DIGITAL MODULATIONS 1999 Bijan Mobasseri

Why digital modulation? If our goal was to design a digital baseband communication system. We have done that Problem is baseband communication won’t takes us far, literally and figuratively Digital modulation to a square pulse is what analog modulation was to messages ©2000 Bijan Mobasseri

A block diagram Messsage source Source coder 1011 Line coder Pulse shaping modulator channel decision detector demodulator ©2000 Bijan Mobasseri

GEOMETRIC REPRESENTATION OF SIGNALS

The idea We are used to seeing signals expressed either in time or frequency domain There is another representation space that portrays signals in more intuitive format In this section we develop the idea of signals as multidimensional vectors ©2000 Bijan Mobasseri

Have we seen this before? Why yes! Remember the beloved ej2πfct which can be written as ej2πfct=cos(2πfct)+jsin(2πfct) quadrature inphase ©2000 Bijan Mobasseri

Expressing signals as a weighted sum Suppose a signal set consists of M signals si(t),I=1,…,M. Each signal can be represented by a linear sum of basis functions ©2000 Bijan Mobasseri

Conditions on basis functions For the expansion to hold, basis functions must be orthonormal to each other Mathematically: Geometrically: j i k ©2000 Bijan Mobasseri

Components of the signal vector Each signal needs N numbers to be represented by a vector. These N numbers are given by projecting each signal onto the individual basis functions: sij means projection of si (t)on j(t) si j sij ©2000 Bijan Mobasseri

Signal space dimension How many basis functions does it take to express a signal? It depends on the dimensionality of the signal Some need just 1 some need an infinite number. The number of dimensions is N and is always less than the number of signals in the set N<=M ©2000 Bijan Mobasseri

Example: Fourier series Remember Fouirer series? A signal was expanded as a linear sum of sines and cosines of different frequencies. Sounds familiar? Sines and cosines are the basis functions and are in fact orthogonal to each other ©2000 Bijan Mobasseri

Example: four signal set A communication system sends one of 4 possible signals. Expand each signal in terms of two given basis functions 1 2 1 -0.5 1 1 1 1 2 ©2000 Bijan Mobasseri

Components of s1(t) s=(1,-0.5) This is a 2-Dsignal space. Therefore, each signal can be represented by a pair of numbers. Let’s find them For s1(t) s1(t) 1 t -0.5 1 1 s=(1,-0.5) t 1 2 ©2000 Bijan Mobasseri

Interpretation s1(t) is now condensed into just two numbers. We can “reconstruct” s1(t) like this s1(t)=(1)1(t)+(-0.5)2(t) Another way of looking at it is this 2 1 1 -0.5 ©2000 Bijan Mobasseri

Signal constellation Finding individual components of each signal along the two dimensions gets us the constellation 2 s4 s2 1 -0.5 0.5 -0.5 s1 s3 ©2000 Bijan Mobasseri

Learning from the constellation So many signal properties can be inferred by simple visual inspection or simple math Orthogonality: s1 and s4 or orthogonal. To show that, simply find their inner product, < s1, s4> < s1, s4>=s11xs41+s12xs42(1)(0.5)+(1)(-0.5)=0 ©2000 Bijan Mobasseri

Finding the energy from the constellation This is a simple matter. Remember, Replace the signal by its expansion ©2000 Bijan Mobasseri

Exploiting the orthogonality of basis functions Expanding the summation, all cross product terms integrate to zero. What remains are N terms where j=k ©2000 Bijan Mobasseri

Energy in simple language What we just saw says that the energy of a signal is simply the square of the length of its corresponding constellation vector 2 E=9+4=13 3 ©2000 Bijan Mobasseri

Constrained energy signals Let’s say you are under peak energy Ep constraint in your application. Just make sure all your signals are inside a circle of radius sqrt(Ep ) ©2000 Bijan Mobasseri

Correlation of two signals A very desirable situation in is to have signals that are mutually orthogonal. How do we test this? Find the angle between them transpose s1 s2  ©2000 Bijan Mobasseri

Find the angle between s1 and s2 Given that s1=(1,2)T and s2=(2,1)T, what is the angle between the two? ©2000 Bijan Mobasseri

Distance between two signals The closer signals are together the more chances of detection error. Here is how we can find their separation 2 1 1 2 ©2000 Bijan Mobasseri

Constellation building using correlator banks We can decompose the signal into its components as follows s1 1 s2 N components s(t) 2 sN N ©2000 Bijan Mobasseri

Detection in the constellation space Received signal is put through the filter bank below and mapped to a point s1 1 s2 s(t) 2 components mapped to a single point sN N ©2000 Bijan Mobasseri

Constellation recovery in noise Assume signal is contaminated with noise. All N components will also be affected. The original position of si(t) will be disturbed ©2000 Bijan Mobasseri

Actual example Here is a 16-level constellation which is reconstructed in the presence of noise ©2000 Bijan Mobasseri

Detection in signal space One of the M allowable signals is transmitted, processed through the bank of correlators and mapped onto constellation question is based on what we see , what was the transmitted signal? received signal which of the four did it come from ©2000 Bijan Mobasseri

Minimum distance decision rule It can be shown that the optimum decision, in the sense of lowest BER, is to pick the signal that is closest to the received vector. This is called maximum likelihood decision making this is the most likely transmitted signal received ©2000 Bijan Mobasseri

Defining decision regions An easy detection method, is to compute “decision regions” offline. Here are a few examples decide s2 decide s1 decide s1 s2 s1 decide s1 measurement s2 s1 s1 s4 s3 decide s2 decide s3 decide s4 ©2000 Bijan Mobasseri

if XZi si was transmitted More formally... Partition the decision space into M decision regions Zi, i=1,…,M. Let X be the measurement vector extracted from the received signal. Then if XZi si was transmitted ©2000 Bijan Mobasseri

How does detection error occur? Detection error occurs when X lands in Zi but it wasn’t si that was transmitted. Noise, among others, may be the culprit X si departure from transmitted position due to noise ©2000 Bijan Mobasseri

P{error|si}=P{X does not lie in Zi|si was transmitted} Error probability we can write an expression for error like this P{error|si}=P{X does not lie in Zi|si was transmitted} Generally ©2000 Bijan Mobasseri

Example: BPSK (Binary Phase Shift Keying) BPSK is a well known digital modulation obtained by carrier modulating a polar NRZ signal. The rule is 1: s1=Acos(2πfct) 0:s2= - Acos(2πfct) 1’s and 0’s are identified by 180 degree phase reversal at bit transitions ©2000 Bijan Mobasseri

Signal space for BPSK Look at s1 and s2. What is the basis function for them? Both signals can be uniquely written as a scalar multiple of a cosine. So a single cosine is the sole basis function. We have a 1-D constellation cos(2pifct) -A A ©2000 Bijan Mobasseri

Eb= A2Tb/2 --->A=sqrt(2Eb/Tb) Bringing in Eb We want each bit to have an energy Eb. Bits in BPSK are RF pulses of amplitude A and duration Tb. Their energy is A2Tb/2 . Therefore Eb= A2Tb/2 --->A=sqrt(2Eb/Tb) We can write the two bits as follows ©2000 Bijan Mobasseri

BPSK basis function As a 1-D signal, there is one basis function. We also know that basis functions must have unit energy. Using a normalization factor E=1 ©2000 Bijan Mobasseri

Formulating BER BPSK constellation looks like this -√Eb √Eb received if noise is negative enough, it will push X to the left of the boundary, deciding 0 instead X|1=[√Eb+n,n] noise -√Eb √Eb noise transmitted ©2000 Bijan Mobasseri

Finding BER Let’s rewrite BER But n is gaussian with mean 0 and variance No/2 -sqrt(Eb) ©2000 Bijan Mobasseri

BER for BPSK Using the trick to find the area under a gaussian density(after normalization with respect to variance) BER=Q[(2Eb/No)0.5] or BER=0.5erfc[(Eb/No)0.5] ©2000 Bijan Mobasseri

BPSK Example Data is transmitted at Rb=106 b/s. Noise PSD is 10-6 and pulses are rectangular with amplitude 0.2 volt. What is the BER? First we need energy per bit, Eb. 1’s and 0’s are sent by ©2000 Bijan Mobasseri

Solving for Eb Since bit rate is 106, bit length must be 1/Rb=10-6 Therefore, Eb=20x10-6=20 w-sec Remember, this is the received energy. What was transmitted is probably several orders of magnitude bigger ©2000 Bijan Mobasseri

Solving for BER Noise PSD is No/2 =10-6. We know for BPSK BER=0.5erfc[(Eb/No)0.5] What we have is then Finish this using erf tables ©2000 Bijan Mobasseri

Binary FSK (Frequency Shift Keying) Another method to transmit 1’s and 0’s is to use two distinct tones, f1 and f2 of the form below But what is the requirements on the tones? Can they be any tones? ©2000 Bijan Mobasseri

Picking the right tones It is desirable to keep the tones orthogonal Since tones are sinusoids, it is sufficient for the tones to be separated by an integer multiple of inverse duration, i.e. ©2000 Bijan Mobasseri

Example tones Let’s say we are sending data at the rate of 1 Mb/sec in BFSK. What are a few orthogonal tones? Bit length is 10-6 sec. Therefore, possible tones are (use nc=0) f1=1/Tb=1 MHz f2=2/Tb=2MHz ©2000 Bijan Mobasseri

BFSK dimensionality What does the constellation of BFSK look like? We first have to find its dimension s1 and s2 can be represented by two orthonormal basis functions: Notice f1 and f2 are selected to make them orthogonal ©2000 Bijan Mobasseri

BFSK constellation There are two dimensions. Find the components of signals along each dimension using ©2000 Bijan Mobasseri

Decision regions in BFSK Decisions are made based on distances. Signals closer to s1 will be classified as s1 and vice versa 45 degree line ©2000 Bijan Mobasseri

Detection error in BFSK Let the received signal land where shown. Assume s1 is sent. How would a detection error occur? x2>x1 puts X in the s2 partition s2 Pe1=P{x2>x1|s1 was sent} x2 X=received s1 x1 ©2000 Bijan Mobasseri

Where do (x1,x2) come from? Use the correlator bank to extract signal components x1(gaussian) 1 x= s1(t)+noise x2(gaussian) 2 ©2000 Bijan Mobasseri

Finding BER We have to answer this question: what is the probability of one random variable exceeding another random variable? To cast P(x2>x1) into like of P(x>2), rewrite P(x2>x1|x1) x1 is now treated as constant. Then, integrate out x1 to eliminate it ©2000 Bijan Mobasseri

BER for BFSK Skipping the details of derivation, we get ©2000 Bijan Mobasseri

BPSK and BFSK comparison: energy efficiency Let’s compare their BER’s What does it take to have the same BER? Eb in BFSK must be twice as big as BPSK Conclusion: energy per bit must be twice as large in BFSK to achieve the same BER; a 3dB disadvantage ©2000 Bijan Mobasseri

Comparison in the constellation space Distances determine BER’s. Let’s compare Both have the same Eb, but BPSK’s are farther apart, hence lower BER ©2000 Bijan Mobasseri

Differential PSK Concept of differential encoding is very powerful Take the the bit sequence 11001001 Differentially encoding of this stream means that we start we a reference bit and then record changes ©2000 Bijan Mobasseri

Differential encoding example Data to be encoded 1 0 0 1 0 0 1 1 Set the reference bit to 1, then use the following rule Generate a 1 if no change Generate a 0 if change 1 1 0 1 1 0 1 1 1 ©2000 Bijan Mobasseri

Detection logic Detecting a differentially encoded signal is based on the comparison of two adjacent bits If two coded bits are the same, that means data bit must have been a 1, otherwise 0 ? ? ? ? ? ? ? ? 1 1 0 1 1 0 1 1 1 unknown transmitted bits Encoded received bits ©2000 Bijan Mobasseri

DPSK: generation Once data is differentially encoded, carrier modulation can be carried out by a straight BPSK encoding Digit 1:phase 0 Digit 0:phase 180 1 1 0 1 1 0 1 1 1 0 0 π 0 0 π 0 0 0 Differentially encoded data Phase encoded(BPSK) ©2000 Bijan Mobasseri

DPSK detection Data is detected by a phase comparison of two adjacent pulses No phase change: data bit is 1 Phase change: data bit is 0 0 0 π 0 0 π 0 0 0 Detected data 1 0 0 1 0 0 1 1 ©2000 Bijan Mobasseri

Bit errors in DPSK Bit errors happen in an interesting way Since detection is done by comparing adjacent bits, errors have the potential of propagating Allow a single detection error in DPSK 0 0 π π 0 π 0 0 0 Incoming phases Detected bits 1 0 1 0 0 0 1 1 1 0 0 1 0 0 1 1 Transmitted bits Back on track:no errors 2 errors ©2000 Bijan Mobasseri

Conclusion In DPSK, if the phase of the RF pulse is detected in error, error propagates However, error propagation stops quickly. Only two bit errors are misdetected. The rest are correctly recovered ©2000 Bijan Mobasseri

Why DPSK? Detecting regular BPSK needs a coherent detector, requiring a phase reference DPSK needs no such thing. The only reference is the previous bit which is readily available ©2000 Bijan Mobasseri

M-ary signaling Binary communications sends one of only 2 levels; 0 or 1 There is another way: combine several bits into symbols 1 0 1 1 0 1 1 0 1 1 1 0 0 1 1 Combining two bits at a time gives rise to 4 symbols; a 4-ary signaling ©2000 Bijan Mobasseri

8-level PAM Here is an example of 8-level signaling binary 0 1 0 1 0 0 0 0 0 0 0 1 1 1 0 1 0 0 1 1 1 7 5 3 2 1 -1 -3 -5 -7 ©2000 Bijan Mobasseri

A few definitions We used to work with bit length Tb. Now we have a new parameter which we call symbol length,T 1 1 Tb T ©2000 Bijan Mobasseri

Bit length-symbol length relationship When we combine n bits into one symbol; the following relationships hold T=nTb- symbol length n=logM bits/symbol T=TbxlogM- symbol length All logarithms are base 2 ©2000 Bijan Mobasseri

Example If 8 bits are combined into one symbol, the resulting symbol is 8 times wider Using n=8, we have M=28=256 symbols to pick from Symbol length T=nTb=8Tb ©2000 Bijan Mobasseri

R=symbol rate=Rb/n=Rb/logM Defining baud When we combine n bits into one symbol, numerical data rate goes down by a factor of n We define baud as the number of symbols/sec Symbol rate is a fraction of bit rate R=symbol rate=Rb/n=Rb/logM For 8-level signaling, baud rate is 1/3 of bit rate ©2000 Bijan Mobasseri

Why M-ary? Remember Nyquist bandwidth? It takes a minimum of R/2 Hz to transmit R pulses/sec. If we can reduce the pulse rate, required bandwidth goes down too M-ary does just that. It takes Rb bits/sec and turns it into Rb/logM pulses sec. ©2000 Bijan Mobasseri

Issues in transmitting 9600 bits/sec Want to transmit 9600 bits/sec. Options: Nyquist’s minimum bandwidth:9600/2=4800 Hz Full roll off raised cosine:9600 Hz None of them fit inside the 4 KHz wide phone lines Go to a 16 - level signaling, M=16. Pulse rate is reduced to R=Rb/logM=9600/4=2400 baud ©2000 Bijan Mobasseri

Using 16-level signaling Go to a 16-level signaling, M=16. Pulse rate is then cut down to R=Rb/logM=9600/4=2400 pulses/sec To accommodate 2400 pulses /sec, we have several options. Using sinc we need only 1200 Hz. Full roll-off needs 2400Hz Both fit within the 4 KHz phone line bandwidth ©2000 Bijan Mobasseri

Bandwidth efficiency Bandwidth efficiency is defined as the number of bits that can be transmitted within 1 Hz of bandwidth =Rb/BT bits/sec/Hz In binary communication using sincs, BT=Rb/2--> =2 bits/sec/Hz ©2000 Bijan Mobasseri

M-ary bandwidth efficiency In M-ary signaling , pulse rate is given by R=Rb/logM. Full roll-off raised cosine bandwidth is BT=R= Rb/logM. Bandwidth efficiency is then given by =Rb/BT=logM bits/sec/Hz For M=2, binary we have 1 bit/sec/Hz. For M=16, we have 4 bits/sec/Hz ©2000 Bijan Mobasseri

M-ary bandwidth Summarizing, M-ary and binary bandwidth are related by BM-ary=Bbinary/logM Clearly , M-ary bandwidth is reduced by a factor of logM compared to the binary bandwidth ©2000 Bijan Mobasseri

BM-ary=Bbinary/logM=9600/log8=3200 Hz 8-ary bandwidth Let the bit rate be 9600 bits/sec. Binary bandwidth is nominally equal to the bit rate, 9600 Hz We then go to 8-level modulation (3 bits/symbol) M-ary bandwidth is given by BM-ary=Bbinary/logM=9600/log8=3200 Hz ©2000 Bijan Mobasseri

Bandwidth efficiency numbers Here are some numbers n(bits/symbol) M(levels) (bits/sec/Hz) 1 2 1 2 4 2 3 8 3 4 16 4 8 256 8 ©2000 Bijan Mobasseri

Symbol energy vs. bit energy Each symbol is made up of n bits. It is not therefore surprising for a symbol to have n times the energy of a bit E(symbol)=nEb Eb E ©2000 Bijan Mobasseri

QPSK quadrature phase shift keying This is a 4 level modulation. Every two bits is combined and mapped to one of 4 phases of an RF signal These phases are 45o,135o,225o,315o Symbol energy Symbol width ©2000 Bijan Mobasseri

QPSK constellation 00 01 45o √E 10 11 Basis functions S=[0.7 √E,- 0.7 √E] ©2000 Bijan Mobasseri

QPSK decision regions 00 01 10 11 Decision regions re color-coded ©2000 Bijan Mobasseri

QPSK error rate Symbol error rate for QPSK is given by This brings up the distinction between symbol error and bit error. They are not the same! ©2000 Bijan Mobasseri

Symbol error Symbol error occurs when received vector is assigned to the wrong partition in the constellation When s1 is mistaken for s2, 00 is mistaken for 11 s2 s1 11 00 ©2000 Bijan Mobasseri

Symbol error vs. bit error When a symbol error occurs, we might suffer more than one bit error such as mistaking 00 for 11. It is however unlikely to have more than one bit error when a symbol error occurs 00 10 10 11 10 Sym.error=1/10 Bit error=1/20 00 11 10 11 10 10 symbols = 20 bits ©2000 Bijan Mobasseri

Interpreting symbol error Numerically, symbol error is larger than bit error but in fact they are describing the same situation; 1 error in 20 bits In general, if Pe is symbol error ©2000 Bijan Mobasseri

Symbol error and bit error for QPSK We saw that symbol error for QPSK was Assuming no more than 1 bit error for each symbol error, BER is half of symbol error Remember symbol energy E=2Eb ©2000 Bijan Mobasseri

QPSK vs. BPSK Let’s compare the two based on BER and bandwidth BER Bandwidth BPSK QPSK BPSK QPSK Rb Rb/2 EQUAL ©2000 Bijan Mobasseri

M-phase PSK (MPSK) If you combine 3 bits into one symbol, we have to realize 23=8 states. We can accomplish this with a single RF pulse taking 8 different phases 45o apart ©2000 Bijan Mobasseri

8-PSK constellation Distribute 8 phasors uniformly around a circle of radius √E 45o Decision region ©2000 Bijan Mobasseri

Symbol error for MPSK We can have M phases around the circle separated by 2π/M radians. It can be shown that symbol error probability is approximately given by ©2000 Bijan Mobasseri

Quadrature Amplitude Modulation (QAM) MPSK was a phase modulation scheme. All amplitudes are the same QAM is described by a constellation consisting of combination of phase and amplitudes The rule governing bits-to-symbols are the same, i.e. n bits are mapped to M=2n symbols ©2000 Bijan Mobasseri

16-QAM constellation using Gray coding 16-QAM has the following constellation Note gray coding where adjacent symbols differ by only 1 bit 0000 0001 0011 0010 1000 1001 1011 1010 1100 1101 1111 1110 0100 0101 0111 0110 ©2000 Bijan Mobasseri

Vector representation of 16-QAM There are 16 vectors, each defined by a pair of coordinates. The following 4x4 matrix describes the 16-QAM constellation ©2000 Bijan Mobasseri

What is energy per symbol in QAM? We had no trouble defining energy per symbol E for MPSK. For QAM, there is no single symbol energy. There are many We therefore need to define average symbol energy Eavg ©2000 Bijan Mobasseri

Eavg for 16-QAM Using the [ai,bi] matrix and using E=ai^2+bi^2 we get one energy per signal Eavg=10 ©2000 Bijan Mobasseri

Symbol error for M-ary QAM With the definition of energy in mind, symbol error is approximated by ©2000 Bijan Mobasseri

Familiar constellations Here are a few golden oldies V.22 600 baud 1200 bps V.22 bis 600 baud 2400 bps V.32 bis 2400 baud 9600 bps ©2000 Bijan Mobasseri

M-ary FSK Using M tones, instead of M phases/amplitudes is a fundamentally different way of M-ary modulation The idea is to use M RF pulses. The frequencies chosen must be orthogonal ©2000 Bijan Mobasseri

MFSK constellation: 3-dimensions MFSK is different from MPSK in that each signal sits on an orthogonal axis(basis) 3 s3 √E s1=[√E ,0, 0] s2=[0,√E, 0] s3=[0,0,√E] √E s1 1 √E s2 2 ©2000 Bijan Mobasseri

Orthogonal signals: How many dimensions, how many signals? We just saw that in a 3 dimensional space, we can have no more than 3 orthogonal signals Equivalently, 3 orthogonal signals don’t need more than 3 dimensions because each can sit on one dimension Therefore, number of dimensions is always less than or equal to number of signals ©2000 Bijan Mobasseri

How to pick the tones? Orthogonal FSK requires tones that are orthogonal. Two carrier frequencies separated by integer multiples of period are orthogonal ©2000 Bijan Mobasseri

Example Take two tones one at f1 the other at f2. T must cover one or more periods for the integral to be zero Take f1=1000 and T=1/1000. Then if f2=2000 , the two are orthogonal so will f2=3000,4000 etc ©2000 Bijan Mobasseri

MFSK symbol error Here is the error expression with the usual notations ©2000 Bijan Mobasseri

Spectrum of M-ary signals So far Eb/No, i.e. power, has been our main concern. The flip side of the coin is bandwidth. Frequently the two move in opposite directions Let’s first look at binary modulation bandwidth ©2000 Bijan Mobasseri

BPSK bandwidth Remember BPSK was obtained from a polar signal by carrier modulation We know the bandwidth of polar NRZ using square pulses was BT=Rb. It doesn’t take much to realize that carrier modulation doubles this bandwidth ©2000 Bijan Mobasseri

Illustrating BPSK bandwidth The expression for baseband BPSK (polar) bandwidth is SB(f)=2Ebsinc2(Tbf) BT=2Rb 2/Tb=2Rb BPSK 1/Tb f fc-/Tb fc fc+/Tb ©2000 Bijan Mobasseri

BFSK as a sum of two RF streams BFSK can be thought of superposition of two unipolar signals, one at f1 and the other at f2 + ©2000 Bijan Mobasseri

Modeling of BFSK bandwidth Each stream is just a carrier modulated unipolar signal. Each has a sinc spectrum 1/Tb=Rb f BT=2 f+2Rb f= (f2-f1)/2 f1 f2 fc fc=(f1+f2)/2 ©2000 Bijan Mobasseri

Example: 1200 bps bandwidth The old 1200 bps standard used BFSK modulation using 1200 Hz for mark and 2200 Hz for space. What is the bandwidth? Use BT=2f+2Rb f=(f2-f1)/2=(2200-1200)/2=500 Hz BT=2x500+2x1200=3400 Hz This is more than BPSK of 2Rb=2400 Hz ©2000 Bijan Mobasseri

Sunde’s FSK: f2-f1=Rb We might have to pick tones f1 and f2 that are not orthogonal. In such a case there will be a finite correlation between the tones  Good points,zero correlation 1 2 3 2(f2-f1)Tb Sunde’s ©2000 Bijan Mobasseri

Picking the 2nd zero crossing: Sunde’s FSK If we pick the second zc term (the first term puts the tones too close) we get 2(f2-f1)Tb=2--> f=1/2Tb=Rb/2 remember f is (f2-f1)/2 Sunde’s FSK bandwidth is then given by BT=2f+2Rb=Rb+2Rb=3Rb The practical bandwidth is a lot smaller ©2000 Bijan Mobasseri

Sunde’s FSK bandwidth Due to sidelobe cancellation, practical bandwidth is just BT=2f=Rb 1/Tb=Rb f f BT=2 f+2Rb f= (f2-f1)/2 f1 f2 fc fc=(f1+f2)/2 ©2000 Bijan Mobasseri

BFSK example A BFSK system operates at the 3rd zero crossing of -Tb plane. If the bit rate is 1 Mbps, what is the frequency separation of the tones? The 3rd zc is for 2(f2-f1)Tb=3. Recalling that f=(f2-f1)/2 then f =0.75/Tb Then f =0.75/Tb=0.75x106=750 KHz And BT=2(f +Rb)=2(0.75+1)106=3.5 MHz ©2000 Bijan Mobasseri

Point to remember FSK is not a particularly bandwidth-efficient modulation. In this example, to transmit 1 Mbps, we needed 3.5 MHz. Of course, working at the 3rd zero crossing is responsible Original Sunde’s FSK requires BT=Rb=1 MHz ©2000 Bijan Mobasseri

Bandwidth of MPSK modulation

MPSK bandwidth review In MPSK we used pulses that are log2M times wider tan binary hence bandwidth goes down by the same factor. T=symbol width=Tblog2M For example, in a 16-phase modulation, M=16, T=4Tb. Bqpsk=Bbpsk/log2M= Bbpsk/4 ©2000 Bijan Mobasseri

SB(f)=(2Eblog2M)sinc2(Tbflog2M) MPSK bandwidth MPSK spectrum is given by SB(f)=(2Eblog2M)sinc2(Tbflog2M) Set to 1 for zero crossing BW Tbflog2M=1 -->f=1/ Tbflog2M =Rb/log2M BT= Rb/log2M 1/logM f/Rb Notice normalized frequency ©2000 Bijan Mobasseri

Bandwidth after carrier modulation What we just saw is MPSK bandwidth in baseband A true MPSK is carrier modulated. This will only double the bandwidth. Therefore, Bmpsk=2Rb/log2M ©2000 Bijan Mobasseri

QPSK bandwidth QPSK is a special case of MPSK with M=4 phases. It’s baseband spectrum is given by SB(f)=2Esinc2(2Tbf) B=0.5Rb--> half of BPSK After modulation: Bqpsk=Rb 0.5 f/Rb 1 ©2000 Bijan Mobasseri

Some numbers Take a 9600 bits/sec data stream Using BPSK: B=2Rb=19,200 Hz (too much for 4KHz analog phone lines) QPSK: B=19200/log24=9600Hz, still high Use 8PSK:B= 19200/log28=6400Hz Use 16PSK:B=19200/ log216=4800 Hz. This may barely fit ©2000 Bijan Mobasseri

MPSK vs.BPSK Let’s say we fix BER at some level. How do bandwidth and power levels compare? M Bm-ary/Bbinary (Avg.power)M/(Avg.power)bin 4 0.5 0.34 dB 8 1/3 3.91 dB 16 1/4 8.52 dB 32 1/5 13.52 dB Lesson: By going to multiphase modulation, we save bandwidth but have to pay in increased power, But why? ©2000 Bijan Mobasseri

Power-bandwidth tradeoff The goal is to keep BER fixed as we increase M. Consider an 8PSK set. What happens if you go to 16PSK? Signals get closer hence higher BER Solution: go to a larger circle-->higher energy ©2000 Bijan Mobasseri

Additional comparisons Take a 28.8 Kb/sec data rate and let’s compare the required bandwidths BPSK: BT=2(Rb)=57.6 KHz BFSK: BT = Rb =28.8 KHz ...Sunde’s FSK QPSK: BT=half of BPSK=28.8 KHz 16-PSK: BT=quarter of BPSK=14.4 KHz 64-PSK: BT=1/6 of BPSK=9.6 KHz ©2000 Bijan Mobasseri

Power-limited systems Modulations that are power-limited achieve their goals with minimum expenditure of power at the expense of bandwidth. Examples are MFSK and other orthogonal signaling ©2000 Bijan Mobasseri

Bandwidth-limited systems Modulations that achieve error rates at a minimum expenditure of bandwidth but possibly at the expense of too high a power are bandwidth-limited Examples are variations of MPSK and many QAM Check BER rate curves for BFSK and BPSK/QAM cases ©2000 Bijan Mobasseri

Bandwidth efficiency index A while back we defined the following ratio as a bandwidth efficiency measure in bits/sec/HZ =Rb/BT bits/sec/Hz Every digital modulation has its own  ©2000 Bijan Mobasseri

=Rb/BT= log2M/2 bits/sec/Hz  for MPSK At a bit rate of Rb, BPSK bandwidth is 2Rb When we go to MPSK, bandwidth goes down by a factor of log2M BT=2Rb/ log2M Then =Rb/BT= log2M/2 bits/sec/Hz ©2000 Bijan Mobasseri

Some numbers Let’s evaluate  vs. M for MPSK M 2 4 8 16 32 64  .5 1 1.5 2 2.5 3 Notice that bits/sec/Hz goes up by a factor of 6 from M=2 and M=64 The price we pay is that if power level is fixed (constellation radius fixed) BER will go up. We need more power to keep BER the same ©2000 Bijan Mobasseri

Defining MFSK: In MFSK we transmit one of M frequencies for every symbol duration T These frequencies must be orthogonal. One way to do that is to space them 1/2T apart. They could also be spaced 1/T apart. Following textbook we choose the former (this corresponds to using the first zero crossing of the correlation curve) ©2000 Bijan Mobasseri

T=Tblog2M symbol length MFSK bandwidth Symbol duration in MFSK is M times longer than binary T=Tblog2M symbol length Each pair of tones are separated by 1/2T. If there are M of them, BT=M/2T=M/2Tblog2M -->BT=MRb/2log2M ©2000 Bijan Mobasseri

BT=2Rb/log2M BT=MRb/2log2M Contrast with MPSK Variation of bandwidth with M differs drastically compared to MPSK MPSK MFSK BT=2Rb/log2M BT=MRb/2log2M As M goes up, MFSK eats up more bandwidth but MPSK save bandwidth ©2000 Bijan Mobasseri

MFSK bandwidth efficiency Let’s compute ’s for MFSK =Rb/BT=2log2M/M bits/sec/Hz…MFSK M 2 4 8 16 32 64  1 1 .75 .5 .3 .18 Notice bandwidth efficiency drop. We are sending fewer and fewer bits per 1 Hz of bandwidth ©2000 Bijan Mobasseri

COMPARISON OF DIGITAL MODULATIONS* *B. Sklar, “ Defining, Designing and Evaluating Digital Communication Systems,” IEEE Communication Magazine, vol. 31, no.11, November 1993, pp. 92-101

Notations Bandwidth efficiency measure ©2000 Bijan Mobasseri

Bandwidth-limited Systems There are situations where bandwidth is at a premium, therefore, we need modulations with large R/W. Hence we need standards with large time-bandwidth product The GSM standard uses Gaussian minimum shift keying(GMSK) with WTb=0.3 ©2000 Bijan Mobasseri

Case of MPSK In MPSK, symbols are m times as wide as binary. Nyquist bandwidth is W=Rs/2=1/2Ts. However, the bandpass bandwidth is twice that, W=1/Ts Then ©2000 Bijan Mobasseri

Cost of Bandwidth Efficiency As M increases, modulation becomes more bandwidth efficient. Let’s fix BER. To maintain this BER while increasing M requires an increase in Eb/No. ©2000 Bijan Mobasseri

Power-Limited Systems There are cases that bandwidth is available but power is limited In these cases as M goes up, the bandwidth increases but required power levels to meet a specified BER remains stable ©2000 Bijan Mobasseri

Case of MFSK MFSK is an orthogonal modulation scheme. Nyquist bandwidth is M-times the binary case because of using M orthogonal frequencies, W=M/Ts=MRs Then ©2000 Bijan Mobasseri

Select an Appropriate Modulation We have a channel of 4KHz with an available S/No=53 dB-Hz Required data rate R=9600 bits/sec. Required BER=10-5. Choose a modulation scheme to meet these requirements ©2000 Bijan Mobasseri

Minimum Number of Phases To conserve power, we should pick the minimum number of phases that still meets the 4KHz bandwidth A 9600 bits/sec if encoded as 8-PSK results in 3200 symbols/sec needing 3200Hz So, M=8 ©2000 Bijan Mobasseri

What is the required Eb/No? ©2000 Bijan Mobasseri

Is BER met? Yes The symbol error probability in 8-PSK is Solve for Es/No Solve for PE ©2000 Bijan Mobasseri

Power-limited uncoded system Same bit rate and BER Available bandwidth W=45 KHz Available S/No=48-dBHz Choose a modulation scheme that yields the required performance ©2000 Bijan Mobasseri

Binary vs. M-ary Model R bits/s M-ary Modulator M-ary demodulator ©2000 Bijan Mobasseri

Choice of Modulation With R=9600 bits/sec and W=45 KHz, the channel is not bandwidth limited Let’s find the available Eb/No ©2000 Bijan Mobasseri

Choose MFSK We have a lot of bandwidth but little power ->orthogonal modulation(MFSK) The larger the M, the more power efficiency but more bandwidth is needed Pick the largest M without going beyond the 45 KHz bandwidth. ©2000 Bijan Mobasseri

MFSK Parameters From Table 1, M=16 for an MFSK modulation requires a bandwidth of 38.4 KHz for 9600 bits/sec data rate We also wanted to have a BER<10^-5. Question is if this is met for a 16FSK modulation. ©2000 Bijan Mobasseri

16-FSK Again from Table 1, to achieve BER of 10^-5 we need Eb/No of 8.1dB. We solved for the available Eb/No and that came to 8.2dB ©2000 Bijan Mobasseri

Symbol error for MFSK For noncoherent orthogonal MFSK, symbol error probability is ©2000 Bijan Mobasseri

BER for MFSK We found out that Eb/No=8.2dB or 6.61 Relating Es/No and Eb/No BER and symbol error are related by ©2000 Bijan Mobasseri

Example Let’s look at the 16FSK case. With 16 levels, we are talking about m=4 bits per symbol. Therefore, With Es/No=26.44, symbol error prob. PE=1.4x10^-5-->PB=7.3x10^-6 ©2000 Bijan Mobasseri

Summary Given: Solution R=9600 bits/s BER=10^-5 Channel bandwith=45 KHz Eb/No=8.2dB Solution 16-FSK required bw=38.4khz required Eb/No=8.1dB ©2000 Bijan Mobasseri