Member Avatar for candysandra87

Hi. I'm a student with two questions that I can't figure out. Can anyone post their opinion, or any additional information? thanks

1) Twenty four voice-signals are multiplexed and transmitted over a twisted pair. What would the bandwidth required be (in bps) if synchronous time division multiplexing was used, and along with the standard analog-to-digital sampling rate, and each sample is converted into 8bit values?

2) 10 computer workstations are to be connected to a synchronous time division multiplexor. And each workstation transmits at 128 kbps. During any point in time, 40 percent of the workstations are not transmitting. What is the minimum necessary speed of the line leaving the multiplexor? Will the answer be different if we use a statistical multiplexor instead?

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Member Avatar for ithelp

I think I have seen these problems in tanentbaum's book.

Member Avatar for sinkrideutan

1) 24 * 64Kbits/sec which is a T1 in SDH

2) 10 * 128k * .6 = 786k
yes

but the better thing to do is to use a regular IP router, because IP in general outperforms statistical multiplexing.

Member Avatar for zabee755

Dear the bandwidth required will be 1000bps:*

Member Avatar for joshua.dumont.5

Ok, like many that try these guys are close. The book is looking for a specific type of answer, also known as the "Right" answer.

Assuming that each signal is running at the same rate,
Each sample is converted into an 8 bit value. The number of samples = 2^8 = 256
Synchronous time division takes a byte from signal for processing. Twenty voice signals transmitted over a twisted pair at an 8 bit value would be represented this way. 202562=10240 bps
The bandwidth required is 10240 bps which can be accommodated with a Cat1 line.

Source
Data Communications and Computer Networks 6e P.140

Edited by joshua.dumont.5
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