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

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

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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 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: A 2 (sin 2 + cos 2 ) = a 1 2 sin 2 1 cos 2 1 1
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