Solution manual for optical fiber communications 4th edition by gerd keiser

Download Full Solution Manual for Optical Fiber Communications 4th Edition by Gerd Keiser https://getbooksolutions.com/download/solution-manual-optical-fibercommunications-4th-edition-by-keiser Gerd Keiser, Optical Fiber Communications, McGraw-Hill, 4th ed., 2011 Problem Solutions for Chapter E  100cos 210 t  30 e x  20cos 210 t  50 e y 2.1 2.2  40cos 2108 t  210 e z The general form is: y = (amplitude) cos(t - kz) = A cos [2(t - z/)] Therefore (a) amplitude = m (b) wavelength: 1/ = 0.8 m-1 so that  = 1.25 m (c)  = 2(2) = 4 (d) At time t = and position z = m we have y = cos [2(-0.8 m-1)(4 m)] = 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: A2 (sin2  + cos2 ) = a 21 sin2 1  cos2 1  + a 22 sin2   cos2   + 2a1a2 (sin 1 sin 2 + cos 1 cos 2) or A2 = a 12  a 22 + 2a1a2 cos (1 - 2) Dividing (2) by (1) gives tan  = a sin1  a sin2 a cos 1  a cos 2 Thus we can write x = x1 + x2 = A cos  cos t + A sin  sin t = A cos(t - ) 2.4 First expand Eq (2.3) as Ey E0 y = cos (t - kz) cos  - sin (t - kz) sin  (2.4-1) Subtract from this the expression Ex cos  = cos (t - kz) cos  E0 x to yield Ey E0 y - Ex cos  = - sin (t - kz) sin  E 0x (2.4-2) Using the relation cos2  + sin2  = 1, we use Eq (2.2) to write   E 2  sin2 (t - kz) = [1 - cos2 (t - kz)] = 1   x    E 0x   (2.4-3) Squaring both sides of Eq (2.4-2) and substituting it into Eq (2.4-3) yields 2 E y  E  x cos  = E   y E 0x    E 2   x  1  E   sin  0x   Expanding the left-hand side and rearranging terms yields 2  E    E x   E y    +   -  E x   y  cos  = sin2  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  90  Glass (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 nglass sin glass = nair sin air with air = 90 and nair = 1.0  critical = arcsin n glass = arcsin = 40.5 1.540 2.8 Air r Water  12 cm Find c from Snell's law n1 sin 1 = n2 sin c = When n2 = 1.33, then c = 48.75 r Find r from tan c = , 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 1.45 nglass = = 2.05 sin 45 n pure 1.450 = 83.3 1.460 2.10 critical = arcsin 2.11 Need to show that n1 cos 2  n cos 1  Use Snell’s Law and the relationship sin  tan   cos  n doped = arcsin 2.12 (a) Use either NA = n12  n22  = 0.242 1/ or NA  n1 2 = n1 2(n1  n ) = 0.243 n1 (b) A = arcsin (NA/n) = arcsin 2.13 0.242  = 14  1.0  n   1.00  (a) From Eq (2.21) the critical angle is  c  sin 1    sin 1    41  1.50   n1    (c) The number of angles (modes) gets larger as the wavelength decreases 2.14 NA = n12  n22  = n12  n12(1 )2  1/ 1/ = n1 2  2  /2 Since 