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,
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- 4 times the Nyquist rate
- 6 times the Nyquist rate
- 12 times the Nyquist rate
- 3 times the Nyquist rate
The Correct Option is B
Solution and Explanation
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.
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