Consider an ideal op-amp circuit shown in the figure below:

The transfer function for the circuit can be given as

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Solution

The correct option is A

$Now, V_{b} = \frac{V_{0}}{4}$

Applying KCL, we get,

$\frac{V_{a} - V_{i}}{1 k Ω} + \frac{V_{a} - V_{1}}{2 k Ω} = 0$

$2 V_{i} + V_{1} - 3 V_{a} = 0$

$\frac{V_{2} - V_{1}}{4 k Ω} + \frac{V_{2} - V_{0}}{8 k Ω} = 0$

$3 V_{2} - 2 V_{1} - V_{0} = 0$

$V_{2} = - 3 V$

$3 (- 3) - 2 V_{1} = V_{0}$

$V_{1} = - \frac{1}{2} (9 + V_{0})$

$2 V_{i} - \frac{1}{2} (9 + V_{0}) - 3 (\frac{V_{0}}{4}) = 0$

$2 V_{i} - 4.5 - \frac{1}{2} V_{0} - 3 \frac{V_{0}}{4} = 0$

$\frac{5}{4} V_{0} = 2 V_{i} - 4.5$

$V_{0} = \frac{4}{5} (2 V_{i} - 4.5)$

$v_{0} = 1.6 V_{i} - 3.6$

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