Wireless Sensor Networks: A Survey of Cross-Layer MAC and Routing Protocols and Their Performance Evaluation

This document provides a comprehensive survey of cross-layer MAC and routing protocols in wireless sensor networks. It discusses their performance evaluation and explores the latest research and advancements in this field. The document also includes tables, graphs, and equations to illustrate the concepts. Suitable for students studying wireless sensor networks or researchers in the field.

Wireless Sensor Networks A SURVEY OF CROSS-LAYER MAC AND ROUTING PROTOCOLS AND THEIR
PERFORMANCE EVA

WIRELESS SENSOR NETWORKS
1
Table of Contents
Question 1.............................................................................................................................................2
Question 2.............................................................................................................................................2
Question 3.............................................................................................................................................2
a. 10𝑆𝑖n(2𝜋(100)𝑡)......................................................................................................................4
b. 20𝑆𝑖n(2𝜋(30)𝑡 + 90)................................................................................................................5
c. 5𝑆𝑖𝑛 (500𝜋𝑡 + 180).................................................................................................................5
d. 8𝑆𝑖(400𝜋𝑡 + 270).....................................................................................................................5
Question 5.............................................................................................................................................6
a. Source rate R.............................................................................................................................6
b. Capacity of channel (bps)...........................................................................................................6
Question 6.............................................................................................................................................6
Question 7.............................................................................................................................................7
Question 8.............................................................................................................................................8
Question 9.............................................................................................................................................8
Question 10...........................................................................................................................................8
References...........................................................................................................................................10

WIRELESS SENSOR NETWORKS
2
Question 1
From the given case it is identified that the visitor place order with the chef and the host
connect the placed order to the clerk. Moreover, the mobile phone provides a platform for
communication between the guest and the clerk. The cock delivers the pizza to the worker
with the order form that is header to the pizza after that the clerk packs the pizza with the
delivery address. However, the distribution process or van encircles all the orders to be
transported. For this situation, it is analysed that the road gives the physical path for the
delivery.
Question 2
It is observed that both the French Prime Minister and Chinese Prime Minister use the
translator process for converting the signals and information into the English language. The
complete process can be analysed with the below figure where the telephonic network link
both French and Chinese Primer Minister together.
From the above process it is identified that when the PM of French speaks anything then he
addresses his remarks to the PM of China by using translators. It is analysed that for
communication purpose like transferring data both Prime Ministers use the telephonic
system. therefore, the French Prime Minister involve a translator process for converting
signals into English language and share these signals to the Chinese PM which also uses a
translator process that converts signals into the Chinese language.
Question 3
Graph 1:

WIRELESS SENSOR NETWORKS
3
From the above waveform, it is analysed that it is a sinusoidal wave represented on time
and amplitude axis. So, after evaluating the above graph it is observed that amplitude= 15,
frequency= 1/3 Hz, time period = 3 seconds and the phase shift is 0 degree.
Parameters Value
Amplitude 15
Time period 3 sec
Frequency 1/3 Hz
Phase 0 degree
Graph 2:
Parameters Value

WIRELESS SENSOR NETWORKS
4
Amplitude 4
Time period 6.5 sec
Frequency 1/6.5 Hz
Phase 0 degree
Graph 3:
Parameters Value
Amplitude 7.8
Time period 2.3 sec
Frequency 1/2.3 Hz
Phase 90 degree
Question 4
a. 10𝑆𝑖n(2𝜋(100)𝑡)
For the above signal values of amplitude, frequency, time period and phase are given below:
Parameters Value
Amplitude 10
Frequency 100 Hz
Time period 1/100 sec.
Phase 0 degree

WIRELESS SENSOR NETWORKS
5
1 7 13 19 25 31 37 43 49 55 61 67 73 79 85 91 97
-15
-10
-5
0
5
10
15
time
Amplitude
b. 20𝑆𝑖n(2𝜋(30)𝑡 + 90)
As compared with the standard equation of sine wave the values of parameters can be
determined which is given as:
X= A*sin(2pi*(f)*+ phase)
Parameters Value
Amplitude 20
Frequency 30 Hz
Time period 1/30 sec.
Phase 90 degree

WIRELESS SENSOR NETWORKS
6
1 24 47 70 93 116 139 162 185 208 231 254 277 300 323
-25
-20
-15
-10
-5
0
5
10
15
20
25
time
Amplitude
c. 5𝑆𝑖𝑛 (500𝜋𝑡 + 180)
For the above equation values of amplitude, time period, frequency and phase are indicated
in the below table:
Parameters Value
Amplitude 5
Frequency 250 Hz
Time period 1/250 sec.
Phase 180 degree
1 27 53 79 105131157183209235261287313339365391
-6
-4
-2
0
2
4
6
time
Amplitude
d. 8(400𝜋𝑡 + 270)
Below table provides all information about question along with their answers:

WIRELESS SENSOR NETWORKS
7
Parameters Value
Amplitude 8
Frequency 200 Hz
Time period 1/200 sec.
Phase 270 degree
1 25 49 73 97 121145169193217241265289313337361385409433457481
-10
-8
-6
-4
-2
0
2
4
6
8
10
Chart Title
time Amplitude
Question 5
a. Source rate R
 After analysing the given scenario it is observed that the sample size of the picture is
480*500 and numbers of pictures sent per second is around 30.
 Now, total sample size for the images is: 480 * 500 * 30= 7.2*106
 Therefore, the value of source rate (R) in bps is calculated on 5 bits which are given
as:
 R= 7.2*106*5
 After calculating the above equation it has found that R= 36 Mbps that means the
value of source rate is 36 Mbps.
b. The capacity of channel (bps)
 According to the provided case channel bandwidth is 4.5 MHz and signal to noise
ratio in dB is 35. Now, convert SNR (dB) in the normal form of SNR which is given as:
 SNR = 1035/10 = 3162

WIRELESS SENSOR NETWORKS
8
 According to Shannon formula, the value of channel capacity can be determined
with the help of the following equation:
 C= B*log2*(1+ SNR)
 Now, put all values in the above equation that can provide the value of channel
capacity in bits per seconds.
 C= 4.5*106*log2*(1+ 3162)
 Therefore, the capacity of the channel is 52.335*106 bps.
Question 6
 Here, frequency (f) = 4GHz
 Distance between satellite and earth (d) = 35,863 KM
 So, the value of free space loss is can be determined by using the following formula:
 20log10 (f) +20log10 (d) - 147.56 dB
 By putting values of f and d in the above equation isotropic free loss can be
determined and which is given as:
 Free space loss= 20log10 (4*109) +20log10 (35.863*106) - 147.56 dB
 So, isotropic free space loss is 195.6 dB.
Question 7
(𝑡) = 5 sin(200𝜋𝑡) + sin(600𝜋𝑡)
The value of frequency can be determined by taking GCD of both frequencies and from the
above equation it is identified that f1= 100 Hz and f2= 300 Hz. So, the overall frequency is
GCD (100, 300)= 100 Hz.
Now, Bandwidth = 300-100= 200 Hz.
The spectrum for the given signal can be determined by identifying the frequency pair of
each signal given in the question. Therefore, frequency pairs are {(100, 2.5), (-100, 2.5),
(300, 0.5), (-300, 0.5)} and below figure shows the frequency spectrum of given signal.

WIRELESS SENSOR NETWORKS
9
-300 -200 -100 0 100 200 300
0
0.5
1
1.5
2
2.5
3
FREQUENCY
MAGNITUDE
Now, channel capacity can be determined with the help of Nyquist criteria that provided a
formula for finding channel capacity that is C= 2*B*log2(M).
By putting the value of M=2, 4, and 8 the value of channel capacity can be calculated and
below table shows the channel capacity for M=2, 4 and 8.
Parameters Values
For M=2 C= 2*200log2 (2)
C= 400 bits/s/Hz
For M= 4 C= 2*200log2 (4)
C= 800 bits/s/Hz
For M=8 C= 2*200log2 (8)
C= 1200 bits/s/Hz
Question 8
According to Nyquist Theorem, the value of data rate over the channel can be increased by
increasing the numbers of levels in the system (Pan, et al., 2013). There are several
disadvantages of this approach such as less efficient, reduce the reliability of the system and
also decrease the overall performance of channels (Song, et al., 2012). Therefore, data rate
over the channel can be increased by increasing numbers of levels but it is not an effective
approach.
Question 9
Circuit switching Packet switching

WIRELESS SENSOR NETWORKS
10
It includes three phases such as connection
establishment, connection released and data
transfer (Perelló, et al., 2013).
In which directly data transfer takes place
(Harai, 2012).
Data is processed at the source system only Information is processed at all intermediate
node involving source system.
The delay between information units is
uniform (Martinelli, 2012).
Delay is not uniform
Less flexible More flexible
It is analysed that packet switching is an advanced version of circuit switching that uses in
communication systems for exchanging information. There are various advantages of packet
switching over circuit switchings such as more flexible, easy to design and implement and it
is a store and forward technique (Martinelli, 2012).
Question 10
It is observed that d= 40 Km and H(1)=2H(2) [as per the given question]
Now, D= 3.57*sqrt (kH1+kH2)
It is assumed that value of K= 4/3
So, putting values in above equation
40= 3.57*1.1547(sqrt(3H2))
Therefore, H2= 31.6 meters and H1= 2*31.6 =63.3 meters

WIRELESS SENSOR NETWORKS
11
References
Harai, H., (2012) Optical packet & circuit integrated network for future networks. IEICE
transactions on communications, 95(3), pp.714-722.
Martinelli, A., (2012) An emerging paradigm or just another trajectory? Understanding the
nature of technological changes using engineering heuristics in the telecommunications
switching industry. Research Policy, 41(2), pp.414-429.
Pan, X., Wang, W., Feng, D., Liu, Y., Fu, Q. and Wang, G., (2013) On deception jamming for
countering bistatic ISAR based on sub-Nyquist sampling. IET Radar, Sonar &
Navigation, 8(3), pp.173-179.
Perelló, J., Spadaro, S., Ricciardi, S., Careglio, D., Peng, S., Nejabati, R., Zervas, G.,
Simeonidou, D., Predieri, A., Biancani, M. and Dorren, H.J., (2013) All-optical packet/circuit
switching-based data center network for enhanced scalability, latency, and throughput. IEEE
Network, 27(6), pp.14-22.
Song, Z., Liu, B., Pang, Y., Hou, C. and Li, X., (2012) An improved Nyquist–Shannon irregular
sampling theorem from local averages. IEEE Transactions on Information Theory, 58(9),
pp.6093-6100.