Discussion
A 50 Hz noise signal of 1 mV peak value is introduced at the input of an op-amp based integrator (RC = 1 s). The output will be :
Answer: B
The output will be about 3.2 \(\mu\)V.
An integrator's gain at angular frequency \(\omega\) is \(1/(\omega RC)\). For the 50 Hz signal:
\[ \begin{aligned} \omega RC &= 2\pi \times 50 \times 1 = 314 \\ V_{out} &= \frac{1\ \text{mV}}{314} \approx 3.2 \times 10^{-6}\ \text{V} \end{aligned} \]
The integrator strongly attenuates the noise because its gain falls as frequency rises, which is the point of the question. 3.2 mV and 3.2 V would mean the gain was 3.2 or 3200 instead of 1/314, and 3.2 nV is a thousand times too small.
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