Slide 1
Log-Normal Shadowing
The average large scale path loss model does not capture the
randomness of the obstacles present in between the transmitter and
receiver. The received signal power at a given distance will
typically experience random variation as the wireless channel will
fluctuate depending on the number of obstacles, changes in
reflecting surfaces and scattering objects. This random behavior of
the environment is usually modeled by log-normal shadowing. 1
Log-Normal Shadowing
Hence in the models for the path loss effects due to shadowing
can be combined to give the variation in the received signal power
with distance as
where is given in Eq. (1.15) and is a zero mean Gaussian
distributed random variable (in dB) with standard deviation (in dB)
2
(1.16)(1.17)(1.18)
Log-Normal Shadowing
The received power with path loss and shadowing
Since PL(d) is a random variable with normal distribution in dB
and mean , therefore Pr(d) is also a random variable with normal
distribution in dB. Let be the distance dependent mean of
Pr(d).
3
(1.19) (1.20)
Log-Normal Shadowing
The probability that the received signal is above a particular
threshold can be obtained as
4
(1.21)
Log-Normal Shadowing
Similarly the probability that the received signal is below a
particular threshold can be obtained as
This probability is also known as outage probability 5
(1.22) Nyquist Pulse Shaping for Linear ModulationsWhen
rectangular pulses are passed through a bandlimited channel, the
pulses will spread in time, and the pulse for each symbol will
smear into the time intervals of succeeding symbols. This causes
intersymbol interference (ISI) and leads to an increased
probability of the receiver making an error in detecting a
symbol.Raised-cosine pulse is used to eliminate ISI while keeping
the transmission bandwidth low.At the correct sampling instant, the
ISI is zero.Typically, the raised cosine frequency response is
split between the transmitter and the receiver meaning the
transmitter and receiver both have square-root raised cosine
spectrum6Nyquist Pulse Shaping for Linear ModulationsIntersymbol
Interference Examples7
*http://www.fiberoptics4sale.com/wordpress/fiber-optics-for-telecommunicationa-complete-and-quick-tutorial-part-1/Nyquist
Pulse Shaping for Linear ModulationsNyquist condition for zero ISI
is
Where GT(f), GR(f) and C(f) denotes the frequency response of
transmitting, receiving and channel filters. T=1/2W is the
theoretically minimum value of T for which transmission with zero
ISI can take place. The corresponding impulse response is
x(t)=sinc(t/T). In other words, a Channel with Bandwidth W= 1/2T,
can support a maximum Symbol rate of Rs=1/T symbols per second.
Hence without ISI, the maximum symbol transmission per Hz is 2
symbols/s/Hz.
8
fX(f)-W=-1/2TW=1/2TT
x(t)x(t-T)
(2.27)(2.28)Nyquist Pulse Shaping for Linear ModulationsOne
commonly used pulse spectrum that has the desirable spectral
properties is the raised cosine spectrum.
where 01 is called the roll-off factor, the bandwidth occupied
by the signal beyond W=1/2T=Rs/2 (also called minimum Nyquist
bandwidth) is called excess bandwidth and is usually expressed as
percentage of W.
9
(2.29)Nyquist Pulse Shaping for Linear ModulationsHence the
total Bandwidth occupied by the signal is B= W+ W = 0.5(1+)Rs If
=0.5, the excess bandwidth is 50% whereas if =1, the excess
bandwidth is 100%. The impulse response for X(f) is
10
(2.31)(2.30)11
X(f)x(t)Bandwidth of Raised Cosine filter (Baseband):
Bandwidth of Raised Cosine filter (Passband):
When =0, Rs=2B (RC pulse reduces to ideal Nyquist filter)When
=1, Rs=B
As the excess bandwidth increases ( 0 ) means faster convergence
and reduction in errors due to mismatch in sampling instants.
However, it reduces the effective symbol rate of transmission.
(2.33)(2.32)
Power Efficient Modulations
Non-linear modulations are power efficient.In non-linear
modulation, the amplitude of the transmitted signal is constant,
regardless of the variation in the modulating digital signal
Constant envelopeThe phase of the transmitted signal is usually
continuous. Use power efficient Class C amplifier.May occupy a
larger bandwidth (than linear modulations)MFSK M-aryFrequency Shift
KeyingMSK Minimum Shift KeyingIt has a modulation index of 0.5 to
allow the minimum frequency spacing that allows two FSK signals to
be coherently orthogonal.MSK implies the minimum frequency
separation that allows orthogonal detection.GMSK Gaussian
MSK12Modulation Index= delta f /Bandwidth= 2*delta f /Rb = 0.5=
1/2T=12Minimum Shift KeyingFeatures of MSKIt has constant
envelope.There is phase continuity in the RF carrier at the bit
transition instants.It is a FSK signal with binary signaling
frequencies of fc+1/4T and fc-1/4T. Hence the frequency deviation
is equal to 1/4T+1/4T=1/2T.Since the minimum frequency spacing is
1/2T hence the two FSK signals are coherently orthogonal.
131490(a)DataSplitter
Delay, Tb
MSK OutputQ OutputI OutputData In(c)(b)(d)
1001101111000111
(e)15
where k is 0 or depending on whether aI(t) is 1 or -1(2.33)
MSK has lower sidelobesIn MSK, 99% of the power is within a
bandwidth B = 1.2/TIn QPSK, 99% of the power is within a bandwidth
B = 8/TMSK has wider main lobe bandwidth.MSK is less bandwidth
efficient in terms of 1st null bandwidthGaussian Pulse shaping
filterEffective when used with power efficient modulation schemes
(e.g. MSK).Unlike Nyquist filters which have zero-crossings at
adjacent symbol peaks and a truncated transfer function, the
Gaussian filter has a smooth transfer function with no
zero-crossings.The transfer function
B= BandwidthThe impulse response
17
(2.34)(2.35)
Gaussian Minimum Shift Keying (GMSK)
Gaussian Minimum Shift Keying (GMSK)
Has constant envelope hence has excellent power efficiency (as
nonlinear Class C power amplifiers can be used) Has good spectral
efficiency (as side-lobes of MSK signal are reduced by passing
through Gaussian low pass filter)Drawback is that it introduces
ISI, however the degradation due to ISI is not severe if the 3dB
bandwidth- bit duration product (BT) of the filter is more than
0.5.
18
19
As the BT product decreases, the sidelobes falls off rapidly,
hence it consumes less bandwidth Reducing BT increases the
irreducible error floor due to ISI. For mobile communication, as
long as the GMSK irreducible error rate is less than that produced
by the mobile channel, there is no penalty in using GMSK. GMSK Bit
error rate:
where is constant related to BT, i.e. = 0.68 for BT= 0.25
(2.30)20
Occupied RF Bandwidth as a fraction of Rb GMSK spectrum becomes
more compact with decreasing BT value but degradation due to ISI
increases. For BT=0.5887, BER degradation is about 0.14 dB compared
to the case without ISI.Gaussian Minimum Shift Keying (GMSK)Example
:Data rate = 270 kbps, BT = 0.25, find the 3-dB bandwidth, 90%
power RF bandwidth.
Thus the 3dB bandwidth is 67.567 kHz. To determine 90% power
bandwidth at pass-band, we use the table to find that 0.57R is the
desired value. Thus, the occupied RF spectrum for a 90% power
bandwidth is given by RF BW=0.57Rb=0.57270103=153.9kHz
21
M-QAMCoherent M-ary QAM For M-ary QAM the signal can be
described as
Where we have assumed a rectangular pulse shaping window and Amc
and Ams are the sets of amplitude levels obtained by mapping k-bit
sequences into signal amplitudes.
22
(2.31)M-QAMFunctional diagram of QAM
Modulator23AmcBinaryDataPulse ShapingFilter
90deg.PhaseShiftAmsSerial to Parallel converterPulse
ShapingFilterOsc
TxQAMSignalSin wctCos wctM-QAMA constellation for 16 QAM24
M-QAMProbability of Error for QAM The probability of symbol
error for the M-ary QAM (when k is even) is
where M=2k, k is even,
and Es is the average energy per symbol. 25
(2.32)(2.33)M-QAMProbability of symbol Error for QAM for any k1
can be tightly upper-bounded as
where Eb is the energy per bit.
26
(2.34)M-QAM27
***MATLAB DEMOModulation performance in Slow Flat fading
channelsFor slow flat fading channel the received signal can be
given as
The BER in a slow flat fading case can be obtained by averaging
the error in AWGN channels over the fading probability density
function
Pe(X) is BER at the specific value of signal-noise-ratio X,
X=2Eb/Nop(X) is the pdf of X due to fading channel.
28
Gain of the channelPhase shift of the channelTransmitted
signalAWGN
(2.35)(2.36)Modulation performance in Slow Flat fading
channelsFor Rayleigh fading channel, the fading envelope has
Rayleigh distribution, so the fading power 2 has exponential
distribution, hence
where = Average SNR where the average channel gain is usually
assumed to be 1. Example: For BPSK, the bit error probability for a
particular value of , will be
Hence the probability of error of BPSK in slow fading
channel
29
(2.37)(2.38)(2.39)Modulation performance in Slow Flat fading
channelsFor Rayleigh fading channel, the fading envelope has
Rayleigh distribution, so the fading power 2 has exponential
distribution, hence
where = Average SNR where the average channel gain is usually
assumed to be 1. Example: For BPSK, the bit error probability for a
particular value of , will be
Hence the probability of error of BPSK in slow fading
channel
30
(2.37)(2.38)(2.39)BER of common digital modulations in Rayleigh
Fading channelFor coherent BPSK:
For coherent BFSK:
Differential BPSK:
Noncoherent Orthogonal BFSK:
GMSK: where 0.68, for BT=0.25, 0.85, for BT=. Please Note :
=Eb/No The approximation only holds for large SNR values31
(2.40)(2.41)(2.42)(2.43)(2.44)
For Fading channel, there is a linear relationship between BER
and SNR, i.e.
whereas for non-fading AWGN channel BER and SNR has exponential
relationship.
To achieve a BER of 10-3 -10-6 in fading channel may require
30-60dB mean SNR whereas for an AWGN channel it requires just 5-7dB
SNR
32Probability of error for Binary DPSK and FSK (non-coherent)
DetectionThe BER for AWGN channel for DPSK and non-coherent FSK
modulation is given by, where c=1 (DPSK) and c=0.5 (FSK) and X=
2Eb/No where X is the instantaneous value and For Rayleigh fading
channel, the pdf of X is given as
where is average signal to noise ratio, i.e. E(X)= =Eb/No Hence
the probability of error in slow flat fading channel can be
obtained by averaging the error in AWGN channel over the fading
probability density function (i.e. averaging over all possible
values of X) 33
Probability of error for Binary DPSK and FSK (non-coherent)
Detection 34
If c=1 (DPSK) and c=1/2(FSK) we have
DPSKFSKExample If the average SNR for a Rayleigh faded DPSK
signal is 30dB what is the probability of error at the receiver?
Also find the probability of error in AWGN channel if the
instantaneous SNR is 30dB. Solution: Average SNR, =30dB
35
Example For BFSK (non-coherent) modulation(assume ideal Nyquist
Filtering), to achieve a BER of 10-3, how much additional signal
power (in dB) is required in a Rayleigh fading environment as
compared to the AWGN channel? Solution: For AWGN ChannelHence, For
Rayleigh faded Channel
Hence, additional power required, = 29.99-10.95=19dB
36
36Prove that if is Rayleigh distributed, then 2 is exponentially
distributed. Let y= 2, hence by transformation theorem of single
random variables, the pdf of y will be
Since , hence 37
38Diversity Techniques for Fading ChannelsThe problem with
Rayleigh FadingFor a Rayleigh fading channel the instantaneous SNR,
X=2Eb/No has an exponential distribution given by (Eq. 2.36)
where is the average SNR of the Rayleigh fading channel.Let be a
desired SNR threshold at which acceptable signal reception is
achieved. Then the outage probability will be
In particular, if = then Pr(SNR< )=1-e-1=0.63. The received
signal is (on average) 63% less than the average SNR.
39
Why Diversity ?Example: Suppose the binary antipodal signals
(BPSK) s(t) are transmitted over a fading channel and the received
signal is
Where n(t) is AWGN. The channel gain a is specified by the pdf
p(a)=0.1(a)+0.9(a-2) Let us determine the BER which employs a
filter matched to s(t). The probability of error for fixed value of
a is given by
Since a takes two values, a=0 and a=2 with probabilities 0.1 and
0.9 40
(3.1)(3.2)Why Diversity ?The average probability of error is
Hence, if we assume that , (known as error floor) Now if we
transmit this signal over two statistically independent fading
channels with gain a1 and a2 where p(ak)=0.1(ak)+0.9(ak-2), k=1,2
The noises of two channels are statistically independent. The
demodulator employs a matched filter for each channel and simply
adds the filter output to form the decision. Lets us determine the
Pe for this case. 41
(3.3)(3.4)Why Diversity ?The probability of error for fixed
values of a1 and a2 is
In this case we have four possible values for the pair (a1,a2),
namely, (0,0), (0,2), (2,0) and (2,2) with probabilities, 0.01,
0.09, 0.09, 0.81. Hence the average BER is
Hence, if , (error floor), which is smaller than the earlier
case. What is the error floor if there are three statistically
independent paths? 42
(3.5)(3.6)Why Diversity ?Communication over a flat fading
channel has poor performance due to significant probability that
channel is in a deep fade
Diversity is a way to protect against deep fades by providing
more than one resolvable signal paths that fade independently.
Diversity helps in reducing the depth & duration of small-scale
fades.
43
Motivation of Diversity TechniquesThe solution: create multiple
channels or branches that have uncorrelated fading
44TTFading of two highly correlated channels Fading of two
uncorrelated channels Motivation of Diversity TechniquesWhen
channel is in a deep fade, many data symbols are lostIf receiver
has several replicas of the same signal obtained from independent
fading channels, the chance of signal recovery increases.If then
having M independent replicas of the same transmitted signal result
in Pr(all M signals having SNR< at the same time)=pM Tm) There
are two types of guard time intervalsNull or zero padding and
Cyclic extension 72OFDM Symbol MOFDM Symbol M+1 M-1 TN TN Tg Tg
OFDM Symbol MOFDM Symbol M+1 M-1 TN TN Tg Tg
MainPathDelayedPath72Cyclic Prefix/ Guard IntervalNull or zero
paddingIn this case no signal is transmitted during the interval
Tg, only the last TN of the TN + Tg is used to recover the data.
However the zero padding may give rise to intercarrier interference
(ICI) and loss of orthogonality between the subcarriers. 73FFT
Integration Time= TN=1/fGuardIntervalSubcarrier 1Delayed Subcarrier
2Part of subcarrier 2 causing ICI on subcarrier 1Cyclic Prefix/
Guard IntervalCyclic ExtensionIn cyclic extension, the last Tg
samples of the OFDM symbol (at the IFFT output) is appended at the
start of the symbol.
At the receiver only the last TN of TN+Tg is used to recover the
data. Cyclic extension eliminates not only inter-carrier
interference but also inter-symbol interference.
74OFDM Symbol MOFDM Symbol M+1 TN TN Tg Tg
Cyclic Prefix/ Guard Interval75Guard Interval(Tg) FFT
Integration Time= TN=1/fOFDM Symbol Duration, T=TN+TgEach
subcarrier has integer number of cycles for the FFT integration
period (for the below waveforms it is 2, 4 and 6 respectively).If
the condition is valid, the spectra of the subcarrier will have
null spectrum nulls at all other subcarrier frequencies. 76
Guard IntervalFFT Integration Time= TN=1/fOFDM Block Duration,
T=TN+TgSubcarrier with CPSubcarrier with zero paddingDelayed
Subcarrier (NO ICI)Delayed Subcarrier (Will cause ICI)Cyclic
Prefix/ Guard IntervalAdding of cyclic prefix ensures that the
delayed replicas of all subcarriers associated with an OFDM symbol
always have an integer number of cycles within an FFT integration
interval as long as length of cyclic prefix is greater than the
channel delay spread. Thus no ISI will occur in this case.
77
Cyclic Prefix/ Guard Interval78GuardIntervalGuardIntervalSymbol
MSymbol M+1Symbol M-1 TN Tg TQAMDemodQAMModparallel to serialinvert
channel
frequencydomainequalizer serial to parallelN-IFFTadd cyclic
prefixparallel to serialDAC andtransmitN-FFTserialtoparallelremove
cyclic prefixTRANSMITTERRECEIVERN subchannelsSyncreceive
andADCchannelOFDM Baseband Block DiagramData OutSymbol Rate= R=1/T
(bits/s)
Data InSymbol Rate= R=1/T (bits/s)
RECEIVERBasic OFDM systemThe whole baseband OFDM system divided
into following blocks:Modulation/ DemodulationSerial to Parallel
Converter/Parallel to Serial ConverterFast Fourier transform/
Inverse Fast Fourier TransformGuard Interval (Cyclic Prefix)
Insertion and RemovalChannelFrequency domain
EqualizationDAC/ADC
80
OFDM Signal generation and transmission
Incoming data from modulator is split into blocks of Nd symbols
each where Nd< N.Usually Np pilot symbols are inserted on each
OFDM block for channel estimation at receiver end.Ng guard
subchannels are also inserted (Ng /2 at each end of the spectrum)
to compensate for non-ideal filter shapes and/or to mitigate
interference to adjacent channels. Guard subchannels are usually
null (zero) symbols. Hence the total number of subcarriers,
N=Nd+Np+Ng Example: In IEEE 802.11a Wireless LAN standard N=64,
Np=4, Ng=12, Nd=4881The reason is that you can remove the DC offset
by applying a DC blocking filter since there is no information
around DC. The DC offset is a very big problem in receiver design
and it is much larger in Direct-Conversion receiver.81The complete
system (Analytically)82
D[N-], D[N-+1], . ,D[N-1] D[0], D[1], D[2], ., D[N--1] D[N-],
D[N-+1]. D[N-1] Cyclic PrefixOriginal length N sequenceAppend last
sequence to beginning(5.7)The complete system (Analytically)83
Linear ConvolutionCircular Convolution(5.8)(5.9)(5.10)Overhead
due to cyclic prefixThe benefits of adding cyclic prefix comes at a
cost. Since symbols are added to the input data block, there is an
overhead of /N and hence the data rate of the OFDM system reduces
by N/(+N).For example in IEEE 802.11a, Number of samples for CP, =
16 and N=64, hence the overhead is 16/64=0.25 and the data rate
reduces by 64/80=0.8.
8484EXAMPLE 1Consider an OFDM system with total passband
bandwidth B= 1MHz and ideal Nyquist filtering. A single carrier
system would have symbol time Ts=1/B=1s. The channel has maximum
delay spread of Tm=5s, so with T= 1s and Tm= 5s there would be
clearly severe ISI. Assume an OFDM system with MQAM modulation
applied to each subchannel. To keep the overhead small, the OFDM
system uses N=128 subcarriers to mitigate ISI. So TN=NTs=128 s. The
length of the cyclic prefix is set to =8 > Tm/Ts to insure no
ISI between OFDM symbols. For these parameters, find the subchannel
bandwidth, the total transmission time associated with each OFDM
symbol. The overhead of the cyclic prefix, and the data rate of the
system assuming M=16. 8586Given: Total Bandwidth, B= 1MHz Input
Symbol Time= 1s Channel Delay spread, Tm= 5s M=16 Length of cyclic
prefix= 8 samples Number of subcarriers= 128 OFDM symbol Duration,
TN=128 s Deduce: Subchannel Bandwidth, BN The total transmission
time, T of OFDM symbol Overhead of cyclic prefix, /N. Data rate of
the system, R
87Solution:Step 1: The subchannel Bandwidth BN=1/TN=7.812KHz or
BN= B/N=7.812KHzStep 2: The total transmission time, T for each
OFDM symbol T=TN+Tg=128+Ts = 128+8=136s Step 3: Overhead with
cyclic prefix /N =8/128=0.0625=6.25%Step 4: Data rate of the system
k=log2M=4bits/symbol per subcarrier every T seconds Hence data
rate, R=4128/13610-6=3.76 Mbps What will be the data rate without
CP?87EXAMPLE 2Consider the design of an OFDM system. The goal is to
transmit data a rate of 2 Mbps using 16-QAM with an available
bandwidth of 600 kHz. It is known that the delay spread of the
channel is upto a maximum of 20s. You can assume a cyclic prefix
interval equal to the max delay spread. Four guard channels at each
end of the signal spectrum are also required. Find the total number
of sub-carriers and total transmission time of OFDM
symbol.8889Given: Required Data Rate, R= 2 Mbps Bandwidth, B= 600
kHz Ng=24=8 Tm=Tg=20s M=16 Deduce: Total Number of subcarriers, N
The total transmission time, T of OFDM symbol 90Let Ng, Nd and N be
the number of guard subchannels, the number of data subcarriers and
total number of subcarriers respectively. Moreover, let Tg, TN and
T be the cyclic prefix interval, OFDM symbol duration and total
block duration respectively. TN=1/ f where f is the inter-carrier
spacing Total transmission time/ block duration, T=Tg+TN For 16 QAM
we have, k=4 bits per symbol. The required bit rate is 2 Mbps,
hence we must have
From the bandwidth constraint we have
(5.11)(5.12)91Combining (5.11) and (5.12) we get,
Hence the total number of subcarriers are,
N=Nd+Ng=100+8=108Substituting Nd=100 in Eq.(5.11) we get the OFDM
symbol duration as TN=108/600=0.18ms and Total Block Duration=
T=0.18ms+0.02ms=0.2ms
92OFDM Symbol MOFDM Symbol M+1 TN TN Tg Tg D[N-], D[N-+1], .
,D[N-1] D[0], D[1], D[2], ., D[N--1] D[N-], D[N-+1]. D[N-1] Cyclic
PrefixOriginal length N sequenceAppend last sequence to
beginning
Spectrum of OFDM Signal
As discussed in previous slides, each OFDM symbol is transmitted
during the time duration TN=NTs, where Ts is the input symbol rate.
Since it is a block by block transmission, the receiver input
during each symbol duration (after removing the samples associated
with cyclic prefix) can be written as
where w(t) is the rectangular function given as 93
(5.13)(5.14)
Spectrum of OFDM Signal
Taking the Fourier Transform
The Fourier transform of w(t) is
Thus the received OFDM symbol has a Fourier spectrum 94
(5.15)(5.16)(5.17)e(jn)=(-1)^n94
Spectrum of OFDM Signal
It can be easily verified that there is no ICI by showing
X(kf)=X(k/TN)=cdk (where c is constant)95
(5.18)e(jk)=(-1)^k
95
Frequency Domain EqualizationSlide 22 in Lecture 15 demonstrated
the frequency domain equalization for the OFDM system. It was shown
that in absence of noise the best estimate of dn was
Equalization in frequency domain consist of scaling the received
sub-carrier by a complex scalar, namely the estimated channel
response at the nth carrier.Channel frequency response are
typically estimated by various means (such as pilots sub-carriers,
training symbols, long preamble etc.)96
Advantages of OFDM By dividing the channel into narrowband flat
fading subchannels OFDM is more resilience to frequency selective
fading thus eliminating the use of expensive and complicated time
domain equalizers which are commonly used in single carrier
systems.Allows carriers to overlap, resulting in lesser wasted
bandwidth without any Inter Carrier Interference (ICI).Eliminates
ISI and ICI through the use of a cyclic prefix.Using adequate
channel coding and interleaving one can recover symbols lost due to
the frequency selectivity of the channel.Channel equalization
becomes simpler by using frequency domain equalization
techniques.97Peak-to-average-Power Ratio (PAPR)One of the major
drawbacks of OFDM as compared to single carrier systems is large
peak to average power ratio (PAPR). An OFDM signal consists of a
number of independently modulated subcarriers which when added
coherently give a large PAPR. Large PAPR causes nonlinear
distortion in the transmitted OFDM signal when it is passed through
a nonlinear High Power Amplifier (HPA)
Drawbacks of OFDM
98
NIDFTx(t) Nonlinear systemd(0)d(1)A-Ad(N-1)ClippingSpectral
RegrowthratioWhen Nsignals are added with the same phase, they
produce a peak power that is N timesthe average power
98Peak-to-average-Power Ratio (PAPR)
Problems with High PAPRCompromises the efficiency of the high
power amplifier as the power amplifier needs to be operated in
linear region to avoid nonlinear distortions.A high PAPR signals
needs a high resolution A/D converter at the receiver, since the
dynamic range of the signal is much larger for high PAPAR
signals.High resolution A/D increases the complexity and power
consumption at the receiver front end. PAPR of continuous time
signal is given asWhat is PAPR of square wave and sine wave?What
can be the maximum value of PAPR for OFDM signal with N
subcarriers?
99
1/0.5=2(3dB)99Inter-Carrier InterferenceOFDM signals are more
sensitive to carrier frequency offset and phase noise than single
carrier systems. Phase noise and frequency offset are two major
phenomena that causes the inter-carrier interference. Frequency
offset is the difference between the transmitter and receiver
frequency. Phase noise is the random phase jitter cause by the
mismatch in oscillator frequencies etc. Example: If the carrier
frequency Oscillator is accurate to 1 part per million then the
frequency offset, For IEEE 802.11a, f0=5 GHz, then fe=5000 Hz which
will degrade the orthogonality of the subcarriers as shown in the
figure.
Drawbacks of OFDM 100
Impairments of OFDM systemThe performance of the OFDM systems
can suffer due to various systems or channel-induced impairments.
These areFrequency Offset Timing MismatchTime Varying
channelNonlinear system behavior 101
Timing Offset
If the cyclic prefix length of the OFDM block duration, , is
greater than the sample timing offset, , i.e. then the effects of
timing offset are negligible and only causes a phase shift in the
data symbol as shown below. Let the received signal x(t) be sampled
at a rate 1/Ts with timing offset
102
(5.19)Delta over equal sign denotes the definition102
Timing Offset
Taking the FFT of xk we get
It is quite obvious from Eq. (5.20) that timing offset only
result in the phase shift of the data symbol (without ISI).
Moreover, the amount of phase shift depends linearly with
sub-carrier index.
103
(5.20)104Frequency OffsetAs mentioned before frequency offset
introduces ICI and affects the orthogonality of the OFDM
subcarriers. This has significant impact on the SNR of the OFDM
system. In order to maintain an Signal-to interference ratio of
20dB or more the offset cannot be more than 4% of the inter-carrier
spacing. The signal corresponding to the subcarrier i can be simply
expressed as ( suppressing the data symbol and carrier
frequency)
An interfering subchannel can be written as
If the signal is demodulated with a frequency offset of /TN,
then this interference becomes
(5.21)(5.22)
(5.23)105The ICI between the subchannel xi and xi+m is simply
the inner product between them
It is quite obvious from Eq. (5.24) that if =0, then Im=0 (hence
no ICI). The total ICI power on the subcarrier i is then
Where Co is some constant. Some interesting observationsAs TN f
hence ICI The ICI grows with frequency offset, ICI is not directly
affected by number of subcarriers, N (How is it indirectly affected
by N?)
(5.24)(5.25)Since exp(-i2pi)=1105
OFDM Synchronization
Synchronization in OFDM is required to find the start of an OFDM
frame and before the demodulation can be done for the subcarriers.
As stated before, since OFDM has a long symbol duration (due to
parallel transmission) it is less sensitive to timing jitters as
compare to single carrier systems. However, frequency offset and
phase noise severely impacts the performance of OFDM system. There
are various levels of synchronization in OFDM as shown below.
106107Coarse timing recovery (Frame/ Packet/Slot
synchronization)Coarse frequency Offset Estimation/ correctionsFine
frequency corrections (after FFT)Fine timing corrections (after
FFT)Coarse timing synchronizationIt provides the frame (packet or
slot) starting point. The coarse timing synchronization is usually
achieved by correlating the incoming signal with a known preamble.
This is performed prior to FFT demodulation. Since timing offset
does not violate the orthogonality of the sub-carriers, fine timing
offset can be done after the FFT.Null symbol type synchronization:
In this case a silence period of a know duration is inserted at the
start of the frame followed by envelope detection. More suitable
for continuous stream traffic like DAB, however ineffective for
packet/burst type of transmission like IEEE 802.11a WLAN. 108
Coarse timing synchronization
Preamble symbols are usually used in coarse timing
synchronization of packet/ burst type of transmission like IEEE
802.11a WLAN. In this case known preamble symbols are repeatedly
transmitted at the start of each frame/packet. The receiver
correlates the received signal with the stored preamble. Repetition
of the stored preamble reduces the false alarm rates and improves
the probability of detection of start of the frame.
109Input BufferCorrelator outputThreshold comparison Stored
preambleRSSIEstimateFrame SynchronizationReceived DataRSSI:
Received signal strength indication109