Question

The transfer function of a Butterworth filter is given by:

• A. \( H(j\Omega) = \frac{6}{1 + \left( \frac{1}{\Omega_c} \right)^N} \)
• B. \( H(j\Omega) = \frac{1}{1 + j \left( \frac{2\Omega}{\Omega_c} \right)^N} \)
• C. \( H(j\Omega) = \frac{1}{1 + j \left( \frac{\Omega}{\Omega_c} \right)^N} \)
• D. \( H(j\Omega) = \frac{N}{1 + \left( \frac{\Omega}{2\Omega_c} \right)^N} \)

Hint:

- The Butterworth filter provides a maximally flat frequency response in the passband. - The standard low-pass Butterworth transfer function is: \[ H(j\Omega) = \frac{1}{1 + j \left( \frac{\Omega}{\Omega_c} \right)^N} \] where \( \Omega_c \) is the cutoff frequency and \( N \) is the order of the filter.

Answer 1

The Correct Option is C

Step 1: Understanding Butterworth Filter
- A Butterworth filter is a type of low-pass filter that provides maximum flatness in the passband.
- The transfer function of an N-order Butterworth filter is given by: \[ H(j\Omega) = \frac{1}{1 + j \left( \frac{\Omega}{\Omega_c} \right)^N} \] where:
- \( \Omega_c \) is the cutoff frequency,
- \( N \) is the filter order.
Step 2: Evaluating the Options
- (A) Incorrect: The denominator should not include \( \left( \frac{1}{\Omega_c} \right)^N \).
- (B) Incorrect: The term \( 2\Omega \) in the denominator is incorrect.
- (C) Correct: Matches the standard Butterworth filter transfer function.
- (D) Incorrect: The factor \( N \) in the numerator and \( 2\Omega_c \) in the denominator are incorrect.