The pole- zero pattern of the active circuit shown below would be:

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Solution

The correct option is A

The given circuit is an active all pass filter.
Applying kVL at node $V_{1}$ . we get
$\frac{V_{1} - V_{i}}{R_{1}} + \frac{V_{1} - V_{0}}{R_{1}} = 0$ $. . . (i)$
Applying kVL at node $V_{2},$ we get
$\frac{V_{2}}{1 / s C} + \frac{V_{2} - V_{i}}{R} = 0$ $. . . (i i)$
Due to virtual short,
$V_{1} = V_{2}$ $. . . (i i i)$
Form equatin (i),
$\frac{2 V_{1}}{R_{1}} = \frac{V_{i} + V_{0}}{R_{1}}$
$V_{1} = \frac{V_{i} + V_{0}}{2}$ $. . . (i v)$
From equation (ii), (iii) and (iv), we get
$V_{2} (\frac{1}{1 / s C} + \frac{1}{R}) = \frac{V_{i}}{R}$
$(\frac{V_{i} + V_{0}}{2}) (s C + \frac{1}{R}) = \frac{V_{i}}{R}$
$\Rightarrow \frac{V_{0}}{V_{i}} (s) = \frac{1 - s C R}{1 + s C R}$
$= - [\frac{s - ω_{0}}{s + ω_{0}}] (W h e r e, ω_{0} = \frac{1}{R C})$
$∴$ It has one zero at $s = ω_{0}$ and one pole at $s = - ω_{0}$

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