1) Consider a digital multimedia transmission system. Separate audio and video c

Question

1) Consider a digital multimedia transmission system. Separate audio and video coders are used to code these signals. For the transmission purpose audio and video bits are packed in packets together and transmitted over a transmission link. Each packet is 1000 bytes long that includes the header bits. The system generates a packet after every 20 ms. Calculate following values: a) Minimum data transmission rate required to support the above system. b) Calculate the minimum SNR required in dB to transmit above audio and video information if a 200 kHz transmission channel bandwidth is used. c) Calculate end to end delay to transmit a 1 MB video file over a 100 km transmission link using a 1 Mbits/sec transmission data rate. The system uses a packet size of 1000 bytes. Assume pipeline packet transmission.

Explanation / Answer

a) Data transmission/transfer rate is the rate at which data is transferred over a given period in time. Hence transmission rate would be 1000/ (20 x 10-3)B/s = 50KBps that is 50 kilobytes per second

b) According to Shannon theorem, C = B log2 (1 + S/N)

where C is the achievable channel capacity, B is the bandwidth of the line, S is the average signal power and N is the average noise power and S/N is signal-to-noise ratio

Also decibel level is dB = 10 log10 S/N

Here, C = 50KBps = 400Kbps, B = 200kHz putting these values -

400 x 103 = 200 x 103  log2 (1 + S/N)

2 =  log2 (1 + S/N)

4 =  (1 + S/N)

S/N = 3

S/N(in dB) = 10 log10 3 = 4.77dB

c) Packet will first experience a transmission delay. That is, all the bits of a packet have to be put out on the link Transmission delay = packet size / Transimission rate

Here transmission delay is (1000 x 8)b / 106 bps = 8ms

Next is propagation delay. That is, the bits have to propagate at the speed of waves in the transmission medium to reach the other end. Propagation delay = length of the link / Transimission rate

Here propagation delay is (100 x 103)km / 106 bps = 100ms that is once a bit is sent, it takes 100ms to reach end of link

I don't kbow properly about end to end delay but i have read somewhere that it is sum of all the delays experienced Hence according to me it is 8+100 = 108ms

Hope i have answered your question satisfactorily.Leave doubts in comment section if any.

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