Digital Signal Processing Basics
A 6 kHz sine wave is sampled at a sampling frequency of 48 kHz. Which of the following is the normalized angular frequency ω (the phase advance per sample) of the resulting discrete-time signal?
- A0.125 rad/sample
- Bπ/8 rad/sample
- Cπ/4 rad/sample
Show answer and explanation
Answer: C. π/4 rad/sample
The normalized angular frequency is ω = 2πf/fs and tells how far the phase advances during one sample. ω = 2π × 6 kHz ÷ 48 kHz = 2π × 0.125 = π/4 rad/sample. f/fs = 0.125 means "0.125 cycles per sample", so one period takes 8 samples. The Nyquist frequency of 24 kHz corresponds to ω = π, and discrete-time frequencies are considered in the range −π to π.
- A. This is f/fs, whose unit is cycles per sample. To express it in radians, multiply by 2π.
- B. This multiplies by π instead of 2π (π × 6/48 = π/8).
- C. Correct. 2π × 6/48 = π/4 rad/sample.