Computer Networks Fundamentals: Homework Assignment

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Table of Contents
Question 1........................................................................................................................................2
Question 3........................................................................................................................................4
Question 4........................................................................................................................................6
Question 5......................................................................................................................................10
Question 6......................................................................................................................................11
Question 7......................................................................................................................................12
Question 8......................................................................................................................................13
Question 9......................................................................................................................................14
References......................................................................................................................................15

Question 1
The IP is a protocol used for the Internet in which the IP address is assigned to each machine and
using this IP address these machines can communicate with each other (Techopedia.com, 2019).
The IP is mainly used for giving identification which helps to uniquely identify the particular
system over the internet. The Network Access Layer is a layer in the TCP/IP model which is at
first position. This layer specifies the accessing of the system to the internet i.e. how the machine
or system is connected with the internet or other types of network. The Network Access layer
also contains the definition of the transmission for the data packets over the physical network
and the medium used for the data transmission. For using of the new hardware, the Network
Access Layer is developed so that the IP datagram can be transmitted by the new hardware
which converts the data into digital signals. The IP protocol help to transmit the data from one
node to another using the IP address whereas the medium used for transmission is defined by the
Network Access Layer. The information is converted or divided into the data packets which is
transfer over the network and these packets contain source and destination address. The IP
decide and maintain the route selection of the packets. The Network Access Layer maps the IP
address into the address which is used for identification that is Ethernet address (O’Reilly, 2019).

Question 2
Figure 1: Telephonic Communication between the Chinese and French Prime Minister
The Following Diagram shows the Communication between the Chinese Prime Minister and the
French Prime minister. The Prime minister uses their Translators for translating their message in
English language. This useful information is transmitted over the Telephonic Communication
layer which uses telephones for communication. This information then is access by the translator
for translating in their language. The translated information is then used by the Prime ministers.

Question 3
(i) As the y-axis shows the amplitude so, Maximum Amplitude = 15.
(ii) The Frequency of the Wave can be given as
F = 1
T where F is frequency and T is Time Period
 F = 1
3 = 0.33 Hz.
(iii) Time Period = 3 sec.
(iv) Phase Angle can be given as
X(t) = A*m*sin (θ) = 0
So, Phase angle = 0∘

(i) As the y-axis shows the amplitude so, Maximum Amplitude = 4.
(ii) The Frequency of the Wave can be given as
F = 1
T where F is frequency and T is Time Period
 F = 1
6.5 = 0.1538 Hz.
(iii) Time Period = 6.5 sec.
(iv) Phase Angle can be given as
X(t) = A*m*sin (θ) = 0
So, Phase angle = 0∘
(i) As the y-axis shows the amplitude so, Maximum Amplitude = 7.5.
(ii) The Frequency of the Wave can be given as
F = 1
T where F is frequency and T is Time Period
 F = 1
2.25 = 0.444 Hz.
(iii) Time Period = 2.25 sec.
(iv) Phase Angle can be given as

X(t) = A*m*sin (θ) = A*m
So, Phase angle = 90∘
Question 4
a. 3 s𝑖𝑛 ( 2 𝜋 (200) 𝑡 )
Amplitude for the given equation = 3
Frequency for the given equation = 200 Hz
Time for the given equation = 0.005 sec.
Phase for the given equation = 0∘
b. 14 s𝑖𝑛 (2 𝜋 (50) 𝑡 + 90)
Amplitude for the given equation = 14

Frequency for the given equation = 50 Hz
Time for the given equation = 0.02 sec.
Phase for the given equation =90∘
c. 4 s𝑖𝑛 (650 𝜋 𝑡 + 180)
Amplitude for the given equation = 4
Frequency for the given equation = 325 Hz
Time for the given equation = 0.003 sec.
Phase for the given equation =180∘

d. 6 s𝑖𝑛 (700 𝜋 𝑡 + 270)
Amplitude for the given equation = 6
Frequency for the given equation = 350 Hz
Time for the given equation = 0.002 sec.
Phase for the given equation = 270∘

Question 5
The selection of the shortest path for the wave can be calculated as 6Hz = 6*109
 Wavelength W = (3∗108)
(6∗109) = 0.495 m
 LdB = 20 log (35.853 * 106) – 20 log (0.495) + 21.98
 LdB = 99.68
The values taken for the antenna are 44dB and 48dB
The lost of free space can be given as:
LdB = 99.68 – 44 - 48
LdB = 7.68

Question 6
The frequency found form the equation 𝑠(𝑡) = 5 sin(100𝜋𝑡) + sin(300𝜋𝑡) + sin(600𝜋𝑡) are:
50, 150 and 300.
Using Fourier transform, the equation we get is
S(f)= 1
2i *{(d*(f-150) – d*(f+150)) + 5*(d*(f-50)-d*(f+50))+d*(f-300)-d*(f+300))}
We know, Fmax = Bandwidth + Fmin
 Bandwidth = 50 + 150 + 300 = 500 MHz
Using the formula given by Nyquist bitrate we get that Bitrate = 2*Bandwidth
When,
M=2
Channel Capacity is 2*500 log2 *2 = 1000 bits/sec.
M=4
Channel Capacity is 4*500 log2 *2 = 2000 bits/sec.
M=8
Channel Capacity is 8*500 log2 *2 = 4000 bits/sec.

Question 7
The data rate is an amount of the data transmitted from one node to another over the channel. It
specifies the Speed of the transmitted data. There are 3 factors on which data rate depends:
The numbers of level available in the digital signals
Bandwidth for a signal
Noise level of a channel
For Noiseless Channels, Nyquist Bit Rate formal is used which can be given as:
Bit Rate = 2 * Bandwidth * log(L) where, base of log is 2, L is Signal Levels.
The Bandwidth is a channel’s Bandwidth which is have fixed quantity value and changes in the
value are not possible. This gives that Data rate is directly propositional to the signal level’s
quantity.
Disadvantages
There is no consideration about the energy loss and the reliability is also reduced for a wave.