Question

A transfer function G(s) with the degree of its numerator polynomial zero and the degree of its denominator polynomial two has a Nyquist plot shown in the figure. The transfer function represents

• A. a stable, type-0 system
• B. a stable, type-1 system
• C. an unstabel, type -0 system
• D. an unstable, type-1 system

Answer 1

The correct option is D an unstable, type-1 system

According to above figure:
$-1+j0$ point is on the right side of the given Nyquist plot. So system is unstable.
$\therefore$Magnitude at $\omega =0$ is infinite so, there is no constant part in the denominator of transfer function of the given system.
Assume,
$G\left(s\right)=\frac{1}{s^2-s}=\frac{1}{s(s-1)}$

$Magnitude=\frac{1}{\sqrt{{\omega }^2({\omega }^2+1)}}=\frac{1}{\sqrt{{\omega }^2+1}}$
$Phase\varphi =-{270}^o+ta{n}^{-1}\left(\omega \right)$
$\omega =0;M=\infty ;\varphi =-{270}^o$
$\omega =\varphi ;M=0;\varphi =-{270}^o+{90}^o=-{180}^o$

So, out assumption is true.
The system is type-1 and is unstable.