Question

3.8 Twenty-four voice signals are sampled uniformly and then time-division multiplexed n sampling operation uses flat-top samples with 1 us duration. The multiplexing oper includes provision for synchronization by adding an extra pulse of sufficient ampli and also 1 us duration. The highest frequency component of each voice signal is 3.4 (a) Assuming a sampling rate of 8 kHz, calculate the spacing between successive pule of the multiplexed signal. (b) Repeat your calculation assuming the use of Nyquist rate sampling.

Answer 1

3.8 to Huttiplexed ynal n=8 Commutator M= 24 * 'In Vorce signal transmission, practically the Sampling rate Vis) = to00 farmples /sec hor of bits/sample No- of Stinals flultiplexed it= 24 gnals Synchronization a' = 1 bit (apulse) extra 8 bits =n 13D "No. of Samples generated in 1 frame is 11- teu HyATs Ts (nMta) bits betueen Suciessive pulies |rampley The spacing of the Multiplexed signal is YR000 Tb = Ts (NMta) = 0.647 Msec [8(ew)+1]

Vorce 1gnal=> Ny quist rate = e Wmox (200 H2 to 3500 H3 ) Given klmas = 3.4 KH NR = 2 (3-4 x1) = 6-8K = 6800 Sarmples/see C) = NR = 6 P00 samples/iee %3D spacing between succesive pules f Multiplexed fignal is Ts Tb = (n1 +a) T = [%200] [8(24) + 1] = 0761 Hsec @3.9 @ FDM- SIB lodulation 1ok fok tok = 120 KH3 BW= 12X 10 *rlinimum BW requered = 120 KH?

TDM- NR= 2hlmar= 2x10K= 20K H2 Commutator (12 Lampling Yate sf Tuttplezed signal is e40 K Samples/lee = 12 x 20K famples/ rec %3D * The minimum channel Band wid th Yequired is Sormpling rate f ultiplered stgnal e40 K %3D BW = - 120KH2. %3D