# Question answers Optical Fiber Communications 4th Edition by Keiser

9. Jul 2020
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### Question answers Optical Fiber Communications 4th Edition by Keiser

• 1. Gerd Keiser, Optical Fiber Communications, McGraw-Hill, 4th ed., 2011 Problem Solutions for Chapter 2 2.1 E  1 00co s 2 1 08 t  3 0 ex  2 0co s 21 08t  5 0 ey  4 0co s 21 08 t  2 10 ez 2.2 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) = 4 (d) At time t = 0 and position z = 4 m we have y = 8 cos [2(-0.8 m -1 )(4 m)] = 8 cos [2(-3.2)] = 2.472 2.3 x1 = a1 cos (t - 1) and x2 = a2 cos (t - 2) Adding x1 and x2 yields x1 + 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 t Since the a's and the 's are constants, we can set a1 cos 1 + a2 cos 2 = A cos  (1) a1 sin 1 + a2 sin 2 = A sin  (2) provided that constant values of A and exist which satisfy these equations. To verify this, first we square both sides and add: 1
• 2. A 2 (sin 2  + cos 2 ) = a 2 2 1  cos 2 1 1 sin + 2 2  cos 2 2 + 2a1a2 (sin 1 sin 2a2 sin 2 or A 2 = a 2  a 2 + 2a a cos ( -  ) 1 2 1 2 1 2 Dividing (2) by (1) gives tan  = a 1 sin   a 2 sin  1 2 a cos   a cos  1 21 2 Thus we can write x = x1 + x2 = A cos  cos t + A sin 2.4 First expand Eq. (2.3) as Ey = cos (t - kz) cos  - sin (t - kz) sin  E0 y Subtract from this the expression E x cos  = cos (t - kz) cos  E0 x to yield Ey - Ex cos  = - sin (t - kz) sin  E 0 y E 0x + cos 1 cos 2)  sin t = A cos(t - ) (2.4-1) (2.4-2) Using the relation cos 2  + sin 2  = 1, we use Eq. (2.2) to write sin 2 (t - kz) = [1 - cos 2 (t - kz)] =  E  1  2 x    E   0x  (2.4-3) Squaring both sides of Eq. (2.4-2) and substituting it into Eq. (2.4-3) yields 2
• 3. E y E 0 y  E x E0x 2 cos   =  E 1  x  E  0x 2       sin 2  Expanding the left-hand side and rearranging terms yields  2  2  E  E  E E    x  + y  - 2 x  y E 0x  E 0y  E 0x E 0y 2.5 Plot of Eq. (2.7). 2.6 Linearly polarized wave. 2.7 Air: n = 1.0 33  33  Glass 90      cos  = sin 2  (a) Apply Snell's law n1 cos 1 = n2 cos 2 where n1 = 1, 1 = 33,and 2 = 90 - 33 = 57  n2 = cos 33 = 1.540 cos 57 (b) The critical angle is found from n glass sin glass = n air sin air with air = 90 and nair = 1.0 critical = arcsin 1 = arcsin 1 = 40.5 n g lass 1.540 3
• 4. 4
• 5. 2.8 Air r Water  12 cm Find c from Snell's law n1 sin 1 = n2 sin c = 1 When n2 = 1.33, then c = 48.75 Find r from tan c = r , which yields r = 13.7 cm. 12 cm 2.9 45  Using Snell's law nglass sin c = nalcohol sin 90 where c = 45 we have n glass = 1.45 = 2.05sin 45 2.10 critical= arcsin n pure = arcsin n doped 1.450 1.460 = 83.3 2.11 Need to show that n1 cos 2  n 2 cos 1  0 . Use Snell’s Law and the relationship tan  sin  cos  5
• 6. 2 2 2.12 (a) Use either NA = n1  n2 or NA  n1 2= n1 2(n1  n2 n 1 1/ 2 = 0.242 ) = 0.243 (b) A = arcsin (NA/n) = arcsin 0.242  1.0  = 14 1  n 2  1 1.00 2.13 (a) From Eq. (2.21) the critical angle is c  sin    sin   41  n  1.50  1    (c) The number of angles (modes) gets larger as the wavelength decreases. 2 2 1/ 2 2 2  (12.14 NA = n1  n2 = n1  n1 = n1 2 1 / 2 2  Since  << 1,  2 << ;  2.15 (a) Solve Eq. (2.34a) for jH: )2 1/ 2 NA  n1  2 jH = j   Er - 1 Hz r  Substituting into Eq. (2.33b) we have j  Er +   E rz =  Solve for Er and let  Er  1 H j z    r  q 2 =  2  -  2 to obtain Eq. (2.35a). (b) Solve Eq. (2.34b) for jHr: jHr = -j  E - 1 Hz Substituting into Eq. (2.33a) we have r 6
• 7. 1 E   j  E + z = - j E r   Solve for E and let q 2 =  2  (c) Solve Eq. (2.34a) for jEr:  1 H z   r -  2 to obtain Eq. (2.35b). 1 1  H z jEr =   jrH Substituting into Eq. (2.33b) we haver    1  EH z z  jrH + = jH  r  r Solve for H and let q 2 =  2  -  2 to obtain Eq. (2.35d). (d) Solve Eq. (2.34b) for jE jE= - 1   H z  jHr r  1 E  H - jHr  z  z r   r Solve for Hr to obtain Eq. (2.35c). Substituting into Eq. (2.33a) we have (e) Substitute Eqs. (2.35c) and (2.35d) into Eq. (2.34c) j 1  Hz Ez   Hz Ez  -  r     = jE z2  q r  r  r r   r   Upon differentiating and multiplying by jq 2 / we obtain Eq. (2.36). 7
• 8. (f) Substitute Eqs. (2.35a) and (2.35b) into Eq. (2.33c) j 1  Ez Hz  Ez Hz     = -jHz- 2   r    q r  r   r r   r     Upon differentiating and multiplying by jq 2 / we obtain Eq. (2.37). 2.16 For  = 0, from Eqs. (2.42) and (2.43) we have Ez = AJ0(ur) e j(t z ) and Hz = BJ0(ur) e j(t z ) We want to find the coefficients A and B. From Eq. (2.47) and (2.51), respectively, we have C = J K   (ua) (wa) A and D = J K    (ua) (wa) B Substitute these into Eq. (2.50) to find B in terms of A: A j 1  a  2 u  1 w2 = B  J' (ua)  K' (wa)    (ua) wK(wa)uJ    For  = 0, the right-hand side must be zero. Also for  = 0, either Eq. (2.55a) or (2.56a) holds. Suppose Eq. (2.56a) holds, so that the term in square brackets on the right-hand side in the above equation is not zero. Then we must have that B = 0, which from Eq. (2.43) means that Hz = 0. Thus Eq. (2.56) corresponds to TM0m modes. For the other case, substitute Eqs. (2.47) and (2.51) into Eq. (2.52): 0 = 1 B jJ (ua)  A uJ' (ua)  u2   a 1  8
• 9. + 1  jJ (ua)  A wK' (wa)J (ua) B      w 2 a 2 K (wa)  With k 2 1 =  2 1 and k 2 2 =  2 2 rewrite this as B =   ja  1    1 1    w u 2 2  (k1 2 J + k2 2 K) A where J and K are defined in Eq. (2.54). If for  = 0 the term in square brackets on the right-hand side is non-zero, that is, if Eq. (2.56a) does not hold, then we must have that A = 0, which from Eq. (2.42) means that Ez = 0. Thus Eq. (2.55) corresponds to TE0m modes. 2.17 From Eq. (2.23) we have  = n2  n2 1 2n12 2 = 1 1  2 n 2 2 n 2 1    << 1 implies n1  n2 Thus using Eq. (2.46), which states that n2k = k2  k1 = n1k, we have 2 k 2  k 2 2 k 2  k 2 2 n 2  n 2 1 1 2.18 (a) From Eqs. (2.59) and (2.61) we have M  2 2 2 a2 n12  n22  2  2 2 a2 NA2 M 1/ 2  1000 1/ 2 0.85m a        30.25m   NA 2  0.22 9
• 10. Therefore, D = 2a =60.5 m (b) 2 2 M  2 30.25m 2  414  0.2 1.32m 2 (c) At 1550 nm, M = 300 2.19 From Eq. (2.58), V = 2 (25 m) 0.82 m (1.48)2  2 1/ (1.46)  2 = 46.5 Using Eq. (2.61)M  V 2 /2 =1081 at 820 nm. Similarly, M = 417 at 1320 nm and M = 303 at 1550 nm. From Eq. (2.72) Pclad  P   total  4 3 M-1/2 = 4 100% 3 1080 = 4.1% at 820 nm. Similarly, (Pclad/P)total = 6.6% at 1320 nm and 7.8% at 1550 nm. 2.20 (a) At 1320 nm we have from Eqs. (2.23) and (2.57) that V = 25 and M = 312. (b) From Eq. (2.72) the power flow in the cladding is 7.5%. 2.21 (a) For single-mode operation, we need V  2.40. Solving Eq. (2.58) for the core radius a V 2 2 1/ 2 2.4 0(1.3 2m) ma = n  n = = 6.55  1 2  22 2 (1.4 80)  (1.4 78) 2 1/ 2 (b) From Eq. (2.23) NA =  2 n 1 n  2 1/ 2 2 =  2 (1.480) 2 1/ 2  (1.478)  = 0.077 (c) From Eq. (2.23), NA = n sin A. When n = 1.0 then 10
• 11. A = arcsin NA =  n  arcsin 0.077  1.0   = 4.4 2.22 n2 = 2  NA 2 = 2 2 = 1.427n1 (1.458) (0.3) a = V = (1.30)(75) = 52 m 2NA 2(0.3) 2.23 For small values of  we can write V  2 a n1 2  For a = 5 m we have  0.002, so that at 0.82 m V  2 (5 0.82 m) m 1.45 2(0.002) = 3.514 Thus the fiber is no longer single-mode. From Figs. 2.18 and 2.19 we see that the LP01 and the LP11 modes exist in the fiber at 0.82 m. 2.25 From Eq. (2.77), Lp = 2  =  n y  n x For Lp = 10 cm For Lp = 2 m Thus ny - nx = ny - nx = 1.3 10 6 101 m 1.3 106 2 m m m = 1.310 -5 = 6.510 -7 6.510 -7 ny - nx 1.310 -5 2.26 We want to plot n(r) from n2 to n1. From Eq. (2.78) n(r) = n1 1 2(r / a) 1 / 2 = 1.48 1 0.02(r / 25) 1 / 2 11
• 12. n2 is found from Eq. (2.79): n2 = n1(1 - ) = 1.465 2.27 From Eq. (2.81)  2 2 2  2an 2 M    1  a k n 2  2    1 where  = n 1  n 2 n 1 = 0.0135 At  = 820 nm, M = 543 and at  = 1300 nm, M = 216. For a step index fiber we can use Eq. (2.61) M step  V2 2 = 1 2a 2   n2  n2 2    1 2 At  = 820 nm, Mstep = 1078 and at  = 1300 nm, Mstep = 429. Alternatively, we can let  in Eq. (2-81): M step = 2an1 2  =  1086 at 820 nm  432 at 1300 nm 2.28 Using Eq. (2.23) we have (a) NA =  2 n 1  n  2 1/ 2 2 =  2 (1.60)  2 1/ (1.49)  2 = 0.58  2 2  1/ 2 = 0.39(b) NA = (1.458)  (1.405) 2.29 (a) From the Principle of the Conservation of Mass, the volume of a preform rod section of length Lpreform and cross-sectional area A must equal the volume of the fiber drawn from this section. The preform section of length Lpreform is drawn into a fiber of length Lfiber in a time t. If S is the preform feed speed, then Lpreform = St. Similarly, if s is the fiber drawing speed, then Lfiber = st. Thus, if D and d are the preform and fiber diameters, respectively, then Preform volume = Lpreform(D/2) 2 = St (D/2) 2 12
• 13. and Equating these yields D St 2 Fiber volume = Lfiber (d/2) 2 = st (d/2) 2 2 d 2 D 2       = st   or s = S  d  2 (b) S = s d 2 D  = 1.2 m/s 0.125 mm  9 mm 2  = 1.39 cm/min 2.30 Consider the following geometries of the preform and its corresponding fiber: 25 m R 4 mm 62.5 m 3 mm FIBER PREFORM We want to find the thickness of the deposited layer (3 mm - R). This can be done by comparing the ratios of the preform core-to-cladding cross-sectional areas and the fiber core-to-cladding cross-sectional areas: A A preform core preform clad = A A fiber core fiber clad or 2 2 )(3  R 2 2 )(4  3 =  (25)2  (62.5)2  (25)2  from which we have R =   7(25)2 1/ 2 = 2.77 mm9 2 2  (62.5)  (25)  13
• 14. Thus, the thickness = 3 mm - 2.77 mm = 0.23 mm. 2.31 (a) The volume of a 1-km-long 50-m diameter fiber core is V = r 2 L =  (2.510 -3 cm) 2 (10 5 cm) = 1.96 cm 3 The mass M equals the density  times the volume V: M = V = (2.6 gm/cm 3 )(1.96 cm 3 ) = 5.1 gm (b) If R is the deposition rate, then the deposition time t is t = M = 5.1 gm = 10.2 min R 0.5 gm / min 2.32 Solving Eq. (2.82) for  yields  K 2  =  where Y =  for surface flaws.Y Thus  = (20 N / mm 3/ 2 )2 (70 MN/ m 2 2 ) = 2.6010 -4 mm = 0.26 m 2.33 (a) To find the time to failure, we substitute Eq. (2.82) into Eq. (2.86) and integrate (assuming that  is independent of time):  f   b / 2 d i t = AY b  b  0 dt which yields 1 1  or t = b1 f  b / 2 1 i  b/ 2  = AYbbt 2 2 f b i  b)/ 2 2 b)/ 2(2 ( (b  2)A(Y) (b) Rewriting the above expression in terms of K instead of yields 14
• 15. 2  2 b Kf 2 b  t = Ki    (b  2)A(Y) b      Y Y  2K 2 b b 2 b 2  i if K orb i << K f (b  2)A(Y) K 2 b  K 2 b i f 15