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Question

In Nyquist plot of a system on adding a pole at s = 0, then plot will -

The correct answer is

rotate clockwise by 90 °

Nyquist Plot Rotation due to a Pole at s=0

The Nyquist plot is a powerful graphical tool used in control systems to determine the stability of a feedback control system. It plots the frequency response of the open-loop transfer function \(G(s)H(s)\) in the complex plane as the frequency \(\omega\) varies from \(-\infty\) to \(+\infty\). The shape and encirclements of the Nyquist plot around the critical point \((-1, j0)\) provide crucial information about system stability.

Understanding the Effect of Adding a Pole at s=0

Adding a pole at \(s = 0\) to a system's open-loop transfer function means multiplying the original transfer function by \(\frac{1}{s}\). A pole at \(s=0\) represents an integrator in the system. Let's consider the original open-loop transfer function as \(G(s)H(s)\). If we add a pole at \(s=0\), the new open-loop transfer function, let's call it \(G'(s)H'(s)\), becomes:

\[G'(s)H'(s) = \frac{1}{s} G(s)H(s)\]

To understand the effect on the Nyquist plot, we need to analyze the change in the magnitude and phase of the frequency response when we substitute \(s = j\omega\).

The original frequency response is \(G(j\omega)H(j\omega)\). The new frequency response is:

\[G'(j\omega)H'(j\omega) = \frac{1}{j\omega} G(j\omega)H(j\omega)\]

Phase Shift due to the Pole at s=0

The phase angle of a complex number \(Z = |Z|e^{j\phi}\) is \(\phi\). When we multiply two complex numbers, their phases add up. So, the phase of the new transfer function will be the sum of the phase of \(\frac{1}{j\omega}\) and the phase of the original \(G(j\omega)H(j\omega)\).

  • The complex number \(\frac{1}{j\omega}\) can be rewritten as \(\frac{-j}{\omega}\).
  • The phase angle of \(\frac{1}{j\omega}\) is \(\angle\left(\frac{1}{j\omega}\right) = \angle\left(\frac{-j}{\omega}\right)\). Since \(-j\) lies on the negative imaginary axis, its angle is \(-90^\circ\) or \(-\frac{\pi}{2}\) radians.

Therefore, the new phase \(\phi'\) of the system with the added pole at \(s=0\) will be:

\[\phi' = \angle G'(j\omega)H'(j\omega) = \angle\left(\frac{1}{j\omega}\right) + \angle\left(G(j\omega)H(j\omega)\right) = -90^\circ + \phi\]

This means that for every point on the original Nyquist plot, its phase angle is decreased by \(90^\circ\).

Impact on Nyquist Plot Rotation

A decrease in the phase angle by \(90^\circ\) corresponds to a specific rotation of the Nyquist plot in the complex plane:

  • A positive change in phase (e.g., \(+90^\circ\)) results in an anti-clockwise rotation.
  • A negative change in phase (e.g., \(-90^\circ\)) results in a clockwise rotation.

Since adding a pole at \(s=0\) introduces a phase shift of \(-90^\circ\), the Nyquist plot of the system will rotate clockwise by \(90^\circ\).

Example Illustration

Consider a very simple open-loop transfer function: \(G(s)H(s) = K\) (a constant gain, where \(K\) is a positive real number).

Its Nyquist plot is simply a point \(K\) on the positive real axis at \(K \angle 0^\circ\).

Now, let's add a pole at \(s=0\): \(G'(s)H'(s) = \frac{K}{s}\).

The frequency response is \(G'(j\omega)H'(j\omega) = \frac{K}{j\omega} = -j\frac{K}{\omega}\).

  • The magnitude is \(\left|\frac{K}{\omega}\right|\).
  • The phase is \(-90^\circ\).

As \(\omega\) varies from \(0^+\) to \(\infty\):

  • When \(\omega \to 0^+\), the magnitude \(\to \infty\), and the phase is \(-90^\circ\). So, the plot starts at infinity along the negative imaginary axis.
  • When \(\omega \to \infty\), the magnitude \(\to 0\), and the phase is \(-90^\circ\). So, the plot approaches the origin along the negative imaginary axis.

This example clearly illustrates that the entire plot has rotated \(90^\circ\) clockwise from its original position (a point on the positive real axis) to a line along the negative imaginary axis.

Conclusion on Nyquist Plot Modification

In summary, the addition of a pole at \(s=0\) to the open-loop transfer function of a system introduces a constant phase lag of \(90^\circ\) (\(-\frac{\pi}{2}\) radians) at all frequencies. This consistent phase shift causes the entire Nyquist plot to rotate in a clockwise direction by \(90^\circ\). This is a fundamental concept in frequency response analysis and stability assessment in control systems, especially when dealing with system type and steady-state errors.

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Important Questions from Nyquist Plot

  1. ______indicates not only whether a system is stable, but also its degree of stability and how stability may be imposed if necessary.

  2. The Nyquist plot of the transfer function \(G\left( s \right) = \frac{K}{{\left( {{s^2} + 2s + 2} \right)\left( {s + 2} \right)}}\)

    Does not encircle the point (–1 + j0) for K = 10 but does encircle the point (-1 + j0) for K = 100 . Then the closed-loop system (having unity gain feedback) is

  3. A closed-loop control system is stable if the Nyquist plot of the corresponding open-loop transfer function

  4. The Nyquist stability criterion and the Routh criterion both are powerful analysis tools for determining the stability of feedback controllers. Identify which of the following statements is FALSE:

  5. The critical point (-1, j0) is mapped to ________ on the Nichols chart.

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