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UGC NET CSE | June 2016 | Part 2 | Question: 26

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ans should be C

The link carries 50000 frames per sec as each frame contains 2 bits per channel ( 100 kbps/2=50kbps)

frame duration is therefore 1/50000 sec= 20 microseconds

Bit rate =frame rate x no of  bits per frame =50000x 8 =400 kbps hence option C
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Answer C

Total bandwidth of input channels = 100 * 1000 * 4 bits/sec

Transmission time by all four channels in 2 bit slot is = 2/ ( 100 * 1000 *4 ) = 0.000005 mlsc

now question is in one sec how many bits you get are transferring ? 

That is = 1/(0.000005) *2 bits/sec = 400Kbps

 

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