Question

Consider an FM broadcast that employs the pre-emphasis filter with frequency response \[ H_{pe}(\omega) = 1 + \frac{j\omega}{\omega_0}, \] where $\omega_0 = 10^{4}$ rad/s. For the network shown in the figure to act as a corresponding de-emphasis filter, the appropriate pair(s) of $(R, C)$ values is/are ________________.

• A. $R = 1\ \text{k}\Omega,\ C = 0.1\ \mu\text{F}$
• B. $R = 2\ \text{k}\Omega,\ C = 1\ \mu\text{F}$
• C. $R = 1\ \text{k}\Omega,\ C = 2\ \mu\text{F}$
• D. $R = 2\ \text{k}\Omega,\ C = 0.5\ \mu\text{F}$

Hint:

For FM emphasis/de-emphasis, always match $RC$ with the standard time constant $1/\omega_0$. A low-pass filter implements de-emphasis, and a high-pass filter implements pre-emphasis.

Answer 1

The Correct Option is A

The pre-emphasis filter has the frequency response \[ H_{pe}(\omega) = 1 + \frac{j\omega}{\omega_0}. \] The corresponding de-emphasis filter should have the inverse response: \[ H_{de}(\omega) = \frac{1}{1 + j\omega/\omega_0}. \]
The given circuit is a standard RC low-pass filter, whose transfer function is: \[ H(\omega) = \frac{1}{1 + j\omega RC}. \] Step 1: Match the denominator terms.
For equivalence: \[ RC = \frac{1}{\omega_0}. \] Given \[ \omega_0 = 10^{4}\ \text{rad/s}, \] so \[ RC = 10^{-4}. \]
Step 2: Check each option.
(A) \[ R = 1\ \text{k}\Omega = 1000,\quad C = 0.1\ \mu\text{F} = 0.1 \times 10^{-6}. \] \[ RC = 1000 \times (0.1 \times 10^{-6}) = 10^{-4} \quad \text{✓ matches}. \] (B) \[ RC = 2000 \times 10^{-6} = 2 \times 10^{-3} \quad (\neq 10^{-4}) \] (C) \[ RC = 1000 \times 2 \times 10^{-6} = 2 \times 10^{-3} \quad (\neq 10^{-4}) \] (D) \[ RC = 2000 \times 0.5 \times 10^{-6} = 10^{-3} \quad (\neq 10^{-4}) \] Only option (A) satisfies the required time constant.
Final Answer: $R = 1\ \text{k}\Omega,\ C = 0.1\ \mu\text{F}$