Optical Communication solved examples

```QUESTION 1
List three features of an optical fiber and briefly explain how it is useful for the optical fiber
communication system. (15 MARKS)
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.
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&deg;
1.3
b) Core of multimode optical fiber (1.475) → clad (1.46)
= sin
1.46
= sin 0.9898 = 81.8&deg;
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&deg;
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: = &amp;⁄ , 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 &lt;
2+
*
* = &times; *
*=
**
cos &lt;
=
2+
2+
* &gt; 2+ &times; cos cos =
* &gt; 2+ &times; cos = 2+ &times;
* &gt; 2+ &times;
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)
* &gt; 2+ &times; = 9 &times; 10/ &times; 1.45 1.445 = 9 &times; 10/ &times; 0.1203
&gt; 1.08 &times; 10/
&gt; 1.08 01
10
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