Assignment Communication Systems Four channels of 5-bit PCM are multiplexed at 8

Question

Assignment Communication Systems Four channels of 5-bit PCM are multiplexed at 8000 frames/s (with no additional pulses added) Determine the output pulse rate and pulse width. 1. 2. A 48-channel PCM system samples voice at 8 kHz and encodes the samples with 8-bit words ters). If 2 bits are added (for framing) after the 48 channels are multiplexed, determine the following: a. Number of bits/frame. b. Number of frames/second c. Number of bits/second. d. What kind of multiplexing (FDM,TDM, cDM, or FBM) is being used? e. Calculate the absolute minimum (Nyquist) bandwidth required.

Explanation / Answer

Answer:-1) Number of channels, N = 4, Number of bits for PCM coding B = 5, frame rate F = 8000 frames/s. Here inone frame 4 channel's data is being transmitted and data of each channel is of 5-bit. So rate in bit/s is the pulse rate, hence pulse rate = 8000*4*5 Hz = 160000 Hz = 0.16 MHz. So pulse width = 1/160000 s = 6.25 micro-s.

Answer:-2)a) Number of bits per frame = (48*8) + 2 bits = 386 bits
b) Number of frame per second = sampling rate = 8000 frame/s
c) Number of bits per second = frame per second * number of bits per frame = 8000*386 bps = 3.088 Mbps
d) The multiplexing that is being used here is TDM, each channel data is transferred one after another and each channel has it's own time to send.
e) Since bit rate is 3.088 Mbps, hence minimum Nyquist bandwidth required to transfer all bits = 3.088/2 MHz
= 1.544 MHz.

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