fiber optics solution

8/7/2019 fiber optics solution

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CHAPTER 1

FillER OPTIC COMMUNICATIONS SYSTEMS

= 10 ioglO (P 2 /P 1 )

Loss Fractional Power

dB) (P2/P I)

0 1 P2/Pl.P1--ill'-o--. P2

-1 0.8 1

- 3 0.510 -1

-6 0.25

-10 0.1 10 -2

-_0 0.011 0 - 3

-30 0.001

40 0.0001 1 0 -4

--0 0.0000110-5

0 -10 -20 -30 -40 -50 dB

= 10 loglO (P 2/P I)

0= loglO (P2/Pl)

_ ;p = 10 dB/lO_ .J.

dB/lO dB/iO_=PI x 10 = 0.001 x 1 0

=2mW

dB/IO -3 -11110: =P1 10 = 2 x 10 x 10 = 0.159 mW

dB/lO -9: = 1 > ]' 10 = 10 x 10

.J. =P2 lO-dB/lO = lOx 10-9 x 1 0 -(-50)/10

-9 5 6 -9 -3= 10 x 10 x 10 = 10 x 10 = 10 W = 1mW

1



1-5 From the text, we find that RG-19/U weighs 1110 kg/km.

1mile of cable x 1110 kg/krn x 1.609 kmlmile x 2.2 Ibs/kg = 3929 Ibs.

1-6 From the text, we find that RG-19/U has an attenuation of 22.6 dB/km at 100MHz.

Using RG-19/U, the allowed loss is:

Loss = 10 10glOh= 10 10glO10-6

= -40 dBP2 10-2

Maximum coaxial cable length =40/22.6 = 1.8 km

Using a fiber with loss, the maximum length of fiber is:

Length = 40/5 = 8 km

1-7 44.7 x 106 bps xl message/64,000 bps = 698 messages

1-8 With manually operated blinker lights, I would guess about 2 or 3 bps.

1-9 Conducting Cable

900 (pairs/cable) x 24 (messages/pair) =21,600 messages

Fiber

144 (fibers/cable) x 672 (messages /fiber) = 96,768 messages

96,768 (fiber cable)/21,600 (copper cable) = 4.48

About 4.5 copper cables are needed to carry the same amount information as thesingle fiber cable.

At the DS-4 rate, each fiber carries 4032 messages. The comparative messagerates are then: 144 x 4032/21600 = 26.88 or about a factor of 27.

2



~ I y = n ( 1 ~ 7 y = 126.67 mm?

D~pert = n ( 7 2 0 Y =3,848.48 mm-

Wavelength (m)8

A =clf = 3 x 10 If

Region of EM

3 x 107

5 x 106

3 x 105

1.5 x 104

3 x 102

0.3

0.03

3 x 10-6

Power

Power

Radio

Radio

Radio

Radio

Microwave

Infrared

gths range from 0.4 urn to 0.7 urn.

- urn, f = 3 x 108mls = 4.3 X 1014 Hz

. 0.7 x 10-6 m

=..li= (7.5 - 4.3) x 1014 = 3.2 x 1014 Hz

3



1-13

W = hf=hc ::: (6.625 x 10-

34

J. S)(3 X 108

m/s)A A

A(!-tm) W(J)

0.6 3.3 X 10-19

0.82 2.4 X 10-19

1.3 1.5 x 10-19

A visible photon has more energy than an infrared photon.

1-14 p::: W/t::: hfN = hNc/A, where N= number of photons/sec

P =6.625xlO-34 xlOlO x3x108 /0.8xlO-6 = 2.5xlO-9W

I :::2.5 X 10-9 W (0.65 AIW) ::: 1 .6 nA

1-15

N::: PA ::: (1 x 10-9W)(1.3 x 1O-

6m) = 6.5 X 109 photons/second

he (6.625 x 10-34 J. S)(3 X 108 m/s]

1-16

Carrier Bit Rate (bps)

10kHz 102

1MHz 104

100 MHz 10 6

lOGHz 108

1u r n 3 X 10 12

For the A=1 I - L m carrier

f = ~ = 3 x 108m/s

A 1 x 10-6 m

:::3 x 1014Hz

4



CARRIER~~~HH~~++++++~~HH~~~++++++~~~ t

t

MODULATED CARRIER

MODULATION

MODULATED CARRIER

= 3 x 108 rnIs = 2.83 x 1014 Hz, BW = O.Olf = 2.83 x 1012 Hz1.06 x 10-6 m

1012Hz X 1 channe1l4000 Hz =7 X 108 channels

" ' i I r n e there are 10 billion (1010) homes each having one 4000 Hz channel, then

mes) X 4000 (Hz/home) = 4 x 1013 Hz is the required bandwidth. Using

_'U:k..--al beam of frequency

1014 Hz

5

- ~ x 1013 = 0.133.. x 1014



1-23 PR = -34 dBm

The bandwidth 13.3% of the carrier frequency. This might be possible.

1-21

(6.4 x 10

4bPS) 14

1010 messages = 6.4 x 10 bpsmessage

A single optical carrier at f ': 3 x 1014 Hz could not be turned on and off this

fast.

1-22 P= 100nW= 100x 10-9W. LetN = number of photons

W=Nhf=Pt

N/t = P/hf = (P/hc) A

(a) At 800 nm

10-7 (08 x 10-6)N/t = . ( ) = 4 X 1011 photons/second

6.63 x 10-34 3 x 108

(b) At 1550 nm

N/t = 4 x 1011 (1.55/.8) =7.8 x 1011 photons/second

(c) The longer waveJlength requires more photons because the energy per

photon is smaller at the longer wavelength.

PT = Transmitted power in dBm

L = -31 dB, system losses

PT = -34 + 31 = -3 dBm

PT =0.5 mW

6



7

5 x 106) = 45 x 10-3 errors/s

000 bps per voice message, yields

PT=4dBm

- -38) = 42 dBm

dBm = 10 log P1(mW)

mW= 1O-9W

dBm = 10 log P2(mW)

;nW = 103 W

--15+1O=35dB

=3 x 108/1.55 X 106 = 1.93548 X 1014

of one percent is a data rate of:

.93548 x 1014) = 1.9354 X 1010 bps = 19.4 Gbps

= 20 Mbps for each HDTV channel

= 19.4 x 10 9 /20 X 106 =0.9677 X 103 =967 video channels



2000

1-31 OC-768 rate is 39,813.12 Mb/s

Number of voice channels is N: -

N=(39,813.12 x 106)/(64 x 103) = 622,080

The actual number is less than this to accommodate overhead such as signaling

and synchronization.

1-32 Photon energy Wp

Wp= hf = hclA = 6.626 x 10-34 x 3 x 10 81A

Wp = 19.878 x 1O -26

/A Joules

IneV

Wp= 19.878 x 10 ,26/ (1.6 x 10,19 A ) = 1.2423 x 1O,6/A

If the wavelength is in nm, the photon energy in eV becomes:

Wp = 1242.3 I A

Photon Energy

1.2

>-CD. . .

0.8GIcw

0.6c

s0.40

s:1l.

0.2

0

0 500 1000

Wavelength

1500

1-33

Wavelength Frequency Energy

1.55 urn 1.935 x 10 14 Hz 0.802 eV

1.55 x 10- 0 mm 1.935 x 101kHz 1.282 x 10'" J

1.55 x 10 -0 m 1.935 x 10" ~vlHz

1.55 x 1O-~km 1.935 x 100 GHz

193.5 THz

8



3x108

-=1.935x1014Hz--=: 6

- -. 1.55x10

2..t2.3 _ 1242.3 = 0.802 eV

- 1 550

9



CHAPTER 2

OPTICS REVIEW

- - - - - t 1 ~ J 4 - - - - -d o 1

tan (Yo12) =x/d.; tan (Yi /2) =x/d;

Thus,tan(I) =~ '=2t -ll~t IYo)~

( )

d' Y I an d an \ 2tan Yo I I

2

Since M = (do/dirl, and tan a=a for small angles, then YiI Yo= 1 1M .

" I i =8.32° using the exact expression or Yi = 8° using the approximate formula.

2-2 NA =nl sin 8 = sin 8

8 NA

0" 0.0

10" 0.174

20" 0.342

30" 0.500

45" 0.707

0.8

0.7

0.60.5

0.4

0.3

0.2

0.1

o o . 10" 2jf 30· 40· 50" 8

10



f=~=_I-

- do ~_ 1

f·

~

M

~

s42.5

1.67

l.0 1

02 3 4 d o f f

0.51

0.33

- = 20 d/{di - 20)

do (mm)

I d; (mm)

I do (mm)infinity

100

60

30

26.7

20

11



r(mm) I I I o

0 1.0

0.2 0.923

0.4 0.726

0.6 0.486

0.8 0.278

1.0 0.135

1.5 0.011

1.0

2.5 f= 20 mm, D = 10 mm, I e = 0.8 urn, d = 2.44 Af 1 0 ; ; ; ; 3.9 urn

2.6 W = 1mm, I e = 0.8 urn, f= 20 mm

Focused spot size= W o ;; ; ;A t 1 nW = 5.09 u rn

-2.0 -1.0 0 1.0 2.0 r(mrrV

2.8 I e =0.8 )lm, W = 1mm, zmoon =3.8 x 1 08m

Divergence Angle =e = = 2lelnw = = 5.09 x 1 0 -4 r

Spot size on the moon = = wo,moon = 8 zmoon! 2 = 96.7 km

Inmiles, wo,moon = 96.7 km x (1 mi.!1.609 km) =60 miles

If R=1km,

W = = 8 R =0.000509 x 103= 0.255 m

o 2 2

I f R = = 10 km,

W = = 8R = = 0.000509 x 104= 255 m

o 2 2 .

12



-ay satellite delay is about half a second. This is noticeable. The

y is not noticeable.

f i- then sin 8t

> sin 8j and then we must have nl >n2'

13



14

2-11 sin 8t = (nl/n2) sin 8; = (1.46/1.48) sin 8; = 0.986485 sin 8;

8; 8t

0 0

15 14.8

30 29.6

45 44.2

60 58.7

75 72.3

90 80.6

90 ~--------,

TRANSMITTED

ANGLE

oo 30 60 90

INCIDENT ANGLE

2-12 sin 8 t = (nl/n2) sin 8; = (1.48/1.46) sin 8; = 1.01369 sin 8;

8;



- ~~-------------,~

TRANSMITTED

ANGLE

o 30 60 90

INCIDENT ANGLE

Fused silica: n =1.46·

Silicon: n =3.5

All wavelengths in run in the following table:

A (Silica) A(Silicon)

548 229

890 371

4431062

The wavelength shortens when an optical beam enters a material from free

space. The frequency remains the same. The photon energy (hi) remains the

same.

Use a refractive index of 1.5. The wave velocity is;

= C = 3 x 108

= 2 x 108 m / sn 1.5

15



1 6

The distance traveled in one second (and, thus; the fiber length) is:

d = vt = 2 x 108X 1= 2 x 10

8m

2-15

Solve for D=2r, with the result:

0 n 2D= 21'= 2w~2 = 1.177w



CHAPTER 3

LIGHTWAVE FUNDAMENTALS

;= ~ : :V I positive (A < 1.3 urn), the longer wavelength travels faster. .6.(-r/L)< 0

-__inegative (A > 1.3 urn), the shorter wavelength travels faster . .6.(-r/L)> 0

"_= 0.85 urn, .6.A= 30 nm, M =90 ps/nm x km

- ) = - M .6.A= - 30 x 90 = - 2700 ps/km

~ rAA=2 nm,

- : : I L ) = - 2 x 90 = - 180 ps/km

-_= 1.55 urn, M= - 20 ps/nm x km, .6.(-r/L)= -M .6.)"

~ rM=30nm,

- : : ' 1 . ) =-M.6.A =- (- 20) x 30 = 600 ps/km

~ r~=2nm,

-L)=-(-20)x2 =40ps/km

- . o d E x L =O.5/.6.(-r/L), Optical.

- = - . o ~ x L = 0.35/.6.('r/L), Electrical.

RRZ xL =O.35/.6.(rlL)

RNRZ x L=0.7/.6.(-r/L)

_= 0.85 urn: (1).6.A = 30 nm, .6.(-r/L)=2.7 ns/km

(2), .6.A=2 nm, .6.(-r/L)=0.18 ns/km

_= I.55 urn: (1) 6.A=30 nm, .6.(-r/L)= 0.6 ns/km

(2) 6.A= 2 nm, .6.(-r/L)= 0.04 ns/km

17



f 3 d B ( o p t ) f 3 d B ( e l e c t ) ( M H z ) R N R Z ( M b / s )

' A t : : . ' A ( M H z ) R R Z ( M b / s )

(urn) (Ill11) 1 0 0 m I k r n I O k r n 1 0 0 m 1krn 1 0 k I n 1 0 0 m 1 k r n I O k r n

0 . 8 5 3 0 1 8 5 0 1 8 5 1 8 . 5 1 3 0 0 1 3 0 1 3 2 6 0 0 2 6 0 2 6

0 . 8 5 2 2 7 7 8 0 2 7 7 8 2 7 7 . 8 1 9 4 4 4 1 9 4 4 1 9 4 3 8 8 8 0 3 8 8 8 3 8 9

1 . 5 5 3 0 8 3 3 0 8 3 3 8 3 . 3 5 8 3 0 5 8 3 5 8 . 3 1 1 6 6 0 1 1 6 6 1 1 7

1 . 5 5 2 1 2 5 0 0 0 1 2 5 0 0 0 1 2 5 0 0 8 7 5 0 0 8 7 5 0 8 7 5 1 7 5 0 0 0 1 7 5 0 0 1 7 5 0

3-5

? 21r(1) 6

kai r = k, = £lL n ==

7.66X

10 rad I m1..0 .82 X 10-6

kglass =ko n = (7.66 X 106) (1.5) = 1.15 X 10

7rad/m

Aglass = Aglass In = 0.8211.5 = 0.55 urn

3-6

When Ax = 20 nm, M I f= 20 1820 = 0.024, or 2.4%

When Ax = 1 nm.Af I f= 1 1820 = 0.0012, or 0.12%

M= 4.39 x 1011 Hz

R = (n l - n2)2 = 0.32nl + n2

3-7 A1 Ga As

I I I = 3.610 loglO 0.68 =-1.67 dB

32%

Transmission loss = 1.67 dB

18



1.0,-----------------n-----.

20 30 40 50 6) 70

In cid en t A n gle, 8 i. (d eg )

il l = 1.48, il2 = 1.46

2 8 ~2 2.28P = -il? COS i+ ill il? - ill sm i, Rp = ip d 2, parallel

p ? 8 .1 2 2' 28il2 cos i+ti I 'V il2 - n I sin i

tl>(>

0! " I : I

0.5_,

8=~

0.00 10

Set Pp = 0 in Eq. (3-29). Solve for 8j = 8B

n4 cos- 8 + n4 siil2 8· = n2 n22 I I I 1 2

19

80'\. 90

8c



8i(deg) a. (m") e-az

0 0.5 1 2 4 z/'A

8c 0 1 1 1 1 1

9.8 x 105,

82 1 .67 .45 .20 .04

84 1.4 x 106 1 .56 .31 .10 .01

86 1.7 x 106 1 .50 .25 .06 .004

88 1.8 x 106 1 .48 .23 .05 .003

90 1.9 x 106 1 .47 .22 .05 .002- -

n~ (I - sin 2 8J + n1 sin2 8j =ny n~

. 28. ( 4 4 ) - 2 2 4SIll I nl - n2 - nl n2 - n2

sin 8 = n2

I , J n T + n~

Thus, 8j is as shown in the triangle, so that

or

3-10

8 sin-l!!2_ = sin" .l.d!i. = 80.57Jlc nl 1.48

k., =21 t = 211: =7.66 x 106

'A .82 x 10- 6

20



O+-------~------~-----:~~~~~Z/Ao 2 3 4

4= -10 log [exp (-0.693 (f/f3_dB )2)]

Let Lf= 1 dB, solve for f/f3_dB :

flf3_dB=0.58

= 4 + 2 cos (comt+ ~ 1 +MI 2) cos (l1~ I 2)

where l1~ = ~2 - ~l and we used the identity

( A + B ) ( A - B )cos A + cos B =2 cos -2- cos -2-

21



22

p

6

4

t

+ cos (Wmt + ~l + n12)

P

(4) PT = 4 + 2 cos (wmt+ ~l + ll~/2) cos (ll~/2)

ll~ 2 cos (ll~ I 2) Pac ' P - P

0 2 4

nl4 1.85 3.7

nl2 1.414 2.8

3nl2 0.77 1.58

n 0 0

o nl2 n ll~

3-13 nl = 1.48

n2=1.465

A=1.3x10-6j.lm

(a) From Eq. 3-29

p = _ 1.4652 cos 85 + 1 48 Y 1.4652 - 1.482 sin285

1.4652 cos 85 + 1.48 V 1.4652 - 1.482 sin2 85



=- 0.187+ j 0.2456 =0.309 ei127.3 =lei74.6

P 0.187+j 0.2456 0.309ei527

E

z

(b) The attenuation factor is given by Eq. (3-35):

a = ko - J n T sin2 8i -n~ = 2n V 1.482 sin2 85 -1.4652

1.3 x 10-6

a = 2n 0.1659 =0.802 radian /um

1.3 x 10-6

The decay is given by:

e -oz = 0.1

az=ln 10

z =2.3/0.802 =2.87 !lm

_ l-f The proof is outlined using Eq. (3-8) in the text.

E = Eo e-az sin (cot - kz)

The intensity is:

1= E 6 e-2az sin2(cot - kz)

Dividing the average intensity at z =L by the average intensity at z =0 yields

the fractional loss:

23



Converting to dB:

Loss (dB) = 10 log e-2aL = -20 a L log e = -8.685 aL

If a is in units of'km'+, then L must be in km, and

Loss (dB)/ L(km) =- 8.685 a

or

dB/km = -8.685 a

3-15 The power is reduced by the factor

Let a = 2 x 10-5 em-1 L = 1 km = 103 m = 105 em,

e-2(2 x 10-5) (10 5) = e - 4 =0.0183 fractional loss

dB = 10 log 0.0183 = -17.4 dB

Alternatively using the conversion between dls/km and a

dB/km = -8.658 a

a =2 x 10 -5 em -1 = 2 x 10 -5 X 10 -5 em =2 km!em km

dB =- 8.085 x 2 =-17.4 dB

3-16 Using

dB/km =- 8.685 a

-0.2 =- 8.685 a

a = 0.023 km -1

3-17 Using Eq. (3 - 17),

6 =- 10 log {e : .693 ( 2 I f3_dB) 2}

24



-0.6 = log e - 2.772 1 d -dB

20.25 =- e -2.7721[3 -dB

24 = e 2.772 1 [3 - dB

2.772/d _dB=ln4= 1.386

f 3- dB = 1.414 GHz Optical 3-dB bandwidth

From Eq. ( 3- 18)

f3-dB (electrical) =0.71 f3-dB (optical)

f3-dB (electrical) = 0.71 x 1.414 = 1 GHz

M = M L ( A _ A 6 )4 A3

Mo = -0.095 ps/(nm 2 x ian)

A = 1290 nm

M = - 0.g95 (1290 _~) = 0.967 ( ps ). 12903nm x ian

1 1 0 =-0.095 ps xnm- x ian

ls ( l n m ) 2 _ l _ k m _1012 ps 10-9 m 103 m

(a) Mo = -0.095 x 103 s/m3

(b) Mo =-0.095 x 10-3 ns/(nm 2 x ian)

~('tIL) = I M ~ A I = 3 ps/km

25



6.A = 2 run

Thus, M = 1.5 ps/nm x km

Assume

A O = 1300 nm

Solve by iteration

-

A A 4 _ 13004 - 63 A3

1320 3.00 x 1012 - 2.856 x 1012 - 0.144 x 1012

= 0.03 x 1012

1315 2.99 x 1012 - 2.856 x 1012 - 0.143x 1012

=0.009 x 1012

A - Ao = 1315 - 1300 = 15 nm

Plot M vs. A from 1275 to 1325 nm to graphically solve this problem.

Extra for this problem:

At: 1310 run

M=.QJ}25_ (1310 _ D i l i t ) = 0.93 ps/nm x km4 13103

At: 1315 nm

M=.QJ}25_ (1315 _ D i l l t . ) = 1.4 ps/nm x km

4 13153

26



:: _1 T =20 ps, soliton pulse width

(a) The maximum rate is:

R = .l= = 0.05 x 10 12 = 50 Gbps

T 20x10-12

(b) The system losses will limit the length of fiber that can be used.

The slope is the value of dM/dA evaluated at AD.

Evaluating at A= Ao yields

(dMldA) =M,

as was to be shown.

IndB, the loss is:

Loss = 10 log P out lPin := 10 log lOyU10

= 10 (yU10) log i0

Loss =yL, Y in dB/km and L in km.

Compare this with the alternative loss calculation

dB = 10 log e-2aL

:= -8.685a

Loss (dB) = 10 log e-2Lyl-8685 = 10 log e023yL = 10(0.23yL) log e

Loss =2.3yL(0.434) =yL

Thus, the two results are identical.

. = -0.5 dBIkm

27



28

This is much less than calculated in problem 3-2, as the dispersion M is much

less at the longer wavelength.

RxL= 0.35 = 0.35 . 83 109

Ia n bl~(r1L) 6x10-12 =5 . x x s

L(km) PoutlPin Efficiency (%)

1 0.89 89

10 0.316 31.6

100 10-5 0.001

3-25 L =0.2 dB/km, 'A = 1550 nm

Use Beer's Law

y = -0.2 dB/km

L(km) PoutlPin Efficiency (%)

1 0.955 95.5

10 0.631 63.1

100 0.01 1

3-26 'A = 1.55 urn, 1 1 ' A =2 nm

M =-3 ps/(nm.km)

l1(r/L) = - M I 1 'A = 3 x 2 = 6 ps/km

3-27 For RZ coding:

RxL=58.3 Gbls

For NRZ coding:



RxL= 0.7 = 0.712 =116.7xl09lanxbls!J.(rIL) 6x10-

RxL=116.7 Gbls

I)0.5 0.5 G I a n

hdB(optica = = 12=83.3 Hz x!J.(rIL) 6xlO-

hdB(electrical) = 0.35 = 58.3 GHz x I a n!J.(rIL)

100m 1 km 10km

- 58 3 G b/s 58 .3 G b/s 5.8 3 G b/s

- 1167 Gb/s 1 16 Gb/ s 11.6 Gb/s-

optical) 833 GHz 83.3 GHz 8.33 GHz

electrical) 583 GHz 58.3 GHz 5.83 GHz

29

fiber optics solution

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