Lect 05 - Digital modulation.pdf

8/11/2019 Lect 05 - Digital modulation.pdf

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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 havedone that

2

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

Messsagesource

Sourcecoder

Line coder Pulseshaping

modulator

1011

3

demodulatordetector

channel

decision


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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

5

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 asej2fct=cos(2fct)+jsin(2fct)

6

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 berepresented by a linear sum of basisfunctions

7

si t( ) = sijj t( )

j=1

N

i = 1,...,

0 t T


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Conditions on basis functions

For the expansion to hold, basis functions

must be orthogonal to each other Mathematically:

i t( ) t( )dt=0 i j

8

Geometrically:

=

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 si nal onto the individual basis

9

functions:

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

( )dtttssT

jiij =0

)(

sij

si

j


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Signal space dimension

How many basis functions does it take to

express a signal? It depends on thedimensionality of the signal

10

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 Fourier series? A signal was

expanded as a linear sum of sins andcosines of different frequencies. Soundsfamiliar?

11

Sins and cosines are the basis functionsand are in fact orthogonal to each other

cos 2nfot( )To cos 2mf

ot( )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 interms of two given basis functions

12

1 1

1 1 2

1

-0.5

2

1


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Components of s1(t)

This is a 2-D signal space. Therefore,

each signal can be represented by apair of numbers. Lets find them

For s t s1(t)

13

s11 = s1(t)1t( )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= 0.5

t

t1 2

1

1

-0.5

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 this

14

1 = 1 - . 2 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 usthe constellations4

2

15

s1

s2

s3

1-0.5

-0.5 0.5


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Finding the energy from the constellation

This is a simple matter. Remember,

Ei = s i2(t)dt

0

T

17

Ei = sijj (t)j=1

N

0

T

sikk(t)k=1

N

dt


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Exploiting the orthogonality

Expanding the summation, all cross

product terms integrate to zero. Whatremains are Nterms where j=kN

T TN

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Ei = s ij2j2 t( )j=1 0 dt= s

ij2j2 t( )dt0

j=1

sij2 j2 t( )dt0

T

=1

1 24 34j=1

N

= sij2

j=1

N


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Energy in simple language

The energy of a signal is simply the

square of the length of its correspondingconstellation vector

19

3

2

E=9+4=13

E


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Constrained energy signals

Lets say you are under peak energy Ep

constraint. Just make sure all yoursignals are inside a circle of radiussqrt(E )

20

Ep


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Find the angle between s1 and s2

Given that s1=(1,2)Tand s2=(2,1)T, what

is the angle between the two?

T 2

22

1 2

1 s1 = 1+4 = 5

s2 = 4 +1 = 5

cos12( )= 5 5 = 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 howwe can find their separation

23

d12

2 = s1 s22 = s1j s2j( )2

j=1

N

=(1)

2+ (1)

2=2

d12 = 21 2

1

2


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Correlator banks

We can decompose the signal into its

components as followsdt

0

T

s1

24

s(t)

1

2

N

dt0

T

dt0

T

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

T

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s(t)

1

2

N

0

dt0

T

dt0

T

s1

s2

sN

components

mapped to a single point


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Actual example

Here is a 16-level constellation which is

reconstructed in the presence of noise

2

Eb/No=5 dB

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-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


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Detection in signal space

One of the M signals is transmitted,

processed through the correlator banksand mapped onto constellation, what wasthe transmitted signal?

28

received signal

which of the four did it

come from


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Minimum distance decision rule

It can be shown that the optimum decision, in thesense of lowest BER, is to pick the signal that isclosest to the received vector. This is calledmaximum likelihood decision making

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this is the most likely

transmitted signal

received


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Defining decision regions

An easy detection method, is to compute decisionregions offline. Here are a few examples

decide s2

30

decide s1

decide s2

s1s2

measurement

ec e s

decide s3 decide s4

s1s2

s3 s4

decide s1

s1


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More formally...

Partition the decision space into M decision regionsZi, i=1,,M. Let X be the measurement vector

extracted from the received signal. Then

if XZ s was transmitted

31


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How does detection error occur?

Detection error occurs when X lands in Zi but itwasnt si that was transmitted. Noise, amongothers, may be the culprit

X

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departure from transmitted

position due to noise

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 was transmitted}

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enera y

Pe = P XZi|si{ }P{i=1

si}


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Example: BPSK

BPSK is a well known digital modulation obtained bycarrier modulating a polar NRZ signal. The rule is

1: s1=Acos(2fct)

0: s2= - Acos(2fct)

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1s and 0s are identified by 180 degree phasereversal at bit transitions


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Signal space for BPSK

Look at s1 and s2. What is the basis function forthem? Both signals can be uniquely written as ascalar multiple of a cosine. So a single cosine is thesole basis function. We have a 1-D constellation

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A-A

cos(2pifct)


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Bringing in Eb

We want each bit to have an energy Eb. Bits inBPSK are RF pulses of amplitude A and duration Tb.Their energy is A2Tb/2 . Therefore

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

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e can wr e e wo s as o ows

s1t( ) = 2Eb

Tbcos 2fct( )

s2 t( ) = 2Eb

Tbcos 2fct( )


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BPSK basis function

As a 1-D signal, there is one basis function. Wealso know that basis functions must have unitenergy. Using a normalization factor

=

37

1t( ) = 2

Tbcos 2fct( )


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Formulating BER

BPSK constellation looks like this

received

if noise is negative enough, it will push

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Eb-Eb

b ,

transmitted

noise

Pe1 =P Eb + n


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Finding BER

Lets rewrite BER

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

Pe1 =P Eb + n


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BER for BPSK

Using the trick to find the area under a Gaussiandensity (after normalization with respect to

variance)]

bE2

40

=

=

o

b

o

N

Eerfc

N

2

1

BP E l


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BPSK Example

Data is transmitted at Rb=106 b/s. Noise PSD is10-6 and pulses are rectangular with amplitude 0.2

volt. What is the BER?

First we need energy per bit, Eb. 1s and 0s are

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2Eb

Tb

cos(2fct) 2Eb

Tb

=0.2

S l i f E


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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 receivedener . What was

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transmitted are probably several orders ofmagnitude bigger

S l i f BER


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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]

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a we ave s en

Finish this using erftables

BER =1

2erfc

Eb

No

=

1

2erfc

2107

2106

=12

erfc( 0.1) =12

erfc(0.316)


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Pi ki th i ht t


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Picking the right tones

It is desirable to keep the tones orthogonal

Since tones are sinusoids, it is sufficient for thetones to be separated by an integer multiple ofinverse duration, i.e.

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fi=n

cTb

,i =1,2

nc = some integer

Ex m l t s


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Example tones

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

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

46

1=

b= z

f2=2/Tb=2MHz

BFSK dimensionality


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BFSK dimensionality

What does the constellation of BFSK look like? Wefirst have to find its dimension

s1 and s2 can be represented by two orthonormalbasis functions:

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Notice f1 and f2 are selected to make them

orthogonal

i t( ) = Tb cos 2fit( ),0 tTb

BFKS constellation


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BFKS constellation

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

s11 = s1 t( )Tb

1t( )dt= Eb

Eb

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0

s12 = s1 t( )0

Tb

2 t( )dt= 0

s1 = ( Eb ,0)

Eb

Decision regions in BFSK


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Decision regions in BFSK

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

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Detection error in BFSK


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Detection error in BFSK

Let the received signal land where shown.

Assume s1 is sent. How would a detection erroroccur?

P =P x >x |s was sent

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x2>x1 puts X in thes2 partition

s1

s2

X=received

x1

x2


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Finding BER


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Finding BER

We have to answer this question: what is theprobability of one random variable exceeding

another random variable? To cast P(x2>x1) into like of P(x>2), rewrite

52

x1 is now treated as constant. Then, integrate outx1 to eliminate it

BER for BFSK


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BER for BFSK

Skipping the details of derivation, we get

=

o

b

N

EQBER

53

=

o

b

NEerfc22

1

BPSK and BFSK comparison


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BPSK and BFSK comparison

Lets compare their BERs

BPSKN

EQP

o

be ,

2

=

What does it take to have thesame BER?

Eb in BFSK must be twice asbig as BPSK

54

BFSKN

EQP

o

be ,

=

be twice as large in BFSK toachieve the same BER

BPSK has more energyefficiency than BFSK

Comparison in the constellation space


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Comparison in the constellation space

Distances determine BERs. Lets compare

Eb 1.4 Eb

55

Both have the same Eb, but BPSKs are fartherapart, hence lower BER

Eb Eb

b

Eb

Differential PSK


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Differential PSK

Concept of differential encoding is verypowerful

Take the the bit sequence 11001001

56

that we start we a reference bit and thenrecord changes

Differential encoding example


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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 the followingrule

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Generate a 0 if change1 0 0 1 0 0 1 1

1 1 0 1 1 0 1 1 1

Detection logic


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Detection logic

Detecting a differentially encoded signal is based onthe comparison of two adjacent bits

If two coded bits are the same, that means databit must have been a 1, otherwise 0

58

1 1 0 1 1 0 1 1 1 Encodedreceived bits

transmitted bits


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DPSK detection


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Data is detected by a phase comparison of twoadjacent pulses

No phase change: data bit is 1

Phase change: data bit is 0

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1 0 0 1 0 0 1 1Detected data

Bit errors in DPSK


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Bit errors happen in an interesting way

Since detection is done by comparing adjacentbits, errors have the potential of propagating

Allow a single detection error in DPSK

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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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M-ary signaling


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y g g

Binary communications sends one of only 2 levels; 0or 1

There is another way: combine several bits intosymbols

1 0 1 1 0 1 1 0 1 1 1 0 0 1 1

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Combining two bits at a time gives rise to 4symbols; a 4-ary signaling

8-level PAM


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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 1

binary

64

7

53

2

1

-1-3

-5

-7


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Bit length-symbol length relationship


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g y g p

When we combine n bits into one symbol; thefollowing relationships hold

Ts=nTb- symbol length

n=logM bits/symbol

66

Ts=TbxlogM- symbol length All logarithms are base 2


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Defining baud


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When we combine n bits into one symbol, numericaldata 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

68

Rs=symbol rate=Rb/n=Rb/logM For 8-level signaling, baud rate is 1/3 of bit rate


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Issues in transmitting 9600 bits/sec


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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

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Go to a 16 - level signaling, M=16. Pulse rate isreduced to

Rs=Rb/logM=9600/4=2400 Hz


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Bandwidth efficiency


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Bandwidth efficiency is defined as the number ofbits that can be transmitted within 1 Hz of

bandwidth=Rb/BT bits/sec/Hz

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In binary communication using sincs, BT=Rb/2 =2bits/sec/Hz

M-ary bandwidth efficiency


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In M-ary signaling , pulse rate is given byRs=Rb/logM. Full roll-off raised cosine bandwidth is

BT=Rs= Rb/logM. Bandwidth efficiency is then given by

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= b T= og s sec z

For M=2, binary we have 1 bit/sec/Hz. For M=16,we have 4 bits/sec/Hz

M-ary bandwidth


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Summarizing, M-ary and binary bandwidth arerelated by

BM-ary=Bbinary/logM

Clearly , M-ary bandwidth is reduced by a factor

74

o og compare o e nary an w

8-ary bandwidth


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Let the bit rate be 9600 bits/sec. Binarybandwidth is nominally equal to the bit rate, 9600

Hz We then go to 8-level modulation (3 bits/symbol)

-

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BM-ary=Bbinary/logM=9600/log8=3200 Hz

Bandwidth efficiency numbers


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Here are some numbers

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

1 2 12 4 2

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4 16 48 256 8

Symbol energy vs. bit energy


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Each symbol is made up of n bits. It is nottherefore surprising for a symbol to have n times

the energy of a bitEs=nEb

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Eb

E

QPSK


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This is a 4 level modulation.

Every two bits is combined and mapped to one of 4phases of an RF signal

These phases are 45o,135o,225o,315o

78

Ttiitf

T

E

ts cs

i

=

+

= 0,

0

4,3,2,1,4

)12(2cos2

)(

Symbol energy

Symbol width

QPSK constellation


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45o

0001

Es

79

11 10

1t( ) = 2

Tcos2fct

2 t( ) = 2T

sin2fct

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

QPSK decision regions


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0001

80

11 10

Decision regions re color-coded

QPSK error rate


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Symbol error rate for QPSK is given by

=

o

se

NEQP

81

This brings up the distinction between symbol errorand bit error. They are not the same!

Symbol error


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Symbol error occurs when received vector isassigned to the wrong partition in the constellation

0011

s1s2

82

When s1 is mistaken for s2, 00 is mistaken for 11

Symbol error vs. bit error


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When a symbol error occurs, we might suffer morethan one bit error such as mistaking 00 for 11.

It is however unlikely to have more than one biterror when a symbol error occurs

83

10 10 11 1000

11 10 11 1000

10 symbols = 20 bits

Sym.error=1/10

Bit error=1/20

Interpreting symbol error


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Numerically, symbol error is larger than bit errorbut in fact they are describing the same situation;

1 error in 20 bits In general, if Pe is symbol error

84

PelogM

BER Pe

Symbol error and bit error for QPSK


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We saw that symbol error for QPSK was

=

o

se

NEQP 2

85

ssum ng no more an error or eac sym o

error, BER is half of symbol error

Remember symbol energy Es=2Eb

=

o

se

N

EQP


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M-phase PSK (MPSK)


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If you combine 3 bits into one symbol, we haveto realize 23=8 states. We can accomplish this

with a single RF pulse taking 8 differentphases 45o apart

87

Ttiitf

TE

ts cs

i

=

+

= 0,

0

8,...,1,4

)1(2cos2)(

8-PSK constellation


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Distribute 8 phasors uniformly around a circleof radius Es

45o

88

Decision region

Symbol error for MPSK


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We can have M phases around the circleseparated by 2/M radians.

It can be shown that symbol error probabilityis approximately given by

89

4,sin2

2

MMN

EQP

o

se

Quadrature Amplitude Modulation


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MPSK was a phase modulation scheme. Allamplitudes are the same

QAM is described by a constellation consistingof combination of phase and amplitudes

90

The rule governing bits-to-symbols are thesame, i.e. n bits are mapped to M=2n symbols


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Vector representation of 16-QAM


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There are 16 vectors, each defined by a pair ofcoordinates. The following 4x4 matrix describes the

16-QAM constellation

92

[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?


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We had no trouble defining energy per symbol E forMPSK. For QAM, there is no single symbol energy.

There are many We therefore need to define average symbol energy

93

avg

Eavg = 1 ai2+ bi

2( )i=1

M

Eavg for 16-QAM


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

18 10 10 18

10 2 2 10

94

E=10 2 2 10

18 10 10 18

Eavg=10

Symbol error for M-ary QAM


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With the definition of energy in mind, symbol erroris approximated by

avgE31

95

( )

oe

NMM 1

Familiar constellations


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Here are a few golden oldies

96

V.22

600 baud1200 bps

V.22 bis

600 baud2400 bps

V.32 bis2400 baud

9600 bps

M-ary FSK


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Using M tones, instead of M phases/amplitudes is afundamentally different way of M-ary modulation

The idea is to use M RF pulses. The frequencieschosen must be orthogonal

97

( ) ( )

Mi

TttfTEts i

si

,...,1

0,2cos2

=

=

MFSK constellation: 3-dimensions


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

s3

3

2s1=[Es, 0, 0]

s2=[0, Es, 0]

E

98

s1

s2

1

2

iT

i ,

0 t T

i = 1,...,

s3=[0, 0, Es]

Es

Es

Orthogonal signals


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We just saw that in a 3 dimensional space, we canhave no more than 3 orthogonal signals

Equivalently, 3 orthogonal signals dont need morethan 3 dimensions because each can sit on one

99

Therefore, number of dimensions is always less thanor equal to number of signals

How to pick the tones?


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Orthogonal FSK requires tones that are orthogonal.

Two carrier frequencies separated by integer

multiples of period are orthogonal

100

Example


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

zero

=

TT

101

0

averages to zero

1 2444 34440

+ cos2 f1 f2( )dt0

T

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

1 24 44 3444

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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Here is the error expression with the usualnotations

( )

sE

QP 1

102

o

Bandwidth of digital


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Bandwidth of digital

modulations

Spectrum of M-ary signals


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So far Eb/No, i.e. power, has been our mainconcern. The flip side of the coin is bandwidth.

Frequently the two move in opposite directions Lets first look at binary modulation bandwidth

104

BPSK bandwidth


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Remember BPSK was obtained from a polar signal bycarrier modulation

We know the bandwidth of polar NRZ using squarepulses was BT=Rb.

105

oesn a e muc o rea ze a carr er

modulation doubles this bandwidth

Illustrating BPSK bandwidth


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The expression for baseband BPSK (polar)bandwidth is

SB(f)=2Ebsinc2

(Tbf)2/Tb=2Rb

106

BT=2Rbf1/T

b

BPSK

fc+/Tbfc-/Tb fc

BFSK as a sum of two RF streams


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BFSK can be thought of superposition of twounipolar signals, one at f1 and the other at f2

0.5

1

1

BFSK for 1 0 0 1 0 1 1

107

0 1000 2000 3000 4000 5000 6000 7000 8000-1

-0.5

0

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

+

Modeling of BFSK bandwidth

E h d l d l


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Each stream is just a carrier modulated unipolarsignal. Each has a sinc spectrum

108

f1 f2

1/Tb=Rb

fc

fc=(f1+f2)/2

f

BT=2 f+2Rb

f= (f2-f1)/2

Example: 1200 bps bandwidth

Th ld 1200 b d d d BFSK d l i


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The old 1200 bps standard used BFSK modulationusing 1200 Hz for mark and 2200 Hz for space.

What is the bandwidth? Use

109

T= + b

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

This is more than BPSK of 2Rb=2400 Hz


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Picking the 2nd zero crossing

If i k th d t (th fi t t t


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If we pick the second zc term (the first term putsthe tones too close) we get

2(f2-f1)Tb=2--> f=1/2Tb=Rb/2remember f is (f2-f1)/2

111

Sundes FSK bandwidth is then given by

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

The practical bandwidth is a lot smaller

Sundes FSK bandwidth

Due to sidelobe cancellation practical bandwidth is


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

112

f1 f2

b b

fc

fc=(f1+f2)/2

f

BT=2 f+2Rb

f= (f2-f1)/2

f

BFSK example

A BFSK system perates at the 3rd zer


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A BFSK system operates at the 3rd zerocrossing of -Tb plane. If the bit rate is 1

Mbps, what is the frequency separation of thetones?

113

The 3rd zc is for 2(f2-f1)Tb=3. Recalling thatf=(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


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FSK is not a particularly bandwidth-friendlymodulation. In this example, to transmit 1

Mbps, we needed 3.5 MHz. Of course, it is working at the 3rd zero

114

crossing t at is responsi e

Original Sundes FSK requires BT=Rb=1 MHz

MPSK bandwidth review

In MPSK we used pulses that are log M times


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In MPSK we used pulses that are log2M timeswider tan binary hence bandwidth goes down

by the same factor.T=symbol width=Tblog2M

115

For example, in a 16-phase modulation, M=16,

T=4Tb.Bqpsk=Bbpsk/log2M= Bbpsk/4

MPSK bandwidth

MPSK spectrum is given by


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MPSK spectrum is given by

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

116

f/Rb

Notice normalized frequency

1/logM

Set to 1 for zero crossingBW: T

bflog

2M=1

f=1/ Tbflog2M =Rb/log2M

BT= Rb/log2M

Bandwidth after carrier modulation

What we just saw is MPSK bandwidth in


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What we just saw is MPSK bandwidth inbaseband

A true MPSK is carrier modulated. This willonly double the bandwidth. Therefore,

117

Bmpsk=2Rb/log2M

QPSK bandwidth

QPSK is a special case of MPSK with M=4 phases


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QPSK is a special case of MPSK with M 4 phases.Its baseband spectrum is given by

SB(f)=2Esinc2

(2Tbf)

118

f/Rb0.5

B=0.5Rb-->half of BPSK

1

After modulation:Bqpsk=Rb


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MPSK vs.BPSK

Lets say we fix BER at some level. How do


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Let s say we f x BE at some le el. How dobandwidth 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

120

.

32 1/5 13.52 dB

Lesson: By going to multiphase modulation, we save bandwidth buthave to pay in increased power, But why?

Power-bandwidth tradeoff

The goal is to keep BER fixed as we increase M.


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g pConsider an 8PSK set.

121

What happens if you go to 16PSK? Signals getcloser hence higher BER

Solution: go to a larger circle-->higher energy

Additional comparisons

Take a 28.8 Kb/sec data rate and lets


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Take a 28.8 Kb/sec data rate and let scompare the required bandwidths:

BPSK: BT=2(Rb)=57.6 KHz BFSK: BT= Rb =28.8 KHz ...Sundes FSK

122

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


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p mtheir goals with minimum expenditure of power

at the expense of bandwidth. Examples are MFSK and other orthogonal

123

signa ing

Bandwidth-limited systems

Modulations that achieve error rates at a


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minimum expenditure of bandwidth but

possibly at the expense of too high a powerare bandwidth-limited

124

Examp es are variations o MPSK an many

QAM Check BER rate curves for BFSK and

BPSK/QAM cases

Bandwidth efficiency index

A while back we defined the following ratio as


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ga bandwidth efficiency measure in

bits/sec/HZ=Rb/BTbits/sec/Hz

125

Every digital modulation has its own

for MPSK

At a bit rate of Rb, BPSK bandwidth is 2Rb


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When we go to MPSK, bandwidth goes down by a

factor of log2M BT=2Rb/ log2M

126

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


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Defining MFSK:

In MFSK we transmit one of M frequencies forb l d ti T


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every symbol duration T

These frequencies must be orthogonal. One way todo that is to space them 1/2T apart. They could

128

.

Following The textbook we choose the former (thiscorresponds to using the first zero crossing ofcorrelation curve)

MFSK bandwidth

Symbol duration in MFSK is M times longer thanbi


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binary

T=Tblog2M symbol length Each pair of tones are separated by 1/2T. If there

129

are o em,

BT=M/2T=M/2Tblog2MBT=MRb/2log2M

Contrast with MPSK

Variation of bandwidth with M differs drasticallymp d t MPSK


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compared to MPSK

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

130

,MPSK save bandwidth

MFSK bandwidth efficiency

Lets compute s for MFSK


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

M 2 4 8 16 32 64

131

1 1 .75 .5 .3 .18

Notice bandwidth efficiency drop. We are sendingfewer and fewer bits per 1 Hz of bandwidth

COMPARISON OF DIGITAL


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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

/l

symbolsof#2n

b lbitM

M=


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sec/log=

/log=

2

2

ss

b bitsT

MTnR

symbolbitsMn

=

Bandwidth efficiency

measure

133

lengthbit11

ratesymbol

durationsym ol

s

s

b

b

b

s

s

nRn

T

R

T

data rateRR

T

===

==

=

bTsTT

b

TBTBB

2 ==

Bandwidth-limited Systems

There are situations where bandwidth is at ami m th f d m d l ti s ith


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premium, therefore, we need modulations with

large Rb/BT. Hence we need standards with large time-

134

an wi t pro uct

The GSM standard uses Gaussian minimumshift keying(GMSK) with BTTb=0.3

Case of MPSK

In MPSK, symbols are mtimes as wide asbinary


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binary.

Nyquist bandwidth is BT=Rs/2=1/2Ts However, the band ass bandwidth is twice

135

that, BT=1/Ts then:

zbits/sec/Hloglog

22 MTB

M

B

R

sTT

b ==

Cost of Bandwidth Efficiency

As M increases, modulation becomes morebandwidth efficient


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bandwidth efficient.

Lets fix BER. To maintain this BER whileincreasing M requires an increase in Eb/No.

136

Power-Limited Systems

There are cases that bandwidth is availablebut power is limited


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but power is limited

In these cases as M goes up, the bandwidthincreasesbut required power levels to meet a

137

speci ie BER remains sta e

Case of MFSK

MFSK is an orthogonal modulation scheme.

N ist b d idth is M ti s th bi s


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Nyquist bandwidth is M-times the binary case

because of using M orthogonal frequencies:BT = M/Ts = MRs then

138

zbits/sec/Hloglog 22M

MTBM

BR

sTT

b ==

Select an Appropriate Modulation

We have a channel of 4KHz with an availableS/No=53 dBHz required data rate R=9600


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S/No=53 dBHz, required data rate R=9600

bits/sec and required BER=10-5

.Choose a modulation scheme to meet these

139

requirements?

Minimum Number of Phases

To conserve power, we should pick the minimumnumber of phases that still meets the 4KHz


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number of phases that still meets the 4KHz

bandwidth A 9600 bits/sec if encoded as 8-PSK results in

140

3200 sym o s sec nee ing 3200Hz

So, M=8

What is the required Eb/No?

== sss R

N

E

TN

E

N

S


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== bo

b

bo

b

osoo

RNE

TNE

NTNN

141

89.2010

2.13

sec)/()()(

10

2.13

==

=

=

o

b

b

oo

b

N

E

dB

dBbitsRdBHzN

dBN

Is BER met? Yes

The symbol error probability in 8-PSK is

P M( ) 2Q 2Es i


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Solve for Es/No

PE M( ) =2Qs

No

sin

M

142

Solve for PE

BER = PE

log2M

=2.2 10

5

3

= 7.3 106

Es

No

= log2M( )Eb

N0

= 3 20.89 = 62.67

Power-limited uncoded system

Same bit rate and BER

Available bandwidth W=45 KHz


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Available bandwidth W=45 KHz

Available S/No=48 dBHz h i h h i h

143

required performance

Binary vs. M-ary Model


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M-ary ModulatorRbbits/s

144

)/(log2

ssymbolM

R bs =

M-ary demodulatorRbbits/s

bo

b

so

s

oRN

E

RN

E

N

S

==

Choice of Modulation

With R=9600 bits/sec and W=45 KHz, the channelis not bandwidth limited


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Lets find the available Eb/No

145

10

2.8

10

2.8)9600log(1048

sec)/()()(

=

==

=

o

b

b

oo

N

E

dB

dBbitsRdBHzN

dBN

Choose MFSK

We have a lot of bandwidth but little power

orthogonal modulation(MFSK)


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orthogonal modulation(MFSK)

The larger the M, the more power efficiencybut more bandwidth is needed

146

Pick the largest M without going beyond the 45

KHz bandwidth.

MFSK Parameters

From Table 1, M=16 for an MFSK modulationrequires a bandwidth of 38.4 KHz for 9600


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q

bits/sec data rate We also wanted to have a BER < 10-5. Question

147

is if this is met for a 16FSK modulation.

16-FSK

Again from Table 1, to achieve BER of 10-5 weneed Eb/No of 8.1dB.


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We solved for the available Eb/No and that came to8.2dB

148

Symbol error for MFSK

For noncoherent orthogonal MFSK, symbol errorprobability is


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PE

M( ) M1

exp E

s

149

o

Es=E

blog

2M

BER for MFSK

We found out that Eb/No=8.2dB or 6.61

Relating Es/No and Eb/No


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g

Es = log2M( )

Eb

150

BER and symbol error are related by

o o

PB=

2m 1

2

m

1

PE

Example

Lets look at the 16FSK case. With 16 levels, weare talking about m=4 bits per symbol. Therefore,


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PB= 2

3

4

PE=8 P

E

151

With Es/No=26.44, symbol error prob.

PE =1.4x10-5 PB=7.3x10-6

Summary

Given:

R=9600 bits/s5


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BER=10-5

Channel bandwith=45 KHz

152

Eb/No=8.2dB

Solution

16-FSK

required bw=38.4khz required Eb/No=8.1dB

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Lect 05 - Digital modulation.pdf