hw9_sol - 6.5 Complete the derivation of Eq. (6.2.3)...

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Unformatted text preview: 6.5 Complete the derivation of Eq. (6.2.3) stamngfrom Eq. (6.2.1). What should be the facet reflectiviries to ensure traveling-wave operation of a semiconductor optical amplifier designed Io provide ZOmdB gain. Assume that R: : 21%;. We begin wizh Eq. (6.2.1): q _ WWW GMV) m {1 MGM)2+4G Rle Sin2[fl(V_Vm}/AVL]- To oiatain Eq. (6.2.3), we need the maximum and minimum vaiues of Gpp. The maximum value occurs at v : vm and is given by m (1 ~R1)(1—R2)G(Vm) (z qu/RERQP The minimum value occurs at v 2 vm iAvL/Q and is given by (1—R1)(1——R2}G(vm) (1+GM)?‘ where we assumed thai GUI,” k mag/2) z G(vm). Thus, Gmax (l-i—GVR;R2)2 AG: 2 ——-—--——~—-—-—- . Gmin EWGVRIRz To ensure traveling—wave operation of a semiconductor opticai ampiifier, AG < 2 0r GVR1R2 < 0.17. Using G = 100 Md R1 = 2,123, we Obtain R1 m 0.24% and R3 2 G.E2%. Gmax Gmin m 6.10 A Raman amplifier is pumped in the backward direction using I W Opr-‘WEIZ Find the oufpm power when a. I-nW Signal is injected into she 5-km-long ampiéfien Assume losses of 0.2 and 0.25 dB/km ar the signal andpump wavelengths, respectively, A312; : 50 pmz, and gR : 6 x 10”14 WW Negiect gain saturation. For backward; pumping, we can use Eqs. (6.3.2) and (6.3.3) after changing the Sign of de/dz. Ignor- ing pump depletion, pump equation beeemes dP, E;— m apPp. It should be iniegrated with the boundary condition PAL) : P9. The resufit is Pp(z) = Psexp[-ap(L w 1)]- Plugged ii imo the sigma} equaEion, we obtain 6313‘ 5 = [—415 + fiPgE—QP(LWZ}] dz ap Integrating {his equation over the ampiifier length L, we obtain PAL) L 8R wg 1:5: 1 t W 7+ ‘P PE “’ d . n [Pb-(0)] f0 [ 0; GP 08 Z Afier some simphfication, the final resulE is PAL} 2 Pia-(0) €Xp(gRPOLeff/ap m OESL}, Where Leg 2 (1 — [WEB/a is the effecfive fiber length. We can now calculate the output power using 135(0) m i {,1 W, L m 5 km, cap m ASH : 50 Jumz, 3R m 6 x10"m {El/W, and _0K2dB/km _5 M1 _{).25dB/km as.“ 4.343 m4.605x10 m , ap— 4343 x 5.7564 ><18fl5 m“? The finai aaswer is PAL) 2 146 pw. MI film-am 1.1:;- gm mrnamm m tIJFM Um £91 will] MI I’d-1.3} In drrlIrt' an rmumj’wn'lr mum-using gain in rh: “mill-1mm. EDFhS. U51: uh: ll'rree-lm'el pumping Schfl'fll' Em realizing gain. Elna can Drum cn'rplny l Him-pl»: um lewl mom-I n—sluming mu m: [4:11 lcm1-Jlll'ulhrfl-lfl'c'l ayslum rpump 12ml] min: man-I1.- amply Maui: n1 :1 n-ITIirl mfcrm’ 1h: pumpuJ pupujauun w- :h: cunt-L1 sum- Taking irru-J an:an differ-2n: minim and abmrpliun truss seam-u flu: ll'i: pump uni signal field; It}: ntnmiu pnpulmium N. and H: “My Eqs. “3.4.2; and. (ii-1.3]. The: mall-signal gam turnip-1nd Lu the cast nr a weak signal. Nuglming 1h: :11 Len-mm no:ng IhHr N. I H;- = h]. m can lindfl'; 11-3: sud-Irina 6% .‘II; F " lfl‘nl-V- - H2] - EFF-z HI;- — Fl = 1}. “'11:”? I}... E FF.’I:n.J|1'_:;-.'.|1+r :nluII-m n fwnd In I}: h; = dfiNr-ip |+ta§+n§1fi¢pl The inmll-Hyl'fll gain can l1|T|-'-' hr ubluil'lud mini: F="f'~"'3' {ff-NI = 'Uf+0ff'IV:—¢11N.. ...
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hw9_sol - 6.5 Complete the derivation of Eq. (6.2.3)...

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