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Computer Networks Chapter 3: Digital transmissions fundamentals Part 1.

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Presentation on theme: "Computer Networks Chapter 3: Digital transmissions fundamentals Part 1."— Presentation transcript:

1 Computer Networks Chapter 3: Digital transmissions fundamentals Part 1

2 Digital representation of information ● Physical layer deals with Digital comm. ● Information is inherently redundant, ex.: picture of sky, english text... ● Compression takes advantage to reduce info to be transmitted. ● Actual compression level depends on material: – Zip, compress (2-6) – JPEG (5-30)

3 H W =++ H W H W H W Color Image Red Component Image Green Component Image Blue Component Image Total bits before compression = 3xHxW pixels x B bits/pixel = 3HWB ● Ex: 8x10 in image, 8 bits/color/pix, 400x400 pix /in^2 resolution => 38.4 MB ● Compression approaches: * GIF – Graphics Interchange Format, lossless compression, limited ratios * JPEG – Joint Photographic Experts Group, lossy compression high ratios 15

4 Streamed info: voice, music, video ● Unlike previous form, compression has to be done on the fly. ● First: Sampling, rate depends on signal variation, highest freq component (PCM: 8000 samples/s) ● Second: Resolution: how many levels (PCM: 8 bits) ● May be combined with compression

5                 (a) (b)

6 Network requirements: ● Transmission errors: Compressed data more susceptible to errors. ● Delay: Interactive applications are more sensitive to delay. ● Jitter: Real time applications can tolerate little jitter, but can tolerate fixed delay. ● Recently minimum measures of these requirements define a QoS.

7 (a) QCIF Videoconferencing (b) Broadcast TV (c) HDTV @ 30 frames/sec = 760,000 pixels/sec @ 30 frames/sec = 10.4 x 10 6 pixels/sec @ 30 frames/sec = 67 x 10 6 pixels/sec 720 480 1080 1920 144 176

8 Playout delay Jitter due to variable delay (b) (c) 123456789 123456 Original sequence (a) 123456789

9 Digital vs Analog comm: ● Analog: Continuous waveform, to regenerate – repeaters => can't remove inband noise ● Digital: Discrete pulses, to regenerate – digital repeaters => signal is perfectly regenerated every time. ● Passage thtough multiple analog repeaters damages signal, not the case for digital repeaters ● Digital networks can handle multiple services

10 (a) Analog transmission: all details must be reproduced accurately Sent Received e.g digital telephone, CD Audio (b) Digital transmission: only discrete levels need to be reproduced e.g. AM, FM, TV transmission

11 Attenuated & distorted signal + noise Equalizer Recovered signal + residual noise Repeater Amp. Analog repeater Amplifier Equalizer Timing Recovery Decision Circuit. & Signal Regenerator Digital repeater

12 Amplitude response: ● Fundamental limit: How fast can we transmit data reliably over a given channel? Depends on: – Energy put into every signal – Distance to be traversed – Noise in the channel – Bandwidth of medium ● The fastest signaling rate over a channel of bandwidth W is r=2W signal per second.

13 f 0 W A(f)A(f) (a) Lowpass and idealized lowpass channel (b) Maximum pulse transmission rate is 2W pulses/second 0W f A(f)A(f) 1 Channel t t

14 signal noise signal + noise signal noise signal + noise High SNR Low SNR SNR = Average Signal Power Average Noise Power SNR (dB) = 10 log 10 SNR t t t t t t

15 f 1 A(f) = 1 1+4  2 f 2 f 0  (f) = tan -1 2  f -45 o -90 o 1/ 2  Frequency response of channel Phase response of channel

16 Channel characterization: ● Any signal can be written as a sum of sinusoids ● If we know the channel response to all frequencies, we can deduce the channel response to any input channel ● Time characterization: impule response. A sinc function will produce a zero- intereference at sampling times. ● Impulse response is inverse Fourrier transform freq. Response.

17 Channel t 0 t h(t) tdtd Sin(2 pi W t)/(2 pi W t)

18 +A+A -A-A 0 T 2T2T 3T3T 4T4T5T5T 111100 Transmitter Filter Comm. Channel Receiver Filter Receiver r(t) Received signal t

19 Fundamental limits in digital comm. ● For a Gaussian distributed noise channel, the capacity of the channel is C=Wlog(1+SNR) ● The probability of error can be made arbitrarily small as long as signaling rate is less than C. ● For twisted copper pair (telephone system) at 30dB the capacity is 33.88kbps. ● How come we have 56kbps modems?

20 (a) 3 separate pulses for sequence 110 (b) Combined signal for sequence 110 t t TTTTTT TTTTTT

21 4 signal levels8 signal levels typical noise


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