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19/02/06 CS6242 : Intro & overview1
University of Manchester
School of Computer Science
CS6242 Mobile Computing
Introduction & layer 1 overview
Barry Cheetham
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Personnel
1. Physical layer & radio channel (Barry & Karim)
2. Medium access control (Nick)
3. Error control (Barry)
4. Network transport layer issues (Nick)
5. WLAN security (Aleksandra)
6. Higher layer issues (Barry + Nick + guest))
Demonstrators: Karim, Ian Featherstone, Basab Sen
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CS624: Mobile Computing
Level:MSc Credit Rating: 15 Degrees: ACS/CS
Pre-requisites: Basic maths Pre-course work: 40 hrs
Taught week: 40 hours: lectures & laboratories
Post-course work: 40 hours assessed practical
Assessment: 67 % practical & 33 % exam
Staff: B. Cheetham, N Filer, K Nasr, A Nenadic, guest,
I. Featherstone, B. Sen
www.cs.man.ac.uk/Study_subweb/Ugrad/coursenotes/CS6242
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Aims & Learning outcomes
Aims: Understanding of concepts underlying current developments
in mobile comms & wireless computer networks.
Learning Outcomes
1) Understanding of radio propagation & interference
2) Understanding of digital transmission systems 3) Understanding of MAC protocols for wireless networks 4) Understanding of the systems, protocols and mechanisms to support mobility for mobile internet users5) Ability to investigate transmission & modulation using MATLAB. & experience of using a network simulation package (OPNET)
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Reading list
J.Schiller, Mobile communications, Addison-Wesley, 2003
Supplemental books
T.S. Rappaport, Wireless communications; Principle and Practice,
A S. Tanenbaum, Computer Networks (4th Edition), Prentice Hall,
2003.
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Detailed Syllabus
Intro to wireless networking & digital trans.
Characteristics of radio propagation.
MAC & error control.
Network & transport layers
WLAN security
Protocols supporting mobility
MATLAB tutorials & assignment
OPNET tutorials & assignment
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Aims for Barry & Karim
1. Up-to-date overview of wired & wireless
telephone & computer networks
2. Principles of digital transmission (i) at base-band
(ii) by single carrier modulation
(iii) by multi-carrier modulation
3. Progagation of radio waves
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1.1. Principles of digital transmission for wired &
wireless telephone & computer networks
ChannelDAC ADC10110 10111
* Transmitter like DAC & receiver like ADC
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Receive symbols
& map to bit-stream
Map to base-band
tvolts
1.1.2 (i) Base-band transmission
Channel10110.. 10110
Ethernet with Manchester coding uses base-band signalling
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Base-band signalling
Map to base-band
1011110t
volts
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Manchester coding
Map to base-band
1011110t
volts
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1.1.2 (ii) Modulation of single carrier
Modulatecarrier
Map to base-band
10110
tvolts
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Amplitude modulation of single carrier
Map to base-band
10110
tvolts
Multiply
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Phase modulation of single carrier
Map to base-band
10110
t
volts
Multiply
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Amplitude & phase modulatn of single carrier
Map to base-band
10110
t
volts
Multiply
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Effect of single carrier modulation on frequency spectrum
carrier
frequency
Power spectral density
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Message * carrier
A cos(Mt) * cos(Ct)
= 0.5A cos(Ct + Mt) + 0.5A cos(Ct - Mt)
= 0.5A cos( (C + M) t ) + 0.5 A cos((C - M)t)
Where do we get those ‘side-bands’ from?
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1.1.3 ASK, FSK, PSK
r(t)
cos(2ct)
b(t)
t
b(t)r(t)
t
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ASK spectrum
frequency
-fC fC
Power spectral density
PSD
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FSK_ frequency shift keying
FM Modulator
(VCO)1
0 0 1 0
Simple generator using voltage controlled oscillor:
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FSK : another generation method
1
0
FSK
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FSK and GMSK
Advantages : constant envelope,
insensitivity to frequency shifts Disadvantage : spectral inefficiency Gaussian minimum shift keying:
2 bits/s /Hz
Spectrum similar to ASK
Used for GSM
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PSK_ phase-shift keying
cos(2ct)Carrier
t
cos(2ct)
Voltage b(t)
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PSK waveform
t
V
1 1 0 0 1 1 0
DPSK more commonly used
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1.1.4 Coherent demodulation of ASK
10110
tvolts
Multiply Threshold detector
Lowpass
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1.1.5 Non-coherent detection of ASK
Rectify & smooth
Thresholddetector
t
10110
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1.1.6. Vector modulator for single carrier
Mult
Mult
ADD
Map
Map
Cos(2fCt)
Sin(2fCt)
bR(t)
bI(t)10110
11011
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Vector de-modulator
Mult
Mult Detect
Detect
Cos(2fCt)
Sin(2fCt)
bR(t)
bI(t) 10110
11011
Lowpass
Lowpass
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Complex base-band
* Transmit:
bR(t)cos(2fCt) + bI(t) sin(2fCt)
= Real { [bR(t) + jbI(t)] . [cos (2fCt) – j sin(2fCt)] }
= Real { [ b(t) ] . exp(-2fCt) }
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Remember:
cos 2 (2fCt) = 0.5 + 0.5 cos(4fCt)
sin 2 (2fCt) = 0.5 - 0.5 cos(4fCt)
sin(2fCt) cos (2fCt) = 0.5sin(4fCt)
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Vector modulator (again)
MultMap
exp(-2fCt)
b(t)10110
11011 Complexsignal
Complxbase-band
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1.1.7 QPSK _ quaternary PSK
Where bR(t) & bI(t) are bipolar, bR(t)cos(2fCt) & bI(t) sin(2fCt) are PSK. ‘2-channel’ modulation process is QPSK.
Bandwidth efficiency twice that of PSK. 2 bits/second per Hz. Widely used.
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1.1.8 Pulse shaping & bandwidth efficiency
Spectra of pulses must be adapted to channel. Channel will have limited bandwidth. Rectangular pulses totally unsuitable in practice. Ideal pulse shaping causes each pulse to ‘ring on’ forever
once it has started. Risk bit-errors due to ‘inter-symbol’ interference (ISI). Can eliminate ISI even when pulses do run into each other, At bb only when symbol rate < 2 x bandwidth of pulse. Max bandwidth efficiency at bb: 2 symbols/s per Hz.
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1.1.9 Bandwidth efficiency with binary
With binary signalling, each symbol represents one bit. With a pulse-shape carefully chosen to eliminate ISI
binary can achieve up to 2 b/s per Hz at base-band. Multiplication by sinusoidal carrier doubles bandwidth Max of 1 b/s per Hz with real base-band signal. With vector-modulation, 2 channels each giving 1 b/s per
Hz possible, Brings achievable bandwidth effic back to 2 b/s per Hz. At expense of complexity of coherent detection. Binary PSK can achieve up to 1 bit/s per Hz, QPSK can achieve 2 bit/s per Hz.
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1.1.10 Multi-level signalling
With 2 b/s/Hz, a modem could achieve 6 kb/s over 3 kHz. 10% of what we know to be possible. Must use multi-level schemes where each symbol represents
more than one bit. Could have pulses of amplitude 0, 1, 2, 3, 4, 5, 6, 7 volts. Each pulse represents 3 bits at once (000, 001, 010, 011,
100, 101, 110, 111). Bandwidth efficiency increases to 6 bit/s per Hz. Cost is increased sensitivity to effects of noise. For 56k modem, more than 6 b/s per Hz required Combined PSK &ASK used to produce range of symbols.
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1.1.11 Single carrier digital modulation schemes
•ASK, FSK, PSK, DPSK, QPSK
•Differential QPSK
•Gaussian FSK & MSK
•Combined ASK & PSK (QAM, APK)
•etc.
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Other modulation techniques
•Direct sequence spread spectrum techniques (DSSS)
•Frequency hopping (FHSS)
•Complementary code keying (CCK)
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1.1.12: Multi-carrier modulation schemes
Advantages with respect to multipath fading. OFDM used for digital radio, TV , WLANs & ADSL. Radio/TV use 1024 carriers & WLANs use 64. Modulation achieved on all carriers by one inverse FFT. Use of ‘cyclic extension’ simplifies pulse shaping &
matched filtering as required with single carrier systems. Equalisation is greatly simplified. More on this later
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1.1.13 Shannon Hartley Law
Channel capacity: C = B log2(1 + S/N) bits/s Max bit-rate achievable with arbitrarily small BER. Bandwidth B Hz & AWGN. S/N is signal/noise power ratio (not in dB), C = B log10(1+S/N)/ log10(2) 0.332 log10(1+S/N) If S/N >>1, C 0.332 x B x SNR in dB. Valid when SNR = 10log10(S/N) >>0 Ex: What is C for 3kHz channel with 50dB SNR? Ex: SNR needed to convey 54 Mb/s over 20MHz ?
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1.2. Telephone networks
•1.2.1 Introduction
•POTs : Analogue using twisted pairs for “last mile”
Digital exchange to exchange.
300 - 3.4kHz speech sampled at 8kHz
ITU-T-G711 – 64-bit log-PCM (8 x 8kHz)
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1.2.2 Wireless Telephone networks
• Cordless (DECT) & cellular (GSM) mobile phones.
• Wireless local loop (then wired)
• Radio medium shared
• Cellular base-stations can “hand off”
• Fading due to multi-path (flat or frequency selective)
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1.2.3 Effects of multi-path in wireless telephony
•Line of sight paths rare in cities
•Fading caused by multi-path – flat or frequency-selective
•Affect gain & phase-delay of channel.
•Coherence b/w BC is largest b/w for which fading appears flat.
•In a city, BC 30 kHz, allowing analogue mobile phones with 30 kHz channels to work without equalisers.
•900 MHz GSM phones with 200 MHz b/w need equalisation.
•Equaliser is filter which reverses filtering effect of channel.
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1.2.4 Effect of multipath on bit-errors & ISI
Reductions in gain due to fading cause small signals to be received, so noise will have greater effect & produce more bit-errors.
Frequency selective fading alters shapes of pulses and thus causes ISI.
Gain and phase-delay affected
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1.3. Wired computer networks
LAN is equiv of local telephone loop Ethernet (IEEE802.3) Orig coax, hubs & contention mode (thin) Now twisted pairs & switched Ethernet Manchester coding MAC protocols
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CSMA for wired networks
Collision detection and avoidance Orig same ‘collision domain’ like radio (When one transmits all hear it). Hubs obsolete now Switches operate at data link layer
(examine headers to decide forwarding)
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1.3.3. Bridges, routers and gateways:
Connections from LAN to outside world Bridge is switch to interconnect LANs Router reads TCP header for destination &
chooses best way to send packet on its way. Like local telephone exchange.
Gateways connect devices with different protocols. E.g TCP/IP & telephone networks for VoIP.
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1.4. Protocols and Layers
Protocols are defined in layers. Well-known description of computer-to-
computer communication is:
‘OSI Reference Model’. “Open Systems Interconnection”, Term invented by “International Standards
Organisation” in 1983.
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1.4.2. ‘7-layer’ OSI reference model
7) Application Layer
6) Presentation Layer
5) Session Layer
2) Data Link Layer
3) Network Layer
4) Transport Layer
1) Physical Layer
7) Application Layer
6) Presentation Layer
5) Session Layer
2) Data Link Layer
3) Network Layer
4) Transport Layer
1) Physical Layer
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1.4.3. TCP/IP Reference Model
Similar to ‘7 layer OSI' model & pre-dates it.
Layer OSI Reference Model TCP/IP Reference Model
765
Application LayerPresentation LayerSession Layer
Application Layer
4 Transport Layer Transport Layer
3 Network Layer Internet Layer
21
Data Link LayerPhysical Layer
Host-to-Network Layer
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1.5. Wireless computer networks
Convergence of: Telephony (with expensive radio access) WLANs (with free radio access
in 2.4 & 5 GHz bands) Widely used for data via hot-spots etc Soon for telephony, multimedia etc.
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1.5.2 IEEE802.11 & other standards
Wi-fi : IEEE802.11 a, b, g & e. Hiperlan 1 & 2 (prob obsolete) Bluetooth WIMAX: IEEE802.15 Phy layer has same problems as telephony: multi-path & AWGN
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1.5.3. Orig IEEE 802.11 phy-layer for WLANs
First in 1997 : 1 & 2 Mb/s in 2.4 GHz band Spectral spreading mandatory Two SS methods:
Freq-hopping (around 80 MHz wide bands)
Direct sequence (XOR with 11-bit ‘chipping
sequence’ [1 0 1 1 0 1 1 1 0 0 0] )
DHSS multiplies bandwidth required by 11.
Reduces effect of noise less power needed.
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1.5.3. Original IEEE 802.11
• Released in1997 1 or 2 Mb/s
• Must use spread spectrum in 2.4 GHz band
•Two versions: FHSS & DSSS
• FHSS version hops around 80 carriers: .4 s dwell
•DSSS uses chipping sequence: {10110111000}
•Each bit => 11 chips. 1Mb/s => 11 Mb/s
• Spreads to 22 MHz.
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1.5.4. IEEE 802.11 frame structure
Preamble Header Payload
80 or 144 32 or 48 variable
Modulation technique:
FHSS : 2 or 4-level Gaussian FSK at 1 Mbaud
DSSS : 2 or 4 level DPSK at 11 Mbaud
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1.5.5. Latest IEEE802.11 standards
IEEE 802.11a : OFDM in 5.17-5.8 GHz band
64 carriers each modulated with PSK etc.
Up to 54 Mb/s. Great for multi-path.
IEEE 802.11b : Operates in 2.4-2.48GHz band
Same as 802.11 for preamble / header
Replaces 11-chip Barker sequence by codes.
1, 2, 5.5 or 11 Mb/s for payload (CCK)
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IEEE802.11g standard (Nov 2001)
* Extension to IEEE802.11b in 2.4 GHz band
* OFDM payload option at up to 54Mb/s
*Same preamble/header as IEEE802.11 orig DSSS & b
* OFDM classified as a spread spectrum technique
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“Bluetooth”
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1.5.6. Intro to IEEE802.11 MAC layer
Contention mode (CSMA/CA) Non-contention mode (central control via
PCF but never implemented) 802.11e standard has EDCA & HCCA QoS standard EDCA is enhanced CSMA/CA HCCA is new centrally controlled MAC
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1.5.7 IEEE802.11 MAC in contention mode
Based on a DCF mechanism. WLAN devices can sample medium & determine
whether any device is currently transmitting. Collision avoidance strategies then employed to
ensure, that a device transmits only when radio channel is likely to be free of other traffic.
Nick Filer will deal with this
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1.5.8.IEEE802.11 MAC in non-contention mode
Possible when there is a central device, which can act as a controller by informing all other devices when they are allowed to transmit or receive data.
Does this by periodically sending "beacons" to enable or disable non-contention mode & "polling" devices by sending further control packets to request data from each device.
A WLAN with central controller capable of fulfilling this co-ordination role is termed an "infrastructure" network
In most cases, the central controller also provides access to the outside world, e.g. via a telephone connection, and it is then termed an "access point".
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1.6. Bluetooth
Short range “piconet” for computer peripherals etc.
Not originally an IEEE standard; Operates in 2.4 GHz band over 10 metres. FHSS over 80 carriers: 160 hops /s Binary FSK at 1Mb/s. Problems with 802.11b transmissions in
range.
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1.7 Telephone & computer networks
•Much commonality physically & conceptually
•Connection oriented/ connectionless?
•Technologies merging: VoIP
ATM
PPP
•Issue is “quality of service” (QoS)
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1.8. Digital transmission in more detail
•Data carried by shaped waveforms or symbols.
•A symbol can carry 1 bit (binary signalling) or more.
•Bit-rate not same as symbol (baud) rate.
•With QPSK, each symbol carries 2 bits.
•For long term or continuous transmission receiver must synchronise to the transmission
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1.8.2. Manchester coding for baseband
+V
V
'one'
T T
'zero'
+V
V
t Manchestercoding
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1.8.3 Digital communication system model
Transmitter
Channel
Receiver..10110.. ..10110..
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Transmitter for synchronous communication
•Analogue waveform suited to the channel.
• If base-band send suitably shaped pulses: e.g. Manchester
•Otherwise modulate carrier with shaped pulse.
•For radio, bandlimited pulses needed.
not time limited.
carrier recovery must be possible
•May need to limit power of transmission.
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Receiver for synchronous communications
•Symbol rate and carrier frequency known but maybe not exactly
•Carrier phase not known.
•Receiver must:
* synchronise carrier, symbols and frames.
* then sample waveform to detect the data.
•Made difficult by attenuation, delay, Doppler & noise.
•Improved by matched filter and equaliser
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Receiver components
Vector Demod
Matched filter
Channel equaliser
Sample & detect
bR(t)
bI(t)
101.
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1.8.4. Digital filter
•Gain G(f) Phase (f)
•Gain in dB: 20 log10(G(f) )
•Phase delay: -(f) / (2f) seconds
•Linear phase: constant phase delay
•Channel acts like a filter
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Example of frequency response of a channel - gain
f
Gain (dB)0
-10
-20
fU fL
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Example of frequency response of a channel - phase
f
-(f)
fU fL
good
bad
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Example of frequency response of a channel - delay
f
-(f) / f
fU fL
good
bad
Different delay at different frequencies
same delay
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Frequency response of a band-pass filter - gain & phase
f
-(f) Gain (dB)
0
-10
-20
fU fL
0
90O
180O
Channel acts like a filter
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Low-pass filter - gain & phase
f
-(f) Gain (dB)
0
-10
-20
fC
0
90O
180O
Remove high frequency parts of signal
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High-pass filter - gain & phase
f
-(f) Gain (dB)
0
-10
-20
fC
0
90O
180O
Remove low frequency parts of signal
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Band-stop filter - gain & phase
f
-(f) Gain (dB)
0
-10
-20
fU
0
90O
180O Remove mid frequency parts of signal
fL
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Channel acts like a filter - not a nice one
t
t
Transmitter sends this
Receiver may get this becauseof different gains & delays at different frequencies
Volt
Volt
+3
+0.1
Not nice
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1.8.5 Frequency shifts and noise
t
t
Transmitter sends this
Receiver may get this because of noise
Volt
Volt
+3
+0.1
Not nice at all
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And still more bad news (for commuters)
… with mobile phones.
Doppler shift:
due to movement of receiver of transmitter
Different frequencies received from transmitted.
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Something about noise
Noise is an unwanted signal.
Can be a sine wave (sounds horrible)
But very often a ‘random signal’.
tVolt
Sounds like
waterfall or the sea.
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Random signal
When examining a signal x(t) with no obvious structure,
it is useful to assume that at any time t,
x(t) is a sample of a random variable X,
with statistical properties we can discuss.
In this case, x(t) is considered to be a “random” signal.
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Random variable
Because it has no structure, you can’t predict its value.
But you can say something about its
* mean (average)
* variance (mean-square value when mean = 0)
* distribution of values
‘PDF’ tells you about distribution of values.
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Probability density function (PDF) - 2 examples
1. Gaussian (normal) with mean m & standard deviation :
m
PDF(x)
m- m+
0.4/
0.24/
x
variance is 2
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2. Uniform between x = A & x = B:
A B m
x
1/(b-a)
From PDF(x) can deduce mean, mean-square value & variance.
mean: m = (B+A)/2
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‘Casino’ analogy0
0
0.1
-0.2-0.3
0.5
0.5
0
0
-0.7
1
-1.5
3
-4
6
9
Gaussian PDF
Assume voltagesgenerated byRoulette wheel
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‘Casino’ analogy 7
-1
-7
1-2
6
2
-8
0
-6
3
-5
5
-4
4
-3
Uniform PDF
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Statistical & electrical properties of noise
Statistical Electrical
mean average voltage
variance power (if mean=0
PDF distribution of voltages
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Exact meaning of PDF
PDF(x)
A B
0.4/
x
Probability of getting a value
between A & B is:
B
AdxxPDF )(
(Area under curve)
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Gaussian PDF continued
m Z
0.4/
x
Probability of getting a value >Z :
ZdxxPDF )(
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Gaussian PDF with zero mean & =1.
Z
0.4
x
Probability of getting a value >Z:
ZdxxPDF )(
= Q(Z) _ see graph
Power =1
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Gaussian PDF with zero mean & any .
Z
0.4/
x
Probability of getting a value >Z:
ZdxxPDF )(
= Q( Z / )
Power= 2
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Spectral properties of a random signal
Although x(t) random, can try to predict next valueFor example, if the values are: -2 , 10, 21, 33, 41, 55, 62, 69, …we may predict that the next sample is around 80.
Considering a second example with no predictability-2, 44, -4, -17, 9, 61, 2, -19, 3, -16, 1, -7, 30, -1, …No correlation between signal & signal delayed for any .No periodicity. Power spectrum will be flat or “white”.
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Power spectral density (PSD)
PSD(f) = power (Watts) per Hz at frequency f
f
f
Power in 1Hz band centred on f Hz is PSD(f) Watts
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Example: Power signal x(t) is white & has ‘2-sided’ PSD N0 /2 Watts/Hz.
What is its power in the bandwidth -B to B Hz?PSD(f)
f
B-B
Answer: N0B Watts
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Example: Power signal x(t) is white & has ‘1-sided’ PSD N0 Watts/Hz.
What is its power in the bandwidth 0 to B Hz?PSD(f)
fB
Answer: N0B Watts
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Example : Considering the sequences of samples of x(t), which one is more likely to be Gaussian ?
Time properties of x(t) (governing shape of power spectrum) & statistical properties are independent.
Can have white or spectrally coloured signal with same PDF
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Additive white Gaussian noise
The noise gets added to your nice clean signal.
Causes bit-errors - mistaking 1 or 0 or vice-versa.
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Additive white Gaussian noise (AWGN)
One parameter needed:
“1-sided” power spectral density (PSD): N0 Watts per Hz
“2-sided” PSD : N0/2 Watts/Hz
If ”1-sided” bandwidth B Hz, power = B . N0 Watts
Variance 2 = power (zero mean)
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Error performance
•Bit-error probability (PB):
probability of a single bit being wrong at receiver (eg 10-3)
•Bit-error rate (BER) :
if bit-error probability is 10-3, BER is ‘1 in 103 ’
(average of 1 bit-error for each 1000 bits).
Not ‘bit-errors per second’.
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1.8.7. Estimating effect of AWGN on BER
t
+V
+V/2
Voltage
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Complementary error function Q(z)
Prob of WGN sample with zero mean & variance 2=1 being greater than z is
z
dttpzQ y)probabiliterror d(normalise )()(
t
p(t)
z
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When 2 1
For WGN of variance 2, probability of a sample exceeding z is Q(z/),
probability that sample > A/2 is Q(A/(2)). Q(z) may be obtained from a graph, ‘erfc in
MATLAB or an approximation valid for z > 3: Q(z) = 0.5 erfc( z / 2 ) (0.4 / z) exp (z2 / 2)
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Example 1.3
After the demodulation, a radio receiver receives +1 volt & 0 volt pulses with AWGN of variance 2 = 0.01. Estimate the bit-error probability assuming a 0.5 volt threshold & an equal number of 1s and 0s.
Solution: 0.5 x Prob (noise > 0.5 volts) when "0" transmitted, plus
0.5 x prob (noise < -0.5) when "1" transmitted, is simply Q(0.5/0.1) = 3 x 10-7.
Bit-error probability PB = 3 x 107 Bit-error rate (BER) is 1 in 0.33 x 107
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Example 1.4 :
A receiver receives +1 & 0 volt pulses and has a bit-error probability of 10-3. What is the variance of the received noise?
Solution: Q(0.5/) = 10-3.
From graph, 0.5/ 3.2
Therefore, = 0.16 and 2 =0.026.
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Example 1.5:
A receiver receives 1 volt bipolar pulses with AWGN of ‘1-sided’ PSD N0 = 0.00002 Watts/Hz. Signal is passed thro’ a low-pass filter with 500 Hz cut-off & is then fed to threshold detector with threshold of zero. Estimate bit-error probability assuming equal numbers of ‘1’s & ‘0’s.
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Solution to 1.5:
Variance 2 = N0 B where B = 500 Hz. 2 = 0.01 and = 0.1. PB = 0.5 Q(1/s) + 0.5 Q(1/s) = Q(1/0.1) = Q(10). This is off scale of graph, but we can use MATLAB or
the approximation given earlier to obtain PB = Q(10) = 0.04 e50 = 7.7 x 1024 (with calculator) PB = 0.5 erfc(10/ 2 ) = 7.7e-24 (from MATLAB)
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Example 1.6:
A digital transmission system is affected by AWGN. Received power decreases with increasing distance according to a “inv-square law", i.e. P(d) 1/d2 where P(d) is power at distance d.
If distance is currently 200 metres, how much nearer must we move towards receiver if the bit-error probability is be reduced from its current value of 10-5 to 10-7.
Solution: Exercise for student.
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1.8.8 Waterfall graphs
t
Volts
T t
VV
-VT
Volts
Unipolar
Consider the unipolar & bipolar signals below:
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Energy per bit EB
Tells us how much energy (in Joules) taken from battery to transmit 1 bit.
Could estimate how many bits could be transmitted before battery runs out.
For unipolar at 1/T bits/s with rect pulses of voltage V, EB = V2T/2 Joules/ bit.
For equivalent bipolar, EB = V2T Joules/bit.
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Example 1.6:
A receiver receives a 10 b/s bit stream as V volt bipolar pulses with AWGN of ‘1-sided’ PSD N0 = 0.00002 Watts/Hz. The signal is passed thro’ a low-pass filter with cut-off frequency 500 Hz and then fed to a threshold detector with a threshold of zero.
For a range of values of 10 log10 (EB/N0) dB, say from -1 to 30 dB, estimate the bit-error probability assuming equal numbers of ‘1’s and ‘0’s. and plot a ‘waterfall graph’ of bit-error probability against EB/N0.
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Solution to 1.6:
Variance 2 = 500 N0. 2 = 0.01 and = 0.1. PB = 0.5 Q(V/) + 0.5 Q(V/) = Q(V/0.1) = Q(10V).
Now EB = V2T = V2/10
and EB/N0 = V2 / (2x10-4) = 5000 V2
Therefore, V = ([EB/N0] / 5000) and PB = Q(10 ([EB/N0] / 5000) )
= Q((0.02[EB/N0]) = 0.5 erfc((0.01[EB/N0])
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MATLAB program for PB against E = EB/No
clear; for i = 1 : 32; EdB(i)=i-2; % This is EB/N0 in dB E = 10^(EdB(i) / 10); % This is EB/N0 PBu(i) = 0.5 * erfc ( sqrt(E*0.01) ); end; semilogy( EdB, PBu, 'b'); grid on; % need to label axes
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Waterfall graph for Ex 1.6
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“Waterfall” graphs of PB against Eb/N0
• PB= bit-error probability
• Eb = energy per bit
Power used to transmit data at a given bit-rate
(watts (joules/s) ÷ bit-rate (bits/s) = joules/bit)
N0 = 1-sided PSD of AWGN
Eb/N0 is sort of ‘signal-to-noise ratio’
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Example of a better waterfall graph
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1.8.9 Matched filtering & equalisation
bI(t)
Matchedfilter
Vectordemodulator
Channelequaliser
Sample &
detect
..1100..
Channelsignal + AWGN
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Mapto b-b
Mod-ulate
Carrier
Channel RF filter
De-modulate
Derive local carrier
HAdaptEqualiser
Detectbits
Sync bit samplingpoints
Initialise
* *
channelnoise
t
tr (t)
s(t)
systemnoise
Single carrier digital radio tansmitter/receiver
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Problems & discussion points
1. As this is a course on digital transmission, why need analog FT?
2.Why is lowest 300Hz bandwidth lost on POTs telephones?
3.Why do we need a hybrid & what happens when it works badly?
4. What causes ‘fading’ in a mobile telephone channel?
5. What is meant by the ‘rms delay spread’?
6. From list, what do you consider the main advantage of digital?
7. If IEEE802.11 is ‘wi-fi’ what is IEEE802.3 ?
8. What is the ‘mac-sub-layer’?
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Further problems & discussion points
9. Why is the 11-chip Barker sequence used with IEEE802.11b?
10. Design a telephone modem using a PC sound card.
11. What does FHSS have to do with the first nude actress?
12. What are the ‘ISM’bands & what are they used for?
13. Why are there no bit-errors in emails?
14. Why is TCP/IP not ideal for VoIP?
15. What is the ‘hidden node problem’ & how is it solved?
16. What is the difference between CSMA/CD & CSMA/CA?