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Dispersion in Single-Mode Fibers The impact of dispersion on the input signal can be summarized as follows: 0t 0 z E ( x, y , z , t ) ( x, y ) e j ( ) s (t , z ) (1) Transverse field Carrier Pulse Envelope distributi on ~ s (t , z ) F 1[ A () H ()] (2) ~ A () s (t ,0) where, “F” stands for the Fourier transform. In this section, we mainly focus on the pulse envelope s(t , z ) . Let us assume that the transverse field distribution of the laser output is same as that of the fiber so that we don’t have to worry about it. Let the pulse envelope of the laser output be si (t ) , i.e. ~ si (t ) s(t ,0) and A() F[si (t )] Si () (3) The fiber can be imagined as a linear system with the transfer function H () . Let the fiber be of length L, inverse group speed 1 and dispersion coefficient 2 . Let the pulse envelope of the output end of the fiber be so (t ) , i.e. s(t , L) so (t ) (4) F[ so (t )] S o () H () S i () si (t ) so (t ) H () Si () S o ( ) H ( ) S i ( ) Step 1: Input pulse envelope si (t ) is given. Take its Fourier transform to obtain Si () . Step 2: Multiply Si () by H () to get the output spectrum, So () . Step 3: Take the inverse Fourier transform of So () to obtain the output pulse envelope so (t ) . Total field distribution at the output is obtained by E ( x, y, L, t ) ( x, y ) e j ( 0t 0 L ) s 0 (t ) (5) Example: (Prob.2 of tutorial 1) Screen Free space Laser L Z=0 Z=L Optical fiber Screen Laser L Fig. a The laser shown in the figure operates at 375 THz. It is turned on for 100ps and then turned off. Sketch the pulse envelope at the screen if the medium is (a) free space (b) fiber. From the tutorial 1, we know that the laser output (ignoring the transverse field distribution) can be written as envelope carrier F (t ,0) rect (t / T0 ) A cos(2f 0t ) f 0 375 THz , T0 100 ps (6) Envelope A Carrier t (a) In tutorial 1, we have seen that the electric field intensity at the screen (z=L) is given by envelope t T1 carrier F (t , L) rect ( ) A cos[2f 0 (t T1 )] T0 T1 L / c, c velocity of light in the free space (7) Let the pulse envelope by s0 (t ) , i.e. t T1 s0 (t ) rect ( ) T0 (8) si (t ) so (t ) t 100 ps 100 ps L/c T1 At the laser At the screen We see that the pulse envelope is delayed by T1 L / c which is the propagation delay. (b) In the case of an optical fiber, first let us consider the case 2 0 , si (t ) rect (t / T0 ) Step 1 : S i ( f ) T0 sin c( f T0 ) H ( f ) e j 2f1L (9) So ( f ) Si ( f )H ( f ) Step 2: T0 sin c( fT0 ) e j 2f1L (10) Recall the Fourier transform property g (t T0 ) G ( f ) e j 2fT 0 (11) Step 3: Using (10), (11), the output pulse envelope can be written as: so (t ) si (t 1L) t 1L rect ( ) T0 (12) si (t ) so (t ) t 100 ps 100 ps 1 L At the laser At the screen When 2 0 , step 2 can be written as So ( f ) T0 sin c( fT0 ) e j 2f1L j ( 2f ) 2 L / 2 2 (13) It is not possible to find the inverse Fourier transform of Eq.(13) and one needs to do it numerically using FFT (Fast Fourier Transform). Please look at the matlab code posted on the web. The effect of dispersion on the input pulse is to broaden the pulse. The output pulse might look like this: si (t ) so (t ) t 100 ps 100 ps 1 L At the laser At the screen Relation between field and power The power/unit area (= optical intensity) of an electromagnetic wave is given by E2 I x 2 (1) where E x is the electric field intensity and is the impedance of a medium. The above relation follows from the Poynting vector. The power flowing through any area A normal to the direction of propagation is E x2 A P (2) 2 2 2 E When complex notation is used, E x should be replaced by x . The electric field intensity in a fiber is given by j ( E x ( x, y, z, t ) ( x, y ) e 0t s(t , z ) 0z) Transverse field Carrier Pulse Envelope distributi on (3) Using Eqs. (2) and (3), we have | s (t , z | 2 2 ( x, y ) A or P K | s (t , z ) | 2 P 2 In this course, we set the constant K to unity to obtain P | s (t , z ) | 2 Fiber Design Considerations Cut-off Wavelength For high bit-rate (>2.5GHz) and long-haul (>100Km) applications, it is absolutely essential that the fiber is single-moded. The single mode condition is given by 2a 2 V (n1 n2 )1 / 2 2.405 2 For examples, if 1.55m , a 4m and (n1 n2 ) 0.1 , V 1.62 . Therefore, this 2 2 1/ 2 fiber is single-moded at this wavelength. However, if 0.7 m corresponding to the optical communication in the visible spectrum, V 3.59 and the fiber is not single- moded at this wavelength. For the given fiber parameters, we can define the cut-off wavelength as 2a (n12 n2 )1 / 2 2 C 2.405 Cut-off wavelength is the frequency above which the optical fiber is single-moded. Fiber Loss Fiber Pin Pout L In an optical fiber, the power attenuation is governed by dP P dz (5) where is the attenuation coefficient and P is the optical power. Solving Eq.(5), dP dz P or ln P z k P k e z If Pin is the input power, k can be calculated as P(0) k Pin If the fiber length is L, the output power can be written as Pout Pin exp( L) The loss power in dB unit through the fiber of length L is given by P loss(dB) 10 log 10 out 10 (L) log 10 e 4.343 L Pin The loss per unit length (dB/Km) 4.343 . Here, is the attenuation coefficient in the unit of Km . 1 Example 1 A fiber of length 40Km has a loss coefficient of 0.046Km . Find the total loss. The loss per unit length (dB/Km) = 4.343 0.046 0.2dB/ Km . Total loss = 0.2 40 8dB

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posted: | 11/13/2011 |

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