Consider the feedback system shown in the figure. The Nyquist plo

SOLUTION

Analyzing all the options, we can conclude that the correct answer is Option (4).

Reason:

\(G\left( s \right) = \frac{{\left( {s - 0.1} \right)}}{{s + 1.1}}\)

\(G\left( {j\omega } \right) = \frac{{2\left( {j\omega - 0.1} \right)}}{{\left( {j\omega + 1.1} \right)}}\)

\(\left| {G\left( {j\omega } \right)} \right| = \frac{{2\sqrt {{\omega ^2} + {{0.1}^2}} }}{{\sqrt {{\omega ^2} + {{1.1}^2}} }}\)

\( = 2 \times \sqrt {\frac{{{\omega ^2} + {{0.1}^2}}}{{{\omega ^2} + {{1.1}^2}}}} \)

ω	|G(jω)|	∠G(jω )
0	0.0909	180°
ω	2	0°

 

\(\angle G\left( {j\omega } \right) = {\tan ^{ - 1}}\left( {\frac{\omega }{{ - 0.1}}} \right) - {\tan ^{ - 1}}\left( {\frac{\omega }{{1.1}}} \right)\)

\( = 180^\circ - {\tan ^{ - 1}}\left( {\frac{\omega }{{0.1}}} \right) - {\tan ^{ - 1}}\left( {\frac{\omega }{{1.1}}} \right)\)

This G(s) satisfies lice given Nyquist plot:

 

Now, the characteristic equation is:

1 + K G(s) = 0

\(1 + K\;2\left( {\frac{{S - 0.1}}{{S\; + 1.1\;}}} \right) = 0\)

s + 1.1 +2Ks – 0.2k = 0

s(1 + 2K) + 1.1 – 0.2K = 0

\(s = \frac{{0.2K - 1.1}}{{2K + 1}}\)

For the system to be stable, the roots must lie on the left haft of the s-plane, i.e.

\(\frac{{0.2k - 1.1}}{{2k + 1}} < 0\)

\(K < \frac{{11}}{2}\)

K < 6.5

If K > 6.5, the system will be unstable as the roots will be on the right half of the s-plane.

Hence option D is current.

Important:

Also, in a practical control system, if gain k is increased, then oscillations in the system will be increased which will make the system unstable.

Option A:

G(s) is an all-pass filter

All-pass filter Nyquist plot has a constant Magnitude of 1 for all frequencies, i.e.

 

Hence option A is incorrect

Option B:

From \(G\left( s \right) = \frac{{2\left( {s - 0.1} \right)}}{{\left( {s + 1.1} \right)}},\)  we have already shown that G(s) is not necessarily strictly proper.

∴ Option B is incorrect.

Proper G(s) may also have a similar Nyquist plot as given in the question.

Option C:

\(G\left( s \right) = \frac{{2\left( {s - 0.1} \right)}}{{s + 1.1}}\)

Here, G(s) is stable, but the system is not a minimum phase system, because one of the zeros is at s = - 0.1. This means that one of the zeroes is in the right half of the s-plane.

A minimum phase system means all the finite pole and zeroes of the open-loop transfer function must lie on the left-hand ride of s-plane.

Hence option C is incorrect.