Question

An analog signal is sampled at 100 MHz to generate 1024 samples. Only these samples are used to evaluate 1024-point FFT. The separation between adjacent frequency points (\( \Delta F \)) in FFT is \(\underline{\hspace{2cm}}\) kHz.

• A. 102.16
• B. 97.66
• C. 100.00
• D. 95.63

Hint:

When calculating the frequency resolution of FFT, divide the sampling frequency by the number of FFT points.

Answer 1

The Correct Option is B

The separation between adjacent frequency points in the FFT is given by the formula: \[ \Delta F = \frac{F_s}{N}, \] where \( F_s \) is the sampling frequency and \( N \) is the number of points in the FFT.
For this problem: \[ F_s = 100 \, \text{MHz}, N = 1024. \] Thus, \[ \Delta F = \frac{100 \, \text{MHz}}{1024} = 97.65625 \, \text{kHz}. \] The closest value to this is 97.66 kHz.
Final Answer: 97.66 kHz