Digital Transmission Fundamentals
Chapter 3.Communication Networks
Leon-Garcia, Widjaja
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
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
Block information
D=t p+LR
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.
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
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
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
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
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
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)
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
Channel Bandwidth
f
A(f)
Sawtooth components
Intermediate components
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
Signal to Noise ratio
σn2=E {n2
}
P=1Rn2
Signal to Noise ratio
SNR=P sig
Pnoise
SNR db=10 log10 SNR
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 0.10111001101.... Volt● We need zero bandwidth for this!
Noise and Signal
Multilevel Signals
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)
Example: Telephone lines
● They have 1% noise (amplitude)● SNR is 10,000:1● SNR in db is 10*log 10000 = 10*4 = 40
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
Channel Capacity
C=W log2(1+SNR)
Example Telephone lines
● Bandwidth 3.4 kHz● SNR 10,000● Capacity is:
– 3400 * log(10000)/log(2) = 3400*4*3.3 = 45,178
Fourier Transform Signals
x (t)=∑ αk cos (2 π k f o t+ϕk )
x (t)=∑ αk sin (2π k f ot+ϕ k )
x (t)=∑ αk cos (2 π k f o t)+¿
∑βk sin(2π k f o t)
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
Bandwidth: Examples
1 0
kHz
1 2 3 4
1 2 3 4
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
Sampling
x (t)x (nT )xn= x(nT )
s (t )
xr( t)=∑n
xn s (t−nT )
Samplingxr=∑
n
xn s (t−nT )
s (t )
x1 x3
x1 x3
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
Sampling Function
s (t )=sin (2πW t )
2πW t
W=1T
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
Sampling Function
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
Quantization Error
en= yn− xn
SNR=σ x
2
σ e2
Quantization Error
y
x−Δ2
Δ2
Quantization Error
σ e2=Δ2
12
Δ=V
2m−1
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
Power of the Signal
-VV
V=4 σ x
SNR
SNR =σ x
2
σ e2 =
V 2
16Δ
2
12
=34V 2
Δ2
=34V 222 (m−1)
V 2 =34
22(m−1 )
=316
4m
SNR db
SNRdb = 10 log(4 )m + 10 log(316
)
= 6m−7.27
= 10 log316
4m
SNR - Example
● For 8 bit per sample (telephones)● SNR = 48 – 7.27 = 40.73 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
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
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
Linear Time Invariant
x2( t )→ y2( t)
x1(t )+ x2( t)→ y1(t)+ y2(t)
x1(t+τ)→ y1( t+ τ)
x1(t )→ y1( t)
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
Amplitude Response Function & Fourier
x (t) =∑k
αk cos(2π k f o t)
y (t )=∑k
A(k f o)akcos (2 π k f o t + ϕ (k f o))
Delay
ϕ( f ) =−2π f τ
cos (2π k f o t−2π f τ) = cos (2 π k f o(t−τ))
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
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
Example: Low-pass channel
A( f )=1
ϕ( f )=2π f τ
h( t)=s (t−τ)
s (t )=sin(2 πW t)
2πW t
−W≤f≤W
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?
Nyquist Signaling Rate
The Nyquist Signaling Rate
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
Shannon Channel Capacity
Withoutnoise
Shannon Channel Capacity
With littlenoise
Shannon Channel Capacity
More bits
Shannon Channel Capacity
Too manybits
Shannon Channel Capacity
C =W log2(1+SNR)
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
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)
Non-Return to Zero
● Consecutive 1s have no intermediate zero in between
● Fewer transitions than RZ
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
DC Component Power
P=0.5
P=0.5∗02+0.5∗12
DC Component
● A long series of 1s can cause even more problems
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
Polar NRZ Power
P=0.5∗0.25+0.5∗0.25
P=0.25
P=0.5∗(−0.5)2+0.5∗0.52
Polar NRZ
● Long sequences of either zeros or ones is still a problem– DC component
– Loss of clocking
Bipolar NRZ
● Zero is 0V● One is either -2.5 or +2.5V (alternates)● DC component disappears● Clocking problem for long strings of 0s
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
1 1 1 0 0
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
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)
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
Modulation-Demodulation
● Modems● We have to use them whenever we cannot
use the baseband– Telephone lines
– Cable
– Wireless
– Fiber
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
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
Phase Shift Keying
● Transmit cos(w t) for 0● Transmit cos(w t + ph) for 1
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
Decoding PSK
A cos (ωc t)×cos (ωc t )=A2
(1−cos(2ωc t ))
A2
(1−cos(2ωc t ))∗LPF =
A2
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!
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
QAM
Y (t )= A cos (ωc t) + B sin(ωc t )
(Y (t )×cos(ωc t))∗LPF =
( A2 (1+cos (2ωc t )) +B2
sin(2ωc t))∗LPF =
A2
QAM
Y (t )= A cos (ωc t) + B sin(ωc t )
(Y (t )×sin (ωc t ))∗LPF =
( A2 sin (2ωc t) +B2
(1−cos (2ωc t)))∗LPF =
B2
QAM Mod-Demod
Bk
×
+
sin(ωc t)
Ak
cos (ωc t)
×
Y k
Y k
×
×
2sin(ωc t)
2cos (ωc t)
Bk
Ak
LPF
LPF
Signal ConstellationsB
A A
B
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
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)
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
Properties of Media
● Distortion– Amplitude response function
– Phase shift (response) function
● Signal to noise ratio– Signal attenuation
– Crosstalk and interference
● Signal delay
Speed of Propagation
● Speed of light– Affected by the dielectric medium, geometry
– Modality
● Skew– Multiple paths
– Multiple modes
– Multiple wires
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)
Twisted Pair
● The twists help minimize interference● Many pairs can be bundled together● Significant attenuation w/ distance
– 1-4 db/mile @ 1kHz
– 10-20 db/mile @ 500kHz
● Bandwidth depends on distance
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
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
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
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
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
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
Old Style Ethernet
● 10Base5 and 10BASE2● Aka thick and thin Ethernet● Baseband, Manchester● Thick, was too bulky
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
Transmission Modes
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
Frequency and Bandwidth
f =cλ
f =3⋅108
10−6 = 3⋅1014
∂ f =cλ2
∂λ = f ∂λλ
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)
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
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– 80-160 channels
– 10-40 Gb/s
– .8-.4 nm separation
Backbone Networks
● High capacity of fiber makes them ideal for backbone networks
● Typically 40-1600 Gb/s with WDM● Cost for the last mile is prohibitive
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)
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
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
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
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
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
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
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
Probability of having j Errors
p( j)= (nj) pj(1− p)n− j
(nj)=n!j !(n− j)!
Two Dimensional Parity
● The simplest method● Not the most efficient
0 1 1 11
1 0 1 00
1 0 0 10
1 0 0 01
1 1 0 00
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
Checksum: example● The checksum of two words
– 10, 12: 1010, 1100
– 10+12 = 22; 1010+1100 = (1)0110
– 22 mod 15 = 7; 0111
– -7 mod 15 = 8; 1000
● At the receiving end– 10, 12, 8, 1010, 1100, 1000
– 10+12+8=30; (1)1110
– 30 mod 15 = 0
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!)
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
Information Polynomial
i( x) = ∑m=0
k−1
im xm
i( x) = i k−1 xk−1+ ik−2 x
k−2+…+i1 x1+i0
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!
Polynomial Division
● Similar to integer division● It is not (exactly) a convolution● We use a longhand-like procedure in our
examples
Polynomial Division
x3+ x+1
x3+ x2+x
x6+ x5
x6+ +x 4+x3
x5+ x4+x3
x5+ +x3+ x2
x 4+ + x2
x 4+ + x2+xxremainder
divisor
divident
quotient
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
Error Polynomial
R( x) = g (x )q (x )+e ( x)
e ( x)?= g (x )q ' ( x)
R( x) = b( x)+e (x )
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
Single Error
e ( x)=x i≠g (x )q ' ( x)
e ( x)=x i
Iff...
● The generating polynomial has more than one 1 in it
● Easy to guarantee that single errors are detected
Double errors
x i ≠ g ( x)q i( x)
e ( x) = x i + x j = x i (1+x j−i)
(1+x j−i) ≠ g (x )q j−i
i< j
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 + xm is divisible by the polynomial is m = 2N-1
● If g(x) is of degree n-k will detect double errors if the codeword length is less than 2n-k -1.
● Often the polynomials used in practice have the form– q(x) = ( 1 + x ) p(x)
– Where p(x) is primitive
Burst of Errors
e ( x) = x i d (x )
x i ≠ g ( x)q i( x)
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 1-2k-n.
Shift register for remainder
Reg 0 Reg 1 Reg 2+ +