University WSN Assignment: Wireless Network Communication Analysis

This document provides a comprehensive solution to a Wireless Sensor Networks (WSN) assignment. The solution addresses several key aspects of WSNs, including network communication models, signal analysis (amplitude, frequency, phase, and time period calculations for various waveforms), channel capacity calculations, and isotropic free space loss. It also explores concepts like packet switching versus circuit switching, and antenna height determination. The assignment covers topics such as the ordering and delivery of a pizza using layer models, the communication between two prime ministers using translators, and the analysis of different waveforms. The solution provides detailed calculations, diagrams, and explanations to facilitate a thorough understanding of the concepts involved in WSNs.

WSN 1
Wireless network communication

WSN 2
Question 1
The guest emphatically submits the request with the cook and the host sends the data
perusing request to the agent. The telephone gadget can possibly give the physical way
to the request to be a shared structure host to an assistant in a positive way.
After that, the cook gives the pizza to the assistant as request. In addition, clerk of the
board packs the order in a container and conveyance car takes all requests for
conveying the request to the visitor. For conveying pizza from worker to guest Street
show a huge job that gives a way to conveying order to the visitor.
Question 2
Below diagram features the correspondence between both PMs by utilizing the
telephonic framework. From given contextual analysis, it is discovered that the two PMs
Uses interpreters for changing over flag into the English. The telephone is a champion
among the effective technique for sending information beginning with one location to
another. However, when the French PM debates something then he conveys his
remarks to the Chinese PM. Therefore, by using translators both PMs can exchange
information and convert into a specific language.
Question 3
For figure first:
Amplitude 15
Phase 00
Frequency 0.33 Hz

WSN 3
Time period 3 seconds
For figure second:
Amplitude 4
Phase 00
Frequency 0.15 Hz
Time period 6.5 seconds
For figure third:
Amplitude 7.8
Phase 900
Frequency 0.43 Hz
Time period 2.3 seconds
Question 4
10 Sin(2*pi(100)t)
 Phase= 0 degree
 Time period= 0.01 seconds
 Amplitude= 10
 Frequency= 100 Hz
1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101
-15
-10
-5
0
5
10
15
Chart Title
time Amplitude

WSN 4
20 Sin(2*pi*(30)t + 90)
 Phase= 90 degree
 Amplitude = 20
 Time period= 0.03 seconds
 Frequency= 30 Hz.
1 17 33 49 65 81 97 113129145161177193209225241257273289305321
-25
-20
-15
-10
-5
0
5
10
15
20
25
Chart Title
time Amplitude
5*Sin(500*pi*t + 180)
 Phase= 180 degree
 Time period= 0.004 seconds
 Amplitude= 5
 Frequency= 250 Hz.

WSN 5
1 20 39 58 77 96 115134153172191210229248267286305324343362381
-6
-4
-2
0
2
4
6
Chart Title
time Amplitude
8*Sin(400*pi*t + 270)
 Phase= 270 degree
 Time period= 0.005 seconds
 Amplitude= 8
 Frequency= 200 Hz.
1 27 53 79 105131157183209235261287313339365391417443469495
-10
-8
-6
-4
-2
0
2
4
6
8
10
Chart Title
time Amplitude
Question 5
Value of pixels= 480*500

WSN 6
Images per seconds= 30 per seconds
SNR (dB)= 35= 1035/10 = 3162
Channel capacity= C= B*log2*(1+ SNR)
So, C= 4.5*106*log2*(1+ 3162)
Or, value of channel capacity is 52.3 Mbps
Question 6
Frequency= 4 GHz
D= 35,863 KM
Now, isotropic free space loss= (4*109) +20log10 (35.863*106) - 147.56 dB
Isotropic free space loss is 195.66 dB
Question 7
S(t)= 5*sin(200*pi*𝑡) + sin(600*pi*𝑡)
By taking greatest common divisor of both frequency factors can provide total
frequency that is 100 Hz.
Bandwidth= 300-100 = 200 Hz
Spectrum:

WSN 7
-300 -200 -100 0 100 200
0.5
2.5 2.5
0.5
Chart Title
Series1
Channel capacity: C= 2*B*log2(M)
Here B= B= 200
M=2, 4, and 8
So, for 2 numbers of levels (M= 2)
C= 400 bits per seconds per Hz
for 4 numbers of levels (M= 4)
C= 800 bits/seconds/Hz
For 8 numbers of levels (M= 8)
C= 1200 bits/seconds/Hz
Question 8
The value of data rate can be improved without increasing bandwidth of the given signal
because Nyquist suggested that numbers of levels are directly linked with data rate
(Chen, et al., 2011). So, when numbers of levels increase then the value of data rate
also increase but it reduces dependability of the channel which is very common
drawback of this technique (Shi, et al., 2008).

WSN 8
Question 9
(Figure: Packet switching virtual vs. circuit switching)
(Source: Das, Parulkar, and McKeown, 2009)
Advantages:
 More efficient (Chaillet, and Bicchi, 2008).
 Improve the performance of the system
 Does not require any physical path
Question 10
It is identified that remoteness between transmitter and receiver is 40 KM. Moreover,
height of first antenna is two times of another antenna. Below formula can be adopted
for determining the height of antennas
D= 3.57*sqrt (kA1+kA2)
Put D= 40 KM, K= 4/3 and A1=2A2
40= 3.57*1.1547(sqrt(3A2))
A2= 31.666 meters and A1= 63.34 meters

WSN 9
Therefore, the height of first antenna is 31.66 meters and height of other receiver or
antenna is 63.34 meters.

WSN 10
References
Chaillet, A. and Bicchi, A., (2008) Delay compensation in packet-switching networked
controlled systems. In 2008 47th IEEE Conference on Decision and Control, 12(2), pp.
3620-3625). IEEE.
Chen, X., Yu, Z., Hoyos, S., Sadler, B.M. and Silva-Martinez, J., (2011) A sub-Nyquist
rate sampling receiver exploiting compressive sensing. IEEE Transactions on Circuits
and Systems I: Regular Papers, 58(3), pp.507-520.
Das, S., Parulkar, G. and McKeown, N., (2009) Unifying Packet and Circuit Switched
Networks with OpenFlow. Dec, 7(2), p.10.
Shi, G., Lin, J., Chen, X., Qi, F., Liu, D. and Zhang, L., (2008) UWB echo signal
detection with ultra-low rate sampling based on compressed sensing. IEEE
Transactions on Circuits and Systems II: Express Briefs, 55(4), pp.379-383.