[PPT] - I n f o r m a t i o n T r a n s m i s s i o n PowerPoint Presentation

SLIDE 1

I n f

r

m a t i

n

T r a n s m i s s i

n

C h a p t e r 4 , D i g i t i a l m

d

u l a t i

n

OVE EDFORS Electrical and information technology

SLIDE 2

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 2

L e a r n i n g

u

t c

m

e s

A

f t e r t h e s e l e c t u r e s ( s l i d e s s p a n t w

l

e c t u r e s ) , t h e s t u d e n t s h

u

l d

– u

n d e r s t a n d t h e b a s i c p r i n c i p l e s

f

h

w

d i g i t a l i n f

r

m a t i

n

i s c a r r i e d

n

a n a l

g

s i g n a l s ( d i g i t a l m

d

u l a t i

n

) , i n c l u d i n g a m p l i t u d e , p h a s e a n d f r e q u e n c y m

d

u l a t i

n

/ k e y i n g ,

– u

n d e r s t a n d h

w

t h e m

d

u l a t i

n

p u l s e s h a p e d e t e r m i n e s b a n d w i d t h

f

t h e s i g n a l a n d w h a t t h e n a r r

w

e s t p

s

s i b l e t r a n s m i s s i

n

b a n d w i d t h i s f

r

a c e r t a i n d a t a r a t e ,

– u

n d e r s t a n d h

w
n

e

r

m

r

e b i t s a r e m a p p e d

n

t

s

i g n a l c

n

s t e l l a t i

n

p

i

n t s ,

– b

e a b l e t

p

e r f

r

m b a s i c c a l c u l a t i

n

s u s i n g r e l a t i

n

s b e t w e e n d a t a r a t e s , s i g n a l c

n

s t e l l a t i

n

s , p u l s e c h a p e s a n d t r a n s m i s s i

n

s p e c t r u m / b a n d w i d t h s ,

– u

n d e r s t a n d t h e f u n d a m e n t a l p r i n c i p l e s

f

h

w

d i g i t a l i n f

r

m a t i

n

i s d e t e c t e d a t t h e r e c e i v e r , i n c l u d i n g

p

t i m a l r e c e i v e r s ,

– u

n d e r s t a n d t h e r e l a t i

n

s h i p s b e t w e e n r e c e i v e s s i g n a l q u a l i t y a n d r e s u l t i n g b i t

e

r r

r

r a t e s ,

– b

e a b l e t

p

e r f

r

m b a s i c c a l c u l a t i

n

s

n

r e s u l t i n g r e c e i v e r p e r f

r

m a n c e ( b i t

e

r r

r

r a t e s ) w h e n t h e m

d

u l a t i

n

t y p e a n d t h e r e c e i v e d s i g n a l q u a l i t y a r e g i v e n .

SLIDE 3

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 3

Wh e r e a r e w e i n t h e B I G P I C T U R E ?

Digital modulation/ transmission techniques Lecture relates to pages 127-146 in textbook.

SLIDE 4

O v e E d f

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s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 4

D i fg e r e n t m

d

u l a t i

n

f

r

m a t s

Amplitude modulation, ASK (amplitude shift keying)
Phase modulation, PSK (phase shift keying)
Frequency modulation, FSK (frequency shift keying)

Transmitted signal, with amplitude, phase or frequency carrying the information We will focus primarily on this one!

SLIDE 5

O v e E d f

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s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 5

4ASK

00 01 11 00 10

Amplitude carries information
Phase constant (arbitrary)

4PSK

00 01 11 00 10

Amplitude constant (arbitrary)
Phase carries information

4FSK

00 01 11 00 10

Amplitude constant

(arbitrary)

Phase slope (frequency)

carries information Comment:

A m p l i t u d e , p h a s e a n d f r e q u e n c y m

d

u l a t i

n

 

A t

SLIDE 6

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 6

Tie p u l s e s h a p e d e t e r m i n e s t h e b a n d w i d t h

c

c u p i e d

SLIDE 7

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 7

T r a i n

f

p u l s e s , r e p r e s e n t i n g 1 1 1

Square pulses
Raised cosine

Ones mapped to positive pulses Zeros mapped to negative pulses

SLIDE 8

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 8

Tie m

d

u l a t i

n

p r

c

e s s

Complex domain Mapping PAM

m

c

 

LP

s t

 

exp 2

c

j f t 

Re{ } Radio signal PAM: “Standard” basis pulse criteria (energy norm.) (orthogonality) Complex numbers Bits Symbol time

sLP(t)= ∑

m=−∞ ∞

cm v (t−mT s)

∫

−∞ ∞

|v(t)|

2dt=1 or =T s

∫

−∞ ∞

v(t)v

*(t−mT s)dt=0 for m≠0

SLIDE 9

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 9

B a s i s p u l s e s a n d s p e c t r u m

Assuming that the complex numbers cm representing the data are independent, then the power spectral density of the base band PAM signal becomes: which translates into a radio signal (band pass) with

Many possible pulses

   

2 2 LP

       

   

dt e t v f S

πft j

t t

s

T

       

c c

f f S f f S f S     

LP LP BP

2 1

SLIDE 10

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 1

B a s i s p u l s e s

(Root-) Raised-cosine [in freq.] Rectangular [in time] TIME DOMAIN

FREQ. DOMAIN

s

T f  freq. Normalized

Normalized freq. f ×T s

s

T t/ time Normalized

s

T t/ time Normalized

SLIDE 11

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 1 1

I n t e r p r e t a t i

n

a s I Q

m
d

u l a t

r
90
c

f

   

 

Re

I LP

s t s t 

   

 

Im

Q LP

s t s t 

 

cos 2

c

f t 

 

sin 2

c

f t  

Radio signal

For real valued basis functions v(t) we can view PAM as:

Pulse shaping filters Mapping

m

c

 

Re

m

c

 

Im

m

c

(Both the rectangular and the (root-) raised-cosine pulses are real valued.) In-phase signal Quadrature signal

SLIDE 12

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 1 2

B i n a r y p h a s e

s

h i f t k e y i n g ( B P S K ) R e c t a n g u l a r p u l s e s

Radio signal (band pass) Base-band signal (low pass)

SLIDE 13

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 1 3

B i n a r y p h a s e

s

h i f t k e y i n g ( B P S K ) R e c t a n g u l a r p u l s e s

Complex representation Signal constellation diagram

SLIDE 14

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 1 4

B i n a r y p h a s e

s

h i f t k e y i n g ( B P S K ) R e c t a n g u l a r p u l s e s

Power spectral density for BPSK

Normalized freq. f ×T b

SLIDE 15

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 1 5

B i n a r y p h a s e

s

h i f t k e y i n g ( B P S K ) R a i s e d

c
s

i n e p u l s e s ( r

l

l

fg

. 5 )

Radio signal (band pass) Base-band signal (low pass)

SLIDE 16

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 1 6

B i n a r y p h a s e

s

h i f t k e y i n g ( B P S K ) R a i s e d

c
s

i n e p u l s e s ( r

l

l

fg

. 5 )

Complex representation Signal constellation diagram

SLIDE 17

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 1 7

B i n a r y p h a s e

s

h i f t k e y i n g ( B P S K ) R a i s e d

c
s

i n e p u l s e s ( r

l

l

fg

. 5 )

Power spectral density for BAM/BPSK

Much higher spectral efficiency than BPSK (with rectangular pulses).

Normalized freq. f ×T b

SLIDE 18

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 1 8

Q u a t e r n a r y P S K ( Q P S K

r

4

P

S K ) R e c t a n g u l a r p u l s e s

Complex representation Radio signal (band pass)

SLIDE 19

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 1 9

Q u a t e r n a r y P S K ( Q P S K

r

4

P

S K ) R e c t a n g u l a r p u l s e s

T wice the spectrum efficiency of BPSK (with rect. pulses). TWO bits/pulse instead of one.

Power spectral density for QPSK

SLIDE 20

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 2

A g

l

d e n b a n d w i d t h r u l e

The narrowest bandwidth of any pulses that act independently is [-1/2T, 1/2T] where T is the symbol interval

SLIDE 21

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 2 1

O t h e r c

m

m

n

s i g n a l c

n

s t e l l a t i

n

s

16 QAM

– Less bandwidth but higher SNR required

8 PSK

– All points on a circle

(same energy)

SLIDE 22

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 2 2

Detection, receivers

SLIDE 23

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 2 3

D e t e c t i n g p u l s e w a v e f

r

m s

Find the method that minimizes the error probability in white Gaussian noise

– Correlation detector

Correlate the received signal with a local copy of the ideal pulse alternatives

SLIDE 24

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 2 4

O p t i m a l r e c e i v e r Wh a t d

w

e m e a n b y

p

t i m a l ?

Every receiver is optimal according to some criterion! We would like to use optimal in the sense that we achieve a minimal probability of error. In all calculations, we will assume that the noise is white and Gaussian – unless otherwise stated.

SLIDE 25

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 2 5

O p t i m a l r e c e i v e r T r a n s m i t t e d a n d r e c e i v e d s i g n a l

t t Transmitted signals 1: 0: s1(t) s0(t) t t Received (noisy) signals r(t) r(t) n(t) Channel s(t) r(t)

SLIDE 26

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 2 6

O p t i m a l r e c e i v e r A fj r s t “ i n t u i t i v e ” a p p r

a

c h

“Look” at the received signal and compare it to the possible received noise free signals. Select the one with the best “fit”. t r(t) Assume that the following signal is received: t r(t), s0(t) 0: Comparing it to the two possible noise free received signals: t r(t), s1(t) 1:

This seems to be the best “fit”. We assume that “0” was the transmitted bit.

SLIDE 27

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 2 7

O p t i m a l r e c e i v e r L e t ’ s m a k e i t m

r

e m e a s u r a b l e

To be able to better measure the “fit” we look at the energy of the residual (difference) between received and the possible noise free signals:

t r(t), s0(t) 0: t r(t), s1(t) 1: t s1(t) - r(t) t s0(t) - r(t) This residual energy is much

smaller. We assume that “0”

was transmitted.

SLIDE 28

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 2 8

O p t i m a l r e c e i v e r Tie A WG N c h a n n e l

The additive white Gaussian noise (AWGN) channel

transmitted signal
channel attenuation
white Gaussian noise
received signal

In our digital transmission system, the transmitted signal s(t) would be one of, let’s say M, different alternatives s0(t), s1(t), ... , sM-1(t).

 

s t 

 

n t

 

r t

 

s t 

 

n t

 

r t

   

s t n t   

SLIDE 29

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 2 9

O p t i m a l r e c e i v e r Tie A WG N c h a n n e l , c

n

t .

It can be shown that finding the minimal residual energy (as we did before) is the optimal way of deciding which of s0(t), s1(t), ... , sM-1(t) was transmitted over the AWGN channel (if they are equally probable). For a received r(t), the residual energy ei for each possible transmitted alternative si(t) is calculated as Same for all i Same for all i, if the transmitted signals are of equal energy. The residual energy is minimized by maximizing this part of the expression.

ei=∫∣r t− sit∣

2dt=∫r t− sitrt− sit *dt

=∫∣rt∣

2dt−2Re { *∫r t si *tdt}∣∣ 2∫∣sit∣ 2dt

SLIDE 30

O v e E d f

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s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 3

O p t i m a l r e c e i v e r Tie A WG N c h a n n e l , c

n

t .

The central part of the comparison of different signal alternatives is a correlation, that can be implemented as a correlator:

r a matched filter

where Ts is the symbol time (duration). The real part of the output from either of these is sampled at t = Ts

 

r t

 

* i

s t

*



∫

T s

 

r t

 

* i s

s T t 

*



SLIDE 31

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 3 1

O p t i m a l r e c e i v e r A n t i p

d

a l s i g n a l s

In antipodal signaling, the alternatives (for “0” and “1”) are This means that we only need ONE correlation in the receiver for simplicity:

       

1

s t t s t t     

 

r t

 

* t



*



If the real part at T=Ts is >0 decide “0” <0 decide “1”

∫

T s

SLIDE 32

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 3 2

O p t i m a l r e c e i v e r I n t e r p r e t a t i

n

i n s i g n a l s p a c e

The correlations performed on the previous slides can be seen as inner products between the received signal and a set of basis functions for a signal space. The resulting values are coordinates of the received signal in the signal space.

 

t 

“0” “1”

Antipodal signals

 

s t

“0” “1”

 

1

s t

Orthogonal signals

Decision boundaries

SLIDE 33

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 3 3

O p t i m a l r e c e i v e r Tie n

i

s e c

n

t r i b u t i

n

Noise pdf. Noise-free positions

s

E

s

E

This normalization of axes implies that the noise centered around each alternative is complex Gaussian

   

2 2

N 0, N 0, j   

with variance σ2 = N0/2 in each direction.

Assume a 2-dimensional signal space, here viewed as the complex plane

Re Im sj si

Fundamental question: What is the probability that we end up on the wrong side of the decision boundary?

SLIDE 34

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 3 4

O p t i m a l r e c e i v e r P a i r

w

i s e s y m b

l

e r r

r

p r

b

a b i l i t y

s

E

s

E Re Im sj si

What is the probability of deciding si if sj was transmitted?

ji

d

We need the distance between the two symbols. In this orthogonal case:

2 2

2

ji s s s

d E E E   

The probability of the noise pushing us across the boundary at distance dji / 2 is

Pr s j  si=Q d ji/2

N 0/2=Q

E s N 0 =1 2 erfc E s 2 N 0

SLIDE 35

O v e E d f

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s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 3 5

O p t i m a l r e c e i v e r B i t

e

r r

r

r a t e s ( B E R )

2PAM 4QAM 8PSK 16QAM Bits/symbol 1 Symbol energy Eb BER

Q 2 E b N 0 

2 2Eb

Q 2 E b N 0 

3 3Eb

~ 2 3 Q 0.87 Eb N 0

4 4Eb

~ 3 2 Q E b, max 2.25 N 0

EXAMPLES:

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O v e E d f

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s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 3 6

2 4 6 8 1 1 2 1 4 1 6 1 8 2 1

6

1

5

1

4

1

3

1

2

1

1

1

O p t i m a l r e c e i v e r B i t

e

r r

r

r a t e s ( B E R ) , c

n

t .

Bit-error rate (BER) 2PAM/4QAM 8PSK 16QAM

/ [dB]

b

E N

SLIDE 37

O v e E d f

r

s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 3 7

Q u a d r a t u r e m

d

u l a t i

n

, w h y i s i t w

r

k i n g ?

Any carrier digital modulation can be expressed as The sine and cosine ”channels” are independent/orthogonal Therefore we can send two pulses at the same time without interference

SLIDE 38

O v e E d f

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s E I T A 3

C

h a p t e r 4 ( P a r t 3 ) 3 8

S U MMA R Y

B

i t s / s y m b

l

s a r e c a r r i e d

n

a n a l

g

s i g n a l s b y a l t e r i n g t h e i r a m p l i t u d e / p h a s e / f r e q u e n c y .

Mo

d u l a t i

n

b a s i c s , b a s i s p u l s e s

R

e l a t i

n

b e t w e e n d a t a r a t e a n d b a n d w i d t h

I

Q m

d

u l a t

r
B

a s i c m

d

u l a t i

n

f

r

m a t s

D

e t e c t i

n
f

d a t a a t r e c e i v e r

p

t i m a l r e c e i v e r i n A WG N c h a n n e l s

I

n t e r p r e t a t i

n
f

r e c e i v e d s i g n a l a s a p

i

n t i n a s i g n a l s p a c e

E

u c l i d e a n d i s t a n c e s b e t w e e n s y m b

l

s d e t e r m i n e t h e p r

b

a b i l i t y

f

s y m b

l

e r r

r
B

i t e r r

r

r a t e ( B E R ) c a l c u l a t i

n

s f

r

s

m

e s i g n a l c

n

s t e l l a t i

n

s

SLIDE 39