})$	roots of $s^2+(7+K^{\prime})s+(6+3K^{\prime})$
Terminal points	$-5\to-1$	$-1\to-3$ and $-6\to-\infty$
Stability	Stable for all $K^{\prime}>0$	Stable for all $K^{\prime}>0$

Topic. Root-locus construction and its verification against the characteristic equation.

Key relations. Locus condition $1+K^{\prime}G_1(s)=0$; branches $=\max(n,m)$; real-axis rule (odd count of real poles and zeros to the right); asymptote count $n-m$ with centroid $\sigma_a=\left(\sum p_i-\sum z_j\right)/(n-m)$; breakaway points from $dK^{\prime}/ds=0$, retaining only roots that give a real positive gain.

Why this works. The locus is simply the set of characteristic roots parameterised by gain, so for first- and second-order systems it can be written down in closed form and the graphical rules become a check rather than a necessity. Part (b) shows why the algebraic check matters: the breakaway equation has roots, but they are complex, and only the roots that return a real positive gain lie on the $K^{\prime}>0$ locus. The discriminant argument then proves the same conclusion a second, independent way.

Common pitfall. Assuming that any two-pole system must break away into a complex pair. Placing the zero between the poles, as in part (b), pulls one branch onto the zero and pushes the other to infinity, leaving both roots real at every gain. The related error is accepting every root of $dK^{\prime}/ds=0$ as a breakaway point without checking that the corresponding gain is real and positive.

Source. Ogata, Modern Control Engineering, 5th ed., §§6–2 and 6–3; Nise, Control Systems Engineering, 8th ed., Ch. 8; Dorf & Bishop, Modern Control Systems, 13th ed., Ch. 7.

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