1

Chapter 4 Modulation and Demodulation

Modulation/Demodulation schemes, Receiver design,

Signal-to-noise ratio (SNR), Bit error rate (BER)

2

4.1 Modulation a. Intensity modulation/ Direct detection

i. digital (on- off keying, OOK)

ii. analog

b. Coherent modulation

3

4.1.1 Signal format (digital)

a. Unipolar return-to-zero (RZ, x%)b. Unipolar non-return-to-zero (NRE)i. need good DC balanceii. For time recover circuits, transitions are

needed. (line coding or scrambling) [(8,10) or (4,5) coding needs extra bandwidth]

The NRE format is very popular for high speed communication systems (e.g. 10Gb/s)

4

4.2 Subcarrier Modulation (SCM) and Multiplexing

Data first modulate the microwave carrier (10MHz~10GHz). Then we use the modulated microwave carrier to modulate the optical carrier.

5

6

CATV

S1(t) X1(t)

Sn(t)

fs1

fsn

Xn(t)

multiplexing

LD PD

X1(t)

Xn(t)

Demultiplexing

fiber

f1 f2 fn freq

7

4.2.1 Clipping and Intermodulation productsThe main issues a. power efficiency b. signal fidelitySource of signal distortion a. clipping

8

9

b. nonlinearityP

I2 3

1 2 3 . . .p a I a I a I

10

Provided a number M of simultaneously present sinusoids lying between ,the second-order terms at frequencies appear.

1 = ( - ) + ( +

2

1

2

1 1

1 1

cos(2 )

cos(2 )cos(2 )

cos(2 ) cos(2 ) )

M

i ii

M M

i j i ji j

M M

i j i j i ji j

let I I f t

I I I f t f t

I I f f t f f t

' ''f and f( )i jf f

11

In general a1>> a2>> a3

=> It seems the second-order terms are more harmful than the third terms.

If we make

will lie completely outside the band

For the following two reasons, may not be satisfied

a. Waste about half the available bandwidth

b. Because is higher, a laser which can be modulated at high frequency is needed.

'' ' 2f f

i jf f ' ''( , )f f'

''

. . i j

i j

e g f f f

f f f

'' ' 1f f

''f

12

two-tone third-order terms (because there are only two distinct frequencies)

triple-beat (CTB)Define: Composite second order (CSO) distortion

Composite triple-beat (CTB) distortion

In general CTB distortion is larger than CSO distortion.

3

1 1 1

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

cos 2 ( ) ,

M M M

i j k i j ki j k

i j k i j k

I I I I f t f t f t

f f f t f f f are called

cos 2 ( ) ,i j k i j kf f f t f f f are called composite

( ),i j k i jpower of the intermoudlation of f f f f f

carrier power

( ),i j k i j kpower of the intermoudlation of f f f f f f

carrier power

13

4.2.2 Applications of SCM

a. Transmitting multiple analog video signal using a single optical transmitter. (CATV, Hybrid fiber-coax HFC)

b. Carrying control data along with the actual data steam. For example, the pilot tone in WDM systems.

14

4.3 Optical Duobinary Modulation (controlled ISI)

T

X(nT)Y(nT)

( ) ( ) ( )

( ) 3 : 0, ,2

( )

( ) ( ) ( )

( ) ( ) ( 2 )

( ) ( ) ( 2 ) ( 3 )

( ) ( ) ( 2 )

y nT X nT X nT T class I

y nT has levels P P

compliacted transmission

We can recover the original signal as

z nT y nT z nY T

y nT y nT T z nT T

y nT y nT T y nT T z nT T

y nT y nT T y nT T

1...( 1) ( )

( ) ( ) ( ) ( 2 )...

( ) (4.1)

n y T

x nT x nT T x nT T x nT T

x nT

15

4.3.2 Optical Single sideband Modulation

16

4.3.3 Multilevel Modulation (M-level)M=2R

R: bits T=Tb log2M

T: band duration Tb: bit duration

4.3.4 Capacity Limits of Optical Fiber

Shanon’s TheoremC=B log2(1+S/N)

B: bandwidth,C: capacity

S/N: signal-to-noise ratio

17

4.4 Demodulation

BER ~ 10-9 or 10-12

3R: Regeneration, reshaping and retiming

18

4.4.1 An Ideal Receiver

It can be viewed as photon counting.Direct detection is used with the following

assumptionsa. When any light is seen, the receiver makes in

favor of symbol “1”, otherwise it makes in favor of symbol “o”

b. Photons will generate electron-hole pairs randomly as a Poisson random process.

When “o” was sent, there is no light. For an ideal receiver, it is error free.

When “1” was sent, because of Poisson random process, if no electron-holes are generated the receiver will make in favor of “o” => error occurs

19

The photon arriving rate for a light pulse with power P is P/hf

h: planck’s constant = 6.63 x 10-34 J-sec hf: the energy of a single photon

Let B denote the bit rate, the T = 1/B is the bit durationRecall for Poisson process P(λ) = exp(-λT) (λT)n /n! where λ= P / hf (photon arriving rate)The probability that n electron-hole pairs are generated duri

ng T = 1/B seconds P(λ) = exp(-P/hfB) (P/hfB)n /n!

20

If n 1 we decide that the bit is “1” otherwise ≧it is “o” bit.

The probability the a light pulse does not generate any electron-hole pair is

P(λ) = exp(-P/hfB)

Assuming that “1” and “o” are transmitted with equal probability,

BER = ½ exp (-P/hfB) + ½ x 0

“1” “0”

21

Let M be the number of photons for “1” bit in T seconds (M = P/hfB)

BER = ½ e-M

which is called the quantum limit

BER = 10-12 M = 27 BER = 10-9 M = 21 (BER = 7.6 x 10-10)

Practically the receiver has noise, more light power is needed to achieve the same BER.

22

4.4.2 A Practical Direct Detection Receiver

23

24

The noise sources a. thermal noise b. shot noiseThe thermal noise current in a resistor R at temper

ate T can be modeled as Additive White Gaussian Noise (AWGN) with zero mean and variance 4KBT/R

KB = Boltzmann’s constance =1.38 x 10-23 J/Ko

Let Be be the signal bandwidth The variance of the thermal noise is 2

2

4thermal B e

t e

K T R B

I B

25

It is the current standard deviation in

where T=bit duration, Be=electrical bandwidth

Let B0 be the optical bandwidth (passband)

B0 2 B≧ e (Haykin P.49)

pA Hz1 1 1

,2 2eB Nyquist bandwidthT T T

Bandwidth

Passband

cf cf

26

For a receiver using PIN diode, the photocurrent is given by

where e = the electronic charge e h(t-tk) = the current impulse due to a photon a

rriving at tk

Let p(t) be the optical power and p(t)/hfc be the photon arrival rate, fc be the optical frequency.

The rate of generation of electrons is Poisson process with rate

( ) ( ) ( .1)k

kk

I t eh t t I

( )eh t dt e

( ) ( )

,c

t R p t e

R e hf the responsivity

the quantum efficiency

27

To evaluate (I.1), we break up the time axis into small interval δt.

The current is given by

Nk are Poisson random variables with rate

( ) ( )k kk

I t eN h t t

( ) , ( ) :k t t k t arrival rate in kth interval

t

t

kth (K+1)th

28

Assume that

The mean value of the photocurrent is

0, ( ) ( )t k t t

( ) ( )

( )

( ) ( )

( ) ( ) 0

( ) ( ) ( ) ( )

tk

k

E I t P

eE h t k t

e k t th t k t

e h t d t

RR P h t d P t

e

29

1 2 1 1 2 2 1 2

1 2

21 2

1 2

1 2

21

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) '

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( )

E x t x t E x t x t

The autocorrelation is given by

E I t I t P

e h t h t d campbell s theorem

E I t P E I t P

eR P h t h t d

R P h t

2( ) ( )

,

( ) ( ) , ( ) 0 0

( ) 1

d P h t d

An ideal photodetector generates pure current impulse

for each received photon

h

30

1 2

21 2 1 1 2

21 2 1 2 1 1 2

( ) ( )

( ) ( )

( ) ( )

( )

( ) ( ) ( )

( ) ( ) ( ) ( )

( )

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

E I t P

R P h t d

R P t d

RP t

E I t I t P

eRP t t t R P t P t

Removing the conditioning over p yields

E I t RE P t

and

E I t I t eRE P t t t R E P t P t

The

1 2 1 2 1 2

21 2 1 1 2

21 2

21 2 1 1 2

1 2

1 2 1 2 1 2

( )

, ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ,

, ( )

, ( ) ( ) ( ) ( )

I

P

autocovariance of I t is

L t t E I t I t E I t E I t

eRE P t t t R E P t P t

R E P t E P t

eRE P t t t R L t t

when L t t is the autocovariance of P t

L t t E P t P t E P t E P t

31

Shot noise

1 2 1 2

2 1

( )

( ) 0, ( ) ( ) ( ) ( )

( )

( ) ( ),

( ) ( )

( )

( ) ( )exp( 2 )

( )exp( 2 )

P

I

I

I I

consider the simple case

E P t P constant

L E P t P t E P t E P t

where t t

E I t RP

L eRP

L eRP

The power spectral density PSD is

S f L i f d

eRP i f d

eRP depends on power

PSP

eRP

-Be Be

f

2

:

( ) 2e

e

e

B

shot I eB

B receiver bandwidth

S f df eRPB

32

The photocurrent is given by

is the shot current with zero mean and variance

sI I i

I RP

si2 eeRPB

2 2shot e

I RP

eIB

33

Let the local resistance be RL

The total current in the resistor is

Where is the thermal current with variance

Assume and are independent, the total current has mean and variance

Note that is proportional to there is a trade-off between SNR and For IM/DD receivers

noiseless amplifier

RLI

s tI I i i ti

2 2(4 / )Tthermal B L e t eK R B I B

tisiI 2

2 2 2shot thermal

2 eB

eBthermal shot

34

4.4.3 Front end amplifier noise

Define Noise Figure (Fn) of the amplifier

=(SNR)in / (SNR)out

typically Fn = 3~5dB

The thermal noise contribution is

Similary

2 4 B e nthermal

L

K TB FR

2 2shot e neIB F

35

4.4.3 APD Noise

36

Gm(t): Gm

= Avalanche multiplication gain will be amplified

Let RAPD=responsivity of APDThe average APD photocurrent is

The variance of shot noise is

FA(Gm): the excess noise factor

FA(Gm) = KAGm + (1 - KA)(2 -1 / Gm)

KA: ionization coefficient ratio

For silicon KA<<1

For InGaAs KA=0.7

Note if Gm=1, FA=1 (4.4) is the variance of shot noise of PIN

2shot

APD mI R P G RP

2 22 ( ) (4.4)shot m A m eG F G RPB

37

4.4.5 Optical Amplifiers

The Amplified Spontaneous Emission (ASE) noise is the main noise source of optical amplifiers.

The ASE noise power for each polarization is

Where nsp: the spontaneous emission factor G: the amplifier gain Bo: the optical bandwidth nsp depends of level of population inversion, With complet

e inversion nsp = 1 typically nsp = 2~5

0( 1) (4.5)N sp cP n hf G B

38

There are two independent polarizations in a single mode fiber

The total ASE noise power

P=2PN

Define

If the standard PIN is used

I=RGP (4.6)

R: responsivity

P: received power

0, ( 1)n sp c N nP PP n hf G B

39

Photocurrent I is proportional to the optical power P which is proportional to the square of the electrical field. Thus the noise field beats against the signal and against itself. Thus signal-spontaneous beat noise and spontaneous-spontaneous beat noise are produced.

For Appendix I, we have

2 2

20

2 2

22 20

(4.7)

2 ( 1) (4.8)

4 ( 1) (4.9)

2 ( 1) (2 ) (4.10)

thermal t e

shot n e

signal spont n e

spont spont n e e

I B

eR GP P G B B

R PP G B

and R P G B B B

40

It is the received thermal noise current

If G is large (e.g > 10dB)

If Bo is small ≈ 2Be

Pn= nsphfc << P

is dominant, which can be modeled as a Gaussian process.

2 4 ,Bt

K TI R resistanceR

2 2 2 2, ,thermal shot sig spont spont spont

2 2spont spont sig spont

2sig sopnt

41

Recall (4.2)

At the amplifier input

At the amplifier output, using (4.6) and (4.9) we have

2

2

2 (4.2)

2

shot e

shot e

eIB

I RP

eRPB

2( )2i

e

RPSNR eRPB

2 20

2 2

( ) 4 ( 1)

( ) 4 ( 1)

n e

e sp c

n sp c

SNR RGP R GPP G B

RGP R GP G B n hf

P n hf

42

The noise figure of the amplifier is

In the above derivation, we assume no coupling losses. The input coupling loss will degrade Fn

0

2

2 2

( ) 2, 1

( ) 4 ( 1)

2

n i

e

sp c e

sp c

F SNR SNR

RP RePBG

RGP R PG G n hf B

n hf R e

, 1

2

2

1 2 3

4 ~ 7

c

c

n sp c

sp

sp n

n

eRecall R ifhf

e

hf

F n hf R e

n

If n F dB

typically F dB

43

4.4.6 Bit Error Rates (BER)

BER is a very important measure for digital communication systems.

Other measures are SNR, CNR, error free second, packet loss…

Consider a PIN receiver without optical amplifiers

P1= optical power of bit “1”

P0= optical power of bit “0”

Assume that thermal noise dominates and is Gaussian.

44

For bit “1”, the mean photocurrent is

The variance is

RL: load resistantFor bit “0”, the variance is

For ideal case P0=0 I0=0

1 1I I RP

20 02 4e B e LeI B K TB R

21 12 4e B e LeI B K TB R

signal thermal noise

20 4 B e LK TB R

receiverpower

amplifierAGC

decision cht

timingcht

optical

signalwithfilter

dataI

45

Assume the decision threshold current is Ith If I > Ith decision is made in favor of “1”If I < Ith decision is made in favor of “0”

46

Assume that bits “1” and “0” are transmitted equally likely

Problem 4.7

The threshold current is

1(0) (1)

2p p

(4.12)0 1 1 0

0 1th

I II

Q(x)

Q(x)=2

2

2

2

2

2

2

1

2

1(4.13)

22

( )

2

22 ( )

y

x

yx

z

x

y

x

Define as

e dy

e dy

Note evfc x e dz

e dy

Q x

47

Then we have

P

P

1

1

0

0

0 1

10

th

th

I IQ

I IQ

0 1 1 0

0 1

0 1 1 1 0 1 1 01

0 1

1 1 0

0 1

1 1 0

1 0 1

0 1 1 0 0 0 1 00

0 1

0 1 0

0 1

0 1 0

0 0 1

th

th

th

th

th

Recall

I II

I I I II I

I I

I I I I

I I I II I

I I

I I I I

48

If “1” and “0” are transmitted equally likely

1 0

0 1

1 1(0 1) (10)

2 2

(4.14)

BER P P

I IQ

11 0

0 1

12 9

( )

10 , 7, 10 , 6

I Ilet r Q BER

For BER r BER r

1 0

1 0 1 0

1 0

2

1

2 2 2

th

I IPractically I

I I I IBER Q Q

49

We may specify BER then calculate the received power (BER=10-12)

Define: The receiver sensitivity as

Psens= The minimum average optical power for given BER (e.g 10-12 )

It is also expressed as the number of photons per bit.

Recall page2521

1

1 01

1

: "1"

:

, 02

2

2

c

sens

sens

sens

c

PM hf B

P optical power for

B the bit rate

P PP ideally P

P P

PM

hf B

50

1

1 0

0 1

0

1 0 1

0 1

2 2 2 2 20 1

2 21

( )

0,2

( ) (4.6)

: 1

( )(4.15)

2

,

,

2 ( )

sens

m

m m

sens

m

thermal thermal shot

shot m A m e

Recall r Q BER

I I

PFor I P

I r RPG

G APD gain G for PIN

rP

RG

For APD shot noise is significant

Recall eG F G RPB

2

(4.4)

4 ( ) sensm A m eeG F G RP B

51

0 1

2

2 2 21

22

2

2

(4.15)2

24 ( )

2 44 ( )

( )

(

therm

sens

m

sensmsenstherm m A m e

sens sensm m thermsensm A m e

sensm thermm A m e

thermsens e A m

m

rP

G R

G RPeG F G RP B

r

G RP G RPeG F G RP B

r r

G RPeG F G B

r rr

P eB F GR G

2 22 2

12

) (4.16)

: , 32

100 , 300 , ( )1.24

41.656 10

1 10 , 7

e n

L

Btherm n e

L

m

r

BAssume B Hz B bit rate F dB

AR T K R W

K TF B BA

R

G BER r

52(KA=0.7, Gm=10)

53

For a system with optical amplifier2

0

2 2

1 0

0 1

2

4 ( 1) (4.9)

(4.6)

(4.14)

2 ( 1)

(4.18)2 ( 1)

e sig spont

sig spont n e

n e

n e

B B dominates

R GPP G B

RGP

I IBER Q

RGPQ

R GPP G B

GPQ

P G B

Ｉ

54

12

2

, 1,

2

1 1

2 ( 1) 2

12 ,

2 2

2

, 10 , 72

2( ) 98

~

~

sp n sp c c

e

n e n e

c n e

If G is large n P n hf hf

BB

GP Pr

G P B P B

PPM M hf B P B

M

MBER Q For BER r

M r M photons bit

practically M a few hundred photons bit

If optical amplifier is not used

M a f

ew thousand photons bit

55

If systems with cascades of optical amplifiers, optical signal consists of a lot of optical noise, and can be ignored. The received optical noise power PASE dominates.

Define Optical signal-to-noise ratio (OSNR)=

Based on above equation, (4.6), (4.9), (4.10) and (4.14)We have

For 2.5Gb/s system, Be=2 GHz, Bo= 36GHz, r=7 => OSNR = 4.37 = 6.14 dBA rule of thumb used by designers OSNR ≈ 20dB because of dispersion and nonlinearity

rec

ASE

PP

0

0

2 ( 1) (4.5)

2 ( 1) ( )

ASE sp c

n

P n hf G B

P G B two pols

02

1 1 4e

BOSNR

Br

OSNR

n sp cP n hf

thermal shot

56

4.4.7 Coherent Detection (ASK)

Assume that phase and polarization of two waves are matched θ=0

The optical power at the receiver.

coupler

detector amplifierdecision circuit

timing circuit

data

localoscillator

2 cos(2 )caP f t

2 cos(2 )Lo LoP f t

57

2

2 2

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

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

2 cos 2 ( )

0.1

r c Lo Lo

c Lo c Lo Lo Lo

Lo Lo c Lo

P t aP f t P f t

aP f t aPP f t f t P f t

aP P aPP f f t high order terms

a for ook

1

21 1

20 0 0

"1" , 1

( 2 )

2

"0" , 0

, 2

~ , 0.01 ,

c Lo

Lo Lo

e

Lo e

Lo Lo

For homodyne f f

when is sent a

I R P P PP

eI B

when is sent a

I RP eI B

Usally P a few mw P mw P P

58

0 1

1 0

1 0

0 1

2 ( 2 ) 2 2 2

( 2 )

2 2

2

2 2

2

2

( ), ( .252)

( .194)

1

e Lo Lo e Lo e Lo

Lo Lo Lo

Lo Lo

Lo

e Lo

e

e

c

c

eB R P P PP eB RP eB RP

I I R P P PP RP

RP R PP R PP

I IBER Q

R PPQ

eB RP

RPQ

eB

BIf B

PBER Q M where M phf B

eR phf

59

At BER=10-12 r =7 M =49Recall for the optical amplified system (p.262)

at BER=10-12 M=98, coherent PSK has sensitivity about 4~5dB better than opti

cal amplified ook.Disadvantages: 1. receivers are very complicated 2. phase noises must be very small 3. high frequency stability is needed 4. optical phase locked loop is needed 5. polarization must be matchedAdvantages: 1. very good performance 2. less effect by nonlinearity and dispersion 3. DPSK does not need OPLL

2

MBER Q

60

4.4.8 Timing RecoveryTiming circuit produces timing clock for the decisio

n circuit.1. Inband timing2. Outband timing

Reference: Alain Blancharal “Phase-Locked Loops” John Wiley and Sons., 1976 Chapter3

61

General time domain equation

Consider sinusoidal input and output

0 0

( ) sin ( )

( ) cos ( )

i iy t A wt t

y t B wt t

VCO

iy phase detector

1uloop filter

amplifier gain K2

2u0yclock

signal

03 2

dk u

dt

62

For sinusoidal type phase detector

1uideal

0

1 1 0

( ) ( )

( ) sin ( ) ( )i

i

t t

u t k t t

Sinusoidal

63

Let F(iw) be the loop filter transfer function and f(t) be its impulse response

2

2

22 2

21 1

2 2 1

22 2

22 1

2 ( ) 2

( ) 2

2 ( )

2 ( )

2 2 2

( ) ( 2 )

2 2

i ft

i ft

i ft

i ft

i ft

i ft

F i f f t e dt w f

f t F i f e df

U i f u t e dt

U i f u t e dt

U i f K U i f F i f

u t u i f e df

K U i f F i f e df

64

1

2 22 2 1

2 ( )2 1

2

( ) ( ) ( 2 )

( ) ( 2 )

( ) ( )

i f i ft

i f t

u t K u e d F i f e d

K u F i f e dfd

K u f t d

2 2 1( ) ( ) ( )u t K u t f t where ＊ denotes the convolution operationThe output phase of the oscillator is

03 2

3 2 1

1 2 3 2

1 1 0

( )( )

( ) ( )

sin ( ) ( ) ( )

( ) sin ( ) ( )

i

i

d tK u t

dtK K u t f t

K K K t t f t

u t K t t

convolution integral

65

1 2 3

00

0 0

0

00

20 0

2

0

( )sin ( ) ( ) ( )

sin ( ) ( ) ( ) ( )

( ) ( )

( )( ) ( ) ( )

( 2 ) ( )

( 2 ) ( )

(

i

i i

i

i

ft

fti i

let K K K K loop gain

d tK t t f t

dtLinearized equations

t t t t

when t t is small

d tK t t f t

dt

let i f t e dt

i f t e dt

d t

20

20

0

)( 2 )

2 ( 2 )

2 ( )

ft

ft

di f e df

dt dt

i f i f e df

i f t

66

00

0 0

0

0

( )( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )( )

( )

i

i

i

d tRecall K t t f t

dtTake Fourier Transform of both sides

iw iw K iw iw F iw

iw KF iw iw K iw F iw

The linearized transfer function of a phase locked loop is

iw KF iwH iw

i iw iw

0

( )

( ) ( ) ( )i

KF iw

The instantaneous phase error is

t t t

67

0

0

( ) ( ) ( )

( )( )

( )

( ) ( ) ( ) ( )

1 ( ) ( )

( )1 ( )

( ) ( )

( )( )

( )

( )1 ( )

( ) ( )

i

i

i i

i

i

i

iw iw iw

iwRecall H iw

iw

iw iw H iw iw

H iw iw

iw iwH iw

iw iw KF iw

let s iw

KF sH s

s KF s

The error function is

s sH s

s s KF s

68

0

00

0

00

0

( )

( ) 1

( )

( )( ) ( )

( )( ) ( )

( )1 ( )

( )

( )( ) ( )

( ) ( ) ( )

i

i

i

i

i

For the first order loop all pass filter

F s

KH s

s Ks s K K s

dRecall iw s

dtd t

K t K tdt

ssRecall H s

s K s

d t dK t t

dt dtwhere t t t phase error

69

2

1

1

2

1

1 2

0

1( )

1( )

1

1( )

1

, ,

( ) ( ) ( ) 0

( ) ( )

i

i

For the second order loops

sF s

s

or F ss

sor F s

s

We adjust K to abtain

t t t at steady state

In normal operation

t changes with t

70

71

72

4.4.9 Equalization

The equalizer is used to reduce the effect of intersymble interference (ISI) due to pulse spreading caused by dispersion.