Question

A two-stage RC-coupled amplifier has stage-1 lower cutoff frequency \(f_{L1} = 100\text{ Hz}\) and stage-2 lower cutoff frequency \(f_{L2} = 1\text{ kHz}\). The overall lower cutoff frequency of the amplifier is approximately:

• A. \(50\text{ Hz} \)
• B. \(100\text{ Hz} \)
• C. \(1\text{ kHz} \)
• D. \(1.1\text{ kHz} \)

Hint:

For cascaded systems with non-identical stages, the overall lower cutoff frequency is always dominated by the highest individual lower frequency value (\(f_{L, \text{overall}} \approx \text{max}(f_{Li})\)).

Answer 1

The Correct Option is C

Concept: When multiple amplifier stages are cascaded in series, the overall bandwidth of the system shrinks. The lower cutoff frequency (\(f_L\)) increases, while the upper cutoff frequency (\(f_H\)) decreases. For a multi-stage amplifier with non-identical stages having lower cutoff frequencies \(f_{L1}, f_{L2}, \dots, f_{Ln}\), the overall lower cutoff frequency can be estimated using the dominant pole approximation method. This rule states that the stage with the highest lower cutoff frequency dominates the overall low-frequency response: \[ \text{Overall } f_L \approx \sqrt{f_{L1}^2 + f_{L2}^2 + \dots + f_{Ln}^2} \] Alternatively, if one frequency is significantly larger than the others, it acts as the dominant limit, meaning the overall lower cutoff frequency will be close to or slightly higher than that largest individual cutoff frequency value.

Step 1: Analyzing the given stage frequencies.
We are given two distinct lower cutoff frequencies for the cascaded stages:
• \(f_{L1} = 100\text{ Hz} = 0.1\text{ kHz}\)
• \(f_{L2} = 1\text{ kHz} = 1000\text{ Hz}\) Comparing the two values, \(f_{L2}\) (\(1\text{ kHz}\)) is ten times larger than \(f_{L1}\) (\(100\text{ Hz}\)). This makes \(f_{L2}\) the dominant pole limiting the low-frequency performance of the cascaded system.

Step 2: Applying the approximation formula to determine the precise trend.
Using the standard multi-stage calculation formula: \[ f_{L, \text{overall}} \approx \sqrt{(100)^2 + (1000)^2} \] Calculate the squares of the frequencies: \[ (100)^2 = 10,000 \] \[ (1000)^2 = 1,000,000 \] Sum the values inside the square root: \[ 10,000 + 1,000,000 = 1,010,000 \] Now calculate the square root of the total sum: \[ f_{L, \text{overall}} \approx \sqrt{1,010,000} \approx 1004.99\text{ Hz} \approx 1.005\text{ kHz} \] Evaluating the choices, this calculated value is approximately \(1\text{ kHz}\). This matches Option (C).