MITS5003 Wireless Networks & Communication Assignment 1

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MITS5003 WIRELESS NETWORKS & COMMUNICATION - 2019S2 - ASSIGNMENT
NO 1

Table of Contents
Question 1..................................................................................................................................................3
Question 2..................................................................................................................................................4
Question 3..................................................................................................................................................5
Question 4..................................................................................................................................................9
Question 5................................................................................................................................................13
Question 6................................................................................................................................................14
Question 7................................................................................................................................................16
Question 8................................................................................................................................................18
Question 9................................................................................................................................................20
References................................................................................................................................................22

Question 1
Solution:
Internet protocols
Transmission control protocol or the internet protocol that is IP both are the network protocols. It
is a set of network communication which is used to deliver the data across the several networks.
The packet address can be conducted due to the IP which helps in sending the data or
information at the specific location. IP is describing the address and the route through which the
packets can be transferred up to the destination. It can be easily controlled by the number of
companies and can be modified according to the requirement. It can easily communicate with the
number of companies due to the flexible operating system. This protocol can work with
hardware or software networks (Mitchell, B 2019).
Network Access layer
It is the type of layer which can be used in data transmission and receiver for the physical
networks. It commonly uses adopter, line card for set up the connections with the wires. It can
also communicate with several computer systems. This interface has the address which helps in
addressing the layers to establishing the communication with effective manner. It can easily
check the errors which come with the incoming frames and acknowledge all the frames of the
receiver. The formats of the outgoing data can be described as the electrical pulses (Rouse, M
2019).

Question 2
Solution:
The interaction diagram of this image can be described as:

Question 3
Solution:
In the first figure, it can be found that the
Step 1;
Maximum amplitude will be
Am=15 equation 1
Step 2;
The time period is given as
T=3 sec
As F=(1/T)
F=(1/3)
F=0.3333 equation 2
Step 3;
n(t)=Am Sin(wt+phase)
consider (wt=0) then n(t)=0
= Am Sin(Phase)=0
= Phase=0

In the second figure, it can be found that
Step 1;
Maximum amplitude will be
Am=4 equation 1
Step 2;
The time period is given as
T=6.5 sec
As F=(1/T)
F=(1/6.5)
F=0.1538 equation 2
Step 3;
n(t)=Am Sin(wt+phase)
consider (wt=0) then n(t)=0
= Am Sin(Phase)=0
= Phase=0

In figure 3, it can be found that:
Step 1;
Maximum amplitude will be
Am=7.5 equation 1
Step 2;
The time period is given as
T=2.25 sec
As F=(1/T)
F=(1/2.25)
F=0.444 equation 2
Step 3;
n(t)=Am Sin(wt+phase)
consider (wt=0) then n(t)=maximum=Am
= Am Sin(Phase+wt)=Am
wt will be considered as 0, the phase will be calculated as:
= Phase=90 degree

Question 4
a. 3𝑆𝑖𝑛(2𝜋(200)𝑡)
The equation will be given as
3𝑆𝑖𝑛(2𝜋(200)𝑡)
From the equation, it will be found that
A = 3
F = 200
t = 0.005
phase angle = 0
The respective waveform according to solution values is shown as:
b. 14𝑆𝑖𝑛(2𝜋(50)𝑡 + 90)
The equation will be given as
14𝑆𝑖𝑛(2𝜋(50)𝑡 + 90
From the equation, it will be found that
A = 14
F = 50
t = 0.02
phase angle = 90
The respective waveform according to solution values is shown as:

c. 4𝑆𝑖𝑛(650𝜋𝑡 + 180)
The equation will be given as
14𝑆𝑖𝑛(2𝜋(50)𝑡 + 90
From the equation, it will be found that
A = 14
F = 325
t = 0.003
phase angle = 180
The respective waveform according to solution values is shown as:

d. 6𝑆𝑖𝑛(700𝜋𝑡 + 270)
The equation will be given as
14𝑆𝑖𝑛(2𝜋(50)𝑡 + 90
From the equation, it will be found that
A = 8
F = 350
t = 0.002
phase angle = 270
The respective waveform according to solution values is shown as:

Question 5
Solution 5:
In this, the loss of space or path can be derived as:
Loss(dB)=20log10(4πR/Λ)
It is given in the question that the frequency will be 6 GHz and radius will be 35,863.
Put these values in the equation,
Hence,
Λ=c/f
Λ=(3*108/6*109)
Λ=0.05
The loss(dB) can be derived as
= 20log10(4*3.14*35863*1000/0.05)
= 20log10(6005857067)
= 20*195.571
The resultant loss in dB will be
Loss(dB)=3911.429 dB

Question 6
Solution:
It is given in the equation that:
𝑠(𝑡) = 5 sin(100𝜋𝑡) + sin(300𝜋𝑡) + sin(600𝜋𝑡)
It can be from the equation that
F1=100/2=50
F2=300/2=150
F3=600/2=300
Now, use the Fourier transform, the equation will be written as
S(f)= (1/2i){(d(f-300)-d(f+300))+(d(f-150)-d(f+150))+5(d(f-50)-d(f+50))}
The formula for Fmax is
Fmax=bandwidth=Fmin
The bandwidth can be calculated as
Bandwidth=300+50
Bandwidth=350MHz
By use Nyquist formula:
Bit rate=2*Bandwidth(log)2M
Case 1:
When M=2
Channel capacity will be
=2*350(log)2*2
=700b/s
Case 2:
When M=4
Channel capacity will be
=2*350(log)2*4
=1400b/s