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A sinusoidal signal of 2 kHz frequency is applied to a delta modulator. The sampling rate and step-size Δ of the delta modulator are 20,000 samples per second and 0.1V respectively. To prevent slope overload, the maximum amplitude of the sinusoidal signal (in Volts) is
1. \(\frac{1}{{2{\rm{\pi }}}}\)
2. \(\frac{1}{{\rm{\pi }}}\)
3. \(\frac{2}{{\rm{\pi }}}\)
4. \({\rm{\pi }}\)

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Correct Answer - Option 1 : \(\frac{1}{{2{\rm{\pi }}}}\)

Concept: In Delta modulation Slope Overload error occurs if the step size is small or Step size × sampling frequency  ≤ slope of m(t).

                Therefore optimum step size is required to avoid the Slope overload error  i.e The slope of m(t) ≤ Step size × sampling frequency

Application: For preventing slope overload, we should have:

The slope of m(t) ≤ Step size × sampling frequency

For sinusoidal signal, if \({{\rm{V}}_{{\rm{input}}}} = {{\rm{A}}_{\rm{m}}}\sin \left( {{{\rm{\omega }}_{\rm{m}}}{\rm{t}}} \right)\) 

\({\rm{slop}}{{\rm{e}}_{{\rm{input}}}} = {\left( {\frac{{{\rm{d}}{{\rm{V}}_{{\rm{input}}}}}}{{{\rm{dt}}}}} \right)_{{\rm{max}}}} = {{\rm{A}}_{\rm{m}}}{{\rm{\omega }}_{\rm{m}}}\)

\(\\ {{\rm{A}}_{\rm{m}}}{{\rm{\omega }}_{\rm{m}}} \le \frac{{\rm{\Delta }}}{{{{\rm{T}}_{\rm{s}}}}} = \frac{{{\rm{step\;size}}}}{{{\rm{sampling\;interval}}}} = {\rm{\Delta \;}}{{\rm{f}}_{\rm{s}}}\)

\({{\rm{A}}_{\rm{m}}}2{\rm{\pi }}\left( {2 × {{10}^3}} \right) \le 2 × {10^4} × 0.1\)

\(\therefore {{\rm{A}}_{\rm{m}}} \le \frac{1}{{2{\rm{\pi }}}} \)

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