Wireless Network: Modulation Techniques, Analog and Frequency Modulation, Error Detection, Direct Sequence Spread Spectrum, WLAN Modes, TCP vs OSI Protocols, Hexagonal Cells

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WIRELESS NETWORK
STUDENT NAME
COLLEGE

Q 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
d. Differential Phase shift keying

e. Four level Amplitude Shift Keying
f. Four level Phase Shift Keying
g. Eight level Amplitude Shift Keying

Q 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.
Fi= fc+ (2i -1-M)fd
F1= 125 kHz 0000
F2= 175 kHz 0001
F3= 225 kHz 0011
F4=275 kHz 0111
F5=325 kHz 1000
F6=375 kHz 1100
F7=425 kHz 0100
F8=475 kHz 0110
F9=525 kHz 0101
F10=575 kHz 1010
F11=625 kHz 0010
F12=675 kHz 1001
F13=725 kHz 1011
F14=775 kHz 1101
F15=825 kHz 1110
F16=875 kHz 1111
Q 3 Draw the approximate Analog Modulation and Frequency Modulation waveforms

Q 4 Draw the 16 QAM Constellation Diagram having two different amplitude levels and eight
different phase levels.
Q 5 Explain and draw the Error Detection Process for Cyclic Redundancy Check (CRC).

CRC is based on the binary division. Here, a redundant bit sequence named the CRC bits are
appended to data end unit in order the resulting unit data might become accurately divisible by
second, which is a pre-determined binary number.
At the destination end, the incoming unit data is then divided by the number and if reminder is
not there at this step then unit data is then pre assumed correct and accepted. The reminder
shows that unit data is damaged during transmission and is rejected.
Q 6 Compute the frame check sequence for the following information: Message = 10111100,
Pattern = 11011
d(t) 1 1 0 1 1
x(t): +1 + 1 - 1 + 1 +1
c(t): 1 0 1 1 1 1 0 0
g(t): +1 -1 +1 +1 +1 +1 -1 -1 +1
Spread sequence, p(t) = x(t)g(t):
x(t): +1 +1 +1 +1 +1 +1 -1 -1 -1 +1 +1 +1
g(t): +1 + 1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1
p(t): +1 - 1 -1 +1 +1 -1 -1 -1 -1 -1 -1 +1
The sequence is 100110000001
Q 7 Compute the transmitted signal using Direct Sequence Spread Spectrum for the following
information: Input: 1011, Locally Generated PN bit stream: 101011011010, T = 3Tc
d(t) 1 0 1 1
x(t): +1 - 1 + 1 +1
c(t): 1 0 1 0 1 1 0 1 1 0 1 0

g(t): +1 -1 +1 -1 +1 +1 -1 +1 +1 -1 + 1 -1
Spread sequence, p(t) = x(t)g(t):
x(t): +1 +1 +1 +1 +1 +1 -1 -1 -1 +1 +1 +1
g(t): +1 -1 -1 +1 +1 -1 +1 +1 +1 -1 -1 -1
p(t): +1 -1 -1 +1 [phone]
The Transmitted signal is 100110000000
Q 8
What is the difference between Infrastructure and ad hoc modes in WLAN? Draw their
relative diagrams as well.
Infrastructure Mode
In the infrastructure mode, the wireless LAN networks are composed of the station with more than
one access points. The access point for the device is taken as the base station which is used for
cellular system. Here, each and every communication which occurs between the stations have to get
through the access point.

Ad Hoc Mode
In the adhoc mode, the wireless network is made up of only the 802.11 NIC compliant stations.
There is no existence of any access point in the network. This type of communication is also called
the peer to peer mode where the systems networked communicates with one another directly over
the wireless radio waves which is 802.11 compliant packets.
Q 9
Compare the differences of TCP and OSI protocols for wired and wireless LANs using
diagrams.
OSI is protocol independent standard, TCP is based on the standard protocols which the

generic and acting as gateway of
communication between the end user and
the network
development of the internet took place. Its regarded as
the communication protocol that allows the connection
of the host.
The transport layer in OSI gives a guarantee
of the delivery of the packets
The transport layer does not give any assurance of the
packet delivery
The OSI model follows the vertical
approach
TCP/IP follows the horizontal approach.
The OSI model contains a separated session
and presentation layer
TCP/IP is not at all made up of separate session and
presentation layer.
The Transport Layer of the OSI model is
Connection Oriented.
The Transport Layer of the TCP/IP is both Connection
less and also Connection oriented.
Q 10
Explain why the square and circle shapes cells for cellular communications are not
appropriate as compared to hexagonal shape cells.
The hexagonal shape of the cell of mostly preferred as compared to the square and circular shape
cells because of the following reasons;
a. The hexagonal shape of the cell has its surface area larger as compared to the square and the
circular shape cells, therefore the small number of the cells with be required for covering
larger geographical area.
b. The hexagonal shape cells approximates closely the radiation circular pattern which is
currently used by the omni- directional antennas which are found on the base station like the

transmitters. The square and the circular shape cells are not used since they leave gaps in
between them if they are implemented.
Bibliography
Hác, A. (2017). Wireless ATM network architectures. International Journal of Network
Management, 11(3), pp.161-167.
Renugadevi, R. and Vijayalakshmi, K. (2019). Modeling a Novel Network Coding Aware Routing
Protocol for Enhancement of Network Performance in Wireless Mesh Network. Wireless Personal
Communications.