23-Mechatronics-A1 Systems Dynamics and Controls · Undated paper
Question 6 of 6: Root Locus of Two Open-Loop Transfer Functions
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 16-Mex-A1, System Analysis and Control, a three-hour closed-book National Examination for which candidates may use an approved Casio or Sharp calculator. The cover page states that any four (4) of the printed questions constitute a complete paper, all of equal value; all six printed questions are worked here.
Reference texts. N.S. Nise, Control Systems Engineering, 8th
ed. (Ch. 4 second-order time response; Ch. 6 stability & the Routh-Hurwitz criterion;
Ch. 8 root locus; Ch. 10 frequency-response methods); K. Ogata, Modern Control
Engineering, 5th ed. (Ch. 3 Laplace transform; Ch. 5 stability analysis; Ch. 6
root-locus method; Ch. 7 frequency-response analysis).
Question 6: Root Locus of Two Open-Loop Transfer Functions
Given. (a) one real open-loop pole at $s=-5$ and one real zero at
$s=-1$; (b) two real open-loop poles at $s=-1,-6$ and one real zero at $s=-3$, gain
$K'\ge0$.
Find. The root locus — the path traced by the closed-loop poles
as $K'$ runs $0\to\infty$ — for each system, verified against the exact closed-loop
characteristic equation.
Approach. Apply the standard root-locus construction rules (real-axis
segments, asymptote count, breakaway test), then confirm by solving
$1+K'G(s)H(s)=0$ directly for the closed-loop pole as a function of $K'$.
Part (a) — characteristic equation and direct solution.
$$(s+5)+K'(s+1)=0\ \Longrightarrow\ s=-\frac{K'+5}{K'+1}$$
One pole ($-5$), one zero ($-1$): $\#\text{poles}=\#\text{zeros}=1$, so there are no
asymptotes; $dK'/ds=-(s+5)/(s+1)$ has no real root strictly between $-5$ and $-1$, so
there is no breakaway. The whole segment $-5\le s\le-1$ carries an odd count (the zero)
of real poles/zeros to the right of any test point in it, so it is entirely on the
locus.
Part (a) — check by direct solve. $K'=0\Rightarrow s=-5$
(open-loop pole); $K'\to\infty\Rightarrow s\to-1$ (open-loop zero); $K'=1\Rightarrow
s=-3$ (midpoint): a single real root slides monotonically from the pole to the zero.
$$\boxed{\text{Locus (a): single real branch, } s=-5\ (K'{=}0)\ \to\ s=-1\ (K'\to\infty)}$$
Part (b) — real-axis segments and asymptote. Poles $s=-1,-6$;
zero $s=-3$. Test points show $[-3,-1]$ and $(-\infty,-6]$ carry an odd count of
poles/zeros to their right (on the locus), while $[-6,-3]$ carries an even count (not on
the locus). $\#\text{poles}-\#\text{zeros}=2-1=1$ asymptote at $180^\circ$ along the
negative real axis, so the branch leaving $s=-6$ travels left to $-\infty$ entirely on
the real axis (it IS the asymptote), while the branch leaving $s=-1$ travels left and
terminates on the finite zero at $s=-3$.
Part (b) — breakaway check and confirmation.
$K'=-(s+1)(s+6)/(s+3)$; $dK'/ds=0$ has roots only at $s=-3\pm i\sqrt6$ (complex, off the
real axis), so neither real-axis branch breaks away — both stay real for every
$K'\ge0$, confirmed by direct solve of $(s+1)(s+6)+K'(s+3)=0$ at sample gains.
$$\boxed{\text{Locus (b): branch 1: } s=-1\to-3\ (K'{:}0\to\infty);\ \text{branch 2: }
s=-6\to-\infty\ \text{along the } 180^\circ\ \text{asymptote}}$$
Figure Q6(a) — single real-axis branch from the pole $s=-5$ to
the zero $s=-1$.
Figure Q6(b) — two real-axis branches: pole $-1$ to zero $-3$,
and pole $-6$ to $-\infty$ along the single $180^\circ$ asymptote.
