Transcript Slide 1

Chapter 3 – Data Transmission:
Concepts and Terminology
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Transmission Terminology
 data transmission occurs between a transmitter
& receiver via some medium
 guided medium
 eg. twisted pair, coaxial cable, optical fiber
 unguided / wireless medium
 eg. air, water, vacuum
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Transmission Terminology
 direct link
 no intermediate devices
 point-to-point
 direct link
 only 2 devices share link
 multi-point
 more than two devices share the link
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Transmission Terminology
 Simplex transmission
 one direction
• eg. television
 Half-duplex transmission
 either direction, but only one way at a time
• eg. police radio (walkie-talkie: push-to-talk and
release-to-listen)
 Full-duplex transmission
 both directions at the same time
• eg. telephone
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Time domain concepts of signals
 time domain concepts
 analog signal
• various in a smooth way over time
 digital signal
• maintains a constant level then changes to another
constant level
 periodic signal
• pattern repeated over time
 aperiodic signal
• pattern not repeated over time
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Analog and digital signals
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Periodic signals
 The signal period T is the
inverse of signal frequency f :
1
T
f
T in sec onds ( s)
f in Hertz ( Hz)
 The signal s(t) is periodic if:
s(t  T )  s(t )
  t  
 The signal amplitude is
denoted by A
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Sine wave
 Mathematically, the sine wave is given by :
s(t )  A sin(2 ft   )
 Three parameters :
1. Peak amplitude (A)


maximum strength of signal
usually measured in volts
2. Frequency ( f )




rate of change of signal
measured in Hertz (Hz) or cycles per second
period = time for one repetition ( T )
T = 1/f
3. Phase (  )

relative position in time
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Varying Sine Waves
s(t )  A sin(2 ft   )
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Wavelength (λ)
 is the distance occupied by one cycle
 assuming signal velocity v, then  = vT
 or equivalently f = v, since T=1/f
 for the special case when v=c
 c = 3*108 m/s (speed of light in free space)
 c=λf
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Frequency Domain Concepts
 signal are made up of many frequencies
 components are sine waves
 Fourier analysis can shown that any signal
is made up of component sine waves
 Fourier series of a square wave with
amplitudes A and –A :
sin(2 kft)
s(t )  A 

 k 1, k odd
k
4

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Fourier Transform
Mathematical tool that relates the frequency-domain
description of the signal to its time-domain description
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Time-domain vs frequency-domain
Figure 3.5a: frequency domain function for the
signal of Figure 3.4c.
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Time-domain vs frequency-domain
Time-domain
Frequency- domain
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Spectrum and bandwidth
 Spectrum
 range of frequencies contained in signal
 Absolute bandwidth
 width of spectrum
 effective bandwidth
 often just bandwidth
 narrow band of frequencies containing most energy
 DC Component
 component of zero frequency
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Acoustic Spectrum
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Analog and digital data transmission
 data
– entities that convey meaning
 signals & signalling
– electric or electromagnetic representations of
data, physically propagates along medium
 transmission
– communication of data by propagation and
processing of signals
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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
300-3400Hz
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Digital Data
 as generated by computers etc.
 has two dc components
 bandwidth depends on data rate
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Analog Signals
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Digital signals
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Advantages and disadvantages of digital signals
 cheaper
 less susceptible to noise
 but greater attenuation
 digital now preferred choice
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Transmission Impairments
 signal received may differ from signal
transmitted causing:
 analog - degradation of signal quality
 digital - bit errors
 most significant impairments are
 attenuation and attenuation distortion
 delay distortion
 noise
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Attenuation
where signal strength falls off with distance
depends on medium
received signal strength must be:
 strong enough to be detected
 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
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Delay distortion
 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 intersymbol interference
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Noise
 Additional unwanted signals inserted
between transmitter and receiver
 Thermal
 due to thermal agitation of electrons
 uniformly distributed
 white noise
N 0  kT (W / Hz)
N 0  noise power density in watts per 1Hz of bandwidth
k  Boltzm ann' s const ant  1.3810 23 J / K
T  Tem praturein Kelvins
 Interference from other users in a multi-user
environment (e.g., mobile environment)
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Noise
 crosstalk
 a signal from one line is picked up by another
 impulse
 irregular pulses or spikes
• eg. external electromagnetic interference




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
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Noise: example
0
+5V
1
-5V
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Data-rate
Data rate: is the rate, in bits per second (bps), at
which data can be communicated
1
1
1
data Rate  R 


 50 kbps
bit duration Tb 0.02m sec
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Spectrum, bandwidth and Data-rate
 Spectrum of a signal: is the range of frequencies that it
contains
 Absolute bandwidth: is the width of the spectrum
 Effective bandwidth: is a relatively narrow band that contains
most signal energy
 Any transmission system has a limited bandwidth
 Square wave have infinite components and hence infinite
bandwidth, but most energy in first few components
 Limited bandwidth increases distortion
 Limited bandwidth also limit the data rate that can be carried
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Bandwidth
Bandwidth B  3 f  1 f  2 f
Assum e f  1KHz,
then Bandwidth B  2 KHz
Absolute B  
Effective B  width of m ain lobe
1

X
Assum e X  1 m sec,
then effective B  1KHz
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Data-rate and bandwidth
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Channel Capacity
 Channel Capacity: max possible rate at which data
can be transmitted over a given communication
path, under given conditions
 Channel capacity is a function of :
 data rate - in bits per second [bps]
 bandwidth - in Hertz [Hz]
 noise - on communication link
 error rate - the rate at which errors occur, reception of 1
when 0 is transmitted, and visa versa
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Nyquist Bandwidth
 Consider noise free channels
 If rate of signal transmission is 2B then we can
carry signal with frequencies no greater than B
 i.e., given bandwidth B, highest signal rate is 2B
 For binary signals (0,1), 2B bps need bandwidth B Hz
 Can increase rate by using M signal levels or M
symbols (e.g. M=4, Quaternary: 00, 01, 10,11)
 Nyquist formula is:
C  2B log2 M
[bps]
 So increase rate by increasing signal levels
 at cost of receiver complexity
 limited by noise & other impairments
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Shannon Capacity Formula
 Consider relation of data rate, noise & error rate
 faster data rate shortens each bit so bursts of noise affects more bits
 given noise level, higher rates means higher errors
signal power
 Signal-to-Noise Ratio (SNR): SNR 
noise power
 SNR in decibles (dB):
SNRdB  10log10 SNR
 Shannon’s channel capacity (C) in bits/s is related to the
channel bandwidth (B) in Hertz and SNR by:
C  B log2 (1  SNR)
 theoretical maximum capacity
 get lower in practise
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Nyquit bandwidth and Shannon Capacity
 Example: Suppose that the spectrum of a channel is
between 3MHz and 4MHz and the SNRdB=24dB. Find:
1. The channel bandwidth (B)
2. The channel capacity (C)
3. Based on Nyquist formula, how many signalling levels are
required to achieve the max capacity
Solution:
1. B = 4MHz - 3MHz = 1MHz
2. SNRdB  24dB  10log10 SNR
 SNR  251
C  B log2 (1  SNR)  106 log2 (1  251)  8 106  8Mbps
3. C  2B log2 M
8 106  2 106 log2 M
 M  16
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Decibels and signal strength
 It is customary to express gain or loss (attenuation) in decibels:
 Logarithmic unit (compressed scale)
 Multiplication and division reduce to addition and subtraction
 The decibel power gain (GdB):
Pout
GdB  10log10
Pin
Pin : input power level
Pout : output power level
 The decibel power loss (LdB):
Pout
Pin
LdB  10 log10
 10log10
Pin
Pout
Pin
Vin2 / R
Vin
L

10
log

10
log

20
log
 The decibel voltage loss: dB
10
10
10
Pout
Vout2 / R
Vout
where V is the voltage across resistor R
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Decibels and signal strength
 Example 1: if a signal with a power level of 10mW is inserted
onto a transmission line and the measured power some
distance away is 5mW, then the loss can be expressed as:
LdB  10log10
Pin
10m W
 10log10
 3dB
Pout
5m W
 Example 2: Consider a series of transmission elements in
which the input is at a power level of 4mW, the first element is a
transmission line with 12dB loss, the second element is an
amplifier with 35dB gain, and the third element is a
transmission line with 10dB loss.
1. The net gain is -12 + 35 – 10= 13dB
Pout
2. The output power (Pout): GdB  13dB  10log10
4m W
Pout  4 101.3 m W  79.8m W
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Decibels and signal strength
 The dBW (decibel-Watt):
powerdBW
 powerW 
 10log10 
  10log10 ( powerW )
 1W 
 Example: a power of 1W is 0dBW,
a power of 1000W is 30dBW,
a power of 1mW is –30dBW
 The dBm (decibel-milliWatt):
powerdBm
 powermW 
 10log10 

 1m W 
 Example: a power of 1mW is 0dBm,
a power of 30dBm is 0dBW
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Example
 Given a receiver with an effective noise temperature of 294K
and a 10 MHz bandwidth. Find the thermal noise level (N0) at
the receiver’s output in units of dBW?
N 0  kT
[W / Hz]
N  kTB [W ],
k  Boltzm ann' s const.  1.3810 23 J / K
T  Tem praturein Kelvins
B  Bandwidth
kTB
N dBW 10 log10
 10 log10 kTB  10 log10 k  10 log10 T  10 log10 B
1W
 10 log10 (1.3810 23 )  10 log10 (294)  10 log10 (107 )
 228.6  24.7  70
 133.9dBW
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The expression Eb/N0
 The expression Eb/N0 : is the ratio of signal energy per bit (Eb) to
noise power density per Hz (N0)
Eb  STb ,
where S : signal power
Tb : bit duration
N 0  kT ,
where k : Boltzm annconst.
T :Tem prature
1
data Rate  R 
Tb
Eb S / R
S



N0
N0
kTR
in decibel notation,
 Eb 

  S dB  10 log10 R  10 log10 k  10 log10 T
 N 0  dB
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Example
 For Binary Phase Shift Keying (BPSK) modulation, Eb/N0 = 8.4
dB is required for a bit error rate of 10-4 (one bit error out of
every 10000 bits). If the effective noise temperature is 290 K
(room temperature) and the data rate is 2400 bps, what
received signal power level is required?
 Eb 

  S dB  10 log10 R  10 log10 k  10 log10 T
 N 0  dB
8.4  S dB  10 log10 2400 (228.6)  10 log10 290
8.4  S dB  (10)(3.38)  228.6  (10)(2.46)
 S dB  161.8 dB
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Eb/N0 versus SNR
 We can relate Eb/N0 to the Signal-to-Noise Ratio (SNR):
Eb S / R

N0
N0
Noise power N  N 0 B, where B is the bandwidth
Eb
S/R S B
S


, where
is the Signalto Noise Ratio( SNR)
N0 N / B N R
N
S
The Shannonchannelcapacity: C  B log2 (1  SNR)  B log2 (1  )
N
S

 2C / B  1
N
E
S B
B
B
 b 
 2C / B  1  2C / B  1
N0 N R
R
C
where C / B is the spectral efficiency
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Example
 Suppose we want to find the minimum Eb/N0 required to achieve
a spectral efficiency C/B of 6bps/Hz




Eb
B
1
 2C / B  1  26  1  10.5  10.21dB
N0
C
6
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