Difficulty: Medium
Correct Answer: about 2.38 Hz
Explanation:
Introduction / Context:Negative feedback reshapes an amplifier’s frequency response, typically trading gain for improved bandwidth and more controlled cutoff characteristics. In RC-coupled voltage amplifiers, applying negative feedback reduces the midband gain but extends both the low- and high-frequency bandwidth, often by approximately the desensitivity factor (1 + Aβ) for dominant single-pole regions.
Given Data / Assumptions:
Concept / Approach:For a simple first-order low-frequency pole, negative feedback reduces the effective time constant by approximately the factor (1 + Aβ). Thus the lower cutoff shifts downward by the same factor, improving low-frequency response. Closed-loop gain also reduces to A_cl ≈ A / (1 + Aβ).
Step-by-Step Solution:
Compute desensitivity: 1 + Aβ = 1 + 200 * 0.1 = 21.New lower cutoff: f_L,new = f_L / (1 + Aβ) = 50 / 21 ≈ 2.38 Hz.Closed-loop midband gain (for reference): A_cl ≈ 200 / 21 ≈ 9.52.Verification / Alternative check:
If you sketch Bode plots, the low-frequency pole moves left by a factor of ~21, consistent with the calculated ~2.4 Hz.Why Other Options Are Wrong:
About 50 Hz: ignores bandwidth extension from feedback.About 5 Hz: uses a factor of 10 instead of 21.About 70.5 Hz: would be a narrowing of bandwidth (contrary to negative feedback benefits).About 21 Hz: misapplies (1 + Aβ) as a multiplier rather than a divider.Common Pitfalls:
Forgetting that negative feedback broadens bandwidth while reducing midband gain; mixing up low- and high-frequency corner shifts.Final Answer:
about 2.38 Hz
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