Digital Transmission Fundamentals. Chapter 3. Communication Networks Leon-Garcia, Widjaja

Transcription

1 Digital Transmission Fundamentals Chapter 3. Communication Networks Leon-Garcia, Widjaja

2 The Questions Digital representation Why digital transmission Digital representation of analog signals Characterization of communication channels Fundamental limits Line coding Properties of transmission media Error detection-correction

3 Block vs Stream communication Block: Text, pictures, software Total time of communication Stream: Internet radio, streaming video, midi Restrictions on rate of arrival of packets

4 Block information D=t p + L R

5 Reducing delay Decrease propagation delay Increase speed of light ;-) Optimize routing Avoid satellite channels Increase transmission rate The subject of intense research Reduce data length Compression.

6 Secret to compression In text Some letters or letter pairs are far more common ae is more common than zq We may encode ae with 3 bits and zq with 16 In images, sound, etc Do not transmit non perceivable information (lossy) Exploit statistical correlations Use different encoding

7 Streaming For analog signals, the most important quantity is bandwidth (of the signal) A measure of how fast the signal changes The unit is the Hz It is the frequency of the highest frequency component

8 Streaming Voice (telephony) 4 khz is considered enough We need 8,000 samples per second This is called Pulse Code Modulation Music 20 khz is enough, but use 22 khz We need 44,000 sample per second

9 Streaming PCM A stream of 8-bit (or 16 or...) samples Differential PCM Transmit difference from previous sample. Adaptive DPCM Adapts to variation of voice level Linear Predictive Methods Predict next value and transmit the difference only

10 Why Digital Comunication Many common signals are analog Yet almost all communication is digital now Reasons Cheaper Ability to restore signal after degradation Advanced routing

11 Analog vs Digital Analog: if we transmit.8 Volt and receive.81 Volt the receiver does not know if it is noise Noise is impossible (to a great extend) to remove Digital: If we transmit 1 Volt and receive.8 Volt we know it is 1 Volt All the noise enters the system upon digitization. Can have re-generators at regular intervals Only if the noise is strong and signal weak we lose information (kind of)

12 Channel Bandwidth Amplitude response function The amplification (or attenuation) of every frequency component going through the channel Bandwidth is the frequency after which the components have much more significant attenuation. Frequency after which the ARF dips

13 Channel Bandwidth A(f) f

14 Sawtooth components

15 Intermediate components

16 Signal to Noise Ratio SNR, one of the most useful quantities Ratio of (average) signal power to (average) noise power Power and variance of a signal are sometimes indistinguishable for electrical engineers Commonly reported in decibell (db) It is a logarithmic scale

17 Signal to Noise ratio σ n 2 =E {n 2 } P= 1 R n2

18 Signal to Noise ratio SNR= P sig P noise SNR db =10 log 10 SNR

19 Why is SNR Important? Imagine we had no noise... We could take the entire wikipedia, say 1 TB Treat the whole thing as a number Transmit it as Volt We need zero bandwidth for this!

20 Noise and Signal

21 Multilevel Signals

22 Multilevel Signals The noise should be small enough to allow us to distinguish the levels Increase the power of the signal Decrease the noise Lower the bit rate (if we transmit every bit for a longer time we can average the noise out)

23 Example: Telephone lines They have 1% noise (amplitude) SNR is 10,000:1 SNR in db is 10*log = 10*4 = 40

24 Channel Capacity Depends on: Bandwidth SNR Studied by Claude E. Shannon, father of communication theory Information Theory (1948) It is a probabilistic theory Provides upper limit for the amount of data the receiver can meaningfully determine

25 Channel Capacity C=W log 2 (1+SNR)

26 Example Telephone lines Bandwidth 3.4 khz SNR 10,000 Capacity is: 3400 * log(10000)/log(2) = 3400*4*3.3 = 45,178

27 Fourier Transform Signals x(t)= α k cos(2 π k f o t+ϕ k ) x(t)= α k sin (2π k f o t +ϕ k ) x(t)= α k cos(2 π k f o t)+ β k sin(2 π k f o t)

28 Signal Bandwidth It is the frequency of the highest frequency component... but Components of very high frequency with miniscule weight are always present A better definition is: Bandwidth is the frequency range that contains 99% of the power of the signal

29 Bandwidth: Examples khz

30 Sampling We can approximate an analog signal with a set of discrete measurements Under certain conditions the original analog signal can be reconstructed exactly The Sampling Theorem provides the preconditions The bandwidth of the signal can be no more than half the sampling rate

31 Sampling x(t) x(n T ) x n = x(n T ) s(t ) x r (t)= n x n s(t nt )

32 Sampling x r = n x n s(t nt ) x 1 x 3 s(t ) x 1 x 3

33 Sampling We can do linear interpolation with this little triangle We can do a smoother interpolation with a spline The perfect interpolation is the sampling function

34 Sampling Function sin (2 π W t ) s(t )= 2 π W t W = 1 T

35 Sampling Function Works perfectly... in theory It dies off very slowly (1/t) Cannot be implemented in practice We use various approximations Typically we window it with a Gaussian

36 Sampling Function

37 Quantization In modern applications samples are digitized Digital signals can be transmitted error-free (almost) All the error sneaks in during digitization Simplest is the uniform quantizer

38 Quantization Error e n = y n x n SNR= σ x 2 σ e 2

39 Quantization Error y Δ 2 Δ 2 x

40 Quantization Error σ e 2 = Δ2 12 Δ= V 2 m 1

41 Power of the Signal It is related to V Signal ranges between -V...V V is chosen so that the signal is almost always between -V...V If we assume that the pdf of the signal is Gaussian Signal is almost always within 4 standard deviations

42 Power of the Signal V =4 σ x -V V

43 SNR SNR = σ 2 x σ = 2 e = 3 4 V 2 Δ 2 V 2 16 Δ 2 12 = 3 4 V (m 1) V 2 = (m 1 ) = m

44 SNR db SNR db = 10 log(4 )m + 10 log( 3 16 ) = 6m 7.27 = 10 log m

45 SNR - Example For 8 bit per sample (telephones) SNR = = db Every extra bit adds 6db to the SNR An extra bit halves the quantization error So it reduces the power of the quantization error by 4 The log of 4 is about 6

46 Non-uniform Quantization Depending on the application it might not be optimal to have all intervals same size For audio we prefer larger intervals for larger signal magnitude This may increase The SNR for a particular application The perception of quality by humans

47 Communication Channels We usually assume channels to be Linear Time invariant Neither is perfectly true but works in practice. Summarized as: The response to the sum of two signals is equal to the sum of the responses. The response does not change with time. The response to a sine is a shifted scaled sine

48 Linear Time Invariant x 1 (t ) y 1 (t) x 2 (t) y 2 (t) x 1 (t )+ x 2 (t) y 1 (t)+ y 2 (t) x 1 (t +τ) y 1 (t+ τ)

49 Attenuation The ratio of power going in over the power coming out If we have attenuation 3db per kilometer In one kilometer the power if halved In 15km the power is down about 30,000 times! Voltage is down about 170 times

50 Amplitude Response Function & Fourier x(t) = k y(t ) = k α k cos(2 π k f o t) A(k f o )a k cos(2 π k f o t + ϕ (k f o ))

51 Delay ϕ(f ) = 2π f τ cos(2 π k f o t 2 π f τ) = cos(2 π k f o (t τ))

52 Finding the Frequency Response We could send one cosine after another through the channel and measure the response Or we could send the sum of all the cosines at once It is linear time invariant after all This sum is called impulse or Dirac function or Delta function

53 Impulse Response Function It is the response of the channel to the impulse function It is mathematically equivalent to the amplitude and phase responses of the channel The one can be obtained from the other through a Fourier transform

54 Example: Low-pass channel A(f )=1 W f W ϕ(f )=2 π f τ h(t)=s(t τ) s(t ) = sin(2 π W t) 2 π W t

55 The Nyquist Signaling Rate Let p(t) be the response to a pulse that appears at the other side of the channel What is a good shape for p(t) so that we recover the original signal most easily Reduce the inter-symbol interference. What is the pulse rate under these conditions?

56 Nyquist Signaling Rate

57 The Nyquist Signaling Rate

58 The Nyquist Signaling Rate The peak of a pulse coincides with the zeros of all the other pulses This allows us to pack the pulses closely Pulse rate is twice the bandwidth The sampling function is not implementable There are very good approximations

59 Shannon Channel Capacity Without noise

60 Shannon Channel Capacity With little noise

61 Shannon Channel Capacity More bits

62 Shannon Channel Capacity Too many bits

63 Shannon Channel Capacity C = W log 2 (1+SNR)

64 Line Coding We want to transmit bits over a channel The obvious is to send 5V to transmit 1 and 0V to transmit 0 Sounds good, but can we do better What are the problems with this? Bandwidth Clocking Polarity

65 Return to Zero A transition 0, 5, 0 Volt is an one A constant 0 Volt is a zero Self clocking is easy (as long as we do not have too many zeros in a row)

66 Non-Return to Zero Consecutive 1s have no intermediate zero in between Fewer transitions than RZ

67 DC component This simplest scheme is called Unipolar NRZ Biggest problem: DC component Wastes energy Very low frequencies are tricky Most transmission lines cut-off DC component May lose clock synchronization

68 DC Component Power P= P=0.5

69 DC Component A long series of 1s can cause even more problems

70 Polar NRZ A simple solution is to represent 1: +2.5V 0: -2.5V Same separation Half the energy DC component much smaller 0s and 1s are about 50-50

71 Polar NRZ Power P=0.5 ( 0.5) P= P=0.25

72 Polar NRZ Long sequences of either zeros or ones is still a problem DC component Loss of clocking

73 Bipolar NRZ Zero is 0V One is either -2.5 or +2.5V (alternates) DC component disappears Clocking problem for long strings of 0s

74 NRZ-inverted (differential encoding) It is easy to detect polarity on a cable How do you detect it in wireless? Hard Map bits to transitions No transition: zero Transition: one Still a problem with long strings of zeros

75 Manchester Encoding Like polar encoding (positive-one, negative zero) But: Transmit 10 for one Transmit 01 for zero There is always a transition in the middle of the pulse Clocking easy, little low frequencies Wastes bandwidth

76 Manchester Encoding Variant (differential): Zero: transition at the beginning of pulse One: no transition at the beginning of pulse Variant (mbnb): Straight Manchester uses two bits to encode one (1B2B) 4B5B is used in FDDI (Fiber Distributed Digital Interface)

77 Manchester Encoding Variant 4B3T: Uses 3 ternary bits (0, 1, 2) to represent 4 binary The ternary bits are positive, negative and zero voltage Used in 100Base-T4

78 Modulation-Demodulation Modems We have to use them whenever we cannot use the baseband Telephone lines Cable Wireless Fiber

79 Amplitude Modulation Amplitude Shift Key in digital communication Used in fiber mostly or in conjunction with PSK The carrier frequency is turned on for 1, off for 0

80 Frequency Modulation Frequency Shift Keying We have two carrier frequencies O: transmit one frequency 1: transmit the other Not used much Very similar to PSK

81 Phase Shift Keying Transmit cos(w t) for 0 Transmit cos(w t + ph) for 1

82 Decoding PSK To decode a PSK signal we need A local oscillator at the same frequency (easy) and phase (hard) with the transmitter A signal multiplier We multiply the incoming signal with the cosine produced by the local oscillator

83 Decoding PSK A cos(ω c t) cos(ω c t ) = A 2 (1 cos(2ω ct )) A 2 (1 cos(2ω ct )) LPF = A 2

84 Half the Bandwidth? In baseband line coding the pulse rate was 2W for a channel of bandwidth W It can be shown that the pulse rate is only W for this scheme Somewhere someone is hiding something!

85 Quadrature Amplitude Modulation We can transmit the sum of a sine and a cosine and separate them at the receiver end Mathematicians say that a sine and a cosine are orthogonal to each other Orthogonal means that their dot product is zero The dot product is the integral of their product over their period (or multiple of it) We approximate the integral with a LPF

86 QAM Y (t ) = A cos(ω c t) + B sin(ω c t ) (Y (t ) cos(ω c t)) LPF = ( A 2 (1+cos(2 ω ct)) + B 2 sin(2 ω ct) ) LPF = A 2

87 QAM Y (t ) = A cos(ω c t) + B sin(ω c t ) (Y (t ) sin (ω c t )) LPF = ( A 2 sin (2ω ct) + B 2 (1 cos(2ω ct)) ) LPF = B 2

88 QAM Mod-Demod A k cos(ω c t) + Y k B k LPF A k sin(ω c t) Y k 2cos(ω c t) LPF B k 2sin(ω c t)

89 Signal Constellations B B A A

90 QPSK Same as QAM but A and B take values +1 or -1 Can encode 2 bits per pulse Constellation of 4 points QAM systems fall back to QPSK in the presence of strong noise

91 WiFi-n Combines WiFi-g with Broader channel (40MHz vs 20MHz) MIMO antennas Negotiation between transmitter and receiver Single stream PSK (6.5 Mb/s) 4 stream QAM-64 (5B6B) (600 Mb/s)

92 MIMO Multiple Input Multiple Output Several transmitting and receiving antennas Spatial beam forming E.g. a x b : c Transmit antennas: a Receive antennas: b Spatial streams: c

93 Properties of Media Distortion Amplitude response function Phase shift (response) function Signal to noise ratio Signal attenuation Crosstalk and interference Signal delay

94 Speed of Propagation Speed of light Affected by the dielectric medium, geometry Modality Skew Multiple paths Multiple modes Multiple wires

95 Wired vs Wireless Wired Point to point Needs right of way Can bundle more wires to increase capacity Exponential attenuation Wireless Limited spectrum Heavily regulated Mobile Polynomial attenuation (for small distances)

96 Twisted Pair The twists help minimize interference Many pairs can be bundled together Significant attenuation w/ distance 1-4 1kHz kHz Bandwidth depends on distance

97 Digital Subscriber Lines Asymmetric DSL Uses twisted pairs Depending on the distance can have from 1.5 to 6 Mbps Often use Discrete MultiTone to avoid noisy and distorting areas of the spectrum Subchannels use QAM

98 Local Area Networks Mostly twisted pairs currently Used coaxial cable 20 years ago Cat-3 is Unshielded Twisted Pair Cat-5 (5e, 6) is tightly twisted UTP

99 10BASE-T 10 Mb/s baseband twisted pair Two cat-3 UTP connect between computer and hub (star configuration) Manchester line coding Max distance 100 meters

100 100BASE-T Comes in at least two flavours 100BASE-T4 (defunct) 3 pairs of UTP, cat-3, + one more for collision detection Each pair carries 33 1/3 Mb Uses a 4B3T scheme 100BASE-TX Two cat-5, one per direction 125 Mpulses/s Uses a 4B5B scheme

101 Coaxial Cable Solid center conductor inside a braided cylindrical outer conductor Great immunity against interference and crosstalk (low SNR) High bandwidth; good for backbone networks Costly to make, install and handle Fiber (mostly) took over

102 Cable Modem Original cable was for downstream only Was modified to carry a narrow band of upstream Each TV channel can carry 36Mb/s User modem has to listen for packets in a certain channel and compete for time slots on the cable

103 Old Style Ethernet 10Base5 and 10BASE2 Aka thick and thin Ethernet Baseband, Manchester Thick, was too bulky

104 Optical Fiber Thin ultra-transparent glass fiber Transparency varies with wavelength For some bands (1300 nm, 1550 nm) goes down to about.2 db/km

105 Transmission Modes

106 Propagation Different modes travel with different speeds After a long run the two modes will be out of sync and the signal will be heavily distorted Very thin fibers will support only one mode Unimodal fibers can be used for large distances (5-70 km) Unimodal fibers have very little (but non-zero) distortion

107 Frequency and Bandwidth f = c λ f = = f = c λ 2 λ = f λ λ

108 Higher Frequency-Higher Bandwidth Light is EM radiation, just higher frequency (in the order of a million times higher then WiFi) What looks like a narrow band can carry many Gb of information Main limit to distance is power loss Intermediate optical amplifiers (now are cheap) Electronic regeneration (expensive)

109 Wavelength Division Multiplexing Similar to Frequency Division Multiplexing Like tuning to radio stations We do not quote frequencies in optical bands Hence wavelength Lasers of different color carry different signals Color is in the infrared and thus invisible

110 WDM Early systems handled 16 channels, 2.5 Gb/s each Coarse WDM have few channels widely separated (cheap) Dense WDM have dense packing of channels channels Gb/s.8-.4 nm separation

111 Backbone Networks High capacity of fiber makes them ideal for backbone networks Typically Gb/s with WDM Cost for the last mile is prohibitive

112 LAN on Fiber Can easily do 1Gb/s and 10Gb/s For shorter distances copper is competitive For larger distances (.5-5 km) fiber wins Multimode can do up to.5 km Two primary standards 1000BASE-SX shortwave (850nm) 1000BASE-LX longwave (1300nm)

113 Radio Transmission Spectrum ranges from 3 khz to 300 GHz Wavelength is from 10 km to 1 mm Bands are spaced every factor ten LF, MF, HF, VHF, UHF, SHF, EHF Higher frequencies more directional Most digital communication is currently 1-5 GHz Point-to-point, satellite comm in 2-40 GHz

114 Infrared Does not penetrate walls Reduce interference Huge potential bandwidth In theory! IrDA (Infrared Data Association) Various standards to connect devices like keyboards, mice and printers to a computer

115 Error Detection and Correction Probability of error in a transmission is never zero Using various techniques we can make it arbitrarily small Media have error rates from 1.0e-3 to 1.0e-9 Tolerance to error also varies hugely

116 Error Control Techniques Two basic approaches ARQ (Automatic Retransmission request) Needs a return channel Forward Error Correction Enough redundancy to recover missed signal Both require error detection Both trade bandwidth for reliability

117 Error Detection Basic idea simple Specify a pattern Add enough bits to the signal to satisfy the pattern Check if it still satisfies the pattern after transmission Simplest pattern is a parity bit

118 Single Parity Check-code Takes k information bits and produces one check bit Append the check bit to the information bits and form a codeword If the check bit is the modulo-2 sum of the information bits we have an even parity scheme If the receiver sees odd parity, it means an error occurred. We use modulo-2 arithmetic throughout this section

119 Probability of an Error We usually assume that the probability of an error (flipped bit) is small Also assume that the probability or error in two bits is independent Let p is the probability of an error Let n be the number of bits in a packet

120 Probability of having j Errors p( j) = ( n j) p j (1 p) n j ( n j) = n! j!(n j)!

121 Two Dimensional Parity The simplest method Not the most efficient

122 Internet Checksum Most communication systems use some form of checksum Minimizes the probability of error going undetected Has to be simple to compute, portable IP uses 16 bit, 1's complement We will play with 4 bits

123 Checksum: example The checksum of two words 10, 12: 1010, = 22; = (1) mod 15 = 7; mod 15 = 8; 1000 At the receiving end 10, 12, 8, 1010, 1100, =30; (1) mod 15 = 0

124 Polynomial Codes Also known as Cyclic Redundancy Check (or Codes) We use polynomials with modulo-2 coefficients Every bit string is a polynomial In our mind! Every bit is a coefficient A byte is a 7th degree polynomial (not 8!)

125 Modulo-2 Arithmetic 1+1 = 0 1+0=0 0+1=0 1-1 = 0 Addition and subtraction are the same 1*1 = 1 1*0=0 0*1=0 0*0=0

126 Information Polynomial k 1 i( x) = m=0 i m x m i( x) = i k 1 x k 1 +i k 2 x k 2 + +i 1 x 1 +i 0

127 Operations on Polynomials Add two polynomials Pairwise add of the corresponding coefficients Bit-wise XOR Subtract Exactly the same as add! Multiply Use the distributive property Unsurprisingly, polynomial multiplication is a Convolution!

128 Polynomial Division Similar to integer division It is not (exactly) a convolution We use a longhand-like procedure in our examples

129 Polynomial Division x 3 + x 2 +x quotient x 3 + x+1 x 6 + x 5 x 6 + +x 4 +x 3 divident divisor x 5 + x 4 +x 3 x 5 + +x 3 + x 2 remainder x x 2 x x 2 +x x

130 Generator polynomial When we k bits of information embedded in n bit codeword we have n-k check-bits We will use an n-k degree generator poly We multiply our information polynomial by the n-k power of x (stick n-k zeros at the end) Divide by the generator polynomial Add the remainder to the shifted information polynomial

131 Error Polynomial R( x) = b( x)+e(x) R( x) = g (x)q (x)+e( x)? e( x) = g (x)q' ( x)

132 Error Polynomial The error is detectable iff the error polynomial is not divisible by the generating polynomial We examine 3 cases: One error Two errors Burst of errors

133 Single Error e( x)=x i e( x)=x i g (x)q ' ( x)

134 Iff... The generating polynomial has more than one 1 in it Easy to guarantee that single errors are detected

135 Double errors e( x) = x i + x j = x i (1+x j i ) x i g ( x)q i ( x) i< j (1+x j i ) g (x)q j i

136 Double errors The second condition will hold if the polynomial is primitive A polynomial of degree N is primitive iff the smallest value of m for which 1 + x m is divisible by the polynomial is m = 2 N -1 If g(x) is of degree n-k will detect double errors if the codeword length is less than 2 n-k -1. Often the polynomials used in practice have the form q(x) = ( 1 + x ) p(x) Where p(x) is primitive

137 Burst of Errors e( x) = x i d (x) x i g ( x)q i ( x)

138 Burst of Errors If the degree of d(x) is less than g(x), g(x) will never divide d(x) If the length of the burst is less than the degree of the generating polynomial the error will be detected I the error burst is longer it will be detected with probability 1-2 k-n.

139 Shift register for remainder + Reg 0 + Reg 1 Reg 2