Solution Manual for Optical Fiber Communications 4th Edition by Gerd Keiser

Solution Manual for Optical Fiber Communications 4th Edition by Gerd Keiser

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
FOR MORE OF THIS COURSE AND ANY OTHER COURSES, TEST BANKS, FINAL EXAMS, AND SOLUTION MANUALS CONTACT US AT [email protected] Gerd Keiser,  Optical Fiber Communications , McGraw-Hill, 4 th  ed., 2011 Problem Solutions for Chapter 2  2.1 E 100cos 2 10 8 t 30 e x 20 cos 2 10 8 t 50 e y 40cos 2 10 8 t 210 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 x 1  = a 1  cos ( t -  1 )  and  x 2  = a 2  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  1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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: A 2  (sin  + cos ) =  a 1 2 sin 2 1 cos 2 1   a 2 2 sin 2 2 cos 2 2  + 2a 1 a 2  (sin  1  sin  2  + cos  1  cos  2 ) or A 2  =  a 1 2 a 2 2   + 2a 1 a 2  cos ( 1  -  2 ) Dividing (2) by (1) gives tan   =  a 1 sin 1 a 2 sin 2 a 1 cos 1 a 2 cos 2 Thus we can write x = x 1  + x 2  = A cos   cos  t + A sin   sin  t = A cos( t -  ) 2.4 First expand Eq. (2.3) as  E y E 0 y = cos ( t - kz) cos   - sin ( t - kz) sin    (2.4-1) Subtract from this the expression  E x E 0 x cos   = cos ( t - kz) cos  to yield E y E 0 y - E x E 0x  cos   = - sin ( t - kz) sin    (2.4-2) 2
Image of page 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)] =  1 E x E 0x 2       (2.4-3) Squaring both sides of Eq. (2.4-2) and substituting it into Eq. (2.4-3) yields E y E 0 y E x E 0x cos 2 1 E x E 0x 2  sin 2   Expanding the left-hand side and rearranging terms yields E x E 0x 2 E y E 0y 2 - 2  E x E 0x E y E 0y
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern