Question

Addition of a pole at the origin to a transfer function rotates the polar plot at zero and infinite frequencies by a further angle of

• A. \( 90^\circ \)
• B. \( -90^\circ \)
• C. \( 45^\circ \)
• D. \( -45^\circ \)

Hint:

Adding a pole at the origin (\(1/s\)) to \(G(s)H(s)\) contributes a phase of \(-90^\circ\) at all frequencies \(\omega\).
Adding a zero at the origin (\(s\)) to \(G(s)H(s)\) contributes a phase of \(+90^\circ\) at all frequencies \(\omega\).
This rotation affects both the starting point (\(\omega \to 0^+\)) and the ending point (\(\omega \to \infty\)) of the polar plot.

Answer 1

The Correct Option is B

In control systems, consider a transfer function \( G(s) \). When a pole is added at the origin, the transfer function becomes \( G(s) = \frac{k}{s} \times \text{(original terms)} \). In polar plots, the phase of a transfer function is determined by the angle it makes on the complex plane. For a pole at the origin (i.e., \( s=0 \)), the phase contributed by this pole is given by:

\(\phi = \angle(-s) = -90^\circ\)

This is because in the complex plane, a point at \(-90^\circ\) corresponds to a pole at the origin, hence the plot is rotated by \(-90^\circ\) at zero and infinite frequencies. Thus, the correct answer is:

\(-90^\circ\)