Wireless Networks & Communication Answer 2022

WIRELESS NETWORKS &
COMMUNICATION

Answer 1
Difference between Network access layer and IP layer of TCP protocol
Network Access layer is the first layer in the TCP/IP model that determines flow of the data to
the network physically. It defines signaling of bits optically or electrically by the hardware
devices such as optical fiber, twisted pair copper wire, coaxial cable etc. On the other hand
Internet layer is just another layer after the network access layer in the TCP model which packs
the data into IP datagrams. It performs the task of routing this IP datagrams across network
between participant nodes. The datagram contains the source and destination address which
helps in the transfer of these packets from source to destination.
Answer 2
The following is the flow of interaction between the Chinese and French Prime Minister:
Both the Chinese and French PM would carry the telephone discussion as they are talking in
front of one another. When the Chinese PM converse his message will be passes legitimately to
French PM on the other side. The communication will be passed through the translators at each
end. When the message reaches the other end, the translator at the French PM Telephone
translates the message into French which can be heard and understand by the French Prime
minister.
Answer 3
In figure 1
Maximum Amplitude is 15
Time Period is 8.3-5.3= 3
Frequency 1 / t = 1 / 3= 0.333 Hz
Phase is zero.

In Figure 2
Maximum Amplitude is 4
Time Period is 8.3-1.7= 6.6 sec
Frequency 1 / t= 1 / 6.6 = 0.15 Hz
Phase is zero.
In figure 3
Maximum Amplitude is 7.8
Time Period is 3.5-1.2 = 2.3 sec
Frequency 1 / t = 1 / 2.3 = 0.43 Hz

Phase is zero.
Answer 4
a. 3𝑆𝑖𝑛 (2(200) 𝑡)
Comparing the above equation with
The amplitude is 35
Frequency ɷ = 2π 200 = 1,257
Time Period T = 2π / ɷ = 0.005 sec
Frequency f = 1 / T= 200
Phase is 0
b. 14𝑆(2𝜋(50)𝑡 + 90)
Comparing the above equation with
The amplitude is 14
Frequency ɷ = 2π 500 = 3142
Time Period T = 2π / ɷ = 0.002 sec
Frequency f = 1 / T = 500
Phase is 90
c. 4𝑆(650𝜋𝑡 + 180)
Comparing the above equation with
The amplitude is 4
Frequency ɷ = 650π = 2042
Time Period T = 2π / ɷ = 0.003 sec

Frequency f = 1 / T = 333.3
Phase is π
d. 6𝑆(700𝜋𝑡 + 270)
Comparing the above equation with
The amplitude is 6
Frequency ɷ = 700π = 2200
Time Period T = 2π / ɷ = 0.003 sec
Frequency f = 1 / T = 333.3
Phase is 3π/2
Answer 5
Given Frequency 6 GHz = (6*109) hz
The distance given is 35,863 km = (35,863 * 106) m
LDb= 20 log(f) + 20 log(d) – 147.56 dB
LDb= 20 log(6*109) + 20 log(35,863 * 106) - 147.56 dB
LDb = 195.56 + 211- 147.56 = 259.09 dB
Answer 6
Solution: To identify the period T, the frequency f= 1/T, or the angular Frequency ɷ = 2 π f = 2
π/T of the sinusoidal signal given in the question we will first write the equation in the following
form
As given in the question, the equation is
(𝑡) = 5sin(100𝜋𝑡) + sin(300𝜋𝑡) + sin(600𝜋𝑡)
ɷ1 =100𝜋, f1=50, T1 = 1/50, ɷ2 = 300𝜋, f2= 150, T2 = 1/150 ɷ3= 600 , f3= 300, T3 = 1/300

To calculate the fundamental frequency we will determine the greatest common divisor of all the
frequency components of the given signal.
Fundamental frequency= f0= GCD {50, 150, 300} = 50
Also we can give the bandwidth of the signal as 300 – 50= 250
We know the Channel Capacity given by the Nyquist criteria is
C = 2 B log2 M
Where B stands for the Bandwidth and M is the number of level
For M= 2
C= 2 * 250 log2 2 =500 Mbps
For M=4
C= 2 * 250 log2 3= 790 Mbps
For M=8
C= 2 * 250 log2 8 = 1500 Mbps
Answer 7
The maximum bit rate as per the Nyquist bit rate criteria can be calculated as
BitRate = 2BLog
Where M is the number of levels in signal and B is the bandwidth
It is clear from the equation that data rate is dependent on the number of signal levels as the
bandwidth is constant for a given wave. This implies that for increasing the data rate the signal
levels should be incremented. However, as the number of levels increases in the signal the
reliability decreases, making it a big disadvantage of this approach.
Answer 8
Circuit Switching vs Packet Switching:
To connect multiple devices communicating with one another two of the most common methods
of switching that is Circuit switching and Packet switching are utilized. The main difference
between these two switching technologies is that Circuit switching requires channels or it can be
said that it is connection oriented whereas packet switching does not require establishing
channels and is connection-less. The output of communication is the major advantage of packet
switching over circuit switching. Without establishing devoted circuits

Other difference are described in the table below
Circuit Switching Packet Switching
It is implemented at physical layer This is implemented at network layer.
It establishes direct connection between the
sender and the receiver.
There is no direct connection between the sender and
the receiver.
It works on the pre-defined path in three stages
that are establishing connection, transfer of data
and closing connection
There is no pre-defined path in this switching so the
data is transferred directly.
The complete message is transferred in one go
and there is no additional header required
Since the message is sent in segmented form hence
header is attached with every piece to identify the
packet at the destination.
Data is transferred in the same order The data might reach the destination in out of order.
Data processing takes place at the sender’s end
only.
Data processing takes place at every intermediate
network end along with the sender’s end
This technology is reliable as it ensures complete
delivery of data and in the same order.
It may lose data packet during transmission so the
communication is not that much reliable.
Even a little communication consumes much
resources
It consumes very less resources.
In circuit switching if connection is lost in the
middle of communication, it has to be establish
the entire communication once again.
In packet switching if a path is lost only that packets
that are travelling from the path is lost and not the
entire communication needs to be started.
Answer 9
The maximum distance of LOS propagation between two antennas is given as
Where
As per the question h1=4h2,

Putting the values in the formula, we get
60 = 3.57 (√ 16/3 h2 + √ 4/3 h2)
16.81= 2.58 √ h2
h2 = 42.45 m
h1= 4* 42.45
h1= 169.79 m