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Four channels are multiplexed using TDM. If each channel sends 100 bytes/second and we multiplex 1 byte per channel, then the bit rate for the link is ……………bps
Correct answer is '3200'. Can you explain this answer?
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Verified Answer
Four channels are multiplexed using TDM. If each channel sends 100 byt...
Firstly we understand the working of Multiplexer using TDM.
Each frame contain 1 byte from each channel; { 1 Byte = 8 Bit}
we have 4 channels
Then ,the size of each frame is
1x4= 4 bytes = =~ 32 bits.
Each channel is sending 100 bytes/second
A frame carries 1 byte from each channel,
The frame rate must be 100 frames per second. { its given}
Then the total rate is = 32*100 = 3200 bits/second
Each frame contain 1 byte from each channel; { 1 Byte = 8 Bit}
we have 4 channels
Then ,the size of each frame is
1x4= 4 bytes = =~ 32 bits.
Each channel is sending 100 bytes/second
A frame carries 1 byte from each channel,
The frame rate must be 100 frames per second. { its given}
Then the total rate is = 32*100 = 3200 bits/second
Most Upvoted Answer
Four channels are multiplexed using TDM. If each channel sends 100 byt...
800 bits/second.
Explanation:
- Each channel sends 100 bytes/second, which is equal to 800 bits/second (since 1 byte = 8 bits).
- We are multiplexing 4 channels, each sending 1 byte per time slot. Therefore, in each time slot, we are transmitting 4 bytes, which is equal to 32 bits (since 1 byte = 8 bits).
- Since we are transmitting 32 bits in each time slot, and there are 4 time slots per cycle, the total bit rate for the link is 32 x 4 = 128 bits/cycle.
- However, since we are only transmitting 1 byte per cycle (i.e., 8 bits), we need to divide the total bit rate by 8. Therefore, the final bit rate for the link is 128/8 = 16 bytes/second = 800 bits/second.
Explanation:
- Each channel sends 100 bytes/second, which is equal to 800 bits/second (since 1 byte = 8 bits).
- We are multiplexing 4 channels, each sending 1 byte per time slot. Therefore, in each time slot, we are transmitting 4 bytes, which is equal to 32 bits (since 1 byte = 8 bits).
- Since we are transmitting 32 bits in each time slot, and there are 4 time slots per cycle, the total bit rate for the link is 32 x 4 = 128 bits/cycle.
- However, since we are only transmitting 1 byte per cycle (i.e., 8 bits), we need to divide the total bit rate by 8. Therefore, the final bit rate for the link is 128/8 = 16 bytes/second = 800 bits/second.
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Four channels are multiplexed using TDM. If each channel sends 100 bytes/second and we multiplex 1 byte per channel, then the bit rate for the link is ……………bpsCorrect answer is '3200'. Can you explain this answer?
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