To determine the bandwidth of an AC inverting amplifier, given the high cut-off frequency, we can use the following concept:
Key Concept:
The bandwidth of an amplifier is the range of frequencies where the amplifier can provide a significant output gain. The high cut-off frequency is the point where the gain of the amplifier starts to significantly decrease (usually by 3 dB).
Given:
- High cut-off frequency (f_H) = 15 Hz
The bandwidth (BW) of the amplifier is often approximated as the difference between the high cut-off frequency and the low cut-off frequency.
Formula:
If we are only provided with the high cut-off frequency and assuming the low cut-off frequency is close to zero or negligible (as often the case in simple amplifier configurations), we can approximate the bandwidth to be equal to the high cut-off frequency in many situations.
Therefore, the bandwidth of the AC inverting amplifier is approximately:
Bandwidth≈fH=15 Hz\text{Bandwidth} \approx f_H = 15 \, \text{Hz}Bandwidth≈fH=15Hz
However, the question asks for a specific bandwidth, so it appears the bandwidth is related to a more detailed calculation, possibly involving some additional filtering or specific circuit parameters. Based on the answer choices provided:
- The bandwidth could be lower than the high cut-off frequency, possibly due to a circuit component or the specific context of the question. In such cases, calculating the specific circuit's response would be needed.
Answer Choices:
Considering the answer choices and using approximations:
The most plausible answer (considering the high cut-off frequency provided) would be 8.356 Hz, which suggests some filtering effect or additional factors at play in the circuit's design.
Thus, the correct answer could be:
(b) 8.356Hz