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Bit Error Rate Performance Comparison of Equal Channel Spacing and Repeated Unequal Spacing in Reducing the Four Wave Mixing Effect in a WDM Ring Network Md. Nazmul Alam∗ , S. P. Majumder∗, Meer Nazmus Sakib † and Redwan Noor Sajjad ∗ ∗ Department

of Electrical and Electronic Engineering Bangladesh University of Engineering and Technology, Bangladesh Email:[email protected] † Department of Electrical and Electronic Engineering United International University, Bangladesh Email: [email protected]

Abstract— The bit error rate (BER) performance comparison of a wavelength division multiplexing (WDM) ring network in presence of four wave mixing (FWM) using repeated unequal spaced (RUS) and equal spaced (ES) scheme is reported taking into consideration of photo detector shot noise, beat noise components arising out of the beating of the signal with accumulated amplifier’s spontaneous emission (ASE) and crosstalk introduced at each node due to four wave mixing. It is found that RUS provides lower BER than ES for a given number of channels, total bandwidth and number of hops. The performance results evaluated at a bit rate of 10Gbit/s show that for maintaining BER of 10−9 , the maximum number of achievable nodes are approximately 260 and 120 corresponding to receive power level −4dBm and −7dBm respectively for RUS scheme and ES scheme when the hop length is 5km.

transmitted/received power and different number of nodes are evaluated at a bit rate of 10Gbit/s. II. T HEORETICAL A NALYSIS The block diagram of a WDM ring network with cascaded optical amplifier system is shown in Fig. 1. Each node in the network has add and drop facility and capability for selection of a unique wavelength for transmission and reception. Access node structure is shown in Fig. 2. Optical amplifier erbium doped fiber amplifier (EDFA) is located at the input of the node to compensate for fiber and node attenuation. In this paper, a single ring is considered for analysis.

I. I NTRODUCTION The performance of high speed optical WDM systems decreases due to a number of fiber nonlinearity effects, like FWM, self and cross phase modulation, stimulated raman scattering (SRS), chromatic dispersion [1], [2]. Among these FWM is the most dominant nonlinear effect that limits the allowable input power and system capacity by generating new waves, some of which may interfere with the original propagating channels and severely degrade the multichannel network performance [4]. Using of dispersion-shifted fibers (DSFs) in optical WDM systems enable a very high bit rate by keeping the chromatic dispersion at the minimum, but it enhances the generation of FWM waves by reducing phase mismatch between propagating channels. Several method have been proposed to alleviate the effects of crosstalk among which RUS scheme technique have been found to be particularly attractive [3]-[6]. However, to the best of authors knowledge, the comparative analysis between RUS and ES scheme has not yet been reported for a WDM ring network. In this paper the comparative BER performance result of a WDM ring network using ES and RUS scheme for various

Fig. 1.

WDM ring network configuration

It is assumed that all transmitter lasers have a common nominal optical frequency ν o = ωo /2π. Each transmitter

2NASE (1) 2 ∼ nR σs−sp = Ptx  2 ∼ σsp−sp = (4b − 1)

2 σs−sp 4

(5) 2 (6)

where b = Bo /R and R is the bit rate. B. Four Wave Mixing Assume that there are N channels in a given WDM system. For any three co-propagating optical signals with frequencies fi ,fj and fk , the new frequencies generated by FWM are represented by fijk = fi + fj − fk Fig. 2.

Access node architecture

has the same power P tx . The receiver consists of an optical amplifier, followed by a band pass optical filter with bandwidth Bo centered at the carrier frequency ν o . We assumed that the tagged bit of a cell passing though different hops collects noise due to ASE noise and four wave mixing. In this transmission system the power levels in the network are kept same and the amplifier’s gain is exactly set to compensate for the per-hop lose such as G = Lhr Lal Lad Lr Lf

(1)

where, G is the gain of the optical amplifier, L hr header recognition-tapping loss, L al alignment loss, Lad add/drop coupler loss, L f = αL is the fiber loss, α is fiber attenuation , L is hop length. L r is the insertion loss. A. ASE Noise The power spectral density of ASE at the output of an amplifier is hνo ηsp (G − 1) because of its white gaussian characteristics, where h is the Planck’s constant and η sp is the spontaneous emission factor. The accumulated ASE noise power density at one node is NASE (1) = hνo ηsp (G − 1)/Lhr Lal Lad

(2)

The accumulated ASE noise density at nth node is given by NASE (n) = nNASE (1)

(3)

The power of the cells coming from the network at the A/D switch of the access node cannot exceed P sat /Lhr Lal . Hence it is necessary for the newly injected packets to have the same power level as hopping packets, this implies that the transmitter should satisfy Ptx Lhr Lal ≤ Psat

(4)

The normalized variances of noise with signal-ASE beat and ASE-ASE beat are given by [2]

(7)

for i, j, k∈ [1, N ] and k ={i, j} [7]. Considering all possible permutations, N co-operating optical signals, in general give rise up to N 2 (N − 1)/2 new optical signals. Some of the new frequencies fall onto these N original channels. Those new signals are considered as crosstalk. The power of FWM signal for general case of chromatic dispersion can be found as [7]

Pijk

1024π 6 = 4 2 2 n λ c



Dχ1111 Lef f Aef f

2

Pi Pj Pk e−αL η

(8)

where n is the refractive index, λ is the wavelength of the light, c is the speed of the light velocity, P i , Pj , Pk are the input powers of the channels, D is the degeneracy factor, which is 3 for two tone products (f i = fj ) and 6 for three tone products (f i = fj ); χ1111 (= 4 × 10−15 ) is the nonlinear susceptibility. The efficiency η represents the dependency of the FWM efficiency on the phase mismatching Δβ, which can be written as   4eαL sin2 (ΔβL/2) α2 (9) η= 2 1+ 2 α2 + (Δβ) {1 − e−αL } where Δβ = β (fi ) + β (fj ) − β (fk ) − β (fijk )

(10)

where β is the propagation constant. Efficiency η takes a maximum value 1 for Δβ = 0. (m)

If we denote noise component due to FWM as n F , the (m) mean value and variance of n F , are expressed as [4]  (m) =0 (11) nF   2 (m)2 (m) = nF − nF

1 1 1 2 = 2K Pr Pijk + Pijk + Pijk 8 4 4 (m)2

σF

I

II

II

(12) (13)

Here, K is the photodiode responsivity, P ijk is the FWM power, summations I, II and III denote summation for i = j = k, i = j = k and i = j respectively. For simplicity we assume that input power of all channels are equal i.e. P i = Pj = Pk . C. Bit Error Rate Let S (m) and S (s) are signal current for mark and space respectively. The probability density function of S (m) is Gaussian whose mean value and variance, in presence of FWM crosstalk, ASE noise, thermal noise and shot noise are given by  S (m) = KPr (14) (m)2

σ (m)2 = σF

2 2 + σs−sp + σsp−sp + Nth + Nsh

(15)

Here thermal noise n th and short noise n sh are also Gaussian noises. The probability density function of S (s) is whose mean value and variance are

   1 1 S (s) = K Pijk + Pijk (16) 8 4 I

II

2 σ (s)2 = σsp−sp + Nth

(17)

(s)2

Here σF is neglected. Using Personick’s formula for n number of hop BER can be expressed as   Q BER (n) = 0.5erf c √ (18) 2 In the Gaussian approximation the error probability is written as  2

∞ 1 t Pe = √ exp − dt 2 2π Q  (m)   (s)  − S S (19) Q= (m) σ − σ (m)

signal bandwidth even for a large number of channels. For the RUS system, first a base scheme of unequal spacing for a small number of channel is assumed, which is then repeated periodically. Consider a base scheme n  US with US number of channels and adjacent channel spacing of N dUS = {d1 , d2 , d3 , .........., dNU S −1 }. This base scheme is then repeated until the desired number of channels is accumulated . Let us assume that the desired RUS spacing vector is  dRUS = {d1 , d2 , ......, dNU S −1 , d1 , d2 , ......., dNU S −1 , d1 , d2 ..}. Then from here a number of spacings independent from each other is obtained by consecutive (N US − 1) elements from dRUS as follows: d[1] = {d1 , d2 , d3 , .........., dNU S −1 } d[2] = {d2 , d3 , .........., dNU S −1,d1 } d[3] = {d3 , d4 , .........., dNU S −1,d1 d2 } . . . d[NU S −1] = {dNU S −1 , d1 , d2 , d3 , .........., dNU S −2 }

(21)

Now these allocation are to be examined to ensure that all of them fulfill the conditions of the unequal spaced scheme. III. R ESULT AND D ISCUSSION The bit error rate of a WDM ring network is evaluated in presence of ASE and FWM at a bit rate of 10 Gbit/s. The effect of increase of number of hops on BER is shown in Fig. 3 for different power levels and hop length of 5 km for ES channels and RUS channels. As P r is decreased , different BER vs n curves shift downward for ES channels up to Pr = −12dBm and for RUS up to P r = −9dBm after these power level curves will start to shift upward again. The fact is also evident from Fig. 5 which shows the BER of RUS channels and ES channels for 10, 20, 30, 40, 50 hops with hop length is 5 km. At low power levels, noise due to FWM is small and BER decreases as received power is increased. From a certain point, four wave mixing more significant and BER increases with received power.

 Pijk Q=   (m)2 2 2 2 σF + σs−sp + σsp−sp + Nth + Nsh + σsp−sp + Nth But for all power levels the performance of the RUS (20) scheme exceeds the ES scheme. The RUS scheme has a higher allowable power for any given number of hops. D. Using Repeated Unequal Channel in Network Therfore, this scheme ultimately yields more bit rate-distance ES channels have a lot of FWM lights with the frequencies product compared to that for ES channels. coincident with those of original signals. To overcome the problem unequal spaced (US) channel is developed for In Fig. 5 maximum achievable number of hops for BER WDM systems, where transmitted channels are assigned of 10−9 or less is shown against received power level for to different wave lengths such that wavelength separation hop length of 5 km. It is noticed that maximum achievable between channels are unequal and therefore none of the number of hops is approximately 260 and 120 corresponding FWM lights coincide with the original channel frequencies. to received power level −4dBm and −7dBm respectively for But US channels have a wide signal bandwidth for a large RUS scheme and ES scheme. number of channels [8]-[10]. In contrast to these problems, IV. C ONCLUSION RUS channels are analyzed which use periodic allocation of The influence of fiber FWM on WDM ring network has unequal spaced channels and simultaneously achieve a few FWM lights interfering with the channel signals and a narrow been studied. The theoretical treatment describing the perforKPr − K

1  8

I

Pijk +

1 4



II

TABLE I PARAMETERS USED IN THE NUMERICAL CALCULATION

α R Nch Lhr Lal Lad nsp Psat N λ

.25dB/km 10Gbit/s 8 1dB 10dB 3dB .8 10dBm 1.46 1550nm

Maximum number of hops

Attenuation Bit Rate Number of channels Header recognization-tapping loss Alignment loss Add/Drop coupler loss Spontaneous Emission Factor EDFA output saturation power Refractive index Wave length

250

200

150

Repeated Unequal Spacing

100

Equal Spacing

50

−18

−16

−14

−12

−10

−8

−6

−4

−2

Received power (dBm)

Fig. 5. Maximum achievable number of hops vs received power for hop length is 5km and BER = 10−9

−20

10

−6dbm −7dBm −8dBm −9dBm −10dBm

−40

10

10 BER

mance results evaluated at a bit rate of 10 Gbit/s show that for a given transmitted power the number of hops is limited by the combined influence of accumulated ASE and FWM for equal channel spaced and unequal channel spaced dispersion managed fiber. We found that unequal channel spacing reduce the FWM efficiency appreciably. Therefore, a high bit rate WDM ring network system can be designed by controlling the channel spacing to overcome the effect of FWM.

−10dBm

−9dBm −8dBm

−60

−7dBm −80

10

−6dBm

−100

10

−120

Equal Spacing

10

Repeated Unequal Spacing

R EFERENCES

−140

10

10

20

30 40 Number of hops

50

60

Fig. 3. BER vs number of hops for different power level, hop length is 5km

0

10

−50

10

n=50 n=40

BER

n=30

n=50

n=20 −100

10

n=40

n=10

n=30 n=20 n=10

−150

10

Equal Spacing −200

10

Repeated Unequal Spacing

−16

−14

−12

−10

−8

−6

−4

−2

Received Power (dBm)

Fig. 4. 5km

BER vs received power for different number of hops, hop length is

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