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

Let the original open-loop transfer function be \(G(s)H(s)\). Adding a pole at the origin means the new transfer function becomes \(G'(s)H'(s) = \frac{1}{s} G(s)H(s)\). We are interested in the effect on the polar plot, which is a plot of \(G(j\omega)H(j\omega)\) as \(\omega\) varies from 0 to \(\infty\). The factor introduced is \(\frac{1}{j\omega}\). The term \(\frac{1}{j\omega} = \frac{-j}{\omega} = \frac{1}{\omega} e^{-j90^\circ}\). This term contributes a phase shift of \(-90^\circ\) (or \(-\pi/2\) radians) at all frequencies \(\omega>0\).
At \(\omega \to 0^+\) (zero frequency): The term \(1/s\) introduces an additional phase of \(-90^\circ\). So, the starting point of the polar plot (at \(\omega=0^+\)) will be rotated by \(-90^\circ\) compared to the original plot.
At \(\omega \to \infty\) (infinite frequency): The term \(1/s\) introduces an additional phase of \(-90^\circ\). So, the ending point of the polar plot (as \(\omega \to \infty\)) will also be rotated by \(-90^\circ\) compared to the original plot. Therefore, the addition of a pole at the origin rotates the polar plot at zero and infinite frequencies by a further angle of \(-90^\circ\). \[ \boxed{-90^\circ} \]