Wireless Network and Communication

This document discusses various topics related to wireless network and communication. It covers IP and network access layer, circuit switching vs packet switching, frequency parameters, channel capacity, and more. Find study material and solved assignments on Desklib.

[Wireless Network and communication]

WIRELESS NETWORK AND COMMUNICATION 1
Question 1
IP and network access layer
Both IP and network access are major layers of TCP/IP protocol which helps for performing data
transmission between networks and computing devices. IP refers to the internet protocol later
which is mainly utilized for transporting network packets from one location to another while
network access layer is used for accessing the transmitted data and managing overall
communication process (Davidson, 2012). As compared with the network access layer, the IP
layer does not take benefit of sequencing services which is occurred in the data link layer.
Moreover, the internet layer is not able to fulfil the objective of managing link states between the
communication nodes while network access layer is able to perform such activities in the
communication channels (Goralski, 2017). A recent study reported that the internet layer utilizes
IP based packets while the network access layer includes network-based signals and packets
(Jasin, et al., 2017). In the case of IP layer control information is embedded with the transport
and application layer but in-network access layer control information is embedded to the internet
and network systems.
Question 2
The above diagram includes telephone line, translator and translator used by the French and
Chinese PMs for exchanging data. The provided scenario can be analyzed by using the
connection between the telephone and translator which is highlighted below:

WIRELESS NETWORK AND COMMUNICATION 2
It is stated that both ministers are performing data communication with each other using the
translator process. Now, suppose the French PM speaks to the Chinese PM then the signal
transmits using French translators which pass the main information from the main server to
Chinese translator using telephone lines and system. Moreover, the translator used by Chinese
PM translates the signals or data and passes to the Chinese PM using the telephonic system.
Question 3
From above sinusoidal waveform, the values of time period, frequency and other parameters are
described below:
Amplitude= 15
Time period= 3 second
Phase= zero (0) degree
Frequency= 0.33 Hz

WIRELESS NETWORK AND COMMUNICATION 3
Amplitude= 4
Time period= 6.4 second
Frequency= 0.156 Hz
Phase= 0 degree
Amplitude= 7.7
Time period= 2.3 second
Frequency= 0.434 Hz
Phase= 90 degree

WIRELESS NETWORK AND COMMUNICATION 4
Question 4
Equation 1:
Factors Value
Amplitude 3
Frequency 200 Hz
Time period 1/200 sec.
Phase 00
1 250 499 748 997 1246
-4.00000
-3.00000
-2.00000
-1.00000
0.00000
1.00000
2.00000
3.00000
4.00000
Chart Title
time
y
Axis Title
Axis Title
Equation 2:
Parameters Value
Amplitude 14
Frequency 50 Hz
Time period 1/50 sec.
Phase 900

WIRELESS NETWORK AND COMMUNICATION 5
0.00020.0025 0.0048 0.0071 0.0094 0.01170.0140 0.0163 0.0186
-20.00000
-15.00000
-10.00000
-5.00000
0.00000
5.00000
10.00000
15.00000
20.00000
Y
Y
Equation 3:
Parameters Value
Amplitude 4
Frequency 325 Hz
Time period 1/325 sec.
Phase 1800
1 24 47 70 93 116 139 162 185 208 231 254 277
-5
-4
-3
-2
-1
0
1
2
3
4
5
time
y

WIRELESS NETWORK AND COMMUNICATION 6
Equation 4:
Parameters Value
Amplitude 6
Frequency 350 Hz
Time period 1/350 sec.
Phase 2700
1 16 31 46 61 76 91 106121136151166181196
-8
-6
-4
-2
0
2
4
6
8
time
y
Question 5
Frequency= 6 GHz
Distance= 35,863 kilometer
Total free space loss= 20log10 (F) +20log10 (D) - 147.56 dB
Put all the given values in the above free space loss formula:
FSL= 20log10 (6*109) +20log10 (35.863*106) - 147.56 dB
Therefore, at 6GHz frequency total isotropic free space loss is 199.09 dB.

WIRELESS NETWORK AND COMMUNICATION 7
Question 6
S(𝑡) = 5*sin(100𝜋𝑡) + sin(600𝜋𝑡) + sin(300𝜋𝑡)
From the above sinusoidal equation, it is found that the frequency parameters for the included
signals are 50 Hz, 300 Hz and 150 Hz.
Now, the total fundamental frequency can be calculated using the GCF technique that identifies
the greatest common factor in the frequency parameters.
So, fundamental frequency= 50 Hz
Bandwidth= 300- 50
Or, bandwidth= 250 Hz
-300 -250 -200 -150 -100 -50 0 50 100 150 200 250 300
0
0.5
1
1.5
2
2.5
3
Spectrum
Frequency
Amplitude
Channel capacity can be calculated with the help of following Nyquist equation:
C= 2*B*log2(M)
Put, B= 250 Hz and M= 2, 4, and 8
Below table indicates channel capacity at different numbers of levels used in the communication
systems:
Parameters Values

WIRELESS NETWORK AND COMMUNICATION 8
For M=2 C= 500 bits/s/Hz
For M= 4 C= 1000 bits/s/Hz
For M=8 C= 1500 bits/s/Hz
Question 7
In order to solve the given problem Nyquist theorem can be applied that shows the relation
between channel capacity and bandwidth (C= 2*B*log2(M)).
It is reported that the data rate over the channel can be improved by enhancing numbers of levels
rather than bandwidth which is mainly avoided in the communication systems because of less
effectiveness (Sun, et al., 2012). The key drawback of such process is that increasing numbers of
levels may affect stability, reliability and efficiency of the communication channels (Kipnis, et
al., 2015).
Question 8
Packet switching Vs circuit switching
There are numerous differences between packet switching and circuit switching which are
highlighted in the below tabular form:
Circuit switching Packet switching
Guaranteed capacity No guarantees
It requires a path for sending data Send data immediately
No reordering process Data packets may be reordered
No delay occurs Delay occurs
From the previous study, it is identified that the key benefit of packet switching is that it is more
efficient as compared with circuit switching (Heisswolf, et al., 2013). Moreover, packet
switching has the ability to transmit data immediately from one source to another. In-circuit

WIRELESS NETWORK AND COMMUNICATION 9
switching, data can be transferred with less delay and allow a single path for sending all data to
the destination (Lusala, and Legat, 2012). Another advantage of circuit switching is that it
provides a physical path between source and destination which does not occur in the packet
switching.
Question 9
From the provided scenario there are following facts identified:
Distance= 60 km
H1= 4H2 (as per the given information)
Now, the antenna's height can be calculated using the described equation:
60= 3.57*sqrt (kH1+kH2)
K= 1.3
So, 60= 3.57*sqrt (1.3*(5H2))
Or, H1= 169.476 meter
H2= 42.369 meter

WIRELESS NETWORK AND COMMUNICATION 10
References
Davidson, J., (2012) An introduction to TCP/IP. Springer Science & Business Media.
Goralski, W., (2017) The illustrated network: how TCP/IP works in a modern network. Morgan
Kaufmann.
Heisswolf, J., König, R., Kupper, M. and Becker, J., (2013) Providing multiple hard latency and
throughput guarantees for packet switching networks on-chip. Computers & Electrical
Engineering, 39(8), pp.2603-2622.
Jasin, A., Alsaqour, R., Abdelhaq, M., Alsukour, O. and Saeed, R., (2012) Review on current
transport layer protocols for TCP/IP model. International Journal of Digital Content Technology
and its Applications, 6(14), pp.495-503.
Kipnis, A., Goldsmith, A.J., Eldar, Y.C. and Weissman, T., (2015) Distortion rate function of
sub-Nyquist sampled Gaussian sources. IEEE Transactions on Information Theory, 62(1),
pp.401-429.
Lusala, A.K. and Legat, J.D., (2012) A hybrid NoC combining SDM-TDM based circuit-
switching with packet-switching for real-time applications. In 10th IEEE International NEWCAS
Conference, 12(7), pp. 17-20.
Sun, H., Chiu, W.Y., Jiang, J., Nallanathan, A. and Poor, H.V., (2012) Wideband spectrum
sensing with sub-Nyquist sampling in cognitive radios. IEEE Transactions on Signal
Processing, 60(11), pp.6068-6073.