a) Given BW=100Hz, so the min.sampling rate needed = 200Hz (nyquist sampling theorm).

b) given 0.1% accuracy => (delta / 2) = (1 / 1000) => delta = (2 / 1000)

But delta is ( max range - min range)/L (where L is the number of quantization levels)

c) so ( [ 10-(-10) ] / L ) = 2 / 1000 => L = 10,000

The number of bits needed are log(L) (base 2) . Here log(10000) to the base 2 is 13.278 = 14 (higher number)

Therefore 14 bits are needed for each PCM word

d) Min.bit rate = (no.of bits)*(sampling rate) = 14*200 = 2800 bits/sec