Question

A 4 kHz sinusoidal message signal having amplitude 4 V is fed to a delta modulator (DM) operating at a sampling rate of 32 kHz. The minimum step size required to avoid slope overload noise in the DM (rounded off to two decimal places) is V.

Hint:

To avoid slope overload in delta modulation, the step size must be chosen based on the sampling rate and the message signal frequency.

Answer 1

Correct Answer: 2.8

The minimum step size for delta modulation is given by: \[ \Delta = \frac{2A}{\sqrt{f_s \cdot f_m}} \] where: - \( A = 4 \, \text{V} \) is the amplitude of the message signal, - \( f_s = 32 \, \text{kHz} \) is the sampling rate, - \( f_m = 4 \, \text{kHz} \) is the message signal frequency. Substituting the values: \[ \Delta = \frac{2 \times 4}{\sqrt{32 \times 4}} = \frac{8}{\sqrt{128}} \approx 2.80 \, \text{V} \] Thus, the minimum step size required is \( 2.80 \, \text{V} \).