0 0 votes The minimum step-size required for a Delta-Modulator operating at $32 \mathrm{~K}$ samples/ $\mathrm{sec}$ to track the signal (here $u(t)$ is the unit-step function) $\begin{array}{l}x(t)=125 t(u(t)-u(t-1))+(250-125 t)(u(t-1) -u(t-2)) \end{array}$ so that slope-overload is avoided, would be $2^{-10}$ $2^{-8}$ $2^{-6}$ $2^{-4}$ Continuous-time Signals gate2006-ec analog-communications signals-and-systems numerical-answers + – admin 240 views answer comment Share Follow Add Sync Questions Print 0 reply Please log in or register to add a comment.
