Transmission in the Physical Layer

1
Transmission in the Physical Layer

• In this section
• Transmission using electromagnetic waves
• Digital versus Analog
• Analog channel capacity
• Modulation and Encoding
• digital transmission of digital information
• digital transmission of analog information
• analog transmission of digital information

2
Transmission Basics

• A stream of 0 and 1 bits
• Bits are state values that exist for a detectable
  period of time on a transmission medium.
• Voltage on a wire
• Electromagnetic waves specific amplitude or
  frequency variations
• Presence/absence of photons with specified
  wavelength
• Medium is sampled over time
• Limitation sample interval must be long enough
  to include
• Time required for taking reliable sample
• Time required for a state change to stabilize
  between samples

3
Digital and Analog

• Analog
• Information or transmission medium is of a
  continuous nature.
• Example human voice can produce a continuous
  range of frequencies
• Digital
• Information or transmission medium uses discrete
  values
• Example piano keys can produce only a certain
  set of tones.
• Keep in mind that the information to be sent and
  the

4
Propagation Delay

• Amount of time taken for a transmission to pass
  through a medium
• Electromagnetic waves in vacuum speed of light
• Electromagnetic waves in wires or fibres 2.1 x
  108 metres / second (i.e. about 30 slower than
  in a vacuum)
• Delay distance / vmedium
• Delay is non-trivial for
• Geosynchronous satellites altitude is 35,900
  km delay for one hop transmission is 0.24
  seconds
• Interplanetary communication
• To a robot on Mars 10.2 minutes, one way, at
  Mars closest approach

5
Transmission Time

• To send a quantity of information, the time
  required will be (at least) the amount of
  information divided by the link capacity
• Extra overhead that may be needed
• Synchronization
• Flow control
• Error detection / correction

6
Analog signals
amplitude A volts
phase shift ? seconds
frequency f 1 / P hertz
?
time t seconds
0
period P seconds
7
Modulation

• In analog communication, a carrier wave is sent
  at a specified frequency F, with constant
  amplitude A.
• Information is included by modulating the carrier
  wave
• Amplitude modulation
• Frequency modulation
• Phase modulation

8
Signal Analysis

• Any periodic signal can be represented as a
  (possibly infinite) sum of sine waves the
  Fourier series
• where, for i 0, 1, 2, 3,

9
Example Square waves

• Original function
• s(t) A for 0, 0.5P), P, 1.5P), 2, 2.5P),
  etc.
• s(t) A for 0.5P, P), 1.5P, 2P), 2.5P,
  3P), etc.
• Fourier coefficients
• ai 0 for all i
• bi 0 if i is even
• bi 4/(pi) if i is odd

10
Square waves

• Result
• Approximated to 3 terms
• This means that 3frequencies of radiowaves
  could approximatea 1 and a 0

11
Analog Channel Capacity

• The information capacity or bandwidth that a
  signal can hold is related to the range of
  frequencies that can be transmitted
• FH FL
• where FH is the highest possible frequency and
  FL is the lowest possible frequency
• Note that effective bandwidth is limited by the
  energy of each frequency
• In Fourier series, since the amplitude has a 1 /
  i term, most of the energy is in the first few
  terms of the series.

12
Bandwidth and Data rate

• Example 3 terms of square wave
• Use f 1 MHz 1.0x106 Hz ? P 1.0x10-6 s can
  transmit 2 bits per P
• Bandwidth 5x106-1x106 4.0x106 Hz
• Result with 4MHz of bandwidth, can transmit 2
  Mb/s at 1MHz base frequency.
• Note that higher base frequencies have more
  bandwidth room, and therefore potential data rate
  is higher.

P
13
Nyquist Bandwidth

• Suppose that a channel is free of noise.
• If the signal rate is 2B, then a signal with
  frequencies no greater than B is sufficient to
  carry the signal rate.
• see previous example 1 MHz carrier, 2 Mbps data
  rate
• Converse Given a bandwidth of B, the highest
  signal rate that can be carried is 2B
• More generally, if there are L levels of
  amplitude used to carry information with
  bandwidth B, the capacity C is
• C 2B log2 L

14
Example Phone modems

• Suppose you are sending an analog signal through
  the phone system (which is how modems work).
• Approximate bandwidth of phone system 3,000 Hz.
  (range 300 Hz to 3300Hz)
• If you are using L16 levels of signal
• C 2B log2(L)
• C 2 x 3,000 x 4 24,000 bits / second

15
Transmission Impairments

• Noise
• Distortions of original signal
• Thermal noise due to agitation of electrons
• Varies by temperature
• Uniform across bandwidths
• Intermodulation noise when signals of different
  frequencies share the same transmission medium
• Crosstalk unwanted coupling between signal
  paths
• Impulse noise
• Unlike the above, these are neither predictable
  nor constant
• Irregular pulses of short duration but relatively
  high amplitude
• Causes external electromagnetic disturbances
  (e.g. lightning), communication system faults

16
Transmission Impairments

• Attenuation
• The strength of a signal decreases, generally
  exponentially, by distance.
• Use repeaters to re-generate signal before signal
  becomes indistinguishable from noise
• Delay distortion
• Velocity of signal through guided medium varies
  by frequency
• A signal consisting of several frequencies will
  have the components arrive at slightly different
  times
• Information from one bit period can spill into
  the next bit period

17
Capacity of a Channel with Noise

• Claude Shannon examined the problem of
  communication through a noisy channel.
• Basis of his work
• There are a set of symbols that will be sent
  through a channel.
• There are a set of symbols expected to be
  received from the channel.
• Noise affects the channel. The result is that
  the receiver may detect a symbol from the channel
  that is not what was sent.
• Given a particular input symbol, what is the
  probability that the receiver will interpret the
  what is read as each of the possible symbols?

18
Example

• Suppose that we could send any of symbols A, B, C
  or D through a channel with random noise.
• pA pB pC pD 1.
• For an effective channel, we want pB gtgt pA, pC,
  pD

pA
A
B
Noisy channel
pB
B
C
pC
D
pD
19
Coding

• To minimize the probability that a symbol is
  misread, codes can be used to make symbols more
  distinguishable
• Example
• to send information bit 0, encode this as 000.
• to send information bit 1, encode as 111.
• Receiver reads three bits, and may read any of
  000, 001, 010, 100, 101, 110, or 111.
• At least two of the bits will be the same choose
  that value as the detected information bit

20
The trade-off

• If the probability of misreading any 0 or 1 sent
  through the channel is p, then for this encoding
  method, two of those must be misread to have an
  error in the information passed through the
  channel.
• Probability of information error is reduced
• However, we now have to take three channel
  samples, so that our information bit rate
  capacity is reduced.
• We are trading off channel capacity for
  reliability of information transfer

21
Error probabilities

• Suppose that the probability of an error in bit
  detection is p 0.1 (i.e. 10) and that we send
  information bit 0 through the channel.

22
Reducing the overall error probability

• If we sent the original bits as is, the
  probability of an error in the information
  transfer for each information symbol would be
  0.1, and the probability of correct information
  transfer would be 0.9.
• Using this encoding, the probability of correct
  information transfer is 0.729 0.081 0.081
  0.081 0.972
• The probability of incorrect information transfer
  is 0.009 0.009 0.009 0.001 0.028
• Sum of probabilities 0.972 0.028 1
• Result the effective error rate for the channel
  will be 2.8 instead of 10.

23
Result from Shannons work

• Assuming that there is random noise on a channel,
  and the ratio of the signal power to the noise
  power is known, it is possible to determine the
  theoretical capacity of a channel.
• This is the maximum channel capacity for which an
  encoding scheme is possible that will reduce the
  error probability to an arbitrarily small value.
• That is, what is the maximum reliable information
  transfer capacity?
• The error probability is incorporated into the
  signal to noise ratio the more the signal
  with information can be distinguished from
  background noise, the less likely it is that a
  bit sampling error will be made.

24
Shannons Formula for Channel Capacity

• Shannons formula
• Maximum channel capacity C in bits per second is
• where S is the signal power, N is the noise
  power, and B is the bandwidth
• The ratio S / N is the signal to noise ratio.
• Units are often reported in bels, which is the
  logarithm (base 10) of the ratio.
• Decibels 10 times the above value.
• Although measurements of S and N are needed to
  establish the ratio, in this course, we will use
  the ratio S / N as a single quantity.

25
Can the capacity be reached?

• C represents a maximum possible value. How close
  can we get to the maximum?
• Turbo codes (mobile phones), and Reed-Solomon
  codes (CDs, DVDs) can approach the Shannon limit
  to 0.1 dB.
• These codes are computationally complex, but are
  now usable with increased processor speeds

26
Example Phones, again

• Shannons law for phones
• Bandwidth 3,000 Hz
• Signal to noise ratio 35 decibels (dB)
• 35 10 log10(S/N), so S/N ? 3162
• Bit rate 3000 log2( 1 S/N) ? 3000 11.63
  ? 34,880 bps
• In fact, with fully digital connections at
  Internet service providers (that is, no modem at
  their end), 56 Kbps can be achieved in the
  download direction. Otherwise, 33.6 Kbps is the
  fastest standard modem speed that is used.
• Why 56 Kbps? The phone system samples 8 bits,
  8000 times per second. One sample bit is used
  for control, so the data sample is 7 bits
  namely 56,000 data bits per second.

27
Signal Encoding

• To send digital signals a set of 0 and 1
  values, we need a coding scheme to represent each
  value.
• Non-return to zero (NRZ)
• Use low (negative) voltage to represent a 1 and
  high (positive) voltage to represent a 0

high
low
1
0
1
0
0
1
1
0
P duration of 1 bit (period)
P
28
Signal Encoding

• Non-return to zero inverted (NRZI)
• A 1 is represented by a change in voltage a 0 is
  represented by no change

high
low
1
0
1
0
0
1
1
0
29
Difficulties with NRZ(I)

• How many 0 bits are there in the following
  transmission?
• What is the potential for two clocks to be out of
  synchronization?

high
low
30
Manchester Encoding

• Bits are indicated by changes in voltage at the
  centre of the bit period.
• 0 change from high to low
• 1 change from low to high
• Advantages
• It is easier to detect a change in voltage than
  measure a voltage (especially in the presence of
  noise)
• The bit stream is self-synchronizing there is
  always a voltage change at the centre of the bit
  period.

P
high
low
1
0
1
0
0
1
1
0
31
Differential Manchester Encoding

• Like Manchester encoding, there is always a
  voltage change at the centre of the interval
• Bits are encoded by whether or not there is a
  voltage change at the start of the interval
• 0 change voltage
• 1 keep previous voltage

P
high
low
1
0
1
0
0
1
1
0
32
Block Coding

• General idea groups of m bits are transformed
  to blocks of n bits.
• Notation mB/nB
• Allows for redundancy, and improves reliability
• Example There are 16 possibilities for 4 bits.
  If we use 5 actual bits to send 4 bits of
  information (4B/5B), we can choose 16 out of the
  32 possible 5 bit combinations to send
  information.
• Choose 5 bit combinations where there is never
  more than 3 consecutive 1 or 0 bits.

33
4B/5B Encoding
34
Block Coding

• If we have bandwidth limitations, another
  approach can be taken to reduce bandwidth while
  still increasing redundancy.
• Use three levels of signal instead of two , -,
  0
• trits ternary bits
• Encode an 8 bit block (256 combinations) using 6
  trits (729 combinations) 8B/6T encoding.

35
Sampling Analog Waves (1)
Original analog signal
Sample wave at fixed intervals
36
Sampling Analog Waves (2)
Sampled signal pulse amplitude modulation
Quantization pulse code modulation
10 9 8 7 6 5 4 3 2 1 0
Send (as bits) 6 7 8 9 10 10 10 10 9 8 7
37
Transmission Modes

• Parallel
• Bits are sent in parallel over separate lines.
• Advantage speed
• Disadvantage cost
• Usually used for high-speed over a short distance
  (e.g. computer to printer).
• Example 8-bit parallel sends all bits of an
  octet at the same time.
• Serial
• Bits are sent consecutively, one after another.

38
Serial Transmission

• Asynchronous
• Bit patterns used for synchronization sender can
  start transmission at any time.
• Example send 8 bits and then a gap, to indicate
  octet boundaries.
• Alternative use start/stop bits start each
  octet with a 0 bit and end it with a 1 bit
  followed by a gap.
• Start/stop bits help to synchronize the sender
  and the receiver.
• Advantage cost (less complex, no timer needed)
• Disadvantage speed (transmission gaps)

39
Serial Transmission

• Synchronous
• Timed sending sending must start at a
  particular clock interval.
• No gaps between bits
• i.e. one cannot detect boundaries between octets,
  messages, etc.
• It is up to the receiver to reconstruct bit
  groupings.
• If there is nothing to send, bit sequences are
  still sent indicating no data.
• Block encoding schemes can use leftover
  combinations to indicate idle (and other
  special conditions).
• Example 4B/5B uses 11111 as an idle code it
  does not represent any combination of four bits.

40
Analog Transmission

• Basis is a carrier wave a pure sine wave at a
  specified frequency
• Example Radio station broadcasting at 91.5 MHz
• The radio station is sending a carrier wave at
  91.5 MHz, but modulating this wave to carry the
  information.
• Carrier wave frequencies are allocated for
  various purposes e.g. 88 106 MHz FM radio
  stations
• In Canada regulated by Industry Canada

41
Analog Transmission of Digital Data

• Baud rate versus bit rate
• A baud is a carrier wave consistently modulated
  for a specified period of time.
• Modulation changes between one baud and the next.
• Examples
• Using 2 power levels to indicate 1 and 0 one
  baud carries one bit
• Using 4 power levels to indicate 00 01 10 or 11
  one baud carries 2 bits
• Baud rate rate at which bauds are sent
• Bit rate rate at which information (i.e. bits)
  is transferred

42
Amplitude Shift Keying

• Amplitude shift keying (ASK) different
  amplitude (i.e. power) levels are used to
  modulate the carrier wave.
• Simplest form two amplitude levels representing
  0 and 1.
• Power level is held for the length of one baud
• Can use several levels with bit encoding
• Very susceptible to noise.
• Bandwidth required (one direction) B (1 d)
  x Nbaud
• Nbaud baud rate
• d modulation factor (greater than 0, depends on
  modulation process)

43
ASK
44
Frequency Shift Keying

• Frequency shift keying (FSK) use different
  frequencies to modulate the carrier wave.
• Simplest form two frequencies representing 0
  and 1.
• If fc is the carrier frequency, then 0 could
  bef0 fc d0 and 1 could be f1 fc d1
• Frequency is held for the length of one baud
• Not so susceptible to noise.
• Bandwidth required (one direction) B f0
  f1 Nbaud

45
FSK
46
Phase Shift Keying

• Phase shift keying (PSK) use different phases
  (horizontal shift) of the carrier wave for
  modulation
• Simplest form two phases representing 0 and 1.
• A 0 bit could be that the wave has a phase shift
  of 0 for the length of the baud, and 1 could be
  that the wave has a phase shift of p radians
  (i.e. 180 degrees).
• Phase is consistent is held for the length of one
  baud
• Not so susceptible to noise.
• Bandwith required (one direction) same as ASK

47
PSK 0 and p radians
48
PSK Constellation Diagram

• Various phases can be used to represent bit
  combinations
• Example use 8 phases to represent combinations
  of 3 bits

010
011
001
000
100
101
111
110
49
Quadrature Amplitude Modulation

• Quadrature Amplitude Modulation (QAM) combines
  multiple phase shifts with multiple amplitudes to
  encode bits.
• Example 8-QAM
• 2 amplitudes, 4 phases (0, 90, 180, 270)

011
amplitude 1
010
101
000
001
110
amplitude 2
111
50
Example of 8-QAM signal
51
Telephone Modems

• Modem MOdulator-DEModulator
• Usable bandwidth in the telephone system for data
  transmission 600 Hz to 3000 Hz (out of 300 Hz
  to 3300 Hz for voice transmission).
• i.e. 2400 Hz bandwidth, so 2400 baud data rate
• Modems are standardized by the ITU-T, V-series of
  standards
• V.32 32-QAM with check bit at 2400 baud ? 9600
  b/s
• V.32bis 128-QAM with check bit at 2400 baud ?
  14,400 b/s
• V.34bis 960 point constellation giving 28.8
  Kb/s, or 1664 point constellation giving 33.6
  Kb/s
• V.90 33.6 Kb/s modem-to-modem, 56 Kb/s download
  from digital source.
• V.92 adjustable data rate up to 48 Kb/s upload
  on digital phone lines with better than 35 dB
  signal to noise ratio, 56 Kb/s