1 1 vote A speech signal, band limited to $4 \; \mathrm{kHz}$ and peak voltage varying between $+5 \mathrm{V}$ and $-5 \mathrm{V}$, is sampled at the Nyquist rate. Each sample is quantized and represented by $8$ bits. If the bits $0$ and $1$ are transmitted using bipolar pulses, the minimum bandwidth required for distortion free transmission is $64 \; \mathrm{kHz}$ $32 \; \mathrm{kHz}$ $8 \; \mathrm{kHz}$ $4 \; \mathrm{kHz}$ Digital Communications gate2008-ec analog-communications digital-communications sampling-theorem information-theory + – admin 485 views answer comment Share Follow Add Sync Questions Print 0 reply Please log in or register to add a comment.
0 0 votes Binary sequence represented by bipolar pulses. So, $$ \begin{aligned} \text { B.W. } & =R_{b}=n f_{s} \\ & =8 \times 8 \mathrm{~K}=64 \mathrm{kHz} \end{aligned} $$ GO Classes answered Jun 15, 2025 GO Classes comment Share Follow 0 reply Please log in or register to add a comment.