Solution #6 ECE S490
1.0 A multimode fiber captures rays travelling at up to 0.4 radians with respect to the
fiber axis (inside the fiber). The fiber is made of a material whose index of refraction
in 1.5. What is the value of delay spreading in ns/km?
Solution:
The speed of light in the fiber is [300,000,000 m/s] / 1.5 = 200,000,000 meters per
second. The delay along the axis of the axial ray is 1/[200,000,000 m/s] = 5 ns/meter.
The delay along the axis of the highest angle guided ray is (5 ns/meter) / cos (0.4
radians) = 5/0.92 ns/meter. The delay spreading is 5 ns/meter [(1/0.92) –1]= 0.43
ns/meter = 430 ns/km
2.0 A single mode fiber is used with a laser whose optical output frequency shifts when it
is modulated from off to on. This frequency shift is called “chirp”. The amount of
frequency shift is 20GHz. The laser’s nominal wavelength is 1.5 micrometers. The
dispersion (delay change that results with a wavelength change) of the fiber being
used with this laser is 16 ps/km-nm. The fiber is 40 km long. How much change in
delay down the fiber results when the laser is modulated? If the allowable delay
change is 0.5 times the spacing between pulses, then what is the highest date rate that
the fiber will support with this light source?
Solution:
Lambda = c/f where c is the speed of light in free space and f is the frequency of the light
signal.
If we take the derivative of lambda with respect to frequency (f) we get:
d(lambda)/d(f) = -c/(f**2). Substituting f=c/lambda we get
d(lambda)/d(f) = -[lambda**2]/c; which means that the change in wavelength
[d(lambda)] that results from a change in frequency [d(f)] is given by
d(lambda) = [-c/(f**2)] d(f).
In our case, d(f) is 20 GHz, so
d(lambda) = 20,000,000,000 Hz x [.0000015 meters]**2/ 300,000,000 m/s = 1.5 x 10**10 meters = 0.15 nm.
Since the dispersion is 16 ps/km-nm, the change in delay down the fiber is 0.15 nm x 16
ps/km-nm = 2.4 ps/km. If the fiber is 40 km long, then the delay change is 2.4 ps/km x 40
km= 96 ps. If this must be no more than 0.5 x the pulse spacing, then the pulse spacing
must be 192 ps. This means that the bit rate (= 1/pulse spacing) must be no higher than
5.2 Gbits per second.
3.0 A fiber optic system receiver has a minimum required energy per pulse of 20,000
photons per pulse. The photon energy at the optical wavelength of interest is 1.5 x
10**-19 J. The optical transmitter in this system is a laser with a peak output power
of 1.5 mW. The coupling loss from the laser into the fiber is 3dB. The fiber loss is 0.5
dB/km, including splices. Assuming that we want to allow 6 dB of system margin
(actual received signal level vs. minimum allowable received signal level), what is the
maximum allowable fiber length between the transmitter and the receiver at a data
rate of 2.5 Gbps.
Solution:
The receiver requires 20,000 photons per pulse. The energy of a photon is 1.5 x 10**19J. Therefore the energy of a received pulse must be at least 20,000 x 1.5 x 10**-19 J =
3 x 10**-15 J. If the pulse rate is 2.5 Gbps, then the duration of a single pulse is 1/ bit
rate = 400 ps. The transmitter output power is 1.5 mW. Therefore the energy in a
transmitted pulse is 1.5 x 10**-3 W x 4 x 10**-10 seconds = 6 x 10**-13 J. Thus the
transmitter output pulse enrgy is 200 times larger than the pulse energy required at the
receiver. Thus we can allow a factor of 200 => 23dB of loss between the transmitter and
the receiver. We lose 3dB coupling from the transmitter into the fiber.. We want to hold
back 6 dB of this allowed loss for margin. Thus we can allow 23dB – 3dB –6dB = 12 dB
of loss in the fiber. Since the fiber loss is 0.5 dB/km (including splices), we can allow 24
km of fiber.