Wireless Network and Communication - Assignments and Modulation Techniques
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This document covers assignments related to wireless network and communication including modulation techniques such as Amplitude Shift Keying, Frequency Shift Keying, Phase Shift Keying, Differential Phase shift keying, and more. It also explains the Error Detection Process for Cyclic Redundancy Check (CRC), WLAN infrastructure mode, ad-hoc mode, TCP and OSI protocols for wired and wireless LANs, and the drawbacks of square and circle shapes cells for cellular communications.
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Running head: - WIRELESS NETWORK AND COMMUNICATION
WIRELESS NETWORK AND COMMUNICATION
Name of the Student
Name of the University
Author Note
WIRELESS NETWORK AND COMMUNICATION
Name of the Student
Name of the University
Author Note
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1WIRELESS NETWORK AND COMMUNICATION
Assignment No-2
1. Convert the binary data “011010” into analog waveforms using following modulation
techniques:
a. Two level Amplitude Shift Keying
b. Two level Frequency Shift Keying
c. Two level Phase Shift Keying
Assignment No-2
1. Convert the binary data “011010” into analog waveforms using following modulation
techniques:
a. Two level Amplitude Shift Keying
b. Two level Frequency Shift Keying
c. Two level Phase Shift Keying
2WIRELESS NETWORK AND COMMUNICATION
3WIRELESS NETWORK AND COMMUNICATION
d. Differential Phase shift keying
e. Four level Amplitude Shift Keying
f. Four level Phase Shift Keying
d. Differential Phase shift keying
e. Four level Amplitude Shift Keying
f. Four level Phase Shift Keying
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4WIRELESS NETWORK AND COMMUNICATION
g. Eight level Amplitude Shift Keying
2. With fc = 500 kHz, fd = 25 kHz, and M = 16 (L = 4 bits), compute the frequency
assignments for each of the sixteen possible 4-bit data combinations.
Given: -
fc = 500 kHz, fd = 25 kHz, and M = 16, L = 4 bits
fi=fc + (2i – 1 – M) *fd
M=2^L =2^4 = 16, M=16
0000 where i=1, f1= 125KHz
0001 where i=2, f2= 175KHZ
0010 where i=3, f3= 225KHZ
0011 where i=4, f4= 275KHz
0100 where i=5, f5= 325KHz
0101 where i=6, f6= 375KHz
0110 where i=7, f7= 425KHz
0111 where i=8, f8=475KHz
1000 where i=9, f9=525KHz
1001 where i=10, f10= 575KHz
1010 where i=11, f11= 625KHz
g. Eight level Amplitude Shift Keying
2. With fc = 500 kHz, fd = 25 kHz, and M = 16 (L = 4 bits), compute the frequency
assignments for each of the sixteen possible 4-bit data combinations.
Given: -
fc = 500 kHz, fd = 25 kHz, and M = 16, L = 4 bits
fi=fc + (2i – 1 – M) *fd
M=2^L =2^4 = 16, M=16
0000 where i=1, f1= 125KHz
0001 where i=2, f2= 175KHZ
0010 where i=3, f3= 225KHZ
0011 where i=4, f4= 275KHz
0100 where i=5, f5= 325KHz
0101 where i=6, f6= 375KHz
0110 where i=7, f7= 425KHz
0111 where i=8, f8=475KHz
1000 where i=9, f9=525KHz
1001 where i=10, f10= 575KHz
1010 where i=11, f11= 625KHz
5WIRELESS NETWORK AND COMMUNICATION
1011 where i=12, f12=675KHz
1100 where i=13, f13= 725KHz
1101 where i=14, f14=775KHz
1110 where i=15, f15=825KHz
1111 where i=16, f16=875KHz
This scheme can support data rate of
1/Tb = 2Lfd = 2*4*25 = 200Kbps
3. Draw the approximate Analog Modulation and Frequency Modulation waveforms in
complete steps for the following signal:
1011 where i=12, f12=675KHz
1100 where i=13, f13= 725KHz
1101 where i=14, f14=775KHz
1110 where i=15, f15=825KHz
1111 where i=16, f16=875KHz
This scheme can support data rate of
1/Tb = 2Lfd = 2*4*25 = 200Kbps
3. Draw the approximate Analog Modulation and Frequency Modulation waveforms in
complete steps for the following signal:
6WIRELESS NETWORK AND COMMUNICATION
4. Draw the 16 QAM Constellation Diagram having two different amplitude levels and
eight different phase levels.
5. Explain and draw the Error Detection Process for Cyclic Redundancy Check (CRC).
The error in the communication channel during the process of the data transmission
can be detected with the help of the cyclic redundancy check method. In this method cyclic
generator is present at both the end. The receiver and the transmission end comprises of the
generator. At the end of the sender the data is divided by the method of binary division and
the remainder obtained is known as the CRC (Koopman & Chakravarty, 2014). The CRC
consists of the bit which is less than the divisor by 1. At the receiver end the data received is
divided by the same number. If the resultant outcome is zero than there is no error and if the
resultant outcome is non-zero than there is error present. The entire method is used for the
error detection.
4. Draw the 16 QAM Constellation Diagram having two different amplitude levels and
eight different phase levels.
5. Explain and draw the Error Detection Process for Cyclic Redundancy Check (CRC).
The error in the communication channel during the process of the data transmission
can be detected with the help of the cyclic redundancy check method. In this method cyclic
generator is present at both the end. The receiver and the transmission end comprises of the
generator. At the end of the sender the data is divided by the method of binary division and
the remainder obtained is known as the CRC (Koopman & Chakravarty, 2014). The CRC
consists of the bit which is less than the divisor by 1. At the receiver end the data received is
divided by the same number. If the resultant outcome is zero than there is no error and if the
resultant outcome is non-zero than there is error present. The entire method is used for the
error detection.
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7WIRELESS NETWORK AND COMMUNICATION
6. Compute the frame check sequence for the following information:
Message = 10111100, Pattern = 11011
6. Compute the frame check sequence for the following information:
Message = 10111100, Pattern = 11011
8WIRELESS NETWORK AND COMMUNICATION
7. Compute the transmitted signal using Direct Sequence Spread Spectrum for the
following information:
Input: 1011, Locally Generated PN bit stream: 101011011010, T = 3Tc
8. What is the difference between Infrastructure and ad hoc modes in WLAN? Draw
their relative diagrams as well.
WLAN infrastructure mode
WLAN ad-hoc mode
7. Compute the transmitted signal using Direct Sequence Spread Spectrum for the
following information:
Input: 1011, Locally Generated PN bit stream: 101011011010, T = 3Tc
8. What is the difference between Infrastructure and ad hoc modes in WLAN? Draw
their relative diagrams as well.
WLAN infrastructure mode
WLAN ad-hoc mode
9WIRELESS NETWORK AND COMMUNICATION
The connection in the ad-hoc mode is done in a peer to peer communication structure.
The set-up of the ad-hoc network is simple. This mode is cheap and the networks are fast.
The data is transferred from the network to the wireless networks. In case of the infrastructure
mode an access point is required for the communication prices. The set-up is difficult and
costly. The network is slow as it takes time for the transferring of the signal from the access
point which is the source to the final destination (Zhuang, Baskett & Shang, 2013).
9. Compare the differences of TCP and OSI protocols for wired and wireless LANs
using diagrams.
The TCP protocol comprises of 4 different layer and the OSI model comprises of 7
different layer. The protocol present in the OSI model are easily replaceable, in case of the
TCP model the replacing of the protocol is not easy. The OSI model is independent of the
protocol. The TCP is dependent on the protocol (Coonjah, Catherine & Soyjaudah, 2015).
The network layer present in the OSI model offers both connection and connectionless
service. However, the TCP provides only connectionless service. The approach of the two
protocol is different, it is vertical in OSI model and horizontal in the TCP model.
The connection in the ad-hoc mode is done in a peer to peer communication structure.
The set-up of the ad-hoc network is simple. This mode is cheap and the networks are fast.
The data is transferred from the network to the wireless networks. In case of the infrastructure
mode an access point is required for the communication prices. The set-up is difficult and
costly. The network is slow as it takes time for the transferring of the signal from the access
point which is the source to the final destination (Zhuang, Baskett & Shang, 2013).
9. Compare the differences of TCP and OSI protocols for wired and wireless LANs
using diagrams.
The TCP protocol comprises of 4 different layer and the OSI model comprises of 7
different layer. The protocol present in the OSI model are easily replaceable, in case of the
TCP model the replacing of the protocol is not easy. The OSI model is independent of the
protocol. The TCP is dependent on the protocol (Coonjah, Catherine & Soyjaudah, 2015).
The network layer present in the OSI model offers both connection and connectionless
service. However, the TCP provides only connectionless service. The approach of the two
protocol is different, it is vertical in OSI model and horizontal in the TCP model.
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10WIRELESS NETWORK AND COMMUNICATION
10. Explain why the square and circle shapes cells for cellular communications are not
appropriate as compared to hexagonal shape cells.
The cell is so arranged for the cellular communication that are the regions in the cell
r4eceives equal amount of signals and all the regions are covered within the cell. The regions
which are not covered by the cell and are not able to receive signals is known as the
blackouts. In case of the square cell the equal distribution is not possible as the distance from
the centre with each side is greater. This is the reason it is not preferred for the purpose of
cellular communication. On the other hand, the combination of number of circle give rise to a
region where the signal cannot be transmitted. It can be concluded that both the shape
consists of certain drawbacks. however, the bee-hive structure of the hexagon shape helps it
in overcoming all the disadvantages and is the most preferred cell shape for the purpose of
cellular communication (Zhaojun, 2016).
10. Explain why the square and circle shapes cells for cellular communications are not
appropriate as compared to hexagonal shape cells.
The cell is so arranged for the cellular communication that are the regions in the cell
r4eceives equal amount of signals and all the regions are covered within the cell. The regions
which are not covered by the cell and are not able to receive signals is known as the
blackouts. In case of the square cell the equal distribution is not possible as the distance from
the centre with each side is greater. This is the reason it is not preferred for the purpose of
cellular communication. On the other hand, the combination of number of circle give rise to a
region where the signal cannot be transmitted. It can be concluded that both the shape
consists of certain drawbacks. however, the bee-hive structure of the hexagon shape helps it
in overcoming all the disadvantages and is the most preferred cell shape for the purpose of
cellular communication (Zhaojun, 2016).
11WIRELESS NETWORK AND COMMUNICATION
References
Coonjah, I., Catherine, P. C., & Soyjaudah, K. M. S. (2015, December). Experimental
performance comparison between TCP vs UDP tunnel using OpenVPN. In 2015 International
Conference on Computing, Communication and Security (ICCCS) (pp. 1-5). IEEE.
Koopman, P., & Chakravarty, T. (2014, June). Cyclic redundancy code (CRC) polynomial
selection for embedded networks. In International Conference on Dependable Systems and
Networks, 2004 (pp. 145-154). IEEE.
Zhaojun, L. I. (2016). U.S. Patent Application No. 14/937,378.
Zhuang, T., Baskett, P., & Shang, Y. (2013). Managing ad hoc networks of
smartphones. International Journal of Information and Education Technology, 3(5), 540.
References
Coonjah, I., Catherine, P. C., & Soyjaudah, K. M. S. (2015, December). Experimental
performance comparison between TCP vs UDP tunnel using OpenVPN. In 2015 International
Conference on Computing, Communication and Security (ICCCS) (pp. 1-5). IEEE.
Koopman, P., & Chakravarty, T. (2014, June). Cyclic redundancy code (CRC) polynomial
selection for embedded networks. In International Conference on Dependable Systems and
Networks, 2004 (pp. 145-154). IEEE.
Zhaojun, L. I. (2016). U.S. Patent Application No. 14/937,378.
Zhuang, T., Baskett, P., & Shang, Y. (2013). Managing ad hoc networks of
smartphones. International Journal of Information and Education Technology, 3(5), 540.
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