QPSK, OQPSK, CPM Probability Of Error For AWGN And Flat Fading . - WINLAB

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QPSK, OQPSK, CPMProbability Of Error for AWGN and Flat Fading Channels [4]16:332:546 Wireless Communication Technologies Spring 2005 Department of Electrical Engineering, Rutgers University, Piscataway, NJ 08904Sanjit Krishnan Kaul (sanjit@winlab.rutgers.edu)AbstractThis article discusses QPSK, OQPSK, π/4DQPSK and the trade offs involved. Later wediscuss various CPM schemes and their relativebandwidth efficiencies. Probability of Bit erroris discussed for various schemes assuming anAWGN channel. A model is devised, ignoringany changes in phase, to determine the errorprobabilities for flat fading channels. The articleends with a brief discussion of non-coherentdetection.Figure 1:1QPSK andSignalingOffsetThe figure shows a QPSK constellation. The darkblack lines show all possible phase changes.QPSKQPSK or Quadrature Phase Shift Keying,involves the splitting of a data streammk (t) m0 , m1 , m2 , . . ., into an in-phase streammI (t) m0 , m2 , m4 , . . . and a quadraturestream mQ (t) m1 , m3 , m5 , . . . Both thestreams have half the bit rate of the datastream mk (t), and modulate the cosine and sinefunctions of a carrier wave simultaneously. Asa result, phase changes across intervals of 2Tb ,where Tb is the time interval of a single bit (themk (t)s). The phase transitions can be as largeas π as shown in Figure 1.Sudden phase reversals of π can throw theamplifiers into saturation. As shown in Figure 2[1], the phase reversals of π cause the envelopeto go to zero momentarily. This may makeus susceptible to non-linearities in amplifiercircuitry. The above may be prevented usinglinear amplifiers but they are more expensiveand power consuming. A solution to the abovementioned problem is the use of OQPSK. Taught by Dr.University.Figure 2:The figure shows a QPSK waveform. As is seenacross the dotted line corresponding to a phase shift of π, theenvelope reduces to zero temporarily.OQPSK modulation is such that phasetransitions about the origin are avoided. Thescheme is used in IS-95 handsets. In OQPSKthe pulse streams mI (t) m0 , m2 , m4 , . . . andmQ (t) m1 , m3 , m5 , . . . are offset in alignment,Narayan Mandayam, Rutgers in other words are staggered, by one bit period(half a symbol period). Figure 3 [2], shows the1

staggering of the data streams in time. Figure4 [1], shows the OQPSK waveform undergoing aphase shift of π/2. The result of limiting thephase shifts to π/2 is that the envelope will notgo to zero as it does with QPSK.Figure 5:The figure shows a OQPSK constellation. Thedark black lines show all possible phase changes. The signalspace is the same as in the case of QPSK, though phase changesare restricted to 90.shows an example set of phase values that wemay choose.Figure 3:The figure shows the staggering of the in phaseand quadrature modulated data streams in OQPSK. Thestaggering restricts the phase changes to 90 as shown inFigure 4.In OQPSK, the phase transitions take placeevery Tb seconds. In QPSK the transitions takeplace every 2Tb seconds.Figure 6:The figure shows an allowable table of phasetransitions in π/4 DQPSK. The maximum phase translationallowed is 135.Figure 7 [2] gives two possible constellationsand their all possible phase transitions.Figure 4:The figure shows a QPSK waveform. As is seenacross the dotted lines the phase changes are of π/2.The OQPSK constellation is as shown inFigure 5.2π/4 DQPSK SignalingThe signaling is a compromise between QPSK Figure 7: π/4 DQPSK constellations and all possible phaseand OQPSK in that the maximum transitions phase translations.are allowed to be 3π/4. The scheme is usedin North American TDMA (IS-136). Figure 62

3CPMConstant envelope and very good spectral Two terms that characterize CPM arecharacteristics make CPM, Continuous Phase Average Frequency Deviation k̄f andModulation, a preferred choice in wireless Modulation Index h.communications.The complex basebandZ Tequivalent is given byhf (τ ) dτk̄f (kf /T )Z t X0nov(t) A exp j2πkfxn hf (τ nT ) dτ nh β(τ )/π 2k̄f T A exp{jφ(t)}(1) 3.1 CPFSKkTφ(t) 2πkfZk 1Xxn hf (τ nT ) dτ n t(2)hf (τ kT ) dτ 2πkfkTwhere, kT t (k 1)TIf hf (t) 0 for t T , the CPM signal iscalled full response CPM. If hf (t) 6 0 for t T ,the CPM signal is called partial response CPM.Using the standard form of representation of abaseband signal,v(t) AXb(t kT, xk )(6)(7)A conventional FSK signal is generatedby shifting the carrier by an amountfn 1/2 f In , In 1, 3, . . . , (M 1),to reflect the digital information that is beingtransmitted.The type of FSK signal ismemoryless. Further, the switching from onefrequency to another may be accomplished byhaving M 2k separate oscillators tuned tothe desired frequencies and selecting one of theM frequencies according to the k-bit symbolthat is to be transmitted in a signal intervalof duration T k/R seconds. However, suchabrupt switching from one oscillator outputto another in successive signaling intervalsresults in relatively large spectral side lobesoutside of the main spectral band of the signaland, consequently, this method requires alarge frequency band for transmission of thesignal. To avoid the use of signals having largespectral side lobes, the information bearingsignal frequency modulates a single carrierwhose frequency is changed continuously. Theresulting frequency modulated signal is phasecontinuous and, hence, it is called continuousphase FSK.where A is the amplitude, kf is thepeak frequency deviation, hf (t) is thefrequency shaping pulse and T is the symbolduration.The symbol source sequence is{xn } { 1, 3, 5, . . . , (M 1)}, where Mis the alphabet size.Zthethe(3)kk 1n hioXb(t, xk ) uT (t) exp j β(τ )xn xk β(t)hf (t) uT (t)(8)k̄f kf , h 2kf T(9) (4)PIn the above equation, β(τ ) k 1 xn is theaccumulated excess phase (memory) and xk β(t)is the excess phase for the current symbol. 0Rβ(t) 2πkf 0t hf (τ ) dτ β(T )t 00 t Tt T 0β(t) 2πkf t πht/T πht 00 t Tt T(10)(5)CPM signals are usually described by sketching3

3.1.1the excess phase φ(t).MSKMSK, Minimum shift keying is a special case ofbinary CPFSK with h 0.5.k 1X φ(t) β(T )xn xk β(t kT ) for all {xn } t 0 0n β(t) 2πkf t πt/2T 0 t T(12)(11) 0.5πt Tφ(t)isplottedforBinary{xn } { 1, 1} in Figure 8 [3].CPFSK,Therefore, the carrier phase during the intervalkT t (k 1)T is given by,φ(t) 2πfc t π/2k 1Xxn 0.5π xk ((t kT )/T )n k X 2πfc πxk /2T t π/2xn π/2 xkn (13)The MSK bandpass waveform is then given ashs(t) A cos (2πfc πxk /2T )t π/2k 1Xxn πk/2 xkin Figure 8:where kT t (k 1)Th A cos 2π(fc xk /4T )t Phase trajectory for binary CPFSK.k 1X(14)iThe figure corresponds to a rectangular pulseπ/2xn πk/2 xkn representing a bit. Hence, the phase changes at aconstant rate (straight line). Figure 9 compareswhere xk ( 1, 1)the phase changes between a raised cosine pulseSince, xk can be 1, two different frequenciesand a rectangular pulse shape.are modulated for 1. The frequencies arefc 1/4T .Therefore, the differencebetween the frequencies is 1/2T which isthe minimum frequency separation required toensure orthogonality between two sinusoids ofduration T , assuming coherent demodulation.The above is the reason why the scheme is calledminimum shift keying.The Power Spectral Density of MSK is shownin the Figure 10 for different pulse shapes.Figure 11 compares the spectra of MSK andOQPSK. Note that the main lobe of MSK is50% wider than that for OQPSK. However, theside lobes in MSK fall off considerably faster,Figure 9: Phase trajectory for binary CPFSK using a making MSK more bandwidth efficient. Evenrectangular pulse (dotted) and a raised cosine pulse. The I greater efficiency than MSK can be achieved byis the {xn }further reducing h. However, the FSK signalswill no longer be orthogonal and there will be anincrease in the error probability.4

3.2.1GMSKPass the rectangular pulse hf (t) through a premodulation filter given asnoH(f ) exp (f /B)2 ln 2/2(16)B is the bandwidth of the filter. H(f) is bellshaped about f 0. Therefore, the nameGaussian MSK. BT is used to parameterizeGMSK schemes. B is the bandwidth of thepremodulation filter defined above and T is thesymbol duration. The next two figures makeit amply clear in both the time and frequencydomain, that decreasing BT improves spectraloccupancy.Figure 10:PSD for MSK. The dotted curve correspondsto MSK. The other curves correspond to a raised cosine pulse,partial response CPFSK with h 0.5, lasting for 2T ,3T ,4T .It is clearly seen that as the spreading in time is increased, thebandwidth efficiency increases.Figure 12:GMSK pulses for different BT .From Figure 12, we observe that when BT 0.3, the GMSK pulse may be truncated at t 1.5T with a relatively small error incurred.GMSK with BT 0.3 is used in GSM.Figure 11: PSD comparison of MSK and OQPSK.The decrease in spectral occupancy isaccompanied by increase in ISI, as we no longeradhere to the Nyquist Criterion. To counter theISI, GMSK requires equalization. Thus, we can3.2 Partial Response CPMsay that GMSK is a bandwidth efficient schemeThe idea is to make hf (t) of duration greater but not a power efficient one. The table belowthan T .shows occupied RF bandwidth for GMSK andMSK as a fraction of Rb , the bit rate, containinghf (t) hf (t) ukT (t)a given percentage of power. Notice that GMSKK 1X(15) is spectrally tighter than MSK.where ukT uT (t kT )k 05

vector r [r1 r2 . . . rN ], which contains allthe relevant information in the received signalwaveform. Once the vector r has been received,an optimum decision needs to be made regardingwhich signal sm was transmitted, given thatr has been received where 1 m M . Thedecision criterion is based on selecting thesignal corresponding to the maximum of theset of a posteriori probabilities {P (sm r)}.The decision criterion is called the MAP(maximum a posteriori probability) criterion. Itcan be proved that the criterion minimizes theprobability of error and therefore a detector thatimplements it is known as the optimum detector.Figure 13:Power spectral density for a GMSK signal.BT0.2 GMSK0.25 GMSK0.5 GMSKMSK4Using Bayes’ rule,the a posterioriprobabilities can be expressed as P (sm r) p(r sm )P (sm )/p(r), where p(r sm ) is theconditional PDF of the observed vector giventhat sm was transmitted and P (sm ) is thea priori probability of the mth signal 41.2099.9%0.991.091.332.7699.99%1.221.372.082If we further assume that all M signals areequally probable a priori, the optimum detectionrule reduces to finding the transmitted signalthat maximizes p(r sm ).The criterion isalso known as the ML (maximum likelihood)criterion.M-PSK Bandwidth/PowerThe MPSK waveform is given by,For an AWGN channel,si (t) (2Es /Ts )0.5 cos(2πfc (t) 2π/M (i 1))NihXp(r sm ) (πN0 ) N/2 exp ( 1/N0 )(rk smk )2where 0 t Ts and Ts (log2 M )Tbk 1Es Eb log2 M andTb is the energy per symbol.(17)0 m M(18)The table below highlights the bandwidth and Therefore,power efficiency of M-PSK signals.PowerNefficiency ηp is defined as Eb /No required forX 6(rk smk )2Pe 10 . The bandwidth is the first null ln p(r sm ) ( N/2) ln(πN0 ) 1/N0k 1bandwidth.(19)Themaximumoflnp(r m that minimizesηb0.511.5 22.53PN2k 1 (rk smk ) , which is the same as theEb /No 10.5 10.5 14 18.5 23.4 28.5Euclidean distance between the received vectorand the vector sm .5Optimum Receivers AWGNFurther, if we assume that all the sm haveFor a signal transmitted over an AWGN the same energy, the criteria becomes selectingchannel [3], either a correlation demodulator the vector sm that has the maximum correlationor a matched filter demodulator produces the with the received vector r. Remember that6

the vector was received at the output of the a narrower main lobe and a faster roll-off ofmatched filter/ correlation demodulator.side bands. But from above it is clear thatPe (GM SK) Pe (M SK)All the observations made above are assumingmemoryless modulation. However, for signals6 B/W vs Power efficiencythat have memory, the maximum likelihoodtrade offsequence detection algorithm must be used.The algorithm searches for the minimumEuclidean distance path through the trellis that As BTb decreases, bandwidth efficiencycharacterizes the memory in the transmitted increases, but power efficiency decreases(because Pe increases).signal.5.17Probability Of Error5.1.1Binary FSKpIt is given by Q( Eb /No ).5.1.2Error Probabilities for FlatFading ChannelsConsider the transmitted waveform,psi (t) 2Es /Ts cos(2πfc t 2π/M (i 1)),MSKMSK has one of two possible frequencies overany symbol interval.h7.1s(t) A cos (2πfc πxk /2T )t 0 t Ts(22)Flat fading received modelx(t) g(t)si (t) w(t), where w(t) is AWGN.(20) g(t) is the attenuation in the amplitude ofn the signal due to fading. Assume that thechannel is flat and slow. Therefore Ts στwhere kT t (k 1)Tand Ts Tc . Therefore g(t) is effectively aConsider interval 0 t Tb and denote A constant over the symbol duration. Let g(t) α,pthen x(t) αsi (t) w(t) for 0 t Ts .2E /T . Thenπ/2bk 1Xxn π/2 kxkibFor a constant α the maximum likelihooddecoding rule for optimum detection (assuminginputs are equiprobable) holds true. Therefore,the receiver structure remains the same. Ingeneral it can be shown that,XpPe Q(αdik / 2N0 )(23)(p2Eb /Tb cos(2πf1 t θ(o))β(t) p2Eb /Tb cos(2πf2 t θ(o))symbol 1symbol 0(21)On first guess one may conclude that MSK hasthe same error probability as BFSK, but since ithas memory of phase it does better. Usingp phasetrellis it can be shown that Pe Q( 2Eb /Tb ),the approximation is valid for high SNR values.Therefore, MSK is approximately same in BERperformance as BPSK.k 1k6 iTypically, α is Rayleigh or Rician distributedfor non LOS and LOS situations respectively.5.1.3 GMSKTherefore the probability of error may beThe probabilityoferrorofGMSKcanbeshownwritten as,pto be Pe Q( 2αEb /Tb ) where α is a constantXZ pfor a given BTb . For example, for BTb 0.25,Pe Q(αdik / 2N0 )fα (α) dα (24)α 0.68. Similarly, for BTb , α 0.85k 1 0k6 iwhich is the case of MSK.We had earlier noted that GMSK improvedbandwidth efficiency over MSK since it had Consider M7 2, the SNR is given as

γb α2 Eb /N0 . Let β α2 . If α is Rayleigh, βis exponential.Z pQ( 2βEb /N0 )fβ (β) dβ(25)P̄e It is important to note that in the abovemodel, we assumed that coherent detection waspossible. That is why the phase was completelyignored in the model. For coherent detection tobe possible in a fading channel, we need pilotsignals.0Rewrite, γb βEb /N0 . Then7.2E[γb ] γ̄b Eb /N0 E[β](26)f (γb ) (γ̄b ) 1 exp( γb /γ̄b ) , γb 0Detection of signals with unknownphase(27) If we assume that the phase of the signal is notknown at the receiver we will have to use NonCoherent detection. Let the transmitted signalbe,Therefore,Z pP̄e Q( 2γb )(γb ) 1 exp( γb /γ̄b ) dγb (28)p2E/T cos(2πfi t), 0 t Tsi (t) 0(35)Integrating by parts we get,The received signal may be written as,P̄e 0.5 Z (1/2 π) exp( γb (1 γ̄b )/γ̄b )0px(t) γb 0.5 dγb2E/T cos(2πfi t θ) w(t)(36)(29)where w(t) is the AWGN and θ is the unknownphase. So we can assume that θ is a randomvariable uniformly distributed in the interval(30) [0, 2π]. Rewriting, x(t) asSubstituting,z γb (1 γ̄b )/γ̄b )P̄e 1/2 1/2 (π(1 γ̄b )/γ̄b ) 1Z x(t) e z z 0.5 dz0(31)P̄e 1/2 1/2 (π(1 γ̄b )/γ̄b ) 1 Γ(1/2)P̄e 1/2 1/2 ((1 γ̄b )/γ̄b ) 1n2E/T cos(2πfi t) cos(θ) o(37)sin(2πfi t) sin(θ) w(t)pThe signal may be received using a Quadrature(32) ume that bit 1 is transmitted as s1 (t) and bitFor high SNR (γ̄b ) we can say that Pe 0 is transmitted as s (t). As the modulation is2(SN R) 1 , unlike AWGN where they were orthogonal, s (t) and s (t) are orthogonal. We12exponentially related. Consider the probability further assume that the received signal is x(t).of error, if using Binary FSK on a flat fading(channel as modeled above.g1 (t) w(t) ,for 0 t T , if 1 is Txx(t) g2 (t) w(t) ,for 0 t T , if 0 is Tx(34)Pe 1/2 1/2 ((2 γ̄b )/γ̄b ) 1(38)We can further assume that g1 (t) and g2 (t) areTherefore, coherent PSK is 3dB better than orthogonal. The receiver is shown in Figure 14.coherent FSK.8

Figure 14:Non-Coherent detection.receiver is a Quadrature Receiver.Figure 15:Each ARM of theA Quadrature Receiver.References[1] DigitalCommunicationsFundamentalsandApplications, Dr. Bernard Sklar.[2] Digital Communications, Dr. Proakis.[3] Wireless Communications Principles and Practice,Dr. Rappaport.[4] Wireless communication technologies, lecture notes,Spring 2005, Dr. Narayan Mandayam, RutgersUniversity.9

The phase transitions can be as large as π as shown in Figure 1. Sudden phase reversals of π can throw the amplifiers into saturation. As shown in Figure 2 [1], the phase reversals of π cause the envelope to go to zero momentarily. This may make us susceptible to non-linearities in amplifier circuitry. The above may be prevented using

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