a)
Given data:
The total number of digital sources=20
Each frame output slot carries 1 bit from each digital source, but one extra bit is added to each frame for synchronization.
Then, the size of the an output frame in bits = 20
1+1
= 21 bits
b)
Each frame carries 1 bit from each source.
Then, the frame rate=100Kbps
=100,000 frames/s
c)
The duration of an output frame=
=
=10
d)
Output data rate= size of an output frame
frame rate
= (100000 frames/s)
(21 bits/frame)
=21
105 bits/s
=2.1
106 bits/s
=2.1 Mbps (since, 1Mbps=106bits/s)
e)
In each frame 20 digital sources out of 21 are useful.
Therefore, the efficiency of the system=20/21
=95%
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