# Solution Manual Optical Fiber Communications 4th Edition by Keiser

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Link Download Full :https://testbankservice.com/download/solution-manual-optical-fiber-communications-4th-edition-by-keiser  SOLUTION MANUAL FOR OPTICAL FIBER COMMUNICATIONS 4TH EDITION BY GERD KEISER Gerd Keiser, Optical Fiber Communications , McGraw-Hill, 4 th  ed., 2011 Problem Solutions for Chapter 2 2.1 E   1 00co s  2  1 0 8  t   3 0     e x     2 0co s  2  1 0 8 t   5 0     e y     4 0co s  2  1 0 8  t   2 10     e z   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 x 1  and x 2  yields x 1  + x 2  = a 1  [cos  t cos  1  + sin  t sin  1 ] + a 2  [cos  t cos  2  + sin  t sin  2 ] = [a 1  cos  1  + a 2  cos  2 ] cos  t + [a 1  sin  1  + a 2  sin  2 ] sin  t Since the a's and the  's are constants, we can set a 1  cos  1  + a 2  cos  2  = A cos  (1) a 1  sin  1  + a 2  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     A 2  (sin 2     + cos 2    ) = a 2 2  1    cos 2  1  1  sin + 2 2   cos 2  2    + 2a 1 a 2   (sin    1   sin    2 a 2  sin  2 or  A   = 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 2 1 2 Thus we can write x = x 1  + x 2  = A cos   cos  t + A sin 2.4 First expand Eq. (2.3) as E y = cos (  t - kz) cos   - sin (  t - kz) sin  E 0 y Subtract from this the expression E x cos   = cos (  t - kz) cos  E 0 x to yield E y - E x 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 cos2    + sin2    = 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     E y E 0 y   E x E 0x  2 cos    =   E  x  E  0x 2         sin2   Expanding the left-hand side and rearranging terms yields  2  2        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   = sin2   (a) Apply Snell's law n 1  cos  1  = n 2  cos  2 where n 1  = 1,  1  = 33  ,   and  2 = 90   - 33   = 57    n 2 = 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

Mar 17, 2018

#### f_ava_tri_1_mat4

Mar 17, 2018
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