Question

A delta modulator is tested with a 10 kHz sinusodial signal, 1 V peak to peak at input. The signal applied is sampled 10 times the nyquist rate, then the minimum step size required to prevent slope overload is equal

• A. 0.587 Volts
• B. 0.778 Volts
• C. 0.157 Volts
• D. 0.358 Volts

Answer 1

The correct option is C 0.157 Volts

$f_s=10f_N=10\times 20=200kHz$
$\delta f_s\ge \text{Max}\left|\frac{dm\left(t\right)}{dt}\right|$
$\delta \times 200kHz\ge 2\pi f_m{A}_m$
$\delta \ge \frac{2\pi \times (10\times {10}^3)\times \frac{1}{2}}{200kHz}$
$\delta \ge 0.157\text{Volts}$