1

Geometric Representation of Modulation Signals Digital Modulation involves

Choosing a particular signal waveform for transmission for a particular symbol

For M possible symbols, the set of all signal waveforms are:

)}(),...,(),({ 21 tststsS M

For binary modulation, each bit is mapped to a signal from a signal set S that has two signals.

We can view the elements of S as points in vector space.

2

Geometric Representation of Modulation Signals Vector space

We can represent the elements of S as linear combination of basis signals i (t).

The number of basis signals is the dimension of the vector space.

Basis signals are orthogonal to each-other.

Each basis is normalized to have unit energy.

3

Let {j(t)| j = 1,2,…,N} represent a basis of S such that

si(t)=)(

1ts

N

jjij

(1) Any symbol, si(t)

j i dttt ji

0)()(

(2) Basis signals are orthogonal to each other in time

(3) Each basis signal is normalized to have unit energy

1)(2 dtti

E =

Basis signals Coordinate system for S

Gram-Schmidt process systematic way to obtain basis for S

Geometric Representation of Modulation Signals

4

Example

)(),(

)2cos(2

)(

)2cos(2

)(

)2cos(2

)(

11

1

2

1

tEtES

tfT

t

tfT

Ets

tfT

Ets

bb

cb

cb

b

cb

b

b

b

T t 0

T t 0

The basis signal

Two signal waveforms to be used for transmission

One dimensional Constellation Diagram bE bE

Q

I

5

QPSK Constellation Diagram

Q

I

/4

54 7/4

3/4

sE

4

7,

4

5,

4

3,

4

M2 =

sE20

3/2

/2Q

I M1 =

2

3,,

2,0

Rotation by /4 obtain new QPSK signal set

Es = 2Eb

6

π/4003π/4015π/4117π/410

si2si1grey coded

QPSK signalbinary symbol

bE

bE

bE bE

bE

bE

bE

bE

00000π/201

0π1103π/210

si2si1grey coded

QPSK signalbinary symbol

bE2

bE2

bE2

bE2

Signal Space Characterization of QPSK Signal Constellations

ith QPSK signal, based on message points (si1, si2) defined in tables

for i = 1,2 and 0 ≤ t ≤ Ts

si(t) = si1,1(t) + si22(t)

7

• modulated signal selected from 2 QPSK constellations shifted by /4 • for each symbol switch between constellations –total of 8 symbols

states 4 used alternately

• phase shift between each symbol = nk = /4 , n = 1,2,3

- ensures minimal phase shift, k = /4 between successive symbols

- enables timing recovery & synchronizationQ

Ipossible signal transitions

= possible states for k for k-1 = n/4

= possible states for k for k-1 = n/2

/4 QPSK modulation

8

Constellation Diagram Properties of Modulation Scheme can be

inferred from the Constellation Diagram:

Bandwidth occupied by the modulation increases as the dimension of the modulated signal increases.

Bandwidth occupied by the modulation decreases as the signal_points per dimension increases (getting more dense).

Probability of bit error is proportional to the distance between the closest points in the constellation.

Euclidean Distance Bit error decreases as the distance increases (sparse).

9

Linear Modulation Techniques Digital modulation techniques classified

as: Linear

The amplitude of the transmitted signal varies linearly with the modulating digital signal, m(t).

They usually do not have constant envelope. More spectrally efficient. Poor power efficiency Example: QPSK.

Non-linear / Constant Envelope

10

• constant carrier amplitude - regardless of variations in m(t)Better immunity to fluctuations due to fading. Better random noise immunity.

• improved power efficiency without degrading occupied spectrum- use power efficient class C amplifiers (non-linear)

• low out of band radiation (-60dB to -70dB)

• use limiter-discriminator detection- simplified receiver design-high immunity against random FM noise & fluctuations

from Rayleigh Fading

• larger occupied bandwidth than linear modulation

Constant Envelope Modulation

11

Frequency Shift Keying Minimum Shift Keying

Gaussian Minimum Shift Keying

Constant Envelope Modulation

12

Frequency Shift Keying (FSK)

Binary FSK

Frequency of the constant amplitude carrier is changed according to the message state high (1) or low (0)

Discontinuous / Continuous Phase

0)(bit Tt0

1)(bit Tt0

b

b

tffAts

tffAts

c

c

)22cos()(

)22cos()(

2

1

13

Switching between 2 independent oscillators for binary 1 & 0

sBFSK(t)= vH(t) binary 1bH

b

b Tt tfT

E 0)2cos(

21 =

binary 0 sBFSK(t)= vL(t)

bLb

b TttfT

E 0)2cos(

22 =

switch

cos w2t

cos w1tinput data phase jumps

Discontinuous Phase FSK

• results in phase discontinuities• discontinuities causes spectral spreading & spurious transmission • not suited for tightly designed systems

14

single carrier that is frequency modulated using m(t)

sBFSK(t) = ))(2cos(2

ttfT

Ec

b

b

t

FSKcb

b dmktfT

E )(22cos2

where (t) =

t

FSK dmk )(2

• m(t) = discontinuous bit stream• (t) = continuous phase function proportional to integral of m(t)

=

Continuous Phase FSK

15

FSK Example

1 1 0 1

Data

FSK Signal

0 1 1

VCO

cos wct

x

01

a0

a1

modulated compositesignal

16

• complex envelope of BFSK is nonlinear function of m(t)• spectrum evaluation - difficult - performed using actual time averaged measurements

PSD of BFSK consists of discrete frequency components at• fc

• fc nf , n is an integer

PSD decay rate (inversely proportional to spectrum)

• PSD decay rate for CP-BFSK

• PSD decay rate for non CP-BFSK

f = frequency offset from fc

4

1

f

2

1

f

Spectrum & Bandwidth of BFSK Signals

17

Transmission Bandwidth of BFSK Signals (from Carson’s Rule)

• B = bandwidth of digital baseband signal

• BT = transmission bandwidth of BFSK signal

BT = 2f +2B

• assume 1st null bandwidth used for digital signal, B

- bandwidth for rectangular pulses is given by B = Rb

- bandwidth of BFSK using rectangular pulse becomes

BT = 2(f + Rb)

if RC pulse shaping used, bandwidth reduced to:

BT = 2f +(1+) Rb

Spectrum & Bandwidth of BFSK Signals

18

General FSK signal and orthogonality

• Two FSK signals, VH(t) and VL(t) are orthogonal if

0)()(0

dttVtVT

LH

• interference between VH(t) and VL(t) will average to 0 during demodulation and integration of received symbol

0)()(0

dttVtVT

LH

• received signal will contain VH(t) and VL(t)

• demodulation of VH(t) results in (VH(t) + VL(t))VH(t)

0)()(0

dttVtVT

HH

?

?

19

))2(2cos())2(2cos( tftfT

Ec

b

b =

))(2cos())(2cos(2

tfftffT

Ecc

b

b vH(t) vL(t) =then

=bT

c

c

b

b

f

ft

f

tf

T

E

04

)4sin(

4

)4sin(

f

fT

f

Tf

T

E b

c

bc

b

b

4

)4sin(

4

)4sin(=

dtfttfT

EdttVtV

bT

cb

bT

LH 00

)4cos()4cos()()( and

vH(t) vL(t) are orthogonal if Δf sin(4πfcTb) = -fc(sin(4πΔf Tb)

An FSK signal for 0 ≤ t ≤ Tb

vH(t) = ))(2cos(2

tffT

Ec

b

b vL(t) = ))(2cos(2

tffT

Ec

b

b and

20

CPFSK Modulationelimination of phase discontinuity improves spectral efficiency & noise performance

• consider binary CPFSK signal defined over the interval 0 ≤ t ≤ T

s(t) = )(2cos2

ttfT

Ec

b

0 ≤ t ≤ T

θ(t) = phase of CPFSK signal

θ(t) is continuous s(t) is continuous at bit switching times

θ(t) increases/decreases linearly with t during T

tT

hθ(t) = θ(0) ±

‘+’ corresponds to ‘1’ symbol‘-’ corresponds to ‘0’ symbolh = deviation ratio of CPFSK

21

To determine fc and h by substitution

tT

h2πfct + θ(0) + = 2πf2 t+ θ(0)

tT

h2πfct + θ(0) - = 2πf1t+ θ(0)

thus 212

1ff fc=

h = T(f2 – f1)

• nominal fc = mean of f1 and f2

• h ≡ f2 – f1 normalized by T

T

hfc 2f1 =

T

hfc 2f2 = yields and

22

‘1’ sent increases phase of s(t) by πh

‘0’ sent decreases phase of s(t) by πh

variation of θ(t) with t follows a path consisting of straight lines• slope of lines represent changes in frequency

symbol ‘1’ θ(T) - θ(0) = πh

symbol ‘0’ θ(T) - θ(0) = -πh

θ(T) = θ(0) ± πh At t = T

FSK modulation index = kFSK (similar to FM modulation index)

bR

F2kFSK =

• peak frequency deviation F = |fc-fi | = 122

1ff

23

θ(t) - (0) rads

3πh

2πh

πh

0

-πh

-2πh

-3πh0 T 2T 3T 4T 5T 6T t

depicted from t = 0

• phase transitions across

interval boundaries of

incoming bit sequence

• θ(t) - θ(0) = phase of

CPFSK signal is even or odd

multiple of πh at even or odd

multiples of T

Phase Tree

24

Phase Tree is a manifestation of phase continuity – an inherent characteristic of CPFSK

0 ≤ t ≤ TtT

θ(t) = θ(0) ±

thus change in phase over T is either π or -π

• change in phase of π = change in

phase of -π

e.g. knowing value of bit i doesn’t

help to find the value of bit i+1

θ(t) - (0) 3π

2π

π

0

-π

-2π

-3π 0 T 2T 3T 4T 5T 6T t

1 0 0 0 0 1 1

25

assume fi given by as fi = T

inc nc = fixed integer

CPFSK = continuous phase FSK

• phase continuity during inter-bit switching times

si(t) =

tT

in

T

E cb 2cos2 0 ≤ t ≤ Tfor i = 1, 2

= 0 otherwise

si(t) = tfT

Ei

b 2cos2 0 ≤ t ≤ T

= 0 otherwise

for i = 1, 2

26

BFSK constellation: define two coordinates as

for i = 1, 2 i(t) = )2cos(2

tfT i 0 ≤ t ≤ T

= 0 otherwise

let nc = 2 and T = 1s (1Mbps) then f1 = 3MHz, f2 = 4MHz

1(t) = )3

2cos(2

tTT

0 ≤ t ≤ T

= 0 otherwise

2(t) = )4

2cos(2

tTT

0 ≤ t ≤ T

= 0 otherwise

27

0 ≤ t ≤ Ts2(t) =

t

TT

Eb 42cos

2

= 0 otherwise

)(2 tEb=

s1(t) =

t

TT

Eb 32cos

2 0 ≤ t ≤ T

= 0 otherwise

)(1 tEb=

bE

bE

2(t)

1(t)1

0

bE2BFSK Constellation

28

T

dt0

output+

-r(t)

Decision Circuit

sin wct

cos wct

T

dt0

• 2 correlators fed with local coherent reference signals• difference in correlator outputs compared with threshold to determine binary value

Pe,BFSK =

0N

EQ b

Probability of error in coherent FSK receiver given as:

Coherent BFSK Detector

29

• operates in noisy channel without coherent carrier reference • pair of matched filters followed by envelope detector

- upper path filter matched to fH (binary 1)- lower path filter matched to fL (binary 0)

• envelope detector output sampled at kTb compared to threshold

Pe,BFSK, NC =

02exp

2

1

N

Eb

Average probability of error in non-coherent FSK receiver:

r(t) outputDecision Circuit

+

-

EnvelopeDetector

Matched FilterfL

EnvelopeDetector

Tb

Matched FilterfH

Non-coherent Detection of BFSK