geo rep of signal

DIGITAL MODULATIONS(Chapter 8)

1999 BG Mobasseri 2

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

1999 BG Mobasseri 3

A block diagram

Messsagesource

Source coder Line coder Pulse

shaping

demodulatordetector

channel

modulator

decision

1011

GEOMETRIC REPRESENTATION OF SIGNALS

1999 BG Mobasseri 5

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

1999 BG Mobasseri 6

Have we seen this before?

Why yes! Remember the beloved ej2πfct which can be written as

ej2πfct=cos(2πfct)+jsin(2πfct)

inphase

quadrature

1999 BG Mobasseri 7

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

si t sij j t j1

N

i 1,..., M

0 t T

1999 BG Mobasseri 8

Conditions on basis functions

For the expansion to hold, basis functions must be orthonormal to each other

Mathematically:

Geometrically: i t j t dt 0 i j1 i j

i

j

k

1999 BG Mobasseri 9

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)sij si (t) j t

0

T

dtsij

si

j

1999 BG Mobasseri 10

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


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

cos 2nfot To

cos 2mfot dt 0,m n

fo 1/ To


Example: four signal set

A communication system sends one of 4 possible signals. Expand each signal in terms of two given basis functions

1 1

1 1 2

1

-0.5

21


Components of s1(t)

This is a 2-Dsignal space. Therefore, each signal can be represented by a pair of numbers. Let’s find them

For s1(t) s11 s1(t)1 t

0

2

dt 1 1 0

1

dt 0 1

s12 s1(t)2 t 0

2

dt 0 0.5 1 0

1

dt 0.5t

t

1 2

1

1

-0.5

s1(t)

1

s=(1,-0.5)


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

1

-0.5

1

2


Signal constellation

Finding individual components of each signal along the two dimensions gets us the constellation

s4

s1

s2

s3

1

2

-0.5

-0.5 0.5


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


Finding the energy from the constellation

This is a simple matter. Remember,

Replace the signal by its expansionEi s

i

2(t)dt0

T

Ei sij j (t)j1

N

0

T

sikk (t)k 1

N

dt


Exploiting the orthogonality of basis functions

Expanding the summation, all cross product terms integrate to zero. What remains are N terms where j=k

Ei sij

2j

2 t j1

N

0

T

dt sij2

j

2 t dt0

T

j1

N

sij2

j

2 t dt0

T

1

j1

N

sij2

j1

N


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

3

2 E=9+4=13E


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 )

Ep


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

s1

s2 cos 12 s

1

Ts2

s1 s2

transpose


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?

s1T s2 1 2

21

2 2 4

s1 1 4 5

s2 4 1 5

cos 12 45 5

45

12 36.9o


Distance between two signals

The closer signals are together the more chances of detection error. Here is how we can find their separation

d12

2 s1 s22 s1 j s2 j 2

j1

N

(1)2 (1)2 2 d12 2

1 2

1

2


Constellation building using correlator banks

We can decompose the signal into its components as follows

s(t)

1

2

N

dt0

T

dt0

T

dt0

T

s1

s2

sN

N components


Detection in the constellation space

Received signal is put through the filter bank below and mapped to a point

s(t)

1

2

N

dt0

T

dt0

T

dt0

T

s1

s2

sN

componentsmapped to a single point


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


Actual example

Here is a 16-level constellation which is reconstructed in the presence of noise

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2

-1.5

-1

-0.5

0

0.5

1

1.5

2Eb/No=5 dB


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 signalwhich of the four did it come from


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


Defining decision regions

An easy detection method, is to compute “decision regions” offline. Here are a few examples

decide s1

decide s2

s1s2

measurement

decide s1decide s2

decide s3 decide s4

s1s2

s3 s4

decide s1

s1


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 XZi si was transmitted


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

departure from transmittedposition due to noise

X

si


Error probability

we can write an expression for error like thisP{error|si}=P{X does not lie in Zi|si was

transmitted} Generally

Pe P X Zi | si P{i1

M si}


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


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

A-Acos(2pifct)


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

s1 t 2Eb

Tb

cos 2fct

s2 t 2Eb

Tb

cos 2fct


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=11 t 2Tb

cos 2fct


Formulating BER

BPSK constellation looks like this

√Eb-√Eb

X|1=[√Eb+n,n]

transmitted

received

noise

nois

e

Pe1 P Eb n 0 |1 is transmitted

if noise is negative enough, it will pushX to the left of the boundary, deciding 0instead


Finding BER

Let’s rewrite BER

But n is gaussian with mean 0 and variance No/2

Pe1 P Eb n 0 |1 P n Eb

-sqrt(Eb)


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]


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

2Eb

Tb

cos(2fct) 2Eb

Tb

0.2


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 are probably several orders of magnitude bigger


Solving for BER

Noise PSD is No/2 =10-6. We know for BPSKBER=0.5erfc[(Eb/No)0.5]

What we have is then

Finish this using erf tables

BER 12

erfc Eb

No

12

erfc 2 10 7

2 10 6

12

erfc( 0.1) 12

erfc(0.316)


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?

si t 2Eb

Tb

cos 2fit , 0 t Tb

0


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.

f inc i

Tb

,i 1,2

nc some integer


Example tones

Let’s say we are sending data at the rate of 1 Mb/sec in BFSK, What are some typical tones?

Bit length is 10-6 sec. Therefore, possible tones are (use nc=0)

f1=1/Tb=1 MHzf2=2/Tb=2MHz


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

i t 2Tb

cos 2fit ,0 t Tb


BFKS constellation

There are two dimensions. Find the components of signals along each dimension using

s11 s1 t 0

Tb

1 t dt Eb

s12 s1 t 0

Tb

2 t dt 0

s1 ( Eb ,0)

Eb

Eb


Decision regions in BFSK

Decisions are made based on distances. Signals closer to s1 will be classified as s1 and vice versa

45 de

gree l

ine


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 thes2 partition s1

s2

X=received

x1

x2Pe1=P{x2>x1|s1 was sent}


Where do (x1,x2) come from?

Use the correlator bank to extract signal components

x=s1(t)+noise

1

2

dt0

Tb

dt0

Tb

x1(gaussian)

x2(gaussian)


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), rewriteP(x2>x1|x1)

x1 is now treated as constant. Then, integrate out x1 to eliminate it


BER for BFSK

Skipping the details of derivation, we get

Pe BER 12

erfc Eb

2No


BPSK and BFSK comparison:energy efficiency

Let’s compare their BER’s

Pe 12

erfc Eb

2No

,BFSK

Pe 12

erfc Eb

No

, BPSK

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


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

Eb Eb

2 Eb

Eb

Eb 1.4 Eb


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


Differential encoding example

Data to be encoded1 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 0 0 1 0 0 1 11 1 0 1 1 0 1 1 1


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

Encoded received bits

unknown transmittedbits


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)


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

1 0 0 1 0 0 1 1Detected data


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

1 0 1 0 0 0 1 11 0 0 1 0 0 1 1

Back on track:no errors

Transmitted bits

Incoming phases

Detected bits

2 errors


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


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


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


8-level PAM

Here is an example of 8-level signaling

0 1 0 1 0 0 0 0 0 0 0 1 1 1 0 1 0 0 1 1 1binary

75321

-1-3-5-7


A few definitions

We used to work with bit length Tb. Now we have a new parameter which we call symbol length,T

1 10

T

Tb


Bit length-symbol length relationship

When we combine n bits into one symbol; the following relationships hold

T=nTb- symbol lengthn=logM bits/symbol

T=TbxlogM- symbol length All logarithms are base 2


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


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


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.


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 Hz


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


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


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


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


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


Bandwidth efficiency numbers

Here are some numbersn(bits/symbol) M(levels) (bits/sec/Hz)

1 2 12 4 23 8 34 16 48 256 8


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


QPSKquadrature 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

si(t) 2ET

cos 2fct (2i 1)4

,i 1,2,3, 4

0

, 0 t T

Symbol energy

Symbol width


QPSK constellation

45o

0001

11 10

√E

1 t 2T

cos2fct

2 t 2T

sin2fctBasis functions S=[0.7 √E,- 0.7 √E]


QPSK decision regions

0001

11 10

Decision regions re color-coded


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!

Pe erfc(E

2No

)


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

0011s1s2


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

10 10 11 1000

11 10 11 1000

10 symbols = 20 bits

Sym.error=1/10Bit error=1/20


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

Pe

log MBER Pe


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

Pe erfc(E

2No

)

BER 12

erfc(E

2No

)


QPSK vs. BPSK

Let’s compare the two based on BER and bandwidth

BER BandwidthBPSK QPSK BPSK QPSK

12

erfcEb

No

1

2erfc

Eb

No

Rb Rb/2

EQUAL


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

si(t) 2ET

cos 2fct (i 1)4

, i 1,...,8

0

,0 t T


8-PSK constellation

Distribute 8 phasors uniformly around a circle of radius √E

45o

Decision region


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

Pe erfcENo

sinM

,M 4


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


16-QAM constellation using Gray coding

16-QAM has the following constellation Note gray codingwhere adjacent symbolsdiffer by only 1 bit

0010001100010000

1010

1110

0110

1011

1111

0111

1001

1101

0101

1000

1100

0100


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

[ai ,bi ]

3,3 1,3 1,3 3,3 3,1 1,1 1,1 3,1 3, 1 1, 1 1, 1 3, 1 3, 3 1, 3 1, 3 3, 3


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

Eavg 1M

ai2 bi

2 i1

M


Eavg for 16-QAM

Using the [ai,bi] matrix and using E=ai^2+bi^2 we get one energy per signal

E

18 10 10 1810 2 2 1010 2 2 1018 10 10 18

Eavg=10


Symbol error for M-ary QAM

With the definition of energy in mind, symbol error is approximated by

Pe 2 1 1M

erfc

2Eavg

2 M 1 No


Familiar constellations

Here are a few golden oldies

V.22 600 baud1200 bps

V.22 bis600 baud2400 bps

V.32 bis2400 baud9600 bps


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

si t 2ET

cos 2fit ,0 t T

i 1,...,M


MFSK constellation:3-dimensions

MFSK is different from MPSK in that each signal sits on an orthogonal axis(basis)

s1

s2

s3

1

2

3i t 2T

cos 2fit ,

0 t Ti 1,...,M

s1=[√E ,0, 0]s2=[0,√E, 0]s3=[0,0,√E]

√E

√E

√E


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


How to pick the tones?

Orthogonal FSK requires tones that are orthogonal.

Two carrier frequencies separated by integer multiples of period are orthogonal


Example

Take two tones one at f1 the other at f2. T must cover one or more periods for the integral to be zero

2cos 2f1t cos 2f2t dt cos2 f1 f2 dt0

T

averages to zero

0

T

cos2 f1 f2 dt0

T

averages to zero if T=i/(f1-f2); i=integer

Take f1=1000 and T=1/1000. Thenif f2=2000 , the two are orthogonal so will f2=3000,4000 etc


MFSK symbol error

Here is the error expression with the usual notations

Pe 12

M 1 erfcE

2No


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


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


Illustrating BPSK bandwidth

The expression for baseband BPSK (polar) bandwidth is

SB(f)=2Ebsinc2(Tbf)

BT=2Rbf1/Tb

BPSK

fc+/Tbfc-/Tb fc

2/Tb=2Rb


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

0 1000 2000 3000 4000 5000 6000 7000 8000-1

-0.5

0

0.5

1

0 1000 2000 3000 4000 5000 6000 7000 8000-1

-0.5

0

0.5

1

0 1000 2000 3000 4000 5000 6000 7000 8000-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1BFSK for 1 0 0 1 0 1 1

+


Modeling of BFSK bandwidth

Each stream is just a carrier modulated unipolar signal. Each has a sinc spectrum

f1 f2

1/Tb=Rb

fc

fc=(f1+f2)/2

f

BT=2 f+2Rb

f= (f2-f1)/2


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?

UseBT=2f+2Rb

f=(f2-f1)/2=(2200-1200)/2=500 HzBT=2x500+2x1200=3400 Hz

This is more than BPSK of 2Rb=2400 Hz


Sunde’s FSK

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

2Tb

cos(2f1t)0

Tb

cos(2f2t)dt

1 2 3 2(f2-f1)Tb

Good points,zero correlation


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/2remember f is (f2-f1)/2 Sunde’s FSK bandwidth is then given by

BT=2f+2Rb=Rb+2Rb=3Rb The practical bandwidth is a lot smaller


Sunde’s FSK bandwidth

Due to sidelobe cancellation, practical bandwidth is just BT=2f=Rb

f1 f2

1/Tb=Rb

fc

fc=(f1+f2)/2

f

BT=2 f+2Rb

f= (f2-f1)/2

f


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


Point to remember

FSK is not a particularly bandwidth-friendly modulation. In this example, to transmit 1 Mbps, we needed 3.5 MHz.

Of course, it is working at the 3rd zero crossing that is responsible

Original Sunde’s FSK requires BT=Rb=1 MHz

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


MPSK bandwidth

MPSK spectrum is given by

SB(f)=(2Eblog2M)sinc2(Tbflog2M)

f/Rb

Notice normalized frequency

1/logM

Set to 1 for zero crossing BWTbflog2M=1-->f=1/ Tbflog2M=Rb/log2M

BT= Rb/log2M


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


QPSK bandwidth

QPSK is a special case of MPSK with M=4 phases. It’s baseband spectrum is given by

SB(f)=2Esinc2(2Tbf)

f/Rb0.5

B=0.5Rb-->half of BPSK

1

After modulation:Bqpsk=Rb


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


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)bin4 0.5 0.34 dB8 1/3 3.91 dB16 1/4 8.52 dB32 1/5 13.52 dB Lesson: By going to multiphase modulation, we save

bandwidth but have to pay in increased power, But why?


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


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


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


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


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


for MPSK

At a bit rate of Rb, BPSK bandwidth is 2Rb When we go to MPSK, bandwidth goes down

by a factor of log2MBT=2Rb/ log2M

Then=Rb/BT= log2M/2 bits/sec/Hz


Some numbers

Let’s evaluate vs. M for MPSKM 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


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 The textbook we choose the former (this corresponds to using the first zero crossing of correlation curve)


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


Contrast with MPSK

Variation of bandwidth with M differs drastically compared to MPSK

MPSK MFSKBT=2Rb/log2M BT=MRb/2log2M

As M goes up, MFSK eats up more bandwidth but MPSK save bandwidth


MFSK bandwidth efficiency

Let’s compute ’s for MFSK=Rb/M=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

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

M 2m # of symbolsm = log2 M bits/symbol

R = mTs

log2 M

Ts

bits/ sec

Ts symbol durationRs symbol rate

Tb 1R

Ts

m

1mRs

bit length

Bandwidth efficiency measure

RW

log2 MWTs

1

WTb


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


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

RW

log2 MWTs

log2 M bits/sec/Hz


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.


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


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

RW

log2 MWTs

log2 M

M bits/sec/Hz


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


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


What is the required Eb/No?

SNo

Eb RNo

Eb

No

R

Eb

No

(dB) SNo

(dB Hz) R(dB bits / sec

13.2dB


Is BER met? Yes

The symbol error probability in 8-PSK is

Solve for Es/No

Solve for PE

PE M 2Q 2Es

No

sin M

BER PE

log2 M

2.2 10 5

37.3 10 6

Es

No

log2 M Eb

N0

3 20.89 62.67


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


Binary vs. M-ary Model

M-ary ModulatorR bits/s

Rs R

log2 Msymbols / s

M-ary demodulator

SNo

Eb

No

R Es

No

Rs


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

Eb

No

(dB) SNo

dB Hz R(dB bit / s)

Eb

No

(dB) 48dB Hz

(10 log 9600)dB bits / s8.2dB


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.


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.


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


Symbol error for MFSK

For noncoherent orthogonal MFSK, symbol error probability is

PE M M 12

exp Es

2No

Es Eb log2 M


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

Es

No

log2 M Eb

No

PB 2m 1

2m 1PE


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

PB 23

24 1PE

815

PE


Summary

Given:• 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

geo rep of signal

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