All posts by mutiahenarta

Optical Sources Assignment

1. An engineer has two Ga(1-x)Al(x)As LEDs: one has band gap energy of 1.540 eV and the other has x = 0.015.
a. Find the aluminium mole fraction x and the emission wavelength for the first LED
answer :
11
Eg = 1,424 + 1.266X + 0.266X^2
1.540 = 1.424 + 1.266X+0.266X^2
X = 0.09
b. Find the band gap energy and the emission wavelength of the other LED
answer :
Eg = 1.424 + 1.266x + 0.266×2
Eg = 1.424 + 1.266(0.015) + 0.266(0.015)2
Eg = 1.443 eV
1b

2. A double heterojunction InGaAsP LED emitting at a peak wavelength of 1310 nm has radiative and nonradiative recombination time sof 25 and 90 ns, respectively. The drive current is 35 mA.
answer :
a. Find the internal quantum efficiency and the internal power level.
Internal quantum efficiency :
2a
Internal power level :
fgsfg
b. If the refractive index of the light source material is n = 3,5, find the power emitted from the device.
answer :
Power emitted from device :
Po = n2 = 3.52 = 12.25 Watt

3. a GaAl As laser diode has a 500cavity length which has an effective absorption coeffient of 10 cm-1. For the uncoated facets the reflectives are 0.32
a. What is the optional gain at the lasing threshold?
answer :
3a
b. If one end of the laser is coated with a dielectric reflector so that its reflectivityis now 90%, what is the optical gain at the lasing threshold?
answer :
R = 0.9 x 0.32 = 0.288
3b
c. If the internal quantum efficiency is 0,65, what is the external quantum effiency is cases (a) and (b)?
answer :
3c

4. A GaAs laser emitting at 800 nm has a 400 cavity length with a refractive index n = 3.6. If the gain exceeds the total loss throughout the range 750 nm < < 850 nm, how many modes will exist in the laser?
answer :
4

5. A distributed feedback laser has aBragg wavelength of 1570 nm, a second-order grating with = 460 nm, and a 300 cavity length. Assuming perfecttly symmetrical DFB laser, find the zeroth-, first-, and the second-order lasing wavelength to a tenth of a nanometer. Draw a relative amplitude-versus wavelength plot.
answer :
adfagrwb

Even Number Assignment 2

2.8 A point source of light is 12 cm below the surface of a large body of water (n = 1.33 for water). What is the radius of the largest circle on the water surface through which the light can emerge?
Answers :
74746746
sfhshsfh

2.10 Show that the critical angle at an interafce between doped silica with n1 = 1.460 and pure silica with n2 = 1.450 is 83.3 degree.
Answers :
898989

2.13 Consider a dielectric slab having a thickness d = 10 mm and index of refraction n1 = 1.50. Let the medium above and below the slab be air, in which n2 = 1. Let the wavelength be λ = 10 mm (equal to the thickness of the waveguide).
(a) What is the critical angle for the slab waveguide?
(b) Solve Eq. (2.26b) graphically to show that there are three angles of incidence which satisfy this equation.
(c) What happens to the number of angles as the wavelength is decreased?
Answers :
(a) critical angle for the slab waveguide
rtrtr
(b) The Equation :
vbdn

2.18 A step-index multimode fiber with a numerical aperture of 0.20 supports approximately 1000 modes at an 850-nm wavelength.
(a) What is the diameter of its core?
(b) How many modes does the fiber support at 1320 nm?
(c) How any modes does the fiber support at 1550 nm?
Answers :
(a) Number of modes is given by,
1241415
(b)
4464577
(c)
66585685

Even Number Assignment

EVEN NUMBER ASSIGNMENT
1.2 A WDM optical transmission system is designed so that each channel has aspectral width of 0.8 nm. How many wavelength channels can be used in the C-band?
answer :
optical-fiber-communication-ftth-13-638

Wavelength range of C-Band is 1525-1565 nm or 40 nm bandwidth. So, wavelength channels that we used if each channel has aspectral width of 0.8 nm is :
(40 nm)/(0.8 nm)= 50 wavelength channels

1.4 A sine wave is offset 1/6 of a cycle with respect to time zero. What is its phase in degrees and in radians?
answer :
Complete cycle is 360 degrees.
If A sine wave 1/6 cycle, then the phase is :
1/6 x360=60 degrees
60 x 2π/360=1.047 radian

1.6 What is the duration of a bit for each of the following three signals which have bit rates of 64 kb/s, 5 Mb/s, and 10 Gb/s?
answer :
bitdur

1.8 (a) Convert the following absolute power levels to dBm values: 1pW, 1nW, 1mW, 10 mW, 50 mW.
(b) Convert the following dBm values to power levels in units of mW: -13 dBm, -6 dBm, 6 dBm, 17 dBm.
answers :
dbm1
dbm2

1.10 A signal passes through three cascaded amplifiers, each of which has a 5 dB gain. What is the total gain in dB? By what numerical factor is the signal amplified?
Answers :
cascade amplifier :
Cascadeamplifier
pic

1.12 A transmission line has a bandwidth of 2 MHz. If the signal-to-noise ratio at the receiving end is 20 dB, what is the maximum data rate that this line can support?
Answers :
aaaa

1.14 To insert low-speed signals such as 64 kb/s voice channels into a SONET frame, 84 columns in each SPE are divided into seven groups of 12 columns. Each such group is called a virtual tributary.
(a) What is the bit rate of such a virtual tributary?
(b) How many 64 kb/s voice channels can a virtual tributary accommodate?
(c) What is the payload efficiency?
Answers :
SONET frame
(a) each group has bit rate around 6.912 Mb/s
(b) voice channels that can be accommodate by virtual tributary is :
voice channels = (6.912 x 106) / 64000 = 108 voice channels
(c) payload efficiency :
Payload Efficiency is ratio between the payload (P) and Effectively sent data (D)
PE = P/D

Step Index Power

Fiber a = 25 μm, n1 = 1,48; Δ = 0,01; λ = 0,84 μm;
calculate V; M; Pclad/P;
if Δ = 0,003, calculate M and Pclad/P;
V= (2πan_1)/λ √2Δ
V= (2π.25.〖10〗^(-6))/(0,84.〖10〗^(-6) ) √2.0,01
V= 26.445 m/s
M = V2 / 2 = (26,445)2 / 2 = 349,67
Pclad/P = 4/3 (M)-0.5
= 4/3 (349.67)-0.5
= 0.07
if Δ = 0,003 then,
V= (2πan_1)/λ √2Δ
V= (2π.25.〖10〗^(-6))/(0,84.〖10〗^(-6) ) √2.0,003
V= 14,48 m/s
M = V2 / 2 = (14,48)2 / 2 = 104,8352
Pclad/P = 4/3 (M)-0.5
= 4/3 (104,8352)-0.5
= 0,13

Assignment II : V-Parameter

d = 10 μm
λ1 = 850 nm
λ2 = 1300 nm
λ3 = 1550 nm
If given Numerical Aperture 1 %,2 %, and 3 %, Calculate the V-parameter and define the optical fiber mode!
(Note : d = diameter, λ = wavelength)
Answers :
V = k . a . NA

x
NA1 = 1 %
v
NA2 = 2 %
z
NA3 = 3 %
l
The Requirement of Multimode is : V > 2,405
So, Even Though NA 1%, 2 %, and 3 %, The Value of V are still under 2,405 so the value tha we got is Single mode Fiber Optic.