QUESTION 1 List three features of an optical fiber and briefly explain how it is useful for the optical fiber communication system. (15 MARKS) The example of an answer. It has low loss. It is a loss of 0.5 dB / km or less in a wide wavelength range. Since loss is small, long distance transmission is possible. Bandwidth is wide. Large capacity transmission is possible. The possibility of transmission of about 10 Tbps with one optical fiber is expected. Do not receive electromagnetic induction. Since glass does not conduct electricity, stable and high-quality transmission can be realized without external electromagnetic induction. There is no restriction on the installation place. It is thin and lightweight. The optical fiber is thinner and lighter than other transmission media. Compared with conventional metallic cables, it can accommodate more lines and is lightweight, so wiring work is easy. It is inexpensive. Quartz glass which is a raw material of fiber is inexhaustible on the ground and is cheaper than metal. Advantages of optical communications Chapter 1, Slide 24 Large capacity, longlong-haul transmission system at low cost ! Low loss Loss of 0.5 dB / km or less in a wide wavelength range. Wideband It is expected to realize a transmission capacity of about 10 Tbps. No electromagnetic induction Not affected by electromagnetic wave noise. No cross talk between cables. no restriction on the installation place. Skinny and Light Diameter of optical cable is 0.25mm. Cross section is about 1/30 compared to coaxial cable. Low cost Glass is much cheaper than copper. QUESTION 2 a), b) Calculation of dB. (20 MARKS) a) a) Convert optical power expressed in mW to dBm value. For the logarithmic calculation, you may use the following values. (i) 1mW ⇒ 0dBm (ii) 5mW ⇒ 7dBm (iii)15mW ⇒ 11.8dBm (iv) 0.2mW ⇒ -7dBm (v)0.05mW ⇒ -13dBm The unit of optical power is dBm. There were some people who did not m and just dB. b) Input power 0.25mW output power 50mW Gain = 10 log 50/0.25 =10log200 =23dB 3 QUESTION 2 c), d) c) Input power 10mW output power 0.05mW Total loss = -10 log 0.05/10 =-10log0.005 =23dB Fiber loss in dB/km =23dB/50km = 0.46dB/km d) Line loss = 0.18dB/km x5km+0.2dB+0.21dB/kmx20km+0.2dB+0.18dB/kmx10km = 0.9+0.2+4.2+0.2+1.8=7.3dB I was going to connect Fiber 3 for the last 10 km. In that case, the loss is as follows. Line loss= 0.18dB/km x5km+0.2dB+0.21dB/kmx20km+0.2dB+0.3dB/kmx10km = 0.9+0.2+4.2+0.2+3=8.5dB I will make it 8.5 dB correct. 4 QUESTION 3 Find the critical angle below. Numbers in parentheses are refractive indices. Arrows indicate the direction of light travel. (10 MARKS) a) water (1.3) → air (1.0) = sin 1 = sin 0.7692 = 50.3° 1.3 b) Core of multimode optical fiber (1.475) → clad (1.46) = sin 1.46 = sin 0.9898 = 81.8° 1.475 5 QUESTION 4 Prove that the critical angle θc of the core and cladding boundary in the optical fiber satisfies cos = . (10 MARKS) From Snell’s law, we can get the following formula. sin = sin 90° sin = sin = We know that sin cos = 1 Therefor, cos = = 1 sin = 1 6 Question 5 a) In the figure below, prove that the numerical aperture NA #NA = sin $ of the optical fiber can be expressed as NA = . (20 MARKS) . Hints: = &⁄ , sin cos = 1 Because is the critical angle, And sin = sin = 1 sin = from Snell’s law sin = sin NA = sin = sin = 90 = 90 sin = sin 90 = cos = We know that sin cos = 1 Therefore sin = 1 cos = 1 NA = sin = sin = 1 = 7 Question 5 b), c) b) Find the NA of a single mode fiber with n1 = 1.45 and n2 = 1.445. (5 MARKS) NA = = 1.45 1.445 = 0.12 c) Find the NA of the multimode fiber with n1 = 1.45 and n2 = 1.435. (5 MARKS) NA = = 1.45 1.435 = 0.21 8 Question 6 a) a) * cos < 2+ * * = × * *= ** cos < = 2+ 2+ * > 2+ × cos cos = * > 2+ × cos = 2+ × * > 2+ × 9 Question 6 b) b) Find the cutoff wavelength of a single mode fiber with n1=1.45, n2=1.445, 2a=9μm. (5 MARKS) * > 2+ × = 9 × 10/ × 1.45 1.445 = 9 × 10/ × 0.1203 > 1.08 × 10/ > 1.08 01 10