Nyquist Plot and System Stability Analysis

This problem focuses on determining the stability of a closed-loop system by analyzing the behavior of its Nyquist plot. We are given the open-loop transfer function \(G\left( s \right) = \frac{K}{{\left( {{s^2} + 2s + 2} \right)\left( {s + 2} \right)}}\) and specific information about how its Nyquist plot encircles the critical point (–1 + j0) for different values of the gain parameter \(K\). We will use the Nyquist stability criterion to determine the stability for each case.

Open-Loop Poles of the Transfer Function

The first step in applying the Nyquist stability criterion is to find the poles of the given open-loop transfer function \(G(s)\). The poles are the values of \(s\) that make the denominator of the transfer function equal to zero.

\(\left( {{s^2} + 2s + 2} \right)\left( {s + 2} \right) = 0\)

This equation yields the following poles:

• For the term \((s + 2)\): \(s + 2 = 0 \implies s = -2\). This pole is located in the left-half of the s-plane (LHP).
• For the term \((s^2 + 2s + 2)\): We use the quadratic formula \(s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) for a quadratic equation in the form \(as^2 + bs + c = 0\). Here, \(a=1\), \(b=2\), \(c=2\).
  \(s = \frac{-2 \pm \sqrt{2^2 - 4(1)(2)}}{2(1)}\)
  \(s = \frac{-2 \pm \sqrt{4 - 8}}{2}\)
  \(s = \frac{-2 \pm \sqrt{-4}}{2}\)
  \(s = \frac{-2 \pm j2}{2}\)
  \(s = -1 \pm j1\). Both of these complex conjugate poles have a negative real part (i.e., -1), which means they are also located in the left-half of the s-plane (LHP).

Since all the open-loop poles of \(G(s)\) are in the Left-Half Plane (LHP), the number of open-loop poles in the Right-Half Plane (RHP), denoted by \(P\), is 0.

\(P = 0\)

Nyquist Stability Criterion Defined

The Nyquist stability criterion provides a graphical method to determine the stability of a closed-loop control system from its open-loop frequency response (Nyquist plot). For a system with unity gain feedback, we consider the Nyquist plot of \(G(s)\).

The criterion is expressed by the formula:

\(N = P - Z\)

Where:

• \(N\): Represents the net number of clockwise encirclements of the critical point (–1 + j0) by the Nyquist plot of \(G(s)\).
• \(P\): Represents the number of poles of the open-loop transfer function \(G(s)\) that lie in the Right-Half Plane (RHP).
• \(Z\): Represents the number of zeros of the closed-loop characteristic equation (which correspond to the closed-loop poles) that lie in the Right-Half Plane (RHP).

For a closed-loop system to be stable, all its poles must be in the Left-Half Plane (LHP). This means that the number of closed-loop poles in the RHP, \(Z\), must be zero (\(Z = 0\)).

Given our calculation that \(P = 0\), the stability condition \(Z = 0\) implies that:

\(0 = 0 - N \implies N = 0\)

Therefore, for the closed-loop system to be stable, the Nyquist plot must not encircle the critical point (–1 + j0).

Stability Analysis for K = 10

The problem states that for \(K = 10\), the Nyquist plot of \(G(s)\) does not encircle the critical point (–1 + j0).

• This means that the net number of clockwise encirclements, \(N\), is 0.
• Applying the Nyquist criterion formula \(Z = P - N\):
• Substitute \(P = 0\) and \(N = 0\): \(Z = 0 - 0 = 0\)

Since \(Z = 0\), there are no closed-loop poles in the Right-Half Plane. Consequently, the closed-loop system is stable for K = 10.

Stability Analysis for K = 100

The problem states that for \(K = 100\), the Nyquist plot of \(G(s)\) does encircle the critical point (–1 + j0).

• This indicates that the net number of clockwise encirclements, \(N\), is not 0 (\(N \neq 0\)).
• Applying the Nyquist criterion formula \(Z = P - N\):
• Substitute \(P = 0\): \(Z = 0 - N = -N\)

Since the plot encircles the critical point, \(N\) is a non-zero value. For the system to be unstable, \(Z\) must be greater than 0 (\(Z > 0\)). Given \(Z = -N\), for \(Z\) to be positive, \(N\) must be negative (i.e., there must be a net number of counter-clockwise encirclements). In general, if \(P=0\), any non-zero encirclement implies that \(Z \neq 0\), leading to an unstable system.

Therefore, the closed-loop system is unstable for K = 100.

Conclusion on Closed-Loop System Stability

Based on our analysis using the Nyquist stability criterion and the given information about the Nyquist plot encirclements:

• For \(K = 10\), the system is determined to be stable.
• For \(K = 100\), the system is determined to be unstable.

This aligns with the option stating "stable for K = 10 and unstable for K = 100".