Question 4 of 8: Polar Plot and Nyquist Stability with an Unstable Open-Loop Pole
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Phys-B5 Control, National Examination December 2017
— a three-hour closed-book examination with one double-sided handwritten formula/notes
sheet permitted and an approved calculator. Questions 1 and 2 are compulsory; candidates choose
three of the remaining six (3–8). Every question is nonetheless answered in full below so
the paper remains a complete study resource. All eight questions carry equal value (20 marks
each).
Reference texts. N. S. Nise, Control Systems Engineering, 7th ed.
(root locus, Routh–Hurwitz, frequency-response lag/lead design, Nyquist stability,
state-space representation, steady-state errors); K. Ogata, Modern Control Engineering,
5th ed. (frequency-response compensator design, controllability/observability, PID pole
placement).
Question 4: Polar Plot and Nyquist Stability with an Unstable Open-Loop Pole (20 marks)
Figure — normalized polar plot $G_{open}(j\omega)/K$ for $\omega>0$
(solid) and its mirror for $\omega\lt0$ (dashed). The locus crosses the negative real axis at
exactly $-1$ when $\omega=1$ rad/s, approaching a vertical asymptote at $\text{Re}=-2$ as
$\omega\to0^+$ and spiralling into the origin as $\omega\to\infty$.
Given. $G_{open}(s)=K(s+1)/[s(s-1)]$ — unity negative feedback, ONE
open-loop pole in the right-half plane ($s=+1$) in addition to the pole at the origin.
Find. 1) The real/imaginary crossover coordinates of the normalized polar
plot and the sketch. 2) The range of $K_p>0$ for closed-loop stability via Nyquist.
Approach. Split $G_{open}(j\omega)/K$ into real and imaginary parts as
explicit functions of $\omega$ and solve $\text{Im}=0$ for the real-axis crossover. Because
there is an open-loop RHP pole ($P=1$), simple gain/phase-margin shortcuts do not apply; use
the full Nyquist criterion $Z=N+P$ (closed-loop RHP poles = CW encirclements of $-1$ plus
open-loop RHP poles), then cross-check the resulting $K$-range directly against
Routh–Hurwitz on the closed-loop characteristic equation.
Part 1) — real/imaginary parts and the crossover.
$$\frac{G_{open}(j\omega)}{K}=\frac{1+j\omega}{j\omega(j\omega-1)}
=\frac{-2}{\omega^2+1}+j\,\frac{1-\omega^2}{\omega(\omega^2+1)}.$$
As $\omega\to0^+$: $\text{Re}\to-2$, $\text{Im}\to+\infty$ (a vertical asymptote at
$\text{Re}=-2$). As $\omega\to\infty$: both parts $\to0^-$ (the locus spirals into the
origin). Setting $\text{Im}=0$: $1-\omega^2=0\Rightarrow\omega=1$ rad/s, where
$\text{Re}=-2/(1+1)=-1$. So the normalized plot crosses the negative real axis at exactly
$\boxed{(-1,0)\text{ at }\omega=1\ \text{rad/s}}$ — the ACTUAL (non-normalized) locus of
$G_{open}(j\omega)$ therefore crosses the real axis at $\boxed{-K}$ at that same frequency.
The sketch above shows the branch for $\omega>0$ (solid, arrow showing $\omega$ increasing from
the $-2+j\infty$ asymptote down through $(-1,0)$ into the origin) and its complex-conjugate
mirror for $\omega\lt0$ (dashed).
Part 2) — Nyquist contour and stability range.
The standard $\Gamma$ contour runs up the $j\omega$-axis, is indented by a small
clockwise-avoiding semicircle bulging INTO the right-half-plane around the pole at $s=0$ (so
that pole is excluded from the enclosed region, matching $P=$ count of RHP poles only), and
needs no large arc at infinity since $G_{open}(s)\to0$ as $|s|\to\infty$ (relative degree
$\ge1$). There is $P=1$ open-loop pole in the RHP ($s=+1$); for a stable closed loop we need
$Z=0$ closed-loop RHP poles, i.e. $N=Z-P=-1$: exactly ONE counter-clockwise encirclement of the
$-1$ point. From Part 1, the actual locus crosses the negative real axis at $-K$ (at
$\omega=\pm1$) and is otherwise confined to $\text{Re}\lt0$; the $-1$ point is encircled
(counter-clockwise, satisfying $N=-1$) precisely when the crossing $-K$ lies FARTHER left than
$-1$, i.e. $K>1$. Cross-check via Routh–Hurwitz on the closed-loop
characteristic equation $1+G_{open}(s)=0\Rightarrow s(s-1)+K(s+1)=0\Rightarrow
s^2+(K-1)s+K=0$: for a 2nd-order polynomial, stability needs every coefficient positive, i.e.
$K-1>0$ AND $K>0$, giving the identical result:
$$\boxed{K>1\ \text{for closed-loop stability}}$$
(at exactly $K=1$ the real-axis crossing lands ON $-1$, matching the marginal case
$K-1=0$ in the Routh test).