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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Brief, authoritative clarification and corrected calculations for both parts (assumes standard telephony PCM where the sampling rate is 8 kHz and each sample is 8 bits).

For Q1 (24 voice signals, synchronous TDM): standard PCM (ITU G.711) samples narrowband voice at 8000 samples/sec and encodes each sample as 8 bits, so each voice channel is 64 kb/s. Multiplying by 24 channels gives 1.536 Mb/s. Many texts then note a practical T1/DS1 line adds a framing bit per frame, which yields 1.544 Mb/s on the physical T1. 

24 channels × 8000 samples/s × 8 bits/sample = 1,536,000 bps (1.536 Mb/s)
T1 framing adds overhead → physical DS1 = 1.544 Mb/s

(itu.int)

For Q2 (10 workstations at 128 kb/s each): with a synchronous TDM multiplexer you must provision a timeslot for every workstation regardless of activity, so provision for the full sum of input rates (10 × 128 kb/s = 1.28 Mb/s). With a statistical (asynchronous) multiplexer you can size the outgoing line for the expected active fraction: if 40% are idle at any instant, 60% are active → 6 active links, so 6 × 128 kb/s = 768 kb/s. Keep in mind statistical multiplexing needs addressing/packet headers and buffering, and it cannot protect against the rare event that all users transmit at once — in that worst case you still need the full 1.28 Mb/s or you suffer loss/delay. 

Sync TDM: 10 × 128 kb/s = 1.28 Mb/s
Statistical: 0.6 × 10 × 128 kb/s = 768 kb/s (plus framing/headers/queueing)

(en.wikipedia.org)

Notes on earlier replies: was on the right track about aggregating channels but misstated a few figures; confused quantization levels with the sample rate (8 bits gives 256 quantization levels, but sampling rate is 8000 samples/sec for telephony); ’s “1000bps” is not correct for this problem. For homework, state explicitly whether the question expects raw PCM bandwidth or physical-line framing/packet overhead — textbooks sometimes expect the raw 1.536 Mb/s answer, while carrier framing produces 1.544 Mb/s. (en.wikipedia.org)

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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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