MSIT 413: Wireless Technologies Week 4

MSIT 413: Wireless Technologies Week 4
Michael L. Honig Department of EECS Northwestern University January 2011

Outline Finish radio propagation
Applications: location tracking (radar), handoffs Digital modulation

Radio Channels Troposcatter Microwave LOS Mobile radio Indoor radio
Troposcatter enables “over-the-horizon” communications (e.g., ham radio). T T Mobile radio Indoor radio 3

Power Attenuation In dB: Pr = P0 (dB) – 10 n log (d)
distance d reference distance d0=1 Reference power at reference distance d0 Path loss exponent P0 slope (n=2) = -20 dB per decade In dB: Pr = P0 (dB) – 10 n log (d) Pr (dB) slope = -40 (n=4) log (d) 4

Shadow Fading Random variations in path loss as mobile moves around buildings, trees, etc. Modeled as an additional random variable: “normal” (Gaussian) probability distribution Pr = P0 – 10 n log d + X “log-normal” random variable standard deviation -  received power in dB For cellular:  is about 8 dB 5

Large-Scale Path Loss (Scatter Plot)
Most points are less than  dB from the mean 6

Urban Multipath No direct Line of Sight between mobile and base
What does the received signal look like if we transmit a sinusoidal (narrowband) signal? No direct Line of Sight between mobile and base Radio wave scatters off of buildings, cars, etc. Severe multipath 7

Narrowband Fading Received signal r(t) = h1 s(t - 1 ) + h2 s(t - 2) + h3 s(t - 3 ) + … attenuation for path 1 (random) delay for path 1 (random) If the transmitted signal is sinusoidal (narrowband), s(t) = sin 2f t, then the received signal is also sinusoidal, but with a different (random) amplitude and (random) phase: r(t) = A sin (2f t + ) Each term in the sum is a sinusoidal type of signal. Transmitted s(t) Received r(t) A,  depend on environment, location of transmitter/receiver 8

Rayleigh Fading Can show: A has a “Rayleigh” distribution
 has a “uniform” distribution (all phase shifts are equally likely) Probability (A < a) = 1 – e-a2/P0 where P0 is the average received power (averaged over different locations) Ex: P0 =1, a=1: Pr(A<1) = 1 – e-1 = 0.63 (probability that signal is faded) P0 = 1, a=0.1: Pr(A<0.1) = 1 – e-1/100 ≈ (prob that signal is severely faded) Prob(A < a) 1 Assume received amplitude without fading is 1 (corresponds to P0=1). 1-e-a2/P0 a

Small-Scale Fading Fade rate depends on Mobile speed
Speed of surrounding objects Frequency

Time Variations: Doppler Shift
Audio clip (train station) 11

Time Variations: Doppler Shift
velocity v distance d = v t Propagation delay = distance d / speed of light c = vt/c transmitted signal s(t) delay increases received signal r(t) propagation delay Received signal r(t) = sin 2f (t- vt/c) = sin 2(f – fv/c) t received frequency Doppler shift fd = -fv/c 12

Doppler Shift (Ex) Mobile moving away from base  v > 0, Doppler shift < 0 Mobile moving towards base  v < 0, Doppler shift > 0 Carrier frequency f = 900 MHz, v = 60 miles/hour = meters/sec Mobile  Base: fd = fv/c = (900 × 106) × / (3 × 108) ≈ 80 Hz meters/sec

Doppler (Frequency) Shift
½ Doppler “cycle” 80 Hz Doppler cycle means the phase “turns around” 80 times per second. in phase out of phase Frequency= 1/50 Frequency= 1/45

Application of Doppler Shift: Astronomy
Doppler shift determines relative velocity of distant objects (e.g., stars, galaxies…) Observed “spectral lines” (radiation is emitted at discrete frequencies) Is the cluster moving away or coming closer? “red shift”: object is moving away “blue shift” object is moving closer sun light spectrum spectrum of galaxy supercluster 15

Application of Doppler Shift: Police Radar
What happens if other objects are present? Doppler shift can be used to compute relative speed. 16

Scattering: Doppler Spectrum
distance d = v t transmitted signal s(t) received signal ?? power Received signal is the sum of all scattered waves Doppler shift for each path depends on angle (vf cos /c ) Typically assume that the received energy is the same from all directions (uniform scattering) freq. frequency of s(t)

distance d = v t transmitted signal s(t) Doppler Spectrum (shows relative strengths of Doppler shifts) Doppler shift fd power power 2fd frequency frequency of s(t) frequency frequency of s(t) + Doppler shift fd 18

distance d = v t transmitted signal s(t) power Doppler spectrum frequency power 2fd frequency of s(t) frequency of s(t) + Doppler shift fd

Rayleigh Fading Received waveform Amplitude (dB) deep fade phase shift
20

Combined Fading and Attenuation
Received power Pr (dB) distance attenuation shadowing Large-scale effects Time (mobile is moving away from base)

Combined Fading and Attenuation
Received power Pr (dB) distance attenuation shadowing Rayleigh fading Small-scale effect Time (mobile is moving away from base)

Channel Coherence Time
Coherence Time: Amplitude and phase are nearly constant. How does the coherence time change with velocity? Why is coherence time important? Rate of time variations depends on Doppler shift: (velocity X carrier frequency)/(speed of light) Coherence Time varies as 1/(Doppler shift).

Fast vs. Slow Fading transmitted bits time time received amplitude
coherence time received amplitude transmitted bits time Fast fading: channel changes every few symbols. Coherence time is less than roughly 100 symbols. Slow fading: Coherence time lasts more than a few 100 symbols. 24

Fade Rate (Ex) fc = 900 MHz, v = 60 miles/hour  Doppler shift ≈ 80 Hz. Coherence time is roughly 1/80, or 10 msec Data rate (voice): 10 kbps or 0.1 msec/bit  100 bits within a coherence time (fast fading) GSM data rate: 270 kbps  about 3000 bits within a coherence time (slow fading)

Channel Characterizations: Narrowband vs. Wideband
Narrowband signal (sinusoid) Wideband signal (impulse) Multipath channel Amplitude attenuation, Delay (phase shift) infinite duration, zero bandwidth delay spread The impulse response is analogous to echoes heard when clapping your hands in an auditorium. r(t) s(t) Multipath channel time t time t multipath components zero duration, infinite bandwidth

Pulse Width vs. Bandwidth
Power signal pulse bandwidth = 1/T Narrowband frequency time T signal pulse What happens to the signal as the bandwidth goes to zero? To infinity? Power bandwidth = 1/T Wideband time frequency T

Power-Delay Profile Received power vs. time in response to a transmitted short pulse. delay spread  For cellular systems (outdoors), the delay spread is typically a few microseconds.

Two-Ray Impulse Response
reflection (path 2) direct path (path 1) s(t) reflection is attenuated r(t)  time t time t

Two-Ray Impulse Response
reflection (path 2) direct path (path 1) s(t) reflection is attenuated r(t)  time t time t  = [(length of path 2) – (length of path 1)]/c

Urban Multipath s(t) r(t) time t time t r(t) different location for receiver time t Spacing and attenuation of multipath components depend on location and environment.

Delay Spread and Intersymbol Interference
s(t) r(t) Multipath channel time t time t Time between pulses is >> delay spread, therefore the received pulses do not interfere. r(t) s(t) Multipath channel time t Time between pulses is < delay spread, which causes intersymbol interference. The rate at which symbols can be transmitted without intersymbol interference is 1 / delay spread.

Coherence Bandwidth channel gain frequency f1 f2
coherence bandwidth Bc channel gain Frequencies far outside the coherence bandwidth are affected differently by multipath. frequency f1 f2 Is the coherence bandwidth determined by man or nature? The channel gain is approximately constant within a coherence bandwidth Bc. Frequencies f1 and f2 fade independently if |f1 – f2 | >> Bc.

Coherence Bandwidth and Delay Spread
frequency channel gain coherence bandwidth Bc delay spread  channel gain delay spread  In general, the shape of the frequency response depends on the timing and position of the multipath. coherence bandwidth Bc frequency Coherence bandwidth is inversely proportional to delay spread: Bc ≈ 1/.

Narrowband Signal signal power channel (narrowband) gain frequency f1
coherence bandwidth Bc channel gain Frequencies far outside the coherence bandwidth are affected differently by multipath. frequency f1 f2 The signal power is confined within a coherence band. Flat fading: all signal frequencies are affected the same way.

Wideband Signals signal power channel (wideband) gain frequency f1 f2
coherence bandwidth Bc channel gain Frequencies far outside the coherence bandwidth are affected differently by multipath. frequency f1 f2 What are advantages and disadvantages of frequency-selective fading? A wideband signal spans many coherence bands. Frequency-selective fading: different parts of the signal (in frequency) are affected differently by the channel.

Frequency Diversity signal power channel (wideband) gain frequency f1
coherence bandwidth Bc channel gain Frequencies far outside the coherence bandwidth are affected differently by multipath. frequency f1 f2 Wideband signals exploit frequency diversity. Spreading power across many coherence bands reduces the chances of severe fading. Wideband signals are distorted by the channel fading (distortion causes intersymbol interference). What are advantages and disadvantages of frequency-selective fading? Analogous to investing: coherence bands represent companies/sectors; power is investing budget.

Coherence Bandwidth for Cellular
signal power (wideband) coherence bandwidth Bc channel gain Frequencies far outside the coherence bandwidth are affected differently by multipath. frequency What are advantages and disadvantages of frequency-selective fading? f1 f2 For the cellular band, Bc is around 100 to 300 kHz. How does this compare with the bandwidth of cellular systems?

Fading Experienced by Wireless Systems
Standard Bandwidth Fade rate AMPS kHz (NB) Fast IS kHz Fast GSM kHz Slow IS-95 (CDMA) MHz (WB) Fast 3G MHz Slow to Fast (depends on rate) LTE up to 20 MHz Slow > 20 MHz Slow Bluetooth > 5 MHz (?) Slow

Power signal pulse bandwidth = 1/T Narrowband frequency time T signal pulse Power bandwidth = 1/T Wideband time frequency T

Radar Pulse Bandwidth delay  = 2 x distance/c delay 
reflection delay  = 2 x distance/c delay  s(t) s(t) Narrowband signal used in Doppler gives velocity. Wideband radar signal gives position. r(t) r(t) time t Narrow bandwidth pulse time t High bandwidth pulse

Bandwidth and Resolution
reflection delay  = 2 x distance/c s(t) The resolution of the delay measurement is roughly the width of the pulse. Low bandwidth  wide pulse  low resolution High bandwidth  narrow pulse  high resolution r(t) time t If the delay measurement changes by 1 microsec, the distance error is c x 10-6/2 = 150 meters!

Propagation and Handoff
Received Signal Strength (RSS) from right BST from left BST unacceptable (call is dropped) time

Received Signal Strength (RSS) What determines the handoff threshold? from right BST with handoff handoff threshold from left BST unacceptable (call is dropped) time

Received Signal Strength (RSS) from right BST with handoff handoff threshold from left BST RSS margin unacceptable (call is dropped) time time needed for handoff

Received Signal Strength (RSS) from right BST handoff threshold from left BST RSS margin unacceptable (call is dropped) time time needed for handoff

Handoff Threshold Received Signal Strength (RSS) time
from right BST handoff threshold from left BST RSS margin unacceptable (call is dropped) time time needed for handoff What can cause a call to be dropped during handoff? Handoff threshold too high  too many handoffs (ping pong) Handoff threshold too low  dropped calls are likely Threshold should depend on slope on vehicle speed (Doppler).

Handoff Measurements (3G)
Mobile maintains a list of neighbor cells to monitor. Mobile periodically measures signal strength from BST pilot signals. Mobile sends measurements to network to request handoff. Handoff decision is made by network. B C A D

Mobile maintains a list of neighbor cells to monitor. Mobile periodically measures signal strength from BST pilot signals. Mobile sends measurements to network to request handoff. Handoff decision is made by network. What information should the pilot signal contain? B C Pilot signals (transmitted continuously) A D

Mobile maintains a list of neighbor cells to monitor. Mobile periodically measures signal strength from BST pilot signals. Mobile sends measurements to network to request handoff. Handoff decision is made by network. B C request handoff A active link D

Mobile maintains a list of neighbor cells to monitor. Mobile periodically measures signal strength from BST pilot signals. Mobile sends measurements to network to request handoff. Handoff decision is made by network. Why might the network deny the handoff request? B C A link is broken D network activates link

Mobile maintains a list of neighbor cells to monitor. Mobile periodically measures signal strength from BST pilot signals. Mobile sends measurements to network to request handoff. Handoff decision is made by network. Depends on available resources (e.g., channels/time slots/codes). Handoffs take priority over new requests (why?). Hysteresis needed to avoid handoffs due to rapid variations in signal strength. How should this procedure differ from ? Received Signal Strength (RSS) unacceptable (call is dropped) time handoff threshold

Handoff Decision Depends on RSS, time to execute handoff, hysteresis, and dwell (duration of RSS) Proprietary methods Handoff may also be initiated for balancing traffic. 1G (AMPS): Network Controlled Handoff (NCHO) Handoff is based on measurements at BS, supervised by MSC. 2G, GPRS, 3G: Mobile Assisted Handoff (MAHO) Handoff relies on measurements at mobile Enables faster handoff Mobile data, WLANs (802.11): Mobile Controlled Handoff (MCHO) Handoff controlled by mobile

Example Diagnostic Measurements: 1XEV-DO
drive test measurements drive path

Why Digital Communications?
1G (analog)  2G (digital)  3G (digital) Digitized voice requires about 64 kbps, therefore the required bandwidth is >> the bandwidth of the voice signal (3—4 kHz)!

Why Digital Communications?
1G (analog)  2G (digital)  3G (digital) Digitized voice requires about 64 kbps, therefore the required bandwidth is >> the bandwidth of the voice signal (3—4 kHz)! Can combine with sophisticated signal processing (voice compression) and error protection. Greater immunity to noise/channel impairments. Can multiplex different traffic (voice, data, video). Security through digital encryption. Flexible design possible (software radio). VLSI + special purpose digital signal processing  digital is more cost-effective than analog!

Binary Frequency-Shift Keying (FSK)
Bits: Amplitude What sinusoidal parameters can convey information? time

Quadrature Phase Shift Keying (QPSK)
Bits: Amplitude time

Binary Phase Shift Keying (BPSK)
Bits: Baseband signal Amplitude time

Amplitude Shift Keying (4-Level ASK)
Bits: Baseband signal Amplitude What is the data rate? symbol duration time

Baseband  RF Conversion
Passband (RF) signal Baseband signal sin 2fct time X T Modulate to the carrier frequency fc Why not transmit the baseband signal? Power signal bandwidth is roughly 2/T Power frequency frequency  0 fc

Why Modulate?

Why Modulate? The baseband spectrum is centered around f=0. Without modulation all signals would occupy low frequencies and interfere with each other. It is difficult to build effective antennas at low frequencies since the dimension should be on the order of a wavelength. Low frequencies propagate further, causing more interference.

Selection Criteria How do we decide on which modulation technique to use? Which of these criteria are especially important for wireless channels (as opposed to wired channels)?

Selection Criteria How do we decide on which modulation technique to use? Performance: probability of error Pe. Probability that a 0 (1) is transmitted and the receiver decodes as a 1 (0). Complexity: how difficult is it for the receiver to recover the bits (demodulate)? FSK was used in early voiceband modems because it is simple to implement. Bandwidth or spectral efficiency: bandwidth (B) needed to accommodate data rate R bps, i.e., R/B measured in bits per second per Hz. Power efficiency: energy needed per bit to achieve a satisfactory Pe. Performance in the presence of fading, multipath, and interference. Which of these criteria are especially important for wireless channels (as opposed to wired channels)?

Example: Binary vs. 4-Level ASK
Rate = 1/T symbols/sec Bandwidth is roughly 1/T Hz Bandwidth efficiency = 1 bps/Hz Rate = 2/T symbols/sec Bandwidth is roughly 1/T Hz Bandwidth efficiency = 2 bps/Hz What about power efficiency?

Noisy Baseband Signals
Rate = 1/T symbols/sec Bandwidth is roughly 1/T Hz Bandwidth efficiency = 1 bps/Hz Power =A2 (amplitude squared). Rate = 2/T symbols/sec Bandwidth is roughly 1/T Bandwidth efficiency = 2 bps/Hz Power = (A2 + 9A2)/2 = 5A2 What about probability of error vs transmitted power?

Probability of Error 4-ASK Log of Probability of Error BPSK 7 dB
(factor of 5) Signal-to-Noise Ratio (dB)

How to Increase Bandwidth Efficiency?

How to Increase Bandwidth Efficiency?
Increase number of signal levels. Use more bandwidth efficient modulation scheme (e.g., PSK). Apply coding techniques: protect against errors by adding redundant bits. Note that reducing T increases the symbol rate, but also increases the signal bandwidth. There is a fundamental tradeoff between power efficiency and bandwidth efficiency.

The Fundamental Question
Given: B Hz of bandwidth S Watts of transmitted signal power N Watts per Hz of background noise (or interference) power What is the maximum achievable data rate? (Note: depends on Pe.)

Claude Shannon (1916-2001) Father of “Information Theory”
Shannon’s 1948 paper “A Mathematical Theory of Communications” laid the foundations for modern communications and networking: “The fundamental problem of communication is that of reproducing at one point either exactly or approximately a message selected at another point… Semantics of messages are irrelevant to the engineering problem. transmitter receiver

Shannon’s 1948 paper “A Mathematical Theory of Communications” laid the foundations for modern communications and networking: “The significant aspect is that the actual message is one selected from a set of possible messages. The system must be designed to operate for each possible selection, not just the one which will actually be chosen since this is unknown at the time of design.” transmitter receiver

Shannon’s 1948 paper “A Mathematical Theory of Communications” laid the foundations for modern communications and networking: “The choice of a logarithm base corresponds to the choice of a unit for measuring information. If the base 2 is used the resulting units may be called binary digits, or more briefly bits, a word suggested by J. W. Tukey. log2 M bits Transmitter (M possible messages) receiver

Shannon’s 1948 paper “A Mathematical Theory of Communications” laid the foundations for modern communications and networking. Other contributions and interests: digital circuits, genetics, cryptography, investing, chess-playing computer, roulette prediction, maze-solving, unicycle designs, juggling Juggling video

Shannon’s Channel Coding Theorem (1948)
noise Information Source Encoder Channel Decoder bits input x(t) output y(t) Estimated bits Information rate: R bits/second Channel capacity: C bits/second R < C  There exists an encoder/decoder combination that achieves arbitrarily low error probability. R > C  The error probability cannot be made small.

Shannon Capacity Channel capacity: C = B log(1+S/N) bits/second
noise Information Source Encoder Channel Decoder bits input x(t) output y(t) Estimated bits Channel capacity: C = B log(1+S/N) bits/second B= Bandwidth, S= Signal Power, N= Noise Power No fading

Caveats “There exists” does not address complexity issues.
As the rate approaches Shannon capacity, to achieve small error rates, the transmitter and (especially) the receiver are required to do more and more computations.

As the rate approaches Shannon capacity, to achieve small error rates, the transmitter and (especially) the receiver are required to do more and more computations. The theorem does not say anything about delay. To achieve Shannon capacity the length of the transmitted code words must tend to infinity!

As the rate approaches Shannon capacity, to achieve small error rates, the transmitter and (especially) the receiver are required to do more and more computations. The theorem does not say anything about delay. To achieve Shannon capacity the length of the transmitted code words must tend to infinity! The previous formula does not apply with fading, multipath, frequency-selective attenuation.

As the rate approaches Shannon capacity, to achieve small error rates, the transmitter and (especially) the receiver are required to do more and more computations. The theorem does not say anything about delay. To achieve Shannon capacity the length of the transmitted code words must tend to infinity! The previous formula does not apply with fading, multipath, frequency-selective attenuation. It has taken communications engineers more than 50 years to find practical coding and decoding techniques, which can achieve information rates close to the Shannon capacity.

Example: GSM/EDGE Bandwidth = 200 kHz, S/I = 9 dB = 7.943
C = 200,000 x log(8.943) ≈ 632 kbps This is what would be achievable in the absence of fading, multipath, etc. Currently, EDGE provides throughputs of about 230 kbps. Up to 470 kbps possible using additional “tricks”, such as adapting the modulation and coding format to match the channel Preceding Shannon formula is not directly applicable.

Data Rates for Deep Space Applications
Mariner: 1969 (Mars) Pioneer 10/11: 1972/3 (Jupiter/Saturn fly-by) Voyager: (Jupiter and Saturn) Planetary Standard: 1980’s (military satellite) BVD: “Big Viterbi Decoder” Galileo: 1992 (Jupiter) (uses BVD) Turbo Code: 1993 Signal to Noise Ratio

TYPES ALL PSK (PSK Type Summary Slide)

1000 Symbols Per Second x 1 Bit Per Symbol = 1000 Bits Per Second
BPSK Bitream 0 1000 Bits Per Second 1000 Symbols (Phases) Per Second Now in an actual radio system, the modulator may transmit thousands of symbols each second. 1000 Symbols Per Second x 1 Bit Per Symbol = 1000 Bits Per Second

1000 Symbols Per Second x 1 Bit Per Symbol = 1000 Bits Per Second
BPSK Bitream 1 1000 Bits Per Second 1000 Symbols (Phases) Per Second For example, if the modulator transmits 1000 symbols, or phases, each second, then with Binary PSK, since each symbol represents 1 bit, the data rate is 1000 bits per second. 1000 Symbols Per Second x 1 Bit Per Symbol = 1000 Bits Per Second

1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second
QPSK Bitream 00 2000 Bits Per Second 1000 Symbols (Phases) Per Second But if we use 4-PSK, or QPSK, then each symbol represents 2 bits, and the data rate is 2000 bits per second. 1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second

QPSK Bitream 10 2000 Bits Per Second 1000 Symbols (Phases) Per Second (3-bit bitstream animation continues at random) 1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second

8PSK Bitstream 000 3000 Bits Per Second 1000 Symbols (Phases) Per Second And if we use 8-PSK, which conveys 3 bits for each symbol, then the data rate would be 3000 bits per second. 1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second

8PSK Bitstream 001 3000 Bits Per Second 1000 Symbols (Phases) Per Second (3-bit bitstream animation continues at random) 1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second

8PSK Bitstream 111 3000 Bits Per Second 1000 Symbols (Phases) Per Second (3-bit bitstream animation conlcudes) 1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second

Colored Balls 2 The width of the funnel tube is analogous to the available bandwidth. Now the narrower the funnel, the longer it takes to get the jellybeans through the funnel. So, the less bandwidth we have, the longer it takes to transmit our symbols.

Binary Phase Shift Keying (BPSK)
Bits: Baseband signal What is the effect of the discontinuity between bit periods on the signal bandwidth?

Minimum Bandwidth (Nyquist) Pulse Shape
This pulse has the minimum bandwidth for a given symbol rate. Given bandwidth B, the maximum symbol rate without intersymbol interference (ISI) is B, the “Nyquist rate”.

Power signal pulse bandwidth B = 1/T Narrowband frequency time 2T signal pulse Power bandwidth B = 1/T Wideband time frequency 2T The bandwidth B = 1/T (symbol rate) is often called the Nyquist bandwidth.

Shifted Nyquist Pulses
Bits:

Baseband Waveform (Nyquist Signaling)
. . . Bits: . . .

Baseband  RF Conversion
Passband (RF) signal Baseband signal sin 2fct X time T Modulate to the carrier frequency fc Why not transmit the baseband signal? Power signal bandwidth is roughly 1/T Power frequency  0 fc frequency

Passband Signal with Different Carrier Frequencies

time Perfect synchronization T time Offset causes severe ISI!

“Excess” Bandwidth (Raised Cosine Pulses)
Minimum BW frequency time

Raised Cosine Pulses frequency time
Minimum BW 50% excess BW 100% excess BW frequency time Excess bandwidth= (Total bandwidth – Nyquist bandwidth)/Nyquist bandwidth

“Case” Study: In-Flight vs. AT&T (2003)
AT&T accused of stealing design for air-to-ground communication system System features: 12 kbps for voice service 8 PSK 6 kHz bandwidth Are there other modulation schemes that would be suitable?

Circle Vocoder And for voice communications, this data rate depends on the data rate generated by the vocoder.

Blocks Transition (Transitional Slide)

Blocks Stretched 8000 Bits Per Second
The vocoder selected by the Hughes engineers used the same technology that was selected for the North American digital cell phone standard IS54. That vocoder generates 8000 bits per second. But we actually need to transmit significantly more than this because...

Additional Bits Additional Bits Error Correction Control Information
... we need to insert additional bits to guard against errors and to provide overhead control information. The control information tells the transmitter and receiver operational information like which ground station to use, how much power to use, the identity of the person using the service, billing information, and so forth. So, in reality, the system needs to support a data rate significantly larger than 8,000 bits per second – something in the range of approximately bits per second. Channel Ground Station Power Identity Billing

6000 Hz Available Bandwidth
In our case, the FCC has given us 6000 Hz. We can estimate the maximum number of symbols, which we can transmit every second, by applying one of the basic principles of digital communications, sometimes referred to as Nyquist's Theorem. Nyquist's Theorem says that, roughly speaking, we can count on being able to send about 4000 symbols per second over a 6000 Hz channel. Nyquist’s Theorem: Can transmit 4000 symbols per second through a 6000 Hz channel

Colored Balls 2 5000 symbols per second would be quite difficult, and 4000 symbols per second is relatively easy. Nyquist’s Theorem: Can transmit 4000 symbols per second through a 6000 Hz channel

4000 bps < 8000 bps (Vocoder rate)
BPSK 4000 < 8000 BPSK: 1 Bit Per Symbol 4000 Bits Per Second (bps) 4000 bps < 8000 bps (Vocoder rate) BPSK So, let's assume that we are sending 4000 symbols per second. If we use Binary PSK, then we are transmitting 1 bit for each symbol, so the data rate in that case is 4000 bits per second. This is far below the 8000 bits per second generated by the vocoder, which means that we cannot use BPSK to transmit our voice signal.

QPSK 8000 BPS QPSK: 2 Bits Per Symbol
2 X 4000 = 8000 Bits Per Second (bps) 8000 bps = Vocoder rate Need more bits for error correction and control! QPSK So now let's consider 4-PSK, or QPSK. In that case we are transmitting 2 bits per symbol, and with 4000 symbols per second, that gives a data rate of 2 times 4000, or 8000 bits per second. Recall that we need more than this, because on top of the vocoder rate of 8000 bits per second, we need to add additional bits to correct errors, and for control. This means that we cannot use QPSK either.

8PSK 12000 BPS 8PSK: 3 Bits Per Symbol
3 X 4000 = 12,000 Bits Per Second (bps) 8000 bps bps Vocoder rate + Error Correction and Control 8PSK Moving then to 8-PSK, this gives us 3 bits per symbol, and with 4000 symbols per second, that gives a data rate of 3 times 4000, or 12,000 bits per second. This will give us the 8000 bits per second from the vocoder, plus 4000 bits per second for correcting and detecting errors, and for control. For this type of application, an additional 4000 bits per second beyond the vocoder rate gives us a reasonable margin for errors and control, so that we conclude that a voice service over this 6000 Hz channel can be supported with 8-PSK.

16 phases: 4 Bits Per Symbol 4 x 4000 = 16,000 bps
16 PSK BPS 16 phases: 4 Bits Per Symbol 4 x 4000 = 16,000 bps More than enough for vocoder rate + overhead But couldn't we also transmit more than 8 phases, or equivalently, more than 3 bits for every symbol? For example, suppose that we transmit 4 bits per symbol. This would give us a bit rate of 4 times 4000, or 16,000 bits per second, which is more than we need to support a voice service with our 8000 bit per second vocoder. We could do this, but it would make the system more complicated.

MOTION BLUR As explained in the last video, the problem with increasing the number of bits per symbol is that we have to increase the number of phases or symbols to transmit, and these become harder to distinguish at the receiver. Namely, 4 bits per symbol means that we have to choose from among 16 possible phases, and to avoid confusing these symbols at the receiver, we need to transmit with more power.

90˚ QPSK w/Bit Labels 0˚ 180˚ 270˚ Because we can choose from one of four phases, each phase can be used to represent two bits, instead of one as before. This is shown in the figure by labeling each phase with two bits. Namely, zero phase corresponds to transmitting 00, shown in green, 90 degrees corresponds to transmitting 11, shown in purple, 180 degrees corresponds to 10, shown in orange, and 270 degrees corresponds to 01, shown in red. (Bit labels fade-in one at a time)

90˚ 8PSK 45˚ 135˚ 0˚ 180˚ 225˚ 270˚ 315˚ Notice that the angle of each flag position again corresponds directly to the starting phase of the radio wave. (Each pair of colored dots connected by a line wipe down successively)

Sine Wave Animation

QPSK Signal Constellation
amplitude = 1 x 1 x x x

“Rotated” QPSK Signal Constellation
amplitude = 1 x x 1 x x

In-Phase/Quadrature Components
x (a,b)  a sin 2fct + b cos 2fct 1 x x b is the “in-phase” signal component a is the “quadrature” signal component x

In-Phase/Quadrature Components
x x 1 x x

Example Constellations
quadrature QPSK BPSK in-phase x x x x x x 16-QAM quadrature 8-PSK x x x x For the 16-QAM signal constellation, what signal does a particular point represent? x x x x x x x x x x x x x x x in-phase x x x x x

Quadrature Modulation
in-phase signal even bits Baseband Signal X Split: Even/Odd source bits transmitted (RF) signal + Baseband Signal X odd bits quadrature signal

Modulation for Fading Channels
Problems: 1. Amplitude variations (shadowing, distance, multipath) 2. Phase variations 3. Frequency variations (Doppler) Solution to 1: Avoid amplitude modulation Power control Solution to 2 & 3: Avoid phase modulation (use FSK) “Noncoherent” demodulation: does not use phase reference Differential coding/decoding “Coherent” demodulation: Estimate phase shifts caused by channel. Increase data rate/Doppler shift ratio

Binary Frequency-Shift Keying (FSK)
Bits:

Minimum Shift Keying (MSK)
Bits: Frequencies differ by ½ cycle Used in GSM

Binary Differential Modulation
(i+1)st bit = 0: 0o phase shift waveform for ith symbol (i+1)st bit = 1: 180o phase shift

Binary Differential Modulation
(i+1)st bit = 0: 0o phase shift waveform for ith symbol (i+1)st bit = 1: 180o phase shift Drawback: a detection error for the ith bit propagates to the (i+1)st bit.

Example: DQPSK x x x x x x x x constellation for ith symbol
bits: 00 x 10 x x 01 x x x 11 x x constellation for ith symbol constellation for (i+1)st symbol Used in IS-136

Coherent Phase Modulation
Receiver estimates phase offset More complicated than noncoherent (e.g., differential) modulation. Receiver requires a pilot signal. Transmit known symbols, measure phase. Pilot symbols are overhead (not information bits).

Probability of Error

Probability of Error with Fading

Orthogonal Frequency Division Multiplexing (OFDM)
Modulate Carrier f1 substream 1 Split into M substreams Modulate Carrier f2 substream 2 source bits substream M OFDM Signal … + Modulate Carrier fM

OFDM Spectrum … … f1 f2 f3 f4 f5 f6  0
Total available bandwidth Data spectrum for a single carrier Power … … f1 f2 f3 f4 f5 f6 frequency  0 subchannels M “subcarriers, or subchannels, or tones” “Orthogonal” subcarriers  no cross-channel interference.

OFDM Example: 802.11a 20 MHz bandwidth, M=64 (48 for data payload)
Subchannel bandwidth = 20 MHz / 64 = kHz Symbol rate / subchannel = 250 kilosymbols/sec Total symbol rate = 64 x 250 x 103 = 16 Msymbols/sec Bit rate? 16 QAM/subchannel  4 bits/symbol x 250 x 103 = 1 Mbps/subchannel, or 64 Mbps total 64 QAM/subchannel  6 bits/symbol x 250 x 10^3 = 1.5 Mbps/subchannel, or 96 Mbps total Includes overhead (synchronization, error correction, control) Actual data rate: 36 / 54 Mbps

OFDM Example: 802.11a 20 MHz bandwidth, M=64 (48 for data payload)
Subchannel bandwidth = 20 MHz / 64 = kHz Symbol rate / subchannel = 250 kilosymbols/sec Total symbol rate = 64 x 250 x 103 = 16 Msymbols/sec Bit rate? 16 QAM/subchannel  4 bits/symbol x 250 x 103 = 1 Mbps/subchannel, or 64 Mbps total 64 QAM/subchannel  6 bits/symbol x 250 x 103 = 1.5 Mbps/subchannel, or 96 Mbps total Includes overhead (synchronization, error correction, control) Actual data rate: 36 / 54 Mbps

Why OFDM? Single-carrier transmission also possible: 250 x 10^3 symbols/sec in kHz means 16 Msymbols/sec would be transmitted in 20 MHz.

Why OFDM? Exploits frequency diversity channel gain
Flat fading on each subchannel simplifies receiver (no multipath/ISI) Slower symbol rate on each subchannel simplifies signal processing. Drawback: high peak-to-average power. subcarrier bandwidth < coherence bandwidth Bc signal power (wideband) channel gain Frequencies far outside the coherence bandwidth are affected differently by multipath. f1 frequency f2 bits are coded across subcarriers Single-carrier transmission also possible: 250 x 10^3 symbols/sec in kHz means 16 Msymbols/sec would be transmitted in 20 MHz.

MSIT 413: Wireless Technologies Week 4

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MSIT 413: Wireless Technologies Week 4

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