You have 24 pointtopoint connections that each requires a 64 kbps peak transmission rate. A 64 kbps circuit costs $400 per month. A 384 kbps fractional T1 circuit costs $800 per month. A 1.544...

You have 24 pointtopoint connections that each requires a 64 kbps peak transmission rate. A 64 kbps circuit costs $400 per month. A 384 kbps fractional T1 circuit costs $800 per month. A 1.544 Mbps T1 circuit costs $1,700 per month. Because of burstiness, lines are in use only 10% of the time. Compare the monthly operating costs of transmission with individual connections (no multiplexing), with time division multiplexing (TDM), and with statistical TDM, respectively. A pair of 24port TDM multiplexers lease for $300 a month. A pair of 24port STDM multiplexers lease for $600 a month. Recommend a most economical solution. Hint: Do not consider 3 x 3 = 9 scenarios. For each type of circuit, there is only one (1) sensible solution. (15 points)

You have a modem that can operate at a symbol rate of 3,600. Design a modulation system to allow the modem to transmit at a data rate of 28,800 bps. Your design should vary both the amplitude and the phase of the carrier wave signal. A diagram of your design can be helpful, but it is not required. (10 points)

To get some idea on what is involved in digital transmission of multimedia traffic, complete the following: (15 points)
3.1. In PCM encoding, the general rule is that we need to sample at twice the bandwidth. In
addition, if we use n bits for each sample, we can represent 2n loudness (amplitude) levels. What transmission speed would you need if you wanted to encode and transmit, in real time and without compression, twochannel stereo music with a bandwidth of 20 kHz and 60,000 loudness levels?
3.2. How much disk space would your require to store 40 minutes of stereo music as you calculated above?
3.3. Assume that the 40 minutes of uncompressed stereo music is to be transmitted over a dialup circuit at 56 Kbps. How long will it take to complete the transmission?
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2 Answers
In PCM encoding, sampling at twice the bandwidth is required.
A signal carrying twochannel stereo music with a bandwidth of 20 kHz and 60000 loudness levels has to be transmitted. If n bits are used for each sample it is possible to represent 2^n amplitude levels.
For the given signal, to represent 60000 amplitude levels the number of bits used for each sample is 16 as 2^16 = 65536. With a sample size of 15 bits only 2^15 = 32768 levels could be accommodated.
As a twochannel stereo signal has to be transmitted and assuming that the channels are not interleaved, the number of samples sent per second is 2*2*20*1000 = 80000. The sample size is 16 bits. This gives the required transmission speed as 80000*16 = 1280 Kbps.
To store 40 minutes of stereo music the disc space required is 1280*60*40 = 3072000 Kb = 3072 Mb
At a transmission rate of 56 Kbps, it would take 3072000/56 ~~ 914.28 minutes to complete the transmission.
why did you get "2*2*20*1000 = 80000" ??