Estimation of Optical Link Length for Multi Haul Applications

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1 Estimation of Optical Link Length for Multi Haul Applications M V Raghavendra 1, P L H Vara Prasad 2 Research Scholar Department of Instrument Technology 1, Professor & Chairman (BOS) Department of Instrument Technology 2 College of Engineering, Andhra University 1 College of Engineering, Andhra University 2 Vishakapatnam Andhra Pradesh, India Abstract--In a fiber-optic system at long distances or high data rates, the system can be limited either by the losses (attenuation-limited transmission) or, assuming that the link is not limited by the source or detector speed, by the dispersion of the fiber (dispersion-limited transmission).in this paper we demonstrate how the optimum link length can be determined in different aspects like Dispersion & attenuation. Key Words: Link Length, attenuation limited transmission, & Dispersion limited transmission. I. INTRODUCTION Optical Communication mainly depends on the following aspects. Source Selection Power Budget Dynamic Range Timing Analysis Attenuation-Limited Transmission Length Dispersion-Limited Transmission Distance These are explained in the following sections. II.SOURCE SELECTION The starting point for a link design is choosing the operating wavelength, the type of source (i.e., laser or LED), and the fiber type (single-mode or multimode). In a link design, one usually knows the data rate required to meet the objectives. From this data rate and an estimate of the distance, one chooses the wavelength, the type of source, and the fiber type. A silica-based fiber operating with an LED source in the 800 to 900 nm region has a data rate-distance product of about 150 Mbps.km. The same fiber operating with a laser source in the same region of the spectrum has a product of approximately 2.5 Gbps.km. In the region near 1300 nm, an LED can achieve a product of 1.5 Gbps.km and a laser can achieve products in excess of 25 Gbps.km. These benchmarks are summarized in Table 1 The choice of fiber type involves the decision to use either multimode or single-mode fiber, and, if multimode, whether to use graded-index or step-index profiles. This choice is dependent on the allowable dispersion and the difficulty in coupling the optical power into the fiber. If an LED is chosen, then the obvious choice of fiber is a multimode fiber because the coupling losses into a single-mode fiber are too severe. For a laser source, either a multimode or single-mode fiber can be used. The choice depends on the required data rate, as losses in both types of fiber can be made quite low. TABLE 1 Source data rate-distance performance limits. Type Short λ Long λ LED < 150Mbs -1.km <1.5Gbs -1.km Laser < 2.5Gbs -1.km <25Gbs -1.Km ISSN:

2 III.POWER BUDGET With a tentative choice of source, we know the power P T available to be coupled into the fiber. If receiver power P R is necessary to achieve the required performance, then the ratio P T /P R is the amount of acceptable loss that can be incurred and still meet the specifications. This is expressed by Losses (db) + l M = 10 log (P T /P R ) (1) Where l M is the system margin. The losses can be allocated in any desired fashion by the system designer. Generally, the probable losses will be as follows: The source-to-fiber coupling loss l T (db). The connector insertion loss l C, or the splice insertion loss l S. The fiber-to-receiver loss l R. This loss is usually negligible. Allowance l A for device aging effects and future splicing requirements. Fiber losses, expressed as the unit loss times the link length L. Equation 1 can then be written as 10 log(p T /P R ) = losses + l M = L + l T + nl S + l R + l A + l M. (2) After we solve Eq. 2 for the system margin, we find l M = P T (dbm) - P R (dbm) - L - l T - nl S - l R - l A. (3) A positive system margin ensures proper operation of the circuit; a negative value indicates that insufficient power will reach the detector to achieve the BER. For the source we will select an LED that produces 2 mw (P T (dbm) = 3 dbm) in a spot that is 225 m in diameter. For the fiber, we will use a parabolic graded-index fiber (g = 2) with a 50 m core and a 125 m outer diameter that has a numerical aperture of The effective source radius (r s = m) is larger than the fiber radius (a = 25.0 m), so we use Eq to calculate the coupling efficiency. = [NA(0) 2 ] (a/r s ) 2 [g/(g+2)] = (0.25) 2 (25/112.5) 2 [2/(2+2)] (4) = = % l T = 28.1 db. A considerable loss is incurred in coupling the light from the source into the fiber. This loss is subtracted from the optical power to produce (P T ) fiber = P T - l T = = dbm. (5) The only losses at the receiver are the Fresnel reflection losses at the fiber-air and air-detector interfaces. These losses are approximately 0.2 db per interface for a total loss of 0.4 db at the receiver. Including a representative 6 db allowance to compensate for aging effects, the required power at the receiver is, then, (P R ) fiber = = dbm. (6) A representative loss of graded-index fiber cable might be 5 db/km. The fiber losses are found from L = (P T ) fiber -(P R ) fiber = (-33.6) = 8.5 db. (7) The length of the fiber is L = [(P T ) fiber -(P R ) fiber ]/ = 8.5/5 = 1.70 km. (8) For the 1.7 km distance calculated, three additional joints might be assumed as typical since fiber cables might typically be available in lengths up to 1 km. If we assume splices with a loss of 0.1 db per splice, then the 1.7 km distance is not changed very much since we have L + 3l S =(P T ) fiber - (P R ) fiber (9) L = (P T ) fiber - (P R ) fiber 3l S L = [(P T ) fiber - (P R ) fiber - 3l S ] / = ( )/5 = 1.64 km. For connectors with a loss of 1 db, we have L + 3l C =(P T ) fiber - (P R ) fiber (10) L = (P T ) fiber - (P R ) fiber 3l C ISSN:

3 L = [(P T ) fiber - (P R ) fiber - 3l C ] / = ( )/5 = 1.10 km. We note now that we are below 1 km and only two pairs of connectors are required. Redoing the calculation for two pairs of connectors, we have L + 2l C = (P T ) fiber - (P R ) fiber (11) L = (P T ) fiber - (P R ) fiber 2l C L = [(P T ) fiber - (P R ) fiber 2l C ] / = ( )/5 = 1.30 km. The assumption of two pairs of connectors being needed leads to a length of 1.30 km, a length that requires three pairs of connectors. If we are required to use connectors with 1 db loss per connection, we can achieve 1 km of link length with two connectors. IV.DYNAMIC RANGE Using a "best case/worst case" set of calculations; we can see whether our link has sufficient dynamic range. From Eq. (2), we can write the system margin L M as l M = l TR - l system, (12) where l TR is the ratio of the transmitter power to the required receiver power, expressed in db, and l system is the summation of the system losses, given by l system = L + l T + nl S + l R + l A. (13) The dynamic range of the system is found by computing the maximum and the minimum system margins. The two computations are summarized by (l M ) max = (l TR ) max (l system ) min (14) (l M ) min = (l TR ) min (l system ) max. (15) The system dynamic range DR (db) is given by the difference in these values: DR(dB) = (l M ) max - (l M ) min. (16) The receiver must have an equivalent dynamic range in order for the system to work properly. We are basically concerned with keeping the power at the receiver above the minimum detectable power of the detector (P R ) min and below the maximum-rated power of the detector (P R ) max. From Eq.2, the received power is deduced to be P R (dbm) = P T (dbm) - l system. (17) If me assume that we hane transmitted a logical 1 then Case I: Maximum power output combined with minimum fiber attenuation. The maximum transmitter power allowed in this case is (P T ) max = P R (1) max + min (L - 0.5) (18) = (0.3)( ) = dbm. Case II: Minimum power output combined with maximum fiber attenuation. The minimum transmitter power allowed in this case is (P T ) min = P R (1) min + max (L - 0.5) (19) = (0.63)(2-0.5) = dbm. For the source specified, we can produce a maximum output of -8.4 dbm (144 W) at the maximum drive current of 60 ma. The output power of an LED is linearly dependent on the drive current, P / P max = I / I max (20) To produce the maximum allowed transmitter power of dbm (62.3 W), the required drive current is I = I max (P/P max ) I max = (60 x 10-3 ) (62.3/144) = 26.0 ma. (21) The lowest rated output power at the 60 ma maximum-rated drive current is dbm (33.1 W). Hence, to produce the minimum output of 9.77 W, we require a minimum drive current of I = I min (P/P min ) (22) I min = (60 x 10-3 ) (9.77/33.1) = ma. We conclude that a drive current between the values of and 26.0 ma will ensure proper operation of the 2 m link if the source and receiver meet specifications and if the other sources of system losses are negligible. V.TIMING ANALYSIS The rise time of an fiber-optic system t sys is given by t sys = [ i t i 2 ] 1/2 (23) ISSN:

4 where t i is the rise time of each component in the system. The four components of the system that can contribute to the system rise time are as follows: The rise time of the transmitting source t S. The rise time of the receiver t R. If B 3dB is the 3-dB frequency bandwidth of the receiver, the rise time can be calculated as t R = 0.35/B 3dB. (24) The material-dispersion time of the fiber t mat. Equation 3.24 gives the dispersion relation as t mat = -(L/c).( ).( 2 d 2 n/d 2 ). (25) The modal-dispersion time of the fiber link t modal. For a step-index fiber with length L, the modal-dispersion delay is given by t modal = L (n 1 - n 2 ) / c. (26) The delay time is a function of the index profile g. For a parabolic-index fiber (g = 2), the delay is estimated as t modal = (L/c).(NA 2 (0) / 8n 1 2 ) (27) FIGURE 1 Plot of λ 2 (d 2 n/dλ 2 ) vs. wavelength for typical silica glass. If we consider the 100 Mbps link previously described in the power-budget analysis. The postulated LED might have a rise time of 8 ns and a spectral width of 40 nm. Then the pin diode might have a typical rise time of 10 ns. For a silica fiber operating at 830 nm, the value of 2 (d 2 n/d 2 ) is approximately (from Fig.1). For a link distance of 2.5 km, the material-dispersion delay time is t mat = - (L/c).( ).( 2. d 2 n/d 2 ) = - (2.5 x 10 3 /3.0 x 0 8 )(40/830)(0.024) = x 10-9 s = ns. (28) A typical intermodal dispersion for graded-index fibers is 3.5 ns/km. Hence, a 2.5 km link has t modal = 8.8 ns. (29) Calculating the system's rise time, we have t sys = [( t S ) 2 +( t R ) 2 + ( t mat ) 2 +( t modal ) 2 ] 1/2 (30) = ( ) 1/2 ns = 18.3 ns. If the system rise time of the previous example was 18.3 ns. Using this value, if we calculate the data rate that the system can suppor for NRZ coding as t sys < 0.7 T B (31) T B > t sys / 0.7, B R < 0.7/ t sys < 0.7/(18.3 x 10-9 ) < 38.3 Mbps For RZ coding, we have B R = 0.35/ t sys =[0.35/(18.3x10-9 )]=19.1Mbps (32) Neither coding will support the desired 100 Mbps data rate. Inspection of Eq. 30 reveals that: the receiver speed and the material dispersion are too large; the modal dispersion contribution is small because the distance is so short; and we need to use a faster detector. To reduce the material dispersion, inspection of Eq. 25 reveals that one should reduce. Two methods of doing this would be 1. To use an LED with a longer wavelength (while keeping constant), or 2. To use a laser source with its reduced value of. ISSN:

5 VI.DISPERSION-LIMITED TRANSMISSION DISTANCE (a) Material Dispersion-Limited Transmission Consider material dispersion in a link using RZ coding. We require t mat < 0.35T B. (33) Hence 0.35T B > (L max /c) ( ) ( 2 d 2 n/d 2 ) (34) L max = (0.35T B c) ( )[1/( 2 d 2 n/d 2 )] (b)modal Dispersion-Limited Transmission For modal dispersion in a step-index fiber, we have L max = 0.35cT B / (n 1 - n 2 ) = 0.35c / (n 1 - n 2 )B R = 0.7cn 1 /NA 2 B R (35) For modal dispersion in a graded-index fiber, we have L max = 2.8T B cn 1 2 /[NA(0)] 2 = 2.8cn 1 2 /[NA(0)] 2 B R. (36) These latter three equations are useful for estimating the dispersion-limited transmission distances when waveguide dispersion is not significant. If we calculate the modal-dispersion-limited transmission distance for a 50 Mbps data link using SI and GI fibers with = 1% and n 1 = The coding is return-to-zero, then We begin with n 1 - n 2 = n 1 = 0.01(l.45) = For the SI fiber, (L max ) SI = 0.35c/[(n 1 - n 2 )B R ] = [(0.35)(3.0 x 10 8 )] / [(0.0145)(50 x 10 6 )] =1.448 x 10 2 m. (37) For the GI fiber, (L max ) GI = 2.8cn 1 2 / [NA(0)] 2 B R =2.8cn 1 2 / 2n 1 (n 1 -n 2 )B R = 1.4cn 1 / (n 1 -n 2 )B R (38) = [(1.4)(3.0 x 10 8 )(1.45)] / [(0.0145)(50 x 10 6 )] = 840 m. The small transmission distances are caused by the use of a multimode fiber for a fairly high-data-rate signal instead of a single-mode fiber. (c) Attenuation-Limited Transmission Length For comparison purposes we frequently want to calculate the maximum link distance for a system limited only by the fiber attenuation. The formula for this is L max = [P T (dbm)- P R (dbm)] / fiber (39) Here P T is the power of the transmitter, P R is the power that the receiver requires to maintain the bit-error rate or the signal-to-noise ratio, and is the fiber attenuation value. Let us consider a graded-index fiber with n 1 = 1.45 and = 1% and a loss of 1 db/km. It is used with an 850 nm source that produces an output power (in a fiber) of -10 dbm. The source line width is 60 nm. The receiver is a pin-diode receiver that requires a power given by P R (dbm) = log B R [Mbps] (40) To maintain a BER of 1 x 10-9, where B R is the data rate in Mbps. The coding is RZ coding. We set up the equations to find (a) The material-dispersion-limited distance, The equation for the material-dispersion-limited distance is L max = 0.35c B R 2 d 2 n/d 2. (41) The value of 2 (d 2 n/d 2 is estimated to be from Fig. 1. Hence, we have L max = [(0.35)(3.0 x 10 8 )(850 x 10-9 )] / [(B R )(60 x 10-9 )(0.022)] = (6.76 x ) / B R. (42) (b) The modal-dispersion-limited distance. This distance is L max = 2.8cn 1 2 /[NA(0)] 2 B R = 1.4cn 1 2 /n 1 2 B R (43) = 1.4c/ B R = 1.4(3.0 x 10 8 )/(0.01)(B R ) =4.14 x /B R (c)the attenuation-limited distance as functions of the data rate. This distance is ISSN:

6 L max = [P T (dbm) - P R (dbm)]/ = [-10 - ( log (B R ))]/1.0 (44) = log (B R ). Plotting the results for a data rate range extending from 1 kbps to 1000 Mbps. 2(A) 2(B) 2(C) 2(D) FIGURE 2(A), 2(B), 2(C) & 2(D) Graph between Link Length & Data Rate. VII.RESULTS: Figure 2 shows the plot of the curves & is noted that for data rates below about 17 Mbps, the link length is attenuation-limited. Above 17 Mbps, the link is limited by the modal dispersion. The material-dispersion limit is slightly longer than the modal-dispersion limit. VIII.REFERENCE [1] Gerd Keiser, Optical Fiber Communication, McGraw-Hill International Editions. [2] C K Sarkar, D C Sarkar Optoelectronics & Fiber Optic Communication New Age International Pvt.Ltd. [3] John M. Senior, Optical Fiber Communications Principles and Practice, Second edition, Prentice Hall Publications. [4] Joseph C. Palais, Fiber Optic Communications, Fourth edition, Pearson Education Series, ISSN:

7 [5] Nosu. K, Advanced Coherent Lightwave Technologies, IEEE Communications Magazine, Vol.26, No.2, Feb [6] J.R.Barry and E.A.Lee, Performance of coherent optical receivers, Proceedings of IEEE, Vol. 78, No. 8, August [7] H.R.Burris, A.E.Reed, N.M.Namazi, M.J.Vilcheck, M.Ferraro, Use of kalaman filtering in data detection in optical communication systems with multiplicative noise, Proceedings of IEEE, April About the authors M.V.Raghavendra is a Research Scholar Department of Instrument Technology College of Engineering, Andhra University Vishakapatnam Andhra Pradesh, India. He has received his M.Tech Degree from ECE Dept, College of Engineering, Andhra University. His main research includes signal estimation of optical communication. P.L.H.Vara Prasad received his PhD degree in Instrumentation, & his areas of specialization are Surface Science, Ellipsometry, Control Theory, He is currently working as Professor & Chairman (BOS) Department of Instrument Technology College of Engineering, Andhra University Vishakapatnam Andhra Pradesh, India, He conducted two international conferences & attended nine national conferences. ISSN:

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