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.

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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.

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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

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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.

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(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.

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(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.

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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

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(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

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(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

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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.

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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.

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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.

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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