Question

The midband gain of an amplifier is 100 and the lower cutoff frequency is 1 KHz. Find the gain of the amplifier at a frequency of 20 Hz.

• A. 2
• B. 20
• C. 50
• D. 100

Hint:

For a first-order high-pass filter (which determines the lower cutoff frequency), the gain roll-off below \(f_L\) is 20 dB/decade.
Gain magnitude: \(|A(f)| = A_{mid} / \sqrt{1 + (f_L/f)^2}\).
When \(f \ll f_L\), then \(f_L/f \gg 1\), so \(|A(f)| \approx A_{mid} / (f_L/f) = A_{mid} \cdot (f/f_L)\). In this case: \(100 \cdot (20/1000) = 100 \cdot (1/50) = 2\).

Answer 1

The Correct Option is A

For a single-pole high-pass response (typical for lower cutoff frequency of an RC-coupled amplifier), the gain \(A(f)\) at a frequency \(f\) relative to the midband gain \(A_{mid}\) is given by: \[ |A(f)| = \frac{A_{mid}}{\sqrt{1 + (f_L/f)^2}} \] where \(f_L\) is the lower cutoff frequency. Given: \(A_{mid} = 100\), \(f_L = 1 \text{ KHz} = 1000 \text{ Hz}\), and \(f = 20 \text{ Hz}\). Calculate \(f_L/f\): \[ \frac{f_L}{f} = \frac{1000 \text{ Hz}}{20 \text{ Hz}} = 50 \] Now calculate the gain: \[ |A(20 \text{ Hz})| = \frac{100}{\sqrt{1 + (50)^2}} = \frac{100}{\sqrt{1 + 2500}} = \frac{100}{\sqrt{2501}} \] We know \(\sqrt{2500} = 50\). So, \(\sqrt{2501}\) is slightly greater than 50. \(\sqrt{2501} \approx 50.01\). \[ |A(20 \text{ Hz})| \approx \frac{100}{50.01} \approx \frac{100}{50} = 2 \] The gain at 20 Hz is approximately 2. This matches option (a). \[ \boxed{2} \]