Four channels are multiplexed using TDM. If each channel sends 100 bytes /s and we multiplex 1 byte per channel, show the frame traveling on the link, the size of the frame, the duration of a frame, the frame rate, and the bit rate for the link.
Question
Four channels are multiplexed using TDM. If each channel sends 100 bytes /s and we multiplex 1 byte per channel, show the frame traveling on the link, the size of the frame, the duration of a frame, the frame rate, and the bit rate for the link.
Solution
Sure, let's break this down:
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Frame Traveling on the Link: Since we are multiplexing 1 byte per channel and we have 4 channels, the frame traveling on the link will consist of 4 bytes, each byte from a different channel.
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Size of the Frame: As mentioned above, the size of the frame will be 4 bytes (1 byte from each channel).
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Duration of a Frame: Since each channel sends 100 bytes per second and we are taking 1 byte from each channel to form a frame, it will take 1/100 seconds to form a frame. Therefore, the duration of a frame is 1/100 seconds or 0.01 seconds.
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Frame Rate: The frame rate is the reciprocal of the duration of a frame. Therefore, the frame rate is 100 frames per second.
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Bit Rate for the Link: The bit rate is the number of bits transmitted per second. Since each byte consists of 8 bits and we are transmitting 100 frames per second (with each frame being 4 bytes or 32 bits), the bit rate for the link is 100 frames/second * 32 bits/frame = 3200 bits per second (bps).
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