Errata
Context
Present version in the book
Corrected/changed version
5.27 nm
527 nm
Gd = L/[2(M – 1) + L]
Gd = L/[2(M – 1 + L)]
Page 572, Exercise 14.6.
Γ = 0 24 40 50 24 0 24 40 24 24 0 0 50 0 40 0.
Γ = 0 50 25 60 25 0 50 60 25 30 0 30 25 50 30 0.
Page 593, Exercise 15.7
0.1 µs
0.8 µs
Page 130, Exercise 2.7 Page 248, expression for Gd below Eq. 6.5.
ii
Exercise Problems and Solutions for Chapter 2 (Technologies for Optical Networks) 2.1 A step-index multi-mode optical fiber has a refractive-index difference Δ = 1% and a core refractive index of 1.5. If the core radius is 25 µm, find out the approximate number of propagating modes in the fiber, while operating with a wavelength of 1300 nm. Solution: Δ = 0.01, n1 = 1.5, a = 25 μm, w = 1300 nm, and the number of modes Nmode is given by 𝑉𝑉 2
𝑁𝑁𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 2 , with 𝑉𝑉 =
The numerical aperture NA is obtained as
2𝜋𝜋𝜋𝜋 𝑁𝑁𝑁𝑁. 𝑤𝑤
𝑁𝑁𝑁𝑁 = �𝑛𝑛12 − 𝑛𝑛22 ≈ 𝑛𝑛1 √2∆ = 1.5√0.02. Hence, we obtain V parameter as, 2𝜋𝜋 × 25 × 10−6 × �1.5√0.02� = 25.632, 1300 × 10−9 leading to the number of modes Nmode , given by 𝑉𝑉 =
𝑁𝑁𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 =
25.6322 ≈ 329. 2
2.2 A step-index multi-mode optical fiber has a cladding with the refractive index of 1.45. If it has a limiting intermodal dispersion of 35 ns/km, find its acceptance angle. Also calculate the maximum possible data transmission rate, that the fiber would support over a distance of 5 km. Solution: The cladding refractive index n2 =1.45, and the intermodal dispersion Dmod = 35 ns/km. Dmod is
expressed as 𝐷𝐷𝑚𝑚𝑚𝑚𝑚𝑚 ≈
𝑛𝑛1 Δ 𝑛𝑛1 𝑛𝑛1 − 𝑛𝑛2 𝑛𝑛1 − 𝑛𝑛2 = � �= = 35 ns/km. 𝑐𝑐 𝑛𝑛1 𝑐𝑐 𝑐𝑐
Hence, (n1 – n2) = cDmod = (3 × 105) × (35 × 10-9) = 0.0105, and n1 = n2 + 0.0105 = 1.4605. Therefore, we obtain NA as 𝑁𝑁𝑁𝑁 = �𝑛𝑛12 − 𝑛𝑛22 = �1.46052 − 1.452 = 0.174815,
and the acceptance angle is obtained as θA = sin-1(NA) = sin-1(0.174815) = 10.068o. The pulse spreading due to dispersion should remain ≤ 0.5/r, with r as the data-transmission rate, implying that r ≤ 0.5/(Dmod L). Hence, we obtain the maximum possible data transmission rate rmax over L = 5 km as 𝑟𝑟𝑚𝑚𝑚𝑚𝑚𝑚 =
0.5 = 2.86 Mbps. 35 × 10−9 × 5
2.3 Consider that a step-index multi-mode optical fiber receives optical power from a Lambertian source with the emitted intensity pattern given by I(θ) = I0 cosθ, where θ is the angle subtended by an incident light ray from the source with the fiber axis. The total power emitted by the source is 1 mW while the power coupled into the fiber is found to be - 4 dBm. Derive the relation between the
2.1
launched power and the numerical aperture of the optical fiber. If the refractive index of the core is 1.48, determine the refractive index of the cladding. Solution: Transmit power PT = 1 mW, and the power coupled into fiber PC = - 4 dBm = 10- 0.4 W = 0.3981 mW. For a Lambertian source, the coupled power PC = NA2 × PT X (for derivation, see Cherin 1983). Hence, 𝑃𝑃 NA2 = PC/PT = 0.3981. Further, 𝑁𝑁𝑁𝑁2 = 𝑛𝑛12 − 𝑛𝑛22 = 𝑃𝑃 𝐶𝐶 implying that n22 = n12 – PC/PT . 𝑇𝑇
Thus, we obtain n2 as
𝑛𝑛2 = √1.482 − 0.3981 = 1.34. 2.4 Consider a 20 km single-mode optical fiber with a loss of 0.5 dB/km at 1330 nm and 0.2 dB/km at 1550 nm. Presuming that the optical fiber is fed with an optical power that is large enough to force the fiber towards exhibiting nonlinear effects, determine the effective lengths of the fiber in the two operating conditions. Comment on the results. Solution: With L = 20 km, first we consider the case with fiber loss αdB = 0.5 dB/km. So, the loss α in neper/km is determined from αdB = 10log10[exp(α)] as
α = ln (10αdB/10) = ln(100.05) = 0.1151. Hence, we obtain the effective fiber length as Lef f = [1 – exp(-αL)]/α = [(1 – exp(-0.1151 × 20)]/0.1151 = 7.82 km. With αdB = 0.2 dB/km, we similarly obtain Lef f = 13.06 km, which is expected because with lower attenuation, the power decays slowly along the fiber and thus the fiber nonlinearity effects can take place over longer fiber length. 2.5 Consider an optical communication link operating at 1550 nm over a 60 km optical fiber having a loss of 0.2 dB/km. Determine the threshold power for the onset of SBS in the fiber. Given: SBS gain coefficient gB = 5 ×10-11 m/W, effective area of cross-section of the fiber Aeff = 50 µm2, SBS bandwidth = 20 MHz, laser spectral width = 200 MHz. Solution: With αdB = 0.2 dB/km at 1550 nm, we obtain α = ln (10αdB/10) = ln(100.02) = 0.0461. Hence, for L = 60 km, we obtain Lef f as Lef f = [1 – exp(-αL)]/α = [1 – exp(-0.0461 × 60)]/0.0461 = 20.327 km. With Aeff = 50 μm , gB = 5 × 10 m/W, and assuming the polarization-matching factor to be ηp = 2, we obtain the SBS threshold power as 2
𝑃𝑃𝑡𝑡ℎ (𝑆𝑆𝑆𝑆𝑆𝑆) =
-11
21 𝜂𝜂𝑝𝑝 𝐴𝐴𝑒𝑒𝑒𝑒𝑒𝑒 𝛿𝛿𝛿𝛿 21 × 2 × 50 × 10−12 200 �1 + �= �1 + � = 22.73 mW. −11 3 𝑔𝑔𝐵𝐵 𝐿𝐿𝑒𝑒𝑒𝑒𝑒𝑒 5 × 10 × 20.327 × 10 Δ𝜔𝜔𝐵𝐵 20
2.6 Consider an optical communication link operating at 1550 nm over a 60 km optical fiber having a loss of 0.2 dB/km. The effective area of cross-section of the fiber Aeff = 50 µm2, where an optical power of 0 dBm is launched. Determine the nonlinear phase shift introduced by SPM in the fiber. Given: ñ(ω) = 2.6 × 10-20 m2/W.
2.2
Solution: The transmit power PT = 1 mW and the effective area of cross-section of the fiber is Aeff = 50 μm2 Hence, the intensity of launched light I = PT/Aeff = 10-3/(50 × 10-12) = 2 × 107 W/m2. With αdB = 0.2 dB/km., α = 0.0461 and Leff = 20.327 km (see Exercise 2.5). Further, the nonlinearity parameter 𝜉𝜉of the fiber is given by 𝜉𝜉 =
𝑛𝑛� (𝜔𝜔)𝜔𝜔 𝑐𝑐
=
2𝜋𝜋 �𝑛𝑛 (𝜔𝜔) 𝑤𝑤
with ñ(ω) = 2.6 × 10-20 m2/W and w = 1550 nm.
Therefore, the phase shift Δ𝜙𝜙𝑆𝑆𝑆𝑆𝑆𝑆 introduced by SPM in the optical fiber is obtained as Δ𝜙𝜙𝑆𝑆𝑆𝑆𝑆𝑆 = 𝐿𝐿𝑒𝑒𝑒𝑒𝑒𝑒 𝜉𝜉 𝐼𝐼 = (20.327 × 103 ) ×
2𝜋𝜋×2.6×10 −20 1.55×10 −6
× (2 × 107 ) = 0.04285 radian = 2.455o
2.7 Determine the Bragg wavelength of an integrated-optic grating with a period of 527 nm (earlier it was 5.27 nm in the book) and an effective refractive index of 1.47. Sketch a block schematic using Bragg grating and other relevant components for an optical add-drop multiplexer for three wavelengths at a given node in a WDM ring network. Solution: The period of grating is Λ = 527 nm, and the effective refractive index is neff = 1.47. Hence the Bragg wavelength wb of the integrated-optic grating is obtained as 𝑤𝑤𝑏𝑏 = 2𝑛𝑛𝑒𝑒𝑒𝑒𝑒𝑒 Λ = 2 × 1.47 × 0.527 = 1549.38 nm.
Using Bragg grating one can realize an add-drop multiplexer (OADM) for three wavelengths w1, w2,
w3 (w2 as the drop wavelength) as shown in the following. Circulator Bragg grating
w1, w2, w3 w2
Coupler
w1, w2, w3
w2
OADM using Bragg grating
2.8 A laser cavity (i.e., an active layer in DH configuration) has a length L with a medium loss α dB per unit length, which is pumped with a gain g per unit length using an appropriate forward bias current. The walls on two ends of the cavity are designed with reflectivities r1 and r2, and the cavity has a confinement factor of Γ. Determine the condition to be satisfied by the pumped cavity to function as a laser. Give a typical sketch of g as a function of wavelength, and explain its impact on the laser spectrum. Solution: The DH laser cavity operates as shown in the following diagram with the two reflecting walls (shaded parts).
r1
Cavity (α, g, Γ)
r2
L Total loss-cum-gain ρLG in the cavity must be ≥ 1, with ρLG given by 𝜌𝜌𝐿𝐿𝐿𝐿 = 𝑟𝑟1 𝑟𝑟2 𝑒𝑒𝑒𝑒𝑒𝑒{−(𝛼𝛼 − 𝑔𝑔Γ)𝐿𝐿)}. 2.3
Hence, the condition that the cavity must satisfy to function as a laser is expressed as 𝑟𝑟1 𝑟𝑟2 𝑒𝑒𝑒𝑒𝑒𝑒{−(𝛼𝛼 − 𝑔𝑔Γ)𝐿𝐿)} ≥ 1,
leading to the expression for the minimum cavity gain, given by 𝑔𝑔𝑚𝑚𝑚𝑚𝑚𝑚 =
𝛼𝛼 ln (𝑟𝑟 𝑟𝑟 ) − 𝐿𝐿Γ1 2 . Γ
g (w)
w
g(w) shapes the spectrum and thus reduces the number of modes in a laser. 2.9 What is the fundamental difference between the spectral spreads in lasers due to the phase noise and modulation? Determine the spectral spread of a 1550 nm laser transmitting at 10 Gbps with the unmodulated linewidths of (i) 200 MHz and (ii) 0.08 nm. Solution: The spectral spread due to phase noise is an internal phenomenon in a laser, arising from the underlying spontaneous noise and finite bandwidth of the laser cavity. On the other hand, the spectral spread due to the modulation is an external effect from the phase, frequency, amplitude, intensity modulation (governed by the modulation scheme) by the input data stream and hence is related to the bit rate of the data stream. (i) Laser linewidth BL = 200 MHz, modulation bandwidth BM = 10 GHz (considering intensity modulation). Hence the total spectral spread BT of the laser after modulation (3-dB bandwidth) is obtained as 𝐵𝐵𝑇𝑇 = �𝐵𝐵𝐿𝐿2 + 𝐵𝐵𝑀𝑀2 = �(200 × 106 )2 + (10 × 109 )2 = √100.04 × 1018 ≈ 10 GHz.
(ii) Laser linewidth is given in terms of wavelength spread δw = 0.08 nm, which leads to a frequency spread BL, given by 𝐵𝐵𝐿𝐿 =
𝑐𝑐𝑐𝑐𝑐𝑐 3×10 8 ×0.08×10 −9 = (1.55×10 −6 )2 = 9.99 GHz ≈ 10 GHz. 2 𝑤𝑤
Using this value for BL, we obtain the total laser spectral spread as
𝐵𝐵𝑇𝑇 = �𝐵𝐵𝐿𝐿2 + 𝐵𝐵𝑀𝑀2 = �(10 × 109 )2 + (10 × 109 )2 = 14.14 GHz.
2.10 The threshold current density of a stripe-geometry AlGaAs laser is 3000 A/cm2 at 15oC. Estimate the required threshold current at 50oC, when the laser characteristic temperature T0 = 170oK and the contact stripe of the laser has a size (area) of 20 μm × 100 μm. Solution: The threshold current Ith (T) in laser has a strong dependence on temperature, given by
Let T1 = 50oC and T2 = 15oC.
𝑇𝑇 𝑇𝑇0
𝐼𝐼𝑡𝑡ℎ (𝑇𝑇) ∝ exp � �.
At T = T2, the current density Jth (T2) = 3000 A/cm2 = 3 × 107 A/m2. Using the above relation between Ith with T (and hence between Jth with T), we obtain 𝐼𝐼𝑡𝑡ℎ (𝑇𝑇1 ) 𝐽𝐽 (𝑇𝑇 ) exp (𝑇𝑇 /𝑇𝑇 ) 𝑇𝑇 −𝑇𝑇 = 𝑡𝑡ℎ (𝑇𝑇1 ) = exp (𝑇𝑇1 /𝑇𝑇0 ) = exp � 1𝑇𝑇 2 � 𝐼𝐼𝑡𝑡ℎ (𝑇𝑇2 ) 𝐽𝐽 𝑡𝑡ℎ 2 2 0 0
2.4
Therefore, we obtain Jth (T1) as 𝑇𝑇 −𝑇𝑇
50−15 � A/m2 . 170
𝐽𝐽𝑡𝑡ℎ (𝑇𝑇1 ) = 𝐽𝐽𝑡𝑡ℎ (𝑇𝑇2 )exp � 1𝑇𝑇 2 � = 3 × 107 × exp � 0
Using the above expression, we therefore obtain Ith (T1) as the product of Jth (T1) and the stripe area Ast (= 20 μm × 100 μm), given by 50−15 � × (20 × 100 × 10−12 ) = 73.72 A. 170
𝐼𝐼𝑡𝑡ℎ (𝑇𝑇1 ) = 𝐽𝐽𝑡𝑡ℎ (𝑇𝑇1 )𝐴𝐴𝑠𝑠𝑠𝑠 = 3 × 107 × exp �
2.11 Consider that a binary optical signal is incident from a fiber onto a photodetector. Presuming that the incident light has a duality in its nature (particle and wave), give examples for the manifestation of both the forms of light on the performance of a digital optical receiver. Solution: One of the manifestations of the wave nature of light takes place in the form of fiber dispersion mechanisms leading to the pulse spreading at the receiving end. On the other hand, the particle nature of light is reflected from the fact that, the photocurrent (number of electrons per unit time) is proportional to the number of photons (light particles) arriving per unit time at the receiver. 2.12 An APD operates at a wavelength of 900 nm with 95% quantum efficiency. Consider that an incident light with a power of -30 dBm has produced a photocurrent of 15 μA at the APD output. Determine the mean avalanche gain of the APD. Solution: The APD operates at 900 nm with a quantum efficiency of η = 0.95. The incident power Pin = -30 dBm = 10-6 W, which has generated a photocurrent IAPD = 15 μA. Hence, the primary photocurrent Ip in the APD is obtained as 𝐼𝐼𝑝𝑝 = 𝑅𝑅𝑤𝑤 𝑃𝑃𝑖𝑖𝑖𝑖 =
𝜂𝜂𝜂𝜂 𝜂𝜂𝜂𝜂𝜂𝜂 0.95×1.6×10 −19 ×900×10 −9 𝑃𝑃𝑖𝑖𝑖𝑖 = 𝑃𝑃𝑖𝑖𝑖𝑖 = × 10−6 = 0.6878 μA. 6.63×10 −34 ×3×10 8 ℎ𝑓𝑓 ℎ𝑐𝑐
Using the above result, we obtain the APD mean gain MAPD as 𝑀𝑀𝐴𝐴𝐴𝐴𝐴𝐴 =
𝐼𝐼𝐴𝐴𝐴𝐴𝐴𝐴 15 = = 21.8. 𝐼𝐼𝑝𝑝 0.6878
2.13 If the received optical power in a PIN-based optical receiver is -20 dBm and the preamplifier needs a voltage swing of 1 mV at its output, calculate the value of the feedback resistance needed for the preamplifier. Given: optical transmitter uses a laser with perfect extinction, PIN diode responsivity = 0.8 A/W. Solution: The received optical power PR = -20 dBm = 10-5 W and the PIN responsivity = 0.8 A/W. Hence, the photocurrent Ip = 0.8 × 10-5 A. The preamplifier output voltage swing needs to be Vs = 10-3 V which is the product of the input photocurrent Ip and the transimpedance RTI, implying Vs = Ip RTI. Hence we obtain the transimpedance as 𝑅𝑅𝑇𝑇𝑇𝑇 =
𝑉𝑉𝑠𝑠 10−3 = = 125 ohm. 𝐼𝐼𝑝𝑝 0.8 × 10−5
2.14 Draw a block schematic for a strictly nonblocking 8 × 8 optical switch, employing 2 × 2 electrooptic switching elements as the building block in Spanke's architecture. Estimate the total insertion loss of a similar N × N (with N = 2k) optical switch in terms of its number of stages and the losses incurred in all the passive devices during the traversal of a lightpath from an input port to an output port. Solution: 2.5
Considering the Spanke’s switch architecture in Fig. 2.58, we note that the first (i.e., input) column will have N (1:N) electro-optic switches, with k = log2N. Similarly, the second (output) column will have N (N:1) electro-optic switches. Using this observation, the readers are instructed to draw the full switch configuration. Hence, the insertion loss Lsw incurred by a lightpath in the above switch will be the sum of the losses in the two columns of the switches along with the connector losses Lc the input and output sides of each 1:N and N:1 switch. Thus, we express Lsw as 𝐿𝐿𝑠𝑠𝑠𝑠 = 2𝑘𝑘𝐿𝐿𝑒𝑒𝑒𝑒 + 4𝐿𝐿𝑐𝑐 = 2(log 2 𝑁𝑁 × 𝐿𝐿𝑒𝑒𝑒𝑒 + 2𝐿𝐿𝑐𝑐 ),
with Leo as the insertion loss in each (2 × 2) electro-optic switch, used as (1 × 2) and (2 × 1) switches at the input and output columns, respectively. 2.15 Using the results on gain saturation in EDFA (Eq. 2.119), estimate the decrease in EDFA gain when its input power increases from -20 dBm to -10 dBm. Given: Q = 10 dBm. Discuss how the gain saturation in EDFAs can affect the performance of an optical link in a WDM network and suggest some possible remedy. Solution: Using Eq. 2.119, EDFA gains with the input power levels of -20 dBm and -10 dBm are estimated as 27.5 dB and 22.4 dB respectively, implying that the gain falls by 27.5 – 22.4 = 5.1 dB with the 10 dB increase in the input power level, thereby affecting the optical SNR at the destination receivers. Gain control: By using a separate wavelength in each fiber link and controlling its own power, one can bring down the variation in the total optical power (from all wavelengths) handled by each EDFA along the fiber links across the network.
2.6
Exercise Problems and Solutions for Chapter 3 (Optical LANs/MANs and SANs) 3.1 Discuss with suitable illustrations why the fiber-optic bus or ring topology cannot use traditional one-packet-at-a-time MAC protocols, such as CSMA/CD or token-ring protocol, as used in IEEE standards for copper-based LANs. Also indicate the difficulty in using CSMA/CD protocol in passive-star-based optical LANs. Solution: In optical LANs, the packet sizes shrink with high speed while the network size increases. This leaves a large space (and hence bandwidth) in the optical fiber as unused resource. Further, in bus, the progressive power losses of each packet along the optical fiber, makes collision detection difficult when the packets overlap with dissimilar power levels. Passive-star-based optical LANs don’t suffer from the progressive loss, but the ratio of the packet duration to the round-trip propagation delay becomes much smaller with large network size. 3.2 Obtain an expression for the progressive power loss in an optical fiber bus connecting N nodes to set up a LAN, with each node tapping power from and transmitting (coupling) power into the fiber bus using a transmit-receive coupler (Fig.3.2). Using this formulation, evaluate the dynamic range needed in the receivers used for the network nodes connected to the fiber bus with N = 5 and 10. If the receiver dynamic range cannot be more than 10 dB, comment on the feasibility of the LANs in the two cases. Given: connector loss = 1 dB, power tapping for the receiver in the transmit-receive coupler = 10%, insertion loss in the waveguide of each transmit-receive coupler = 0.3 dB, loss at the transmitter-coupling point in each transmit-receive coupler = 0.2 dB, fiber loss = 0.2 dB/km, distance between two adjacent nodes = 1 km. Solution: RX
TX Each node employs a passive transmit-receive coupler, as shown above, with two taps for transmit and receive operations, and four connectors on input/output and transmit/receive ports. The necessary parameters associated with the nodes and the network are defined in the following for the analysis. • • • • • • • • •
Ptx: transmitted power from each node in mW (including the loss of isolator/circulator). d: distance between two successive nodes in km. All nodes are assumed to be equispaced. lf : loss in a fiber segment (d km) in neper/km between each adjacent node pair with, lf = exp(αf d) and lf (dB) = 10 log10(lf ). lc: connector loss (ratio). lcin: insertion loss (loss) inside the coupler. lrtap: tapping ratio for the receiving port. ltxc: loss (ratio) at the transmitting port. lrxc: loss (ratio) at the receiving port due to power tapping (= 1 – lrtap). ltcp: loss (ratio) while coupling the optical signal from the transmitter into the transmit-receive coupler
The power at node 1 output is given by 𝑃𝑃1 = 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐 = 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐2 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 ,
and the power received at node 2 from node 1 is given by 3.1