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Homework 6
1. If an electric field Eo is applied in the [0,1,0] direction inside ADP, find the principal
axes at Eo=104 V/m (Do not assume that if it is small it is equal to zero.). Use
λ=633nm. What is the propagation direction for maximum birefringence and the
value of the birefringence?
2. A bulk cube of KD*P is placed in one arm of a Mach-Zehnder interferometer to
create a modulator. To couple light through the cube the width of the cube must be
greater than 10mm. Use a wavelength of =546nm.
a. Design a modulator using the KD*P cube to have the lowest voltage for 100%
extinction. What is the required voltage?
b. An integrated MZ modulator is fabricated from the same material. The length
is still L=10mm but the thickness is now d=10m. Design the modulator and
calculate the voltage for 100% extinction. (An integrated modulator has to
operate with the voltage applied perpendicular to the propagation direction.)
3. A bulk modulator is designed to modulate a CO2 laser, which has a wavelength of
λ=10.6μm. The modulator is designed to use polarization filtering and uses a cube of
material with dimension of 10mm. Design the modulator and determine the best cube
orientation, and the required voltage range for entire extinction.
a. What is the best material to use? You can only use materials from Table 9.2
that are listed for a wavelength of =10.6m.
b. What is the V?
For problems 4-6 refer to the Avanex specification sheet and the reference article on
lithium niobate modulators. An x-cut crystal means that the direction normal to the
surface of the wafer is the x-direction of the crystal.
4. A modulator is fabricated using lithium niobate. Calculate the V of the modulator as
a function of d/L, where d is the electrode spacing and L is the length of the
modulator. Assume that the modulator is operated in a push-pull configuration. For
the following cases
a. y-propagating using an x-cut crystal
b. x-propagating using a z-cut crystal
5. Why is the x-cut crystal used for the Avanex modulator?
6. Estimate the d/L ratio for the Avanex modulator.
7. An amplitude modulator is configured in a polarization filter setup. The input

polarization state is given by E  4e j 0.1 xˆ  2e j 0.5 zˆ and the output is passed through a
linear polarizer oriented at 45 degrees. The modulator has a length L and the two
no3 V
ne3 V
indices of refraction are given by n x  no  r13 and n z  ne  r33 , where
2
d
2
d
no=2.286, ne=2.2, r13=9.6pm/V, r33=30.9 pm/V, =1550nm, d/L=1000, and L=2cm.
a. Plot the relative optical power as a function of voltage.
b. What is the extinction ratio (Pmin/Pmax)?
Homework 7
For problems 1-6 use the specification sheet for the PT10XGC pin optical receiver.
Assume that the noise bandwidth is equal to the -3dB high frequency corner. Use the
typical values given in the specification sheet.
http://www.bookham.com/documents/datasheets_rcvr/PT10XGC_14Feb07.pdf
1.
2.
3.
4.
5.
6.
What does PRBS stand for (see note 4)?
What is the quantum efficiency of the pin photodiode?
What is the shot noise if the receiver is operating at the minimum detectable power?
What is the thermal noise?
What is the noise equivalent power (NEP)?
What is the smallest possible minimum detectable power if the thermal noise is zero?
This is called the quantum limited detection.
For problems 7-8 use the specification sheet for the ATV10GC APD.
http://www.bookham.com/documents/datasheets_rcvr/ATV10GC_14Feb07.pdf
7. Bookham also sells a similar receiver that uses an APD. If we assume that the
thermal noise is the same as the pin receiver, what is the APD gain?
8. What the quantum limited detection of the APD receiver? Be sure to multiply the
shot noise by the excess noise factor. See the APD users guide for more information
on excess noise factor.
Homework 8
1. A free-space optical communications link is constructed with a CQF938/500 laser
from JDS Uniphase, and a 264-339730-500 receiver
(http://www.cmcelectronics.ca/Pdfs/CustEle_HyMic_Opti_75-Bro.pdf). Figure 1
shows the optical communications link. The laser beam is a perfect circular Gaussian
beam and is spread to have a 3dB width of 10mm (this is the diameter) at the
transmitter side. The receiver collects a circular diameter of 10mm. The
communications link is designed to operate at B=50Mb/s and have a BER=10-9. The
total loss associated with all of the lenses is 3dB and there is no atmospheric
distortion. What is the maximum distance between the transmitter and receiver?
Laser
Receiver
Figure 1
T
2. (Problem 17.1) The gain ripple in units of dB is given by   10 log 10  max
 Tmin
that =1dB requires GR=0.0575.

 . Show

3. (Problem 17.2)
a. Show that the FWHM gain bandwidth of resonant transmission in a Fabryc 1  GR
Perot amplifier is given approximately by  FW HM 
. The
2neff L  GR
increase in the resonant transmission comes with an decrease of the gain
bandwidth.
b. For a SOA with L=500m, neff=3.6, G=100, and R=0.008, find the maximum
gain at resonance and the gain bandwidth at resonance.
Homework 9
For problems 1-9 assume Nt=8E18 cm-3, A=50μm2, and use the parameters in Table 1
and following figures
Table 1: EDFA parameters
Material
Silicate L22
Fluorozirconate F88
τsp (λ<1400nm)
0.001ms
0.001ms
τsp (λ>1400nm)
14.5ms
9.4ms
1. Why is it impractical to create an EDFA using the doping denoted as Silicate L22 to
amplify an optical signal at 980nm? Be sure to use an equation to prove your answer.
(It has nothing to do with the emission or absorption cross-sections.)
2. What is the minimum pump power required for gain for the following EDFAs?
a. Silicate L22 pumped at a wavelength of 980nm.
b. Fluorozirconate F88 pumped at a wavelength 1480nm.
3. A Fluorozironcate F88 EDFA is pumped at a wavelength of λp=1480nm, with a
pump power equal to 5 times the minimum pump power required for gain. If the
amplifier has a length of L=5m, what is the total gain? Assume the signal wavelength
is λs=1554nm.
4. An amplifier is constructed with a peak gain of G=30dB. Let’s define the amplifier
wavelength range as the wavelength range over which the normalized gain variation
is less than 10dB.
a. Which of the amplifier materials (Silicate L22 or Fluorozirconate F88) has the
largest wavelength range?
b. What is the wavelength range for the best material?
5. The pumping efficiency is defined as the gain divided by the pumping power in units
of dB/mw.
a. What is the EDFA material, pump wavelength, and amplified wavelength that
produce the highest pumping efficiency?
b. What is the pumping efficiency for (a) if the pump power is 5 times the
minimum pump power for gain and the amplifier length is L=5m.
6. What is the gain of a Silicate L22 amplifier if λp=980nm, λs=1550nm, Pp=10mw,
L=5m, Ps=100μW?
7. An optical receiver has the following specifications: R=0.9 A/W, Pmin=-20dBm with
B=10Gbps and BER=10-12. The optical fiber has an attenuation of α=0.2 dB/km (this
attenuation includes all splice losses), the transmitted optical power minus all other
losses (coupling, loss margin, etc.) is 20mW, and the wavelength is λ=1550nm.
Ignore any dispersion limits in this problem.
a. What is the maximum link length?
b. What is the maximum link length if an EDFA is used as a pre-amplifier (right
before the receiver)? The EDFA has the following specifications G=30dB,
NF=7dB.
c. What is the maximum link length if an EDFA is used near the middle of the
optical communications link? The EDFA has the following specifications
G=30dB and NF=7dB.
d. What is the maximum link length if 4 EDFAs are used? The EDFA has the
following specifications G=30dB and NF=7dB.