Wireless Networks & Communications: Encoding, Error Control, Multiplexing, and Wi-Fi

This article discusses encoding and error control techniques, multiplexing and multiple accesses, and Wi-Fi in wireless networks & communications. It covers topics such as calculating data rate, CRC, BSS, ESS sizes, and throughput, designing a network, and recommending security strategies. The article also provides explanations of various techniques such as Differential Manchester, Manchester encoding, Pseudo-ternary encoding, FDMA, TDMA, CDMA, and OFDM.

WIRELESS NETWORKS & COMMUNICATIONS
[Author Name(s), First M. Last, Omit Titles and Degrees]
[Institutional Affiliation(s)]

Part I: Encoding and error control
a) Calculate the data rate required for robot to remote controller communication
Data rate is the number of bits sent per second=16/2=8 mbps
b) Explain 3 types of suitable encoding techniques could be used to encode the status string.
i. Differential Manchester: A binary zero is illustrative of the presence of a
transition at the start of a bit
ii. Manchester encoding: A transition occurs at the middle of each of the bits that
have been transmitted. A binary zero hints a low to high transition that occurs in
the middle while a low to high transition is hinted by a binary one (Lu et al.,
2015).
iii. Pseudo-ternary encoding: In this encoding, the absence of a line illustrated a one
while a zero is indicated using an alternating positive and negative. A long string
of zeros do not lead to any loss in the synchronization
c) i. Write the status string in binary for this instance
Step 1: the initial step is to change the provided parameters to binary notation

Motor functionality = 1111
5m/s = 0101
75%= 0.11
48cm = 110000
Step 2: Combination of all the binary to come up with a single string
0101110000111111
ii. Represent the status string on ASK, FSK, and PSK encoding techniques
FSK: Switching or alteration occurs to the carrier wave frequency upon the transmission of logic
but no variations occur to the carrier on transmission of logic zero.
ASK: In the occurrence of a transmission, logic one undergoes transmission over the media. A
logic zero returns no carrier signal transmission over a network.
PSK: A change from a logic one to zero or zero to one changes the phase angle by 180⁰ while no
lack of a transition in the data bots does not result into a change in the phase.

d) Calculate the CRC for the status string derived in c) with polynomial divisor 11001101
Dividend is given as 0101110000111111
Divisor is provided as = 11001101
Stage 1: Sum up the digits of the dividends
= No. of divisors -1= 8-1= 7 units.
Stage 2: Sum up additional digits to the data word to come up with argument data word
= 01011100001111110000000

e) Briefly explain some other error control and flow control techniques that Sam can use to
ensure the accuracy of robot control
Flow control techniques
1. Stop and wait control: An acknowledgement is sent as a response by the receiver of
a packet to the sender when a frame is sent and this goes on until the transmission is
finished. The receiver must hint readiness to receive a packet (Agiwal, Roy &
Saxena, 2016).
2. Sliding window flow control: The sender and the receiver are both having a window
frame which is number as per the sliding window. Th sender sends as many frames as
can be held by the window and the receiver upon receiving the required number of
frames respond by sending an acknowledgement for all frame sides. The
acknowledgement is followed by the quantity of the subsequent frames that are to be
sent (Bi, Zeng & Zhang, 2016).

Error control techniques
Forward Error Control: Used in the detection and correction of errors that occur in unreliable and
noisy channels. The main underlying idea is encoding of the bits by the sender in a redundant
manner through the use of an error correcting bit with the aid of the Hamming code. The
redundancy gives the receiver an opportunity to attempt correcting a small number of errors
which might have taken place in the message and in most cases does corrections to such errors
without necessarily having to be transmitted by the sender (Saifullah et al., 2015).
II: Multiplexing and multiple accesses
FDMA gives the users an opportunity to have access and use a bandwidth simultaneously and is
composed of a base station that is able to hold numerous users at different frequencies. A guard
band is used in eliminating chances of hindrances due to frequencies being close to each other in
this technique.
TDMA: The idea behind this technique is that users are divided with respect to the time slots
such that every user has access to the bandwidth but at various times. There is just a single
bandwidth channel

CDMA: All the transmissions are carried by a single channel simultaneously. This technique
defines communication using various codes which must meet certain criteria in order to achieve
the desired output in the communication channel.
b) Main air interface in the change from 3G to 4G and WiMAX. Explain how OFDM is
different from above techniques and the special feature of signals used by OFDM
Orthogonal Frequency Division Multiplexing closely resembles the FDMA as it mainly uses the
basic principle of FDMA that permits transmission of multiple messages simultaneously over a
single channel. OFDM is different from the conventional multiplexing strategies in the sense that
it has a number of parallel narrow-band subcarriers as opposed to a single wide-band carrier for
the transmission of information thereby enhancing its efficiency when it comes to wireless
communication. This is only achieved by ensuring that the signals are orthogonal to each other
(Maccartney et al., 2015).

c)
i. For 48 subscribers what should be the subcarrier bandwidth (fb)?
Subcarrier bandwidth
Bandwidth=frequency/Subscribers
=40MHz/48 subscribers=833.3 KHz/subcarrier
ii. Propose a suitable subcarrier bit time of T to achieve orthogonality
Bit time T is determined as the reciprocal of the subcarrier bandwidth in attaining orthogonality
T=1/f
=1/0.8333
=1.2s
iii. Explain how OFDM overcomes the issue of intersymbol interference (ISI)
By ascertaining that each of the subcarrier signals is orthogonal to each other such that zero is
the resultant of the dot multiplication of any two adjacent signals (Yang et al., 2015).

Part III: Wi-Fi
a) Design the network specifying the locations of access points and distributions system
showing the backbone network.
b) Calculate the BSS and ESS sizes
The BSS illustrated the area coverage by each of the Wi-Fi hotspot and is determined using the
formula for each of the rooms which are in this case square in shape.
Area=L*W
Lounge and reception=20*10=200m2
1st Room= L*W=10*10=100m2

2nd Room= L*W=10*10=100m2
3rd Room= L*W=10*10=100m2
4th Room= L*W=10*10=100m2
5th Room= L*W=10*10=100m2
The ESS shows the area covered by all the points of access which have a connection with the
distribution system and is determined through finding the sum of each of the areas under an
access point (Zhang et al., 2015).
ESS=Lounge and reception+1st Room+2nd Room+3rd Room+4th Room+5th Room
200+100+100+100+100+100=700m2
b) Calculate the throughput for the DS
The data that pass through the network system per second is determined by
8*100=800 mbps (for individual room)
5*800=4000 mbps (throughput of the rooms)
Throughput of lounge and reception=25*100 mbps=2500 mbps
Total throughput=4000+2500=6500 mbps
c) Recommend a suitable IEEE Substandard for the network and give a reason for your
choice.
An ideal IEE substandard for the network would be 802.11ac owing to its ability to offers up to
thrice faster connection of the substandard 802.11 b/n. The Wi-Fi also permits multiple users
besides provision of minimized buffering, fast surfing and smooth streaming of video (Loo,
Mauri & Ortiz, 2016).
d) Recommend suitable security strategies for the network
 Ensuring the Wi-Fi is off when not being used

 Having a unique, secure and strong password for all the wireless networks
 Conversion of the name of the fault network to SSD
 Adoption of WPA2 encryption for the network to enhance the security

References
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