Solution:

Steps involved in analog to digital conversion,

1) Sampling CT - DT

2) Quantization CA-DA

3) Encoding

$ \mathrm{f}_{\mathrm{s}} \geq 2 \mathrm{f}_{\mathrm{m}}$ samples / sec Critical rate / Nyquist rate: $f_s=2 f_m \\ $

$ \Rightarrow \mathrm{f}_{\mathrm{s}}=2 \times 3.4 \mathrm{k}=6800 \text { samples / sec }\\ $

Practical rate:

$ \mathrm{f}_{\mathrm{s}}\gt2 \mathrm{f}_{\mathrm{m}}\\ $

$ \Rightarrow \mathrm{f}_{\mathrm{s}}=8000 \text { samples / } \mathbf{c}\\ $

Sampling period: $T=1 / \mathrm{f}_{\mathrm{s}}$

$ \Rightarrow \mathrm{T}=1 / 8000=125 \mu \mathrm{s} \text { samples } / \mathbf{c}\\ $

Bit rate : $\mathrm{R}_{\mathrm{s}}=\mathrm{n} . \mathrm{f}_{\mathrm{s}}$ bits $/ \mathrm{sec}$

$ \begin{aligned}\\ &=8 \text { bits } / \text { sample } \times 8000 \text { samples } / \mathrm{sec} \\\\ &=64000 \text { bits } / \mathrm{sec}\\ \end{aligned}\\ $