Question

The transfer function of the circuit as shown in the figure is expressed as

• A. $\frac{R}{1 + sRC}$
• B. $\frac{s}{1 + sRC}$
• C. $\frac{1}{1 + sRC}$
• D. $\frac{C}{1 + sRC}$

Hint:

This is a standard passive low-pass filter.
At $s = 0$ (DC), the capacitor behaves as an open circuit, so the gain is 1.
At $s \rightarrow \infty$ (high frequency), the capacitor behaves as a short circuit, so the gain is 0. Only option (C) satisfies these limits.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the transfer function $H(s) = \frac{V_{out}(s)}{V_{in}(s)}$ of a passive first-order RC network.

Step 2: Key Formula or Approach:
Using s-domain impedance representations:
- Resistor: $Z_R = R$
- Capacitor: $Z_C = \frac{1}{sC}$
We apply the voltage divider rule to determine the output voltage across the capacitor.

Step 3: Detailed Explanation:

• By voltage division across the series RC circuit:
\[ V_{out}(s) = V_{in}(s) \cdot \frac{Z_C}{Z_R + Z_C} \]

• Substituting the values:
\[ H(s) = \frac{V_{out}(s)}{V_{in}(s)} = \frac{\frac{1}{sC}}{R + \frac{1}{sC}} \]

• Multiplying both the numerator and the denominator by $sC$:
\[ H(s) = \frac{1}{sRC + 1} = \frac{1}{1 + sRC} \]

Step 4: Final Answer:
The transfer function of the circuit is $\frac{1}{1 + sRC}$, which corresponds to Option (C).