NET 434: WIRELESS NETWORKS
Homework 1
Student Name:---------------------------------------Student ID:-------------------------------------------Section: -----------------------------------------------
Networks and communications Department
Homework Submission Rules:
 It most be printed A4 clean and neat or you will lose marks!!!
 All homework will be manually compared copying from another
student will result in a Minus one for all involved students!!
 Submission will be on date due.
 No late submission even if you’re absent.
 Date due: 8.11.2015
Assignment Questions(s)
1. The difference between the Terrestrial and Satellite wireless
communications from a network design point of view.
(Reference: Wireless Communications and networks by William
Stallings)
2. List the applications of three different cellular bands of the
spectrum in KSA.
(Reference:
Cellular
Bands
in
KSA:
http://www.citc.gov.sa/English/RulesandSystems/Bylaws/Docume
nts/SM%20002%20E-NFP.pdf)
3. Solve the Problems
Satellite Networks
a. We wanted to put a satellite in a circular orbit at 500 km
above the surface (LEO Orbit).What will be the speed of
rotation required.
i. G=6.67×10−11 𝑚3 /𝑠 2 𝑘𝑔
ii. R=Radius of the earth=6371 Km=637100m
iii. Mass of earth M=5.98×1024 kg
Solution:
𝐺𝑀
6.67 × 10−11 × 5.985.98 × 1024 𝑚
𝑣=√
=√
( )
(6371000 + 500000)
𝑟
𝑠
b. From the following Figure, What can you deduce about the
relationship between the frequency and the Orbital height?
Solution:
As the orbital height increases the free space loss also
increases for a given frequency. We can also observe from
the graph that as the frequency increases the free space loss
also increases.
c. From the following Figure, Determine the relationship
between angle of elevation and different types of attenuation
i. Atmospheric absorption
ii. For absorption
iii. Rain absorption
The signal attenuation due to atmospheric absorption , Fog
absorption and Rain absorption all decrease with the increase in the
elevation angle.
Wireless Transmission
d. Convert the following Power values to 𝑑𝐵𝑊 ?
i. 100W
ii. 1mW
Solution:
𝑃𝑜𝑤𝑒𝑟
𝑃𝑜𝑤𝑒𝑟
𝑑𝑏𝑊=
100𝑊
1𝑊
10−3 𝑊
𝑑𝑏𝑊=
1𝑊
e. Consider a series in which the input is at a power level of
4mW, the first element is a transmission line with a 12dB
loss, the second element is a an amplifier with a 35dB gain
and the third element is a transmission line with a 10dB loss.
Find the net gain over the the link. Calculate the output
power?
Solution:
f. A signal travels through transmission medium and the power
is reduced by half. Calculate the loss of power in dB’s.
Solution:
g. A signal travels through an amplifier and its power is
increased by 10. Calculate the gain in dB’s.
h. One reason that engineers use the decibel to measure the
changes in the strength of a signal is that decibel numbers
can be added (or subtracted) when we are measuring several
points (cascading) instead of just two.
i. In the following Figure, Has the signal gained or lost power
between Points 1 and 4?
The signal has gained in power.
j. Sometimes the decibel is used to measure signal power in
milliwatts. In this case, it is referred to as dBm and is
calculated as dBm = 10 log10 Pm, where Pm is the power in
milliwatts. Calculate the power of a signal if its dBm = −30.
k. The loss in a cable is usually defined in decibels per
kilometer (dB/km). If the signal at the beginning of a cable
with −0.3 dB/km has a power of 2 mW, what is the power of
the signal at 5 km?
l. The power of a signal is 10 mW and the power of the noise
is 1 μW; what are the values of SNR and SNRdB?
m. Calculate the values of SNR and SNRdB for a noiseless
channel?
n. Consider a noiseless channel with a bandwidth of 3000 Hz
transmitting a signal with two signal levels. Calculate the
maximum channel capacity?
o. Consider the same noiseless channel transmitting a signal
with four signal levels (for each level, we send 2 bits).
Calculate the maximum channel capacity?
p. We need to send 265 kbps over a noiseless channel with a
bandwidth of 20 kHz. How many signal levels do we need?
q. Consider an extremely noisy channel in which the value of
the signal-to-noise ratio is almost zero. In other words, the
noise is so strong that the signal is faint. For this channel
calculate the capacity C ?
r. We can calculate the theoretical highest bit rate of a regular
telephone line. A telephone line normally has a bandwidth of
3000 Hz (300 to 3300 Hz) assigned for data
communications. The signal-to-noise ratio is usually 3162.
Find the channel capacity?
s. Determine the isotropic free space loss at 4 GHz for the
shortest path to a synchronous satellite from earth (35683
km)? Hint: The free space loss is given by the following
expression:
t. For a parabolic reflector antenna with a diameter of 2 meter,
operating at 12 GHz, what is the effective area and the
aperture gain?
CDMA
u. Check to see if the following set of chips can belong to an
orthogonal system. [+1 +1] [+1 -1]
Solution:
[+1 +1].[+1 +1]=+2
[+1 +1].[+1 -1]=0
[+1 -1].[+1 -1]=2
So they belong to an orthogonal set
v. Check to see if the following set of chips can belong to
orthogonal system
Apply the same technique as in the previous question to check Yes
they belong to an orthogonal set.
Spread spectrum
w. What is the minimum number of bits in a PN sequence if we use
FHSS with a channel bandwidth of B =4 KHz and Bss =100 KHz?
The number of hops = 100 KHz/4 KHz = 25. So we need
log2(25) bits=4.64≈ 5
x. An FHSS system uses a 4-bit PN sequence. If the bit rate of
the PN is 64 bits per second, answer the following questions:
i. What is the total number of possible hops?
ii. What is number of finished cycles per time of PN?
The total number of hops=24
Bit rate =64 bits per second, Number of bits in the PN
sequence=4 so the number of finished cycles of PN sequence
would be 64/4=16 cycles per second
y. We have a digital medium with a data rate of 10 Mbps. How
many 64-kbps voice channels can be carried by this medium
if we use DSSS with the Barker sequence?
The Barker chip is 11 bits, which means that it increases the
bit rate 11 times.
A voice channel of 64 kbps needs 11 × 64 kbps = 704 kbps.
This means that the bandpass channel can carry: (10 Mbps) /
(704 kbps) ≈ 14 channels.
z. Consider a CDMA system in which users A and B have
codes (-1 1-11-11-11) and (-1-111-1-111) respectively. (a)
Show the output at the receiver if A transmits a data 1 and B
does not transmit.
Data over the channel=da·ca = +1 * [-1 1-11-11-11] = [-1 1-11-1111]
Data Output at receiver = ca· Data = [-1 1-11-11-11] · [-1 1-11-1111] = 8/8=1