Question

With reference to the Fast Fourier Transform, if \( W_4^1 = W_x^2 \), then what is the value of \( x \)?

• A. 2
• B. 4
• C. 8
• D. 16

Hint:

The twiddle factor in FFT is given by \( W_N = e^{-j\frac{2\pi}{N}} \). When equating twiddle factors, always compare exponents carefully.

Answer 1

The Correct Option is B

Step 1: The twiddle factor in the FFT is defined as: \[ W_N = e^{-j\frac{2\pi}{N}} \] where \( W_N^k \) represents the \( k \)th power of the twiddle factor for an \( N \)-point FFT. 
Step 2: Given: \[ W_4^1 = W_x^2 \] Substituting the definition of the twiddle factor: \[ e^{-j\frac{2\pi}{4} \times 1} = e^{-j\frac{2\pi}{x} \times 2} \] 
Step 3: Simplifying: \[ e^{-j\frac{2\pi}{4}} = e^{-j\frac{4\pi}{x}} \] Equating exponents: \[ \frac{2\pi}{4} = \frac{4\pi}{x} \] \[ \frac{\pi}{2} = \frac{4\pi}{x} \] Solving for \( x \): \[ x = 4 \] 
Step 4: Evaluating options:
- (A) Incorrect: \( x = 2 \) does not satisfy the equation.
- (B) Correct: \( x = 4 \) is the correct answer.
- (C) Incorrect: \( x = 8 \) does not match.
- (D) Incorrect: \( x = 16 \) is incorrect.