Digital Modulation 1

7/29/2019 Digital Modulation 1

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DIGITAL MODULATIONS


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Why digital modulation?

If our goal was to design a digital

baseband communication system, we have

done that Problem is baseband communication wont

takes us far, literally and figuratively

Digital modulation to a square pulse is

what analog modulation was to messages


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A block diagram

Messsage

source

Source

coderLine coder

Pulse

shaping

demodulatordetector

channel

modulator

decision

1011


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GEOMETRIC

REPRESENTATION OF

SIGNALS


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The idea

We are used to seeing signals expressed

either in time or frequency domain

There is another representation space thatportrays signals in more intuitive format

In this section we develop the idea of

signals as multidimensional vectors


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Have we seen this before?

Why yes! Remember the beloved ej2fct

which can be written as

ej2fct=cos(2fct)+jsin(2fct)

inphase

quadrature


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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 ofbasisfunctions

si t sijj t j 1

N

i 1,...,M

0 t T


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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 j

1 i j

i

j

k


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Components of the signal

vector

Each signal needs Nnumbers to be

represented by a vector. These Nnumbers

are given by projecting each signal ontothe individual basis functions:

sij means projection of si (t)on j(t)sij si (t)j t

0

T

dt sij

si

j


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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 Nand is

always less than the number of signals inthe set

N


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Example: Fourier series

Remember Fouirer series? A signal was

expanded as a linear sum of sines and

cosines of different frequencies. Soundsfamiliar?

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


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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


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Components of s1(t)

This is a 2-Dsignal space. Therefore, each

signal can be represented by a pair of

numbers. Lets 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)


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Interpretation

s1(t) is now condensed into just two

numbers. We can reconstruct s1(t) like

thiss1(t)=(1)1(t)+(-0.5)2(t)

Another way of looking at it is this

1

-0.5

1

2


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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


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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


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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 sijj (t)j 1

N

0

T

sikk(t)

k1

N

dt


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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 s ij2

j

2t

j 1

N

0

T

dt sij2j2 t dt0

T

j1

N

sij2

j

2 t dt0

T

1

j 1

N

sij2j 1

N


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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

2E=9+4=13

E


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Constrained energy signals

Lets say you are under peak energy Epconstraint in your application. Just make

sure all your signals are inside a circle ofradius sqrt(Ep )

Ep


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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 betweenthem

s1

s2 cos 12 s

1

Ts2

s1 s2

transpose


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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?

s1Ts2 1 2

2

1

2 2 4

s1 1 4 5

s2 4 1 5

cos 12 4

5 5

4

5

12 36.9o


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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

s1j s2 j 2

j 1

N

(1)2 (1)2 2 d12 2

1 2

1

2


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Constellation building using

correlator banks

We can decompose the signal into its

components as follows

s(t)

1

2

N

dt0

T

dt0

dt0

s1

s2

sN

N components


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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

dt0

T

dt0

T

s1

s2

sN

components

mapped to a single point


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Constellation recovery in

noise

Assume signal is contaminated with noise.

All Ncomponents will also be affected.

The original position of si(t) will bedisturbed


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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


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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 itcome from


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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 makingthis is the most likely

transmitted signal

received


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Defining decision regions

An easy detection method, is to compute

decision regions offline. Here are a fewexamples

decide s1

decide s2

s1s2

measurement

decide s1decide s2

decide s3 decide s4

s1s2

s3 s4

decide s1

s1


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More formally...

Partition the decision space into M

decision regions Zi, i=1,,M. Let X be themeasurement vector extracted from the

received signal. Then

if XZisi was transmitted


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How does detection error

occur?

Detection error occurs when X lands in Zi

but it wasnt si that was transmitted.Noise, among others, may be the culprit

departure from transmitted

position due to noise

X

si


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Error probability

we can write an expression for error like

this

P{error|si}=P{X does not lie in Zi|si wastransmitted}

Generally

Pe P XZi | si P{i 1M

si}


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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(2fct)

0:s2= - Acos(2fct)

1s and 0s are identified by 180 degree

phase reversal at bit transitions


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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-A

cos(2pifct)


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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 A2T

b/2 .

Therefore

Eb= A2Tb/2 --->A=sqrt(2Eb/Tb)

We can write the two bits as follows

s1 t 2Eb

Tbcos 2fct

s2 t 2Eb

Tb

cos 2fct


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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

1 t 2

Tbcos 2fct


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Formulating BER

BPSK constellation looks like this

Eb-Eb

X|1=[Eb+n,n]

transmitted

received

noise

Pe1 P Eb n 0 |1 is transmitted

if noise is negative enough, it will push

X to the left of the boundary, deciding 0

instead


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Finding BER

Lets rewrite BER

But n is gaussian with mean 0 and

variance No/2

Pe1 P Eb n 0 |1 P n Eb

-sqrt(Eb)


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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]


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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. 1s and 0sare sent by

2EbTb

cos(2fct) 2Eb

Tb 0.2


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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 receivedenergy.

What was transmitted are probably several

orders of magnitude bigger


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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 erftables

BER 1

2erfc

EbNo

1

2erfc

2 107

2 106

12

erfc( 0.1) 12

erfc(0.316)


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Binary FSK

(Frequency Shift Keying)

Another method to transmit 1s and 0s isto 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

Tbcos 2fit , 0 t Tb

0


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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 integermultiple of inverse duration, i.e.

finc i

Tb

,i 1,2

nc some integer


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Example tones

Lets say we are sending data at the rateof 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 MHz

f2=2/Tb=2MHz


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BFSK dimensionality

What does the constellation of BFSK look

like? We first have to find its dimension

s1 and s2 can be represented by twoorthonormal basis functions:

Notice f1 and f2 are selected to make themorthogonal

i t 2

Tbcos 2fit ,0 t Tb


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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


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Decision regions in BFSK

Decisions are made based on distances.

Signals closer to s1 will be classified as s1

and vice versa


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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 s1

s2

X=received

x1

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


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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)


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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


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BER for BFSK

Skipping the details of derivation, we get

Pe BER 12erfc Eb

2No

BPSK d BFSK i


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BPSK and BFSK comparison:

energy efficiency

Lets compare theirBERs

Pe 1

2erfc

Eb2No

,BFSK

Pe 12erfc 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

C i i th


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Comparison in the

constellation space

Distances determine BERs. Lets compare

Both have the same Eb, but BPSKs arefarther apart, hence lower BER

Eb Eb

2 Eb

Eb

Eb 1.4 Eb


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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


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Differential encoding example

Data to be encoded

1 0 0 1 0 0 1 1

Set the reference bit to 1, then use thefollowing 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


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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 transmitted

bits


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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 0 0 0 0 0Differentially encoded data

Phase encoded(BPSK)


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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 0 0 0 0 0

1 0 0 1 0 0 1 1Detected data


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Bit errors in DPSK

Bit errors happen in an interesting way

Since detection is done by comparing

adjacent bits, errors have the potential ofpropagating

Allow a single detection errorin DPSK

0 0 0 0 0 0

1 0 1 0 0 0 1 1

1 0 0 1 0 0 1 1

Back on track:no errors

Transmitted bits

Incoming phases

Detected bits

2 errors


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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


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Why DPSK?

Detecting regular BPSK needs a coherent

detector, requiring a phase reference

DPSK needs no such thing. The onlyreference is the previous bit which is

readily available


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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 to4 symbols; a 4-ary signaling


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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

7

5

3

2

1

-1

-3

-5

-7


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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


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Bit length-symbol length

relationship

When we combine nbits into one symbol;

the following relationships hold

T=nTb- symbol length

n=logM bits/symbol

T=TbxlogM- symbol length

All logarithms are base 2


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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


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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=R

b/logM

For 8-level signaling, baud rate is 1/3 of bit

rate


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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


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Issues in transmitting 9600

bits/sec

Want to transmit 9600 bits/sec. Options:

Nyquists 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


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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


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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


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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


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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


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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



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numbers

Here are some numbers

n(bits/symbol) M(levels) (bits/sec/Hz)

1 2 12 4 2

3 8 3

4 16 4

8 256 8


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Symbol energy vs. bit energy

Each symbol is made up ofnbits. It is not

therefore surprising for a symbol to have n

times the energy of a bit

E(symbol)=nEb

Eb

E

QPSK


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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

si(t) 2ET

cos 2fct (2i 1)4 ,i 1,2,3,4

0

, 0 t T

Symbol energy

Symbol width


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QPSK constellation

45o

0001

11 10

E

1 t 2

Tcos2fct

2 t 2

Tsin 2fct

Basis functions S=[0.7 E,- 0.7 E]


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QPSK decision regions

0001

11 10

Decision regions re color-coded


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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)

S


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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

S b l bit


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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

I t ti b l


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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

logM

BER Pe

Symbol error and bit error for


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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 12erfc( E

2No)

QPSK BPSK


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QPSK vs. BPSK

Lets compare the two based on BER andbandwidth

BER Bandwidth

BPSK QPSK BPSK QPSK

1

2

erfcEb

No

1

2erfc

Eb

No

Rb Rb/2

EQUAL

M h PSK (MPSK)


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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) 2E

Tcos 2fct (i 1)

4

,i 1,...,8

0

,0 t T

8 PSK t ll ti


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8-PSK constellation

Distribute 8 phasors uniformly around a

circle of radius E

45o

Decision region

S b l f MPSK


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Symbol error for MPSK

We can have M phases around the circle

separated by 2/M radians.

It can be shown that symbol error

probability is approximately given by

Pe erfc ENosin

M ,M 4

Quadrature Amplitude


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Q p

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


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Q g

Gray coding

16-QAM has the following constellation

Note gray coding

where adjacent symbolsdiffer by only 1 bit

0010001100010000

1010

1110

0110

1011

1111

0111

1001

1101

0101

1000

1100

0100

Vector representation


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p

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


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gy p y

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 bi2 i 1

M

E for 16 QAM


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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 10

10 2 2 10

18 10 10 18

Eavg=10

Symbol error for M ary QAM


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Symbol error for M-ary QAM

With the definition of energy in mind,

symbol error is approximated by

Pe 2 11

M

erfc

2Eavg

2 M1 No

Familiar constellations


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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

M ary FSK


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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 2E

T cos 2fit ,0 t T

i 1,...,M

MFSK constellation:


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3-dimensions

MFSK is different from MPSK in that each

signal sits on an orthogonal axis(basis)

s1

s2

s3

1

2

3

i t 2

Tcos 2fit ,

0 t T

i 1,...,M

s1=[E ,0, 0]s2=[0,E, 0]s3=[0,0,E]

E

E

E

Orthogonal signals:

How many dimensions how many


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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 dontneed 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?


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101

How to pick the tones?

Orthogonal FSK requires tones that are

orthogonal.

Two carrier frequencies separated by

integer multiples of period are orthogonal

Example


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102

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. Then

if f2=2000 , the two are orthogonal

so will f2=3000,4000 etc

MFSK symbol error


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103

MFSK symbol error

Here is the error expression with the usual

notations

Pe 1

2M1 erfc

E

2No

Spectrum of M-ary signals


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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

Lets first look at binary modulationbandwidth

BPSK bandwidth


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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 doesnt take much to realize that carriermodulation doubles this bandwidth

Illustrating BPSK bandwidth


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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


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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

1

BFSK for 1 0 0 1 0 1 1

+

Modeling of BFSK bandwidth


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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


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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=2f+2Rb

f=(f2-f1)/2=(2200-1200)/2=500 Hz

BT=2x500+2x1200=3400 Hz

This is more than BPSK of 2Rb=2400 Hz

Sundes FSK


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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

2

Tb

cos(2f1t)

0

Tb

cos(2f2t)dt

1 2 3 2(f2-f1)Tb

Good points,zero correlation

Picking the 2nd zero crossing:


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Sundes 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

rememberf is (f2-f1)/2

Sundes FSK bandwidth is then given by

BT=2f+2Rb=Rb+2Rb=3Rb

The practical bandwidth is a lot smaller

Sundes FSK bandwidth


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112

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

B FSK example


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113

B FSK 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


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114

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 zerocrossing that is responsible

Original Sundes FSK requires BT=Rb=1 MHz


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Bandwidth of MPSK

modulation

MPSK bandwidth review


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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


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MPSK bandwidth

MPSK spectrum is given by

SB

(f)=(2Eb

log2

M)sinc2(Tb

flog2

M)

f/Rb

Notice normalized frequency

1/logM

Set to 1 for zero crossing BW

Tbflog2M=1

-->f=1/ Tbflog2M

=Rb/log2M

BT= Rb/log2M

Bandwidth after carrier

d l ti


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118

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


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QPSK bandwidth

QPSK is a special case of MPSK with M=4

phases. Its 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


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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


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MPSK vs.BPSK

Lets say we fix BER at some level. How dobandwidth and power levels compare?

M Bm-ary/Bbinary (Avg.power)M/(Avg.power)bin

4 0.5 0.34 dB8 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?

Power-bandwidth tradeoff


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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


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p

Take a 28.8 Kb/sec data rate and letscompare the required bandwidths

BPSK: BT=2(Rb)=57.6 KHz

BFSK: BT = Rb=28.8 KHz ...Sundes 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


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y

Modulations that are power-limited achieve

their goals with minimum expenditure of

power at the expense of bandwidth.

Examples are MFSK and other orthogonalsignaling

Bandwidth-limited systems


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y

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


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y

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


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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

Some numbers


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Lets 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

Defining MFSK:


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g

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 1/2Tapart. 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


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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


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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

MFSK bandwidth efficiency


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y

Lets compute s for MFSK

=Rb/M=2log2M/M bits/sec/HzMFSK

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


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COMPARISON OF DIGITALMODULATIONS*

*B. Sklar, Defining, Designing and Evaluating Digital Communication Systems,

IEEE Communication Magazine, vol. 31, no.11, November 1993, pp. 92-101

Notations


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M 2m # of symbols

m = log2 Mbits/symbol

R = mTs

log2 MTs

bits/sec

Ts symbol duration

Rs symbol rate

Tb 1

RTsm

1

mRsbit length


measure

R

W

log2 M

WTs

1

WTb

Bandwidth-limited Systems


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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


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In MPSK, symbols are m times as wide as

binary.

Nyquist bandwidth is W=Rs/2=1/2Ts.

However, the bandpass bandwidth is twicethat, W=1/Ts

Then

R

W

log2 M

WTs log2 Mbits/sec/Hz

Cost of Bandwidth Efficiency


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137

As M increases, modulation becomes more

bandwidth efficient.

Lets fix BER. To maintain this BER while

increasing M requires an increase in Eb/No.

Power-Limited Systems


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138

There are cases that bandwidth is

available but power is limited

In these cases as M goes up, the

bandwidth increasesbut required powerlevels to meet a specified BER remains

stable

Case of MFSK


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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

R

W log2 M

WTs log2 M

Mbits/sec/Hz

Select an Appropriate

Modulation


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140

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


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141

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 resultsin 3200 symbols/sec needing 3200Hz

So, M=8

What is the required Eb/No?


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S

No

EbR

No

Eb

No

R

EbNo

(dB) S

No(dB Hz) R(dB bits /sec

13.2dB

Is BER met? Yes


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The symbol error probability in 8-PSK is

Solve for Es/No

Solve for PE

PE M 2Q2Es

Nosin

M

BER PE

log2 M

2.2 105

3 7.3 106

EsNo

log2 M EbN0

3 20.89 62.67

Power-limited uncoded

system


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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


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M-ary ModulatorR bits/s

Rs

R

log2

Msymbols / s

M-ary demodulator

S

No

E

b

No

R E

s

No

Rs

Choice of Modulation


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With R=9600 bits/sec and W=45 KHz, the

channel is not bandwidth limited

Lets find the available Eb/No

Eb

No

(dB) S

No

dB Hz R(dB bit/ s)

Eb

No(dB) 48dB Hz

(10log9600)dB bits / s

8.2dB

Choose MFSK


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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


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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


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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


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For noncoherent orthogonal MFSK, symbol

error probability is

PE

M M1

2exp

Es

2No

Es

Eblog

2M

BER for MFSK


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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 EbN

o

PB

2m1

2m 1P

E

Example


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Lets look at the 16FSK case. With 16levels, 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

2

3

24 1

PE

8

15P

E

Summary


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Given:

R=9600 bits/s

BER=10^-5

Channel bandwith=45KHz

Eb/No=8.2dB

Solution

16-FSK

required bw=38.4khz

required Eb/No=8.1dB

Documents

Digital Modulation 1