Data and Computer Communications Data Transmission Transmission Terminology

Data and Computer Communications Data Transmission

Transmission Terminology Ø Data transmission occurs between a transmitter & receiver via some medium Ø Guided medium l eg. twisted pair, coaxial cable, optical fiber Ø Unguided / wireless medium l eg. air, water, vacuum

Transmission Terminology Ø Direct link l no intermediate devices Ø Point-to-point l l direct link only 2 devices share link Ø Multi-point l more than two devices share the link

Transmission Terminology Ø Simplex l one direction • eg. television Ø Half duplex l either direction, but only one way at a time • eg. police radio Ø Full duplex l both directions at the same time • eg. telephone

Frequency, Spectrum and Bandwidth Ø Time domain concepts l Analog signal • Varies in a smooth way over time l Digital signal • Maintains a constant level then changes to another constant level l Periodic signal • Pattern repeated over time l Aperiodic signal • Pattern not repeated over time

Analogue & Digital Signals

Periodic Signals

Sine Wave Ø Peak amplitude (A) l l Ø Frequency (f) l l Ø Maximum strength of signal Volts Rate of change of signal Hertz (Hz) or cycles per second Period = time for one repetition (T) T = 1/f Phase ( ) l Relative position in time

Varying Sine Waves s(t) = A sin(2 ft + )

Wavelength ( ) Ø is distance occupied by one cycle Ø between two points of corresponding phase in two consecutive cycles Ø assuming signal velocity v have = v. T Ø or equivalently f = v § especially when v=c § c = 3*108 ms-1 (speed of light in free space)

Frequency Domain Concepts Ø Signals are made up of many frequencies Ø Components are sine waves Ø Fourier analysis can show that any signal is made up of component sine waves Ø Can plot frequency domain functions

Addition of Frequency Components (T=1/f) p Ø c is sum of f & 3 f p p

Frequency Domain Representations Freq. domain func. of Fig 3. 4 c Ø Freq. domain func. of single square pulse p p p Ø < <

Spectrum & Bandwidth Ø Spectrum l Ø Absolute bandwidth l Ø Width of spectrum Effective bandwidth l Often just bandwidth l Ø Range of frequencies contained in signal Narrow band of frequencies containing most energy DC Component l Component of zero frequency

Figure 3. 7 (a) & (b)

Bit time = T / 2

Data Rate and Bandwidth Ø Any transmission system has a limited band of frequencies Ø This limits the data rate that can be carried Ø Square have infinite components and hence bandwidth Ø But most energy in first few components Ø Limited bandwidth increases distortion Ø Have a direct relationship between data rate & bandwidth

Data Rate Calculation Ø Case 1 l l Ø the period of fundamental frequency 1/f = 10 -6 = 1 us. One wave takes two data bits Data rate = 2 Mbps Case 2 l l l Ø Bandwidth 4 MHz, use the sine wave of Fig. 3 -7 (a) 4 MHz = 5 f – f f = 1 MHz, Bandwidth 8 MHz, use the sine wave of Fig. 3 -7 (a) 8 MHz = 5 f – f f = 2 MHz Data rate = 4 Mbps Case 3 l l Bandwidth 4 MHz, use the sine wave of Fig. 3 -4 (c) 4 MHz = 3 f – f f = 2 MHz Data rate = 4 Mbps

Data Rate vs. Bandwidth Ø Bandwidth ↑ l l Data rate ↑ (compare case 1 & 2) Same signal quality Ø Same bandwidth l l Higher signal quality lower data rate Compare case 1 & 3 Ø Same data rate l l Bandwidth ↑ better signal quality Compare case 2 & 3

Analog and Digital Data Transmission Ø Data l Entities that convey meaning Ø Signals & signaling l Electric or electromagnetic representations of data, physically propagates along medium Ø Transmission l Communication of data by propagation and processing of signals

Acoustic Spectrum (Analog)

Audio Signals Freq. range 20 Hz-20 k. Hz (speech 100 Hz-7 k. Hz) Ø Easily converted into electromagnetic signals Ø Varying volume converted to varying voltage Ø Can limit frequency range for voice channel to 300 -3400 Hz Ø

Video Signals Ø USA - 483 lines per frame, at frames per sec l Ø have 525 lines but 42 lost during vertical retrace 525 lines x 30 scans = 15750 lines per sec l l 63. 5 s per line 11 s for retrace, so 52. 5 s per video line Max frequency if line alternates black and white Ø 70% of 483 is subjective resolution – 338. Ø 4/3 ratio – 450 lines Ø Horizontal resolution is about 450 lines giving 225 cycles of wave in 52. 5 s Ø Max frequency of 4. 2 MHz Ø

Digital Data Ø As generated by computers etc. Ø Has two dc components Ø Bandwidth depends on data rate

Analog Signals

Digital Signals

Advantages & Disadvantages of Digital Signals Ø Cheaper Ø Less susceptible to noise Ø But greater attenuation Ø Digital now preferred choice

Transmission Impairments Ø Signal received may differ from signal transmitted causing: l l Analog - degradation of signal quality Digital - bit errors Ø Most significant impairments are l l l Attenuation and attenuation distortion Delay distortion Noise

Attenuation Where signal strength falls off with distance Ø Depends on medium Ø Received signal strength must be: Ø l l Strong enough to be detected Sufficiently higher than noise to receive without error So increase strength using amplifiers/repeaters Ø Is also an increasing function of frequency Ø So equalize attenuation across band of frequencies used Ø l l E. g. using loading coils or amplifiers N_f = -10 log_10(P_f/P_1000)

Delay Distortion Ø Only occurs in guided media Ø Propagation velocity varies with frequency Ø Hence various frequency components arrive at different times Ø Particularly critical for digital data Ø Since parts of one bit spill over into others Ø Causing inter-symbol interference

Noise Ø additional signals inserted between transmitter and receiver Ø thermal due to thermal agitation of electrons l uniformly distributed over frequency l white noise N= k. TB, k = Boltzman’s constant (1. 38*10^-21), T – temperature, B - Bandwidth In decibel watts – take log l Ø intermodulation l signals that are the sum and difference of

Noise Ø Given a receiver with an effective noise temperature of 294 K and a 10 -MHz bandwidth, thermal noise level at the receiver output is N = 228. 6 + 10 log(294) + 10 log(107) Ø intermodulation l signals that are the sum and difference of original frequencies sharing a medium

EXAMPLE 3. 1 Room temperature is usually specified as T = 17°C, or 290 K. At this temperature, thermal noise power density is -23 -21 N 0 = (1. 38 * 10 ) * 290 = 4 * 10 W/Hz = - 204 d. BW/Hz where d. BW is the decibel-watt, defined in Appendix 3 A.

EXAMPLE 3. 2 Given a receiver with an effective noise temperature of 294 K and a 10 -MHz bandwidth, thermal noise level at the receiver’s output is N = - 228. 6 d. BW + 10 log(294) + 10 log 10 = - 228. 6 + 24. 7 + 70 = - 133. 9 d. BW 7

Noise Ø Crosstalk l A signal from one line is picked up by another Ø Impulse l Irregular pulses or spikes • E. g. external electromagnetic interference l l Short duration High amplitude A minor annoyance for analog signals But a major source of error in digital data • A noise spike could corrupt many bits

Channel Capacity Ø Max possible data rate on comms channel Ø Is a function of l l Data rate - in bits per second Bandwidth - in cycles per second or Hertz Noise - on comms link Error rate - of corrupted bits Ø Limitations due to physical properties Ø Want most efficient use of capacity

Nyquist Bandwidth Consider noise free channels Ø If rate of signal transmission is 2 B then carry signal with frequencies no greater than B Ø l i. e. given bandwidth B, highest signal rate is 2 B For binary signals, 2 B bps needs bandwidth B Hz Ø Can increase rate by using M signal levels Ø Nyquist Formula is: C = 2 B log 2 M Ø So increase rate by increasing signals Ø l l At cost of receiver complexity Limited by noise & other impairments

EXAMPLE 3. 3 Consider a voice channel being used, via modem, to transmit digital data. Assume a bandwidth of 3100 Hz. Then the Nyquist capacity, C, of the channel is 2 B = 6200 bps. For M = 8, a value used with some modems, C becomes 18, 600 bps for a bandwidth of 3100 Hz.

Shannon Capacity Formula Ø Consider relation of data rate, noise & error rate l l Faster data rate shortens each bit so bursts of noise affect more bits Given noise level, higher rates mean higher errors Shannon developed formula relating these to signal to noise ratio (in decibels) Ø SNRdb=10 log 10 (signal/noise) Ø Capacity C=B log 2(1+SNR) Ø l l Theoretical maximum capacity Gets lower in practice

Problem Ø Suppose the spectrum is between 3 Mhz and 4 MHz and SNR_DB = 24. Calculate channel capacity using Shannon and Nyquist.

B = 4 MHz - 3 MHz = 1 MHz SNR = 24 d. B = 10 log SNR d. B 10 SNR = 251 Using Shannon’s formula, C = 10 * log (1 + 251) = 10 * 8 = 8 Mbps 6 6 2 This is a theoretical limit and, as we have said, is unlikely to be reached. But assume we can achieve the limit. Based on Nyquist’s formula, how many signal- ing levels are required? We have C = 2 B log M 2 8 * 10 = 2 * (10^6)* log M 6 2 4 = log M 2 M = 16

Summary Ø Looked at data transmission issues Ø Frequency, spectrum & bandwidth Ø Analog vs digital signals Ø Transmission impairments