Question

If the data rate of delta modulator output is 43.2 kbps, for the input signal of 3.6 kHz, then the sampling rate used is equal to,

• A. 4 times the Nyquist rate
• B. 6 times the Nyquist rate
• C. 12 times the Nyquist rate
• D. 3 times the Nyquist rate

Hint:

Remember the data rate in delta modulation is same as the sampling frequency. Nyquist frequency is twice the signal frequency.

Answer 1

The Correct Option is B

Step 1: First we calculate the Nyquist frequency: \[ f_{nyquist} = 2 \times 3.6 kHz = 7.2 kHz \] 
Step 2: The sampling frequency \(f_s\) is same as the data rate in a delta modulator, that is 43.2 kbps. 
Step 3: To find what multiple of Nyquist rate the sampling frequency is, we calculate: \[ \frac{43.2 kHz}{7.2 kHz} = 6 \] Therefore, the sampling frequency is 6 times the Nyquist rate.