sol_hw1 optoelectronics 4

Optical Communication Components and Subsystems ECE 565 Spring 2005 Homework 1 Solutions

1. Calculate the frequency and photon energy of optical communication systems operating at 0.88, 1.3 and 1.55 µm. This is best done with a short program.

Solution: The frequency of a photon is related to its energy through the relation E=hν, where h is Planks constant, h=6.6262 x 10-34 Js. Also, ν = c / λ, where c is the speed of light in vacuum c = 2.9979x108 m/s, and λ is the photon’s wavelength. λ (µm )

ν (Hz)

E (J)

0.88

3.4067 x1014

0.2257 x10 18

1.4089

1.3

2.3061 x1014

0.1528 x10 18

0.9537

1.55

1.9341 x1014

0.1282 x10 18

0.7999

E (eV) -

Matlab code: h=6.6262e-34; c=2.9979e8; lambda=[0.88 1.3 1.55]*1.0e-6; nu = c ./lambda; E=h*nu; Eev = E/1.6022e-19;

2. Calculate the distances over which the optical power is attenuated by a factor of 10 for fibers with losses 0.2 dB/km and 2000 dB/km. Write a program to compute this distance for any loss quoted in dB/km.

Solution: If we launch an optical power of P0 over distance L (in km), and assume that the power at the destination P, then the loss coefficient in dBm/km is α = - L-110 log10(P/P0). In this problem, P/P0 = 0.1. a=[0.2 2000]; att=0.1; L = -10 *log10(att) ./a;

L is 50 km and 5 m, for the cases of α = 0.2 dB/km and 2000 dB/km, respectively. There is a change by four orders of magnitude!

3. A single channel of an optical communication system has a bit rate of 10 Gb/s [return-tozero (RZ) format, i.e., each on-pulse is on for half of the bit duration and then it drops to zero]. In principle, how many channels can be distributed over the range 1.3 – 1.55 µm?

How many in the range 1.5 – 1.6 µm?

Solution: The bandwidth B required for 10 Gb/s RZ format is approximately 2/Tb (twice that for NRZ). Thus, B = 20 GHz. The range of bandwidth supported light in the specified range is ∆ν = c / 1.3x10-6 - c / 1.55x10-6 = 3.7195 x 1013 Hz. In principle, we can support 1859 10Gb/s RZ channels. If the optical range is 1.5-1.6, then the number of channels supported becomes 624 channels.

4. A 1.55 µm optical carrier operating at 2.5 Gb/s receives an average power of –40dBm at the detector. What is the average received power in mW? Assuming 0 and 1 bits are equally likely, what is the power of a 1 bit? How many photons (on average) does the 1 bit contain?

Solution: P (in dBm) = 10 log10 P (in mW). Therefore, P (in mW) = 10–40/10 = 0.0001 mW = 0.1 µW. The power P1 of a “1” bit is 2x0.1 µW = 0.2 µW. The average number of photons is related to the energy of the pulse E (assuming a monochromatic pulse, which is an approximation) by <n> = E/ hν = P1Tb/ hν = 624.2 photons (on average per “1” bit).

5. A receiver needs a minimum of 1000 photons for adequate signal detection. A transmitter launches 1 mW of peak power using a NRZ format (i.e., an on-pulse is on for the entire duration of the bit). The bit rate is 2.5 Gb/s. For a fiber loss of 0.2 dB/Km at 1.55 µm, what is the longest length of the fiber link?

Solution: The minimum received power (in a “1” bit) required is Pr = (1000 hν)/ Tb = 0.3204 µW. By using the equation α (in dBm/km) = - L-110 log10(Pr/P0) and substituting P0 = 1 mW, we obtain L = 174.7 km.