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 \({{\rm{V}}_{{\rm{input}}}} = {{\rm{A}}_{\rm{m}}}\sin \left( {{{\rm{\omega }}_{\rm{m}}}{\rm{t}}} \right)\)
\({\rm{slop}}{{\rm{e}}_{{\rm{input}}}} = {\left( {\frac{{{\rm{d}}{{\rm{V}}_{{\rm{input}}}}}}{{{\rm{dt}}}}} \right)_{{\rm{max}}}} = {{\rm{A}}_{\rm{m}}}{{\rm{\omega }}_{\rm{m}}}\)
\(\\ {{\rm{A}}_{\rm{m}}}{{\rm{\omega }}_{\rm{m}}} \le \frac{{\rm{\Delta }}}{{{{\rm{T}}_{\rm{s}}}}} = \frac{{{\rm{step\;size}}}}{{{\rm{sampling\;interval}}}} = {\rm{\Delta \;}}{{\rm{f}}_{\rm{s}}}\)
\({{\rm{A}}_{\rm{m}}}2{\rm{\pi }}\left( {2 × {{10}^3}} \right) \le 2 × {10^4} × 0.1\)
\(\therefore {{\rm{A}}_{\rm{m}}} \le \frac{1}{{2{\rm{\pi }}}} \)