Data and Computer Communications Chapter 3 – Data Transmission

TransmissionTerminology Transmission Terminology  data transmission occurs between a transmitter & receiver via some medium  guided medium e.g. twisted pair, coaxial cable, optical fiber e.g. twisted pair, coaxial cable, optical fiber  unguided / wireless medium e.g. air, water, vacuum e.g. air, water, vacuum

TransmissionTerminology Transmission Terminology  direct link (guided & unguided) no intermediate devices no intermediate devices  point-to-point (guided) direct link direct link only 2 devices share link only 2 devices share link  multi-point more than two devices share the link more than two devices share the link

TransmissionTerminology Transmission Terminology  simplex one direction one direction eg. televisioneg. television  half duplex either direction, but only one way at a time either direction, but only one way at a time eg. police radioeg. police radio  full duplex both directions at the same time both directions at the same time eg. telephoneeg. telephone

Frequency, Spectrum and Bandwidth  time domain concepts analog signal analog signal varies in a smooth way over timevaries in a smooth way over time digital signal (discrete) digital signal (discrete) maintains a constant level then changes to another constant levelmaintains a constant level then changes to another constant level periodic signal periodic signal pattern repeated over timepattern repeated over time aperiodic signal aperiodic signal pattern not repeated over timepattern not repeated over time

Analogue & Digital Signals

Periodic Signals

Sine Wave  peak amplitude (A) maximum strength of signal maximum strength of signal volts volts  frequency (f) rate of change of signal rate of change of signal Hertz (Hz) or cycles per second Hertz (Hz) or cycles per second period = time for one repetition (T) period = time for one repetition (T) T = 1/f T = 1/f  phase (  ) relative position in time relative position in time

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

Wavelength ( )  is distance occupied by one cycle  between two points of corresponding phase in two consecutive cycles  assuming signal velocity v, we have = vT  or equivalently f = v  especially when v=c  c = 3*10 8 ms -1 (speed of light in free space)

Problem # In a multipoint configuration, a central control may be used that enables only one device to transmit. What is the merit and demerit of such control as compared to distributed control?

3.12 According to Fourier analysis, any composite signal is a combination of simple sine waves with different frequencies, amplitudes, and phases. Note

Frequency Domain Concepts  Signal is made up of many frequencies  components are sine waves  Fourier analysis can show that any signal is made up of component sine waves  can plot frequency domain functions

Addition of Frequency Components (T=1/f)  c is sum of f & 3f (with different amplitudes)

Frequency Domain Representations  freq domain function of Fig 3.4c  freq domain function of single square pulse  -ve amplitude signify?

3.16 A composite periodic signal

3.17 Decomposition of a composite periodic signal in the time and frequency domains

Spectrum & Bandwidth  spectrum range of frequencies contained in signal range of frequencies contained in signal  absolute bandwidth width of spectrum width of spectrum  effective bandwidth often just bandwidth often just bandwidth narrow band of frequencies containing most energy narrow band of frequencies containing most energy  DC Component component of zero frequency component of zero frequency

Data Rate and Bandwidth  any transmission system can accommodate a limited band of frequencies  this limits the data rate that can be carried  Square wave: infinite components and hence infinite bandwidth  but most energy is in first few components  limited bandwidth increases distortion  has a direct relationship between data rate & bandwidth

Data Rate-Bandwidth Relation  Square Wave transmission  Case 1: Three sinusoidal frequency components – f, 3f, 5f => Bandwidth = (5- 1) f = 4f. Let f = 1 MHz, Bandwidth = 4 MHz,  T = 1µs => 1 bit needs 0.5 µs  Data rate = 2 MBPS

Data Rate-Bandwidth Relation  Square Wave transmission  Case 2: Three sinusoidal frequency components – f, 3f, 5f => Bandwidth = (5- 1) f = 4f. Let f = 2 MHz, Bandwidth = 8 MHz,  T = 0.5µs => 1 bit needs 0.25 µs  Data rate = 4 MBPS

Data Rate-Bandwidth Relation  Square Wave transmission  Case 3: Two sinusoidal frequency components – f, 3f only => Bandwidth = (3-1) f = 2f. Let f = 2 MHz, Bandwidth = 4 MHz,  T = 0.5µs => 1 bit needs 0.25 µs  Data rate = 4 MBPS  Shape of signal?

Analog and Digital Data Transmission  data entities that convey meaning / information entities that convey meaning / information  signals & signaling electric or electromagnetic representations of data, physically propagates along medium electric or electromagnetic representations of data, physically propagates along medium  Transmission propagation and processing of signals propagation and processing of signals

Acoustic Spectrum (Analog)

Audio Signals  freq range 20Hz-20kHz (speech 100Hz-7kHz)  easily converted into electromagnetic signals  varying volume converted to varying voltage  can limit frequency range for voice channel to Hz with acceptable reproduction

Video Signals - Bandwidth  525 lines x 30 scans = lines per sec => 63.5  s per line => 63.5  s per line 11  s for horizontal retrace, so 52.5  s per video line 11  s for horizontal retrace, so 52.5  s per video line  max frequency if line alternates black and white  483 lines per frame USA has 525 lines but 42 lost during vertical retrace USA has 525 lines but 42 lost during vertical retrace  horizontal resolution is about 450 lines giving 225 cycles of wave in 52.5  s  max frequency of 4.2MHz

Digital Data  as generated by computers etc.  has two dc components  bandwidth depends on sequence of 1s & 0s

Analog Signals

Digital Signals

Advantages & Disadvantages of Digital Signals  cheaper  less susceptible to noise  but greater attenuation  digital now preferred choice

Preferred Method  Digital, because:  - Technology support of VLSI  - Security (Encryption)  - Integration (data, audio, video)

Transmission Impairments  signal received may differ from signal transmitted causing: analog - degradation of signal quality analog - degradation of signal quality digital - bit errors (‘1’ as ‘0’ or vice-versa) digital - bit errors (‘1’ as ‘0’ or vice-versa)  most significant impairments are attenuation attenuation delay distortion delay distortion noise noise

Attenuation  where signal strength falls off with distance  depends on medium  received signal strength must be: strong enough to be detected strong enough to be detected sufficiently higher than noise to receive without error sufficiently higher than noise to receive without error  so increase strength using amplifiers/repeaters  is also an increasing function of frequency  so equalize attenuation across band of frequencies used e.g. using loading coils (voice grade) or amplifiers e.g. using loading coils (voice grade) or amplifiers

Problem # A signal has passed through three cascaded amplifiers, each with a 4 dB gain. What is the total gain in dB? How much is the signal amplified? What does a negative dB value signify?

Delay Distortion  only occurs in guided media  propagation velocity varies with frequency  hence various frequency components arrive at different times  particularly critical for digital data  since parts of one bit spill over into others  causing inter-symbol interference

Noise  additional signals inserted between transmitter and receiver  thermal due to thermal agitation of electrons due to thermal agitation of electrons uniformly distributed across typical bandwidth uniformly distributed across typical bandwidth white noise white noise  Inter-modulation signals that are the sum and difference (or multiples) of original frequencies sharing a medium signals that are the sum and difference (or multiples) of original frequencies sharing a medium

Noise  crosstalk a signal from one path / line is picked up by another a signal from one path / line is picked up by another  impulse irregular pulses or spikes irregular pulses or spikes e.g. external electromagnetic interferencee.g. external electromagnetic interference short duration short duration high amplitude high amplitude a minor annoyance for analog signals a minor annoyance for analog signals but a major source of error in digital data but a major source of error in digital data a noise spike could corrupt many bitsa noise spike could corrupt many bits

Channel Capacity  maximum possible data rate on a communication channel data rate - in bits per second data rate - in bits per second bandwidth - in cycles per second or Hertz bandwidth - in cycles per second or Hertz noise - on communication link noise - on communication link error rate - of corrupted bits error rate - of corrupted bits  limitations due to physical properties  want most efficient use of capacity

Nyquist Bandwidth  considers noise free channels  if rate of signal transmission is 2B then it can carry signal with frequencies no greater than B i.e. given bandwidth B, highest data rate is 2B i.e. given bandwidth B, highest data rate is 2B  for binary signals, 2B bps needs bandwidth B Hz  can increase rate by using M signal levels  Nyquist Formula is: C = 2B log 2 M  so increase rate by increasing signal levels at the cost of receiver complexity at the cost of receiver complexity limited by noise & other impairments limited by noise & other impairments

Shannon Capacity Formula  considers relation of data rate, noise & error rate faster data rate shortens each bit so bursts of noise affects more bits faster data rate shortens each bit so bursts of noise affects more bits given noise level, higher signal strength means lower errors given noise level, higher signal strength means lower errors  Shannon developed formula relating these to signal to noise ratio (in decibels)  SNR db = 10 log 10 (signal/noise)  Capacity C=B log 2 (1+SNR) theoretical maximum capacity theoretical maximum capacity get lower in practice get lower in practice

Summary  looked at data transmission issues  frequency, spectrum & bandwidth  analog vs digital signals  transmission impairments

Problems  Q1. For a video signal, what increase in horizontal resolution is possible if a bandwidth of 5 MHz is used?  Q2. For a digitized TV picture matrix of 480×500 pixels, where each pixel can take one of 48 intensity values, assume 30 pictures are sent per second. Find the source rate (bps).

Problems  Q3. For the above problem, use Shannon’s formula to calculate the channel capacity if BW used is 4.5 MHz and SNR is 35 dB.  Q4. A multi level signaling system operates at 9600 bps. A signal element encodes a 4-bit word. Find the minimum BW required of the channel.

Problems  Q5. A telephone line has 4 KHz BW. When signal is 10V, the noise is 5 mV. Find the maximum data rate supported by this line.

Problems  Q6. Consider square wave transmission that involves the fundamental frequency and all odd frequencies upto the 9 th harmonic. If f = 2MHz, then find the bandwidth required and the data rate achieved. If only the fundamental frequency and the 3 rd harmonic are transmitted, find the bandwidth required and the data rate achieved in this case. Comment about the receiver requirements for both the cases.