• date post

28-Mar-2015
• Category

## Documents

• view

4.221

16

Embed Size (px)

### Transcript of Optical Fiber Communication - Solution Manual

1Problem Solutions for Chapter 22-1. E 100cos 2108t + 30 ( ) ex+ 20 cos 2108t 50 ( ) ey+ 40cos 2108t + 210( ) ez2-2. The general form is:y = (amplitude) cos(t - kz) = A cos [2(t - z/)]. Therefore(a) amplitude = 8 m(b) wavelength: 1/ = 0.8 m-1 so that = 1.25 m(c) 2 = 2(2) = 4(d) At t = 0 and z = 4 m we havey = 8 cos [2(-0.8 m-1)(4 m)] = 8 cos [2(-3.2)] = 2.4722-3. For E in electron volts and in m we have E = 1.240(a) At 0.82 m, E = 1.240/0.82 = 1.512 eV At 1.32 m, E = 1.240/1.32 = 0.939 eV At 1.55 m, E = 1.240/1.55 = 0.800 eV(b) At 0.82 m, k = 2/ = 7.662 m-1 At 1.32 m, k = 2/ = 4.760 m-1 At 1.55 m, k = 2/ = 4.054 m-12-4. x1 = a1 cos (t - 1) and x2 = a2 cos (t - 2)Adding x1 and x2 yieldsx1 + x2 = a1 [cos t cos 1 + sin t sin 1]+ a2 [cos t cos 2 + sin t sin 2]= [a1 cos 1 + a2 cos 2] cos t + [a1 sin 1 + a2 sin 2] sin tSince the a's and the 's are constants, we can seta1 cos 1 + a2 cos 2 = A cos (1)2a1 sin 1 + a2 sin 2 = A sin (2)provided that constant values of A and exist which satisfy these equations. Toverify this, first square both sides and add:A2 (sin2 + cos2 ) = a12sin21+ cos21( )+ a22sin22+ cos22( ) + 2a1a2 (sin 1 sin 2 + cos 1 cos 2)orA2 = a12+ a22 + 2a1a2 cos (1 - 2)Dividing (2) by (1) givestan = a1sin1+ a2 sin2a1cos1+ a2 cos2Thus we can writex = x1 + x2 = A cos cos t + A sin sin t = A cos(t - )2-5. First expand Eq. (2-3) asEyE0 y= cos (t - kz) cos - sin (t - kz) sin (2.5-1)Subtract from this the expressionExE0 xcos = cos (t - kz) cos to yield EyE0 y- ExE0x cos = - sin (t - kz) sin (2.5-2)Using the relation cos2 + sin2 = 1, we use Eq. (2-2) to write3sin2 (t - kz) = [1 - cos2 (t - kz)] = 1 ExE0x| . ` , 2 ] ] ] (2.5-3)Squaring both sides of Eq. (2.5-2) and substituting it into Eq. (2.5-3) yieldsEyE0 y ExE0xcos ] ] ] 2= 1 ExE0x| . ` , 2 ] ] ] sin2 Expanding the left-hand side and rearranging terms yieldsExE0x| . ` , 2+ EyE0y| . ` , 2- 2 ExE0x| . ` , EyE0y| . ` , cos = sin2 2-6. Plot of Eq. (2-7).2-7. Linearly polarized wave.2-8. 33 33 90 GlassAir: n = 1.0(a) Apply Snell's lawn1 cos 1 = n2 cos 2where n1 = 1, 1 = 33, and 2 = 90 - 33 = 57 n2 = cos 33cos 57= 1.540(b) The critical angle is found fromnglass sin glass = nair sin air4with air = 90 and nair = 1.0 critical = arcsin 1nglass= arcsin 11.540= 40.52-9AirWater12 cmrFind c from Snell's law n1 sin 1 = n2 sin c = 1When n2 = 1.33, then c = 48.75Find r from tan c = r12 cm , which yields r = 13.7 cm.2-10.45 Using Snell's law nglass sin c = nalcohol sin 90where c = 45 we havenglass = 1.45sin 45= 2.052-11. (a) Use either NA = n12 n22( )1/ 2= 0.242or5NA n1 2 = n12(n1 n2)n1= 0.243(b) 0,max = arcsin (NA/n) = arcsin 0.2421.0| . ` , = 142-13. NA = n12 n22( )1/ 2= n12 n12(1 )2[ ]1/ 2= n1 2 2( )1 / 2Since 0, we havehB(t) = 12C ej 2ft12RC+ jf| . ` , df= 1Ce-t/RC7-3. Part (a): hp (t) dt = 1Tb -Tb/2Tb/2 dt = 1Tb.

|,

` Tb2 + Tb2 = 12Part (b): hp (t) dt = 12 1Tb exp t22(Tb)2 ] ] ] dt= 121Tb Tb 2 = 1 (see Appendix B3 for integral solution)Part (c): hp (t) dt = 1Tb exp tTb ] ] ] dt0= - e e0[ ]= 17-4. The Fourier transform isF[p(t)*q(t)] = p(t)* q(t)e j2 ft dt= q(x) p(t x) e j2ft dt dx= q(x) e j2fx p(t x) e j2f( t x) dt dx= q(x) e j2fx dx p(y) e j 2fy dywhere y = t - x= F[q(t)] F[p(t)] = F[p(t)] F[q(t)] = P(f) Q(f)7-5. From Eq. (7-18) the probability for unbiased data (a = b = 0) isPe = 12 P0(vth) + P1(vth) [ ].Substituting Eq. (7-20) and (7-22) for P0 and P1, respectively, we havePe = 12 122 ev2/ 22 dvV/ 2+ e (vV)2/ 22 dvV / 2 ] ] ] In the first integral, let x = v/ 22so that dv = 22dx.In the second integral, let q = v-V, so that dv = dq. The second integral thenbecomes3eq2/ 22dqV / 2 V= 22 e x2 dxV / 2 22where x = q/ 22ThenPe = 12 2222 e x2 dxV/ 2 22+ e x2 dx V/ 2 22 ] ] ] = 12 ex 2 dx 2 ex2 dx0V / 2 22 ] ] ] Using the following relationships from Appendix B, e p2x 2 dx= pand 2 ex 2 dx0t= erf(t), we havePe = 12 1 erf V2 2| . ` , ] ] ] 7-6. (a) V = 1 volt and = 0.2 volts, so that V2 = 2.5. From Fig. 7-6 for V2 = 2.5,we find Pe 710-3 errors/bit. Thus there are (2105 bits/second)(710-3errors/bit) = 1400 errors/second, so that11400 errors/second = 710-4 seconds/error(b) If V is doubled, then V2 = 5 for which Pe 310-7 errors/bit from Fig. 7-6.Thus1(2 105bits / sec ond)(3 107errors / bit )= 16.7 seconds/error7-7. (a) From Eqs. (7-20) and (7-22) we haveP0(vth) = 122 e v2/ 22 dvV / 2= 12 1 erf V2 2| . ` , ] ] ] and4P1(vth) = 122 e( vV)2/ 22dvV / 2= 12 1 erf V2 2| . ` , ] ] ] Then for V = V1 and = 0.20V1P0(vth) = 12

]]]1 - erf.

|,

`12(.2) 2 = 12

]]]1 - erf.

|,

` 20.8 = 12 [ ]1 - erf( ) 1.768 = 12 ( ) 1 - 0.987 = 0.0065Likewise, for V = V1 and = 0.24V1P1(vth) = 12

]]]1 - erf.

|,

`12(.24) 2 = 12

]]]1 - erf.

|,

` 20.96 = 12 [ ]1 - erf( ) 1.473 = 12 ( ) 1 - 0.963 = 0.0185(b) Pe = 0.65(0.0185) + 0.35(0.0065) = 0.0143(c) Pe = 0.5(0.0185) + 0.5(0.0065) = 0.01257-8. From Eq. (7-1), the average number of electron-hole pairs generated in a time t isN = Eh = Pthc/ = 0.65(25 1010W)(1 109s)(1.3 106m)(6.6256 1034Js)(3 108m / s)= 10.6Then, from Eq. (7-2)P(n) = Nn e-Nn! = (10.6)5 e-10.65! = 133822120 e-10.6 = 0.05 = 5%7-9. vN= vout- voutvN2= vout vout[ ]2= vout2-2 vout2+ vout2= vout2- vout27-10. (a) Letting = fTb and using Eq. (7-40), Eq. (7-30) becomes5Bbae = Hp(0)Hout(0)2 Tb2 Hout'()Hp'()20 dTb= I2Tbsince Hp(0) = 1 and Hout(0) = Tb. Similarly, Eq. (7-33) becomesBe = Hp(0)Hout(0)2 Hout(f)Hp(f) 1 + j2fRC ( )20 df= 1Tb2 Hout(f)Hp(f)20 1 + 42f2R2C2( )df= 1Tb2 Hout(f)Hp(f)20 df + 2RC ( )2Tb2 Hout(f)Hp(f)20 f2 df = I2Tb + (2RC)2T3b I3(b) From Eqs. (7-29), (7-31), (7-32), and (7-34), Eq. (7-28) becomes< >v2N = < >v2s +< >v2R +< >v2I +< >v2E = 2q < >i0 < > m2 BbaeR2A2 + 4kBTRb BbaeR2A2 + SIBbaeR2A2 + SEBeA2= .

|,

`2q < >i0 M2+x + 4kBTRb + SI BbaeR2A2 + SEBeA2= R2A2I2Tb .

|,

`2q < >i0 M2+x + 4kBTRb + SI + SER2 + (2RCA)2T3b SEI37-11. First let x v boff( )/ 2 off( ) with dx dv / 2 off( ) in the first part of Eq. (7-49):Pe = 2off2off exp x2( )vth boff2off dx = 1 exp x2( )Q/ 2 dxSimilarly, let y v + bon( )/ 2 on( ) so thatdy = -dv/ 2on( )in the second part of Eq. (7-49):6Pe = - 1 exp y2( ) vth+bon2on dy = 1 exp y2( )Q/ 2 dy7-12. (a) Let x = V2 2 = K2 2 For K = 10, x = 3.536. ThusPe = e-x22 x = 2.9710-7 errors/bit(b) Given that Pe = 10-5 = e-x22 x then e-x2 = 2 10-5 x.This holds for x 3, so that K = 2 2 x = 8.49.7-13. Differentiating Eq. (7-54) with respect to M and setting dbon/dM = 0, we havedbondM = 0= - Q(h/ )2 M2+ xh| . ` , bonI2+ W ] ] ] 1/ 2+ M2+xh| . ` , bonI2(1 ] ] ] + W ' ' 1/ 2+ Q(h/ )M(h/ )12(2 + x)M1+ x bonI2M2+ xh| . ` , bonI2+ W ] ] ] 1/ 2+12 (2 + x)M1+x bonI2(1 )M2+xh| . ` , bonI2(1 ) + W ] ] ] 1/ 2 ' ' 1 / 2Letting G = M2+x.

|,

` h bonI2 for simplicity, yields''( ) G + W1/2 + [ ] G(1-) + W1/2 = G2 (2 + x)'' 1(G + W)1/2 + (1-)[ ] G(1-) + W1/2 Multiply by (G + W)1/2 [ ] G(1-) + W1/2 and rearrange terms to get(G + W)1/2

]]]W - Gx2 (1-) = [ ] G(1-) + W1/2 .

|,

`Gx2 - W 7Squaring both sides and collecting terms in powers of G, we obtain the quadraticequationG2

]]] x24 (1-) + G

]]] x24 W(2-) - W2(1+x) = 0Solving this equation for G yieldsG = - x24 W(2-) t ''x416 W2(2-)2 + x2(1-)W2(1+x)12x22 (1-) = W(2-)2(1-) ''1 -

]]]1 + 16 .

|,

`1+xx2 1-(2-)212 where we have chosen the "+" sign. Equation (7-55) results by lettingG = M2+xopt bonI2 .

|,

` h 7-14. Substituting Eq. (7-55) for M2+xbon into the square root expressions in Eq. (7-54) and solving Eq. (7-55) for M, Eq. (7-54) becomesbon = Q .

|,

` hb1/(2+x)on

]]] h W(2-)2I2(1-) K1/(2+x) ''

]]] W(2-)2(1-) K + W12 +

]]]W2 (2-)K + W12 Factoring out terms:b(1+x)/(2+x)on = Q.

|,

` h(1+x)/(2+x) Wx/2(2+x) I1/(2+x)2 ''

]]] (2-)2(1-) K + 112 +

]]]12 (2-)K + 112

]]] (2-)K2(1-)1/(2+x) or bon = Q(2+x)/(1+x).

|,

` h Wx/2(1+x) I1/(1+x)2 L87-15. In Eq. (7-59) we want to evaluatelim 1(2 )K2(1 )L ] ] ] 1+x2+x= lim 1(2 )K2(1 ) ] ] ] 1+x2+x lim 11L| . ` , 1+x2+xConsider firstlim 1(2 )K2(1 ) ] ] ] = lim 1 (2 )2(1 )1 + 1 + B (1 )(2 )2 ] ] ] 12 ' ' where B = 16(1+x)/x2 . Since 1, we can expand the square root term in abinomial series, so thatlim 1(2 )K2(1 ) ] ] ] = lim 1 (2 )2(1 )1 + 1 + 12 B (1 )(2 )2 Order(1 )2 ] ] ] ' ' = lim 1 B4(2 )= B4 = 4 1+xx2 Thus lim 1(2 )K2(1 ) ] ] ] 1+x2+x= .

|,

`4 1+xx21+x2+x Next consider, using Eq. (7-58) lim 1 1L| . ` , 1+ x2+x= lim 1(2 )K2(1 ) ] ] ] 1/(2+ x) (2 )2(1 )K+1 ] ] ] 12+ 12(2 )K+1 ] ] 12 ' ' From the above result, the first square root term is

]]] (2-)2(1-) K + 112 =

]]]4 1+xx2 + 112 = .

|,

` x2 + 4x + 4x212 = x + 2x From the expression for K in Eq. (7-55), we have that lim 1 K = 0, so that9lim 112(2 )K+1 ] ] 12= 1 Thuslim 1(2 )2(1 )K+1 ] ] ] 12+ 12(2 )K+1 ] ] 12 ' ' = x + 2x + 1 = 2(1 + x)x Combining the above results yieldslim 1(2 )K2(1 )L ] ] ] 1+x2+x= .

|,

`4 1+xx21+x2+x .

|,

`4 1+xx212+x 2(1 + x)x = 2x so that lim 1 M1+xopt = W1/2QI2 2