MITS5003 Wireless Networks & Communication 2022

MITS5003 Wireless Networks & Communication - 2019SS
Assignment No 1
Process of communication
In this situation we have two hosts namely X and Y. Host X want to communicate with Host Y
and to do so it establishes as virtual circuit. When the virtual circuit has been fully developed the
both the ATM layer and the Transportation should be in agreement with the type of service being
used. The Asynchronous Transfer Mode (ATM) is a switched need network that is different from
the switch Ethernet in the following ways; ATM uses fixed-length packets of 53 bytes, it
prioritizes transmissions based on Quality of Service (QoS), it provides no error correction on
the user data and it uses a very different type of addressing from traditional data link layer
protocols such as Ethernet or token ring. The contact between two hosts is comprised of three
parts which Traffic to be offered, the service agreed upon by the two host and the compliance
requirements. (University, 2019)
The virtual circuit won’t work when both sides of the hosts does not agree on the contact point.
The process of communication;

 The host computer X establishes the quality of service (QOS) parameters first using the
network parameters that are located within the node of computer X and then it set ups a
route of communication for instance. X → M →Y
 Switch connection denoted by X 1 is reserved by node X
 X then sends X1 to M requesting for a reserve connection X2.
 The X2 then send to Y by M to requesting for reserve connection X3.
 Host computer Y finds Host computer X signals and it then send X3 as an
acknowledgement and in turn computer X will receive A1back again.
A summary of the Communication process from host computer X to Y and vice versa with the
reserved connection along the way allowing Qos. When the communication is completed the
communication is broken down and the virtual circuits are the resets ready for the next usage.
X → X 1(X )→ X 2 ( M ) → X 3 (Y ) →Y
Y → X 3 (Y )→ X 2 ( M ) → X 1( X ) → X
The ATM- A dressing
Permanent virtual circuit ( PVC )Y → M →Y
Switched virtual circuit (SVC ) X → M → Y ∨X → N → M → Y .
Routes and switches on the actual path.

solution
a) 8 sin ( 2 π ( 500 ) t )
Thisis a modulation signal of the form y= A sin ( w (t )+C ) +D (Alencar, 2018)
Where Amplitude= A ,
Period= 2 π
w ,
Frequency= 1
Period
Phase=C
Solution
Amplitude=8 ,
Period= 2 π
w = 2 π
(2 π∗500)=2∗10−3 seconds
Frequency= 1
Period = 1
2∗10−3 =500 Hz
Phase=0o
b) 3 sin ( 600 t+ 45 )
Amplitude=3 ,
Period= 2 π
w = 2 π
(600)=0.01047 seconds

Frequency= 1
Period = 1
0.01047 =95.493 Hz
Phase=45o
c) 6 sin ( 400 t +135 )
Amplitude=6 ,
Period= 2 π
w = 2 π
(400)=0.0157 seconds
Frequency= 1
Period = 1
0.0157 =63.662 Hz
Phase=135o
d) 2 sin ( 1000t +180 )
Amplitude=2 ,
Period= 2 π
w = 2 π
(1000)=6.2832∗10−3 seconds
Fr equency= 1
Period = 1
6.2832∗10−3 =159.155 Hz
Phase=180o

a) Maximum Amplitude =15
b) Time period=3 s
c) Frequency= 1
Period = 1
3 =0.333 Hz
d) Phase
¿ the modulation equation y =Asin (w (t)+C)
y=0 , A=15 , w ( t )=0 , C=Phase
substituting the known values intothe equation above
0=sin(C )
C=sin−¿ (0 )=0o ¿
Phase=0o
a) Maximum Amplitud e=4
b) Time period=6.5 s
c) Frequency= 1
Period = 1
6.5 =0.15385 Hz
d) Phase
¿ the modulation equation y =Asin (w (t)+C)
CITATION BLT 06 ¿1033 (B . L .Theraja , 2006)
y=0 , A=4 , w ( t ) =0 , C=Phase
substituting the known values intothe equation above
0=sin(C )
C=sin−¿ (0 )=0o ¿
Phase=0o

a) Maximum Amplitude=7.5
b) Time period=2.25 s
c) Frequency= 1
Period = 1
2.25 =0.44444 Hz
d) Phase
¿ the modulation equation y =Asin (w (t)+C)
y=1 , A=4 , w (t )=0 ,C=Phase
substituting the known values intothe equation above
1=sin(C)
C=sin−¿ (1 )=90o ¿
Phase=90o
Solution
Fundamental Frequency f 0
¿ is thelowest frequency
Calculatingthe frequencies∈the signal

f 1=100 π
2 π =50 Hz , f 2= 300 π
2 π =150 Hz f 3= 600 π
2 π =300 Hz
Fundamental Frequency f 0=f 1=50 Hz
Spectrum
The spectrumof a sine signal wave f r omthe tables is given by the following equation
t= e jωt−e− jωt
2
Ont aking ℱ Transfor m ationon the avove equationtwice we get :
¿ 2 π (δ ( w−wo )−δ ( w+ wo ) )
2 j
¿ πj¿
¿ the sine wave we have the wo being the period ¿ be plotted on the Xaxis
¿ πj values being the Amplitude ¿ be plotted on theY axis .
The general graphical spectrum of the signal is:
Using the Amp litude∧period data¿ the sine wave ¿ plot thewave spectrum

Bandwidth
Bandwidth=2∗Maximum Frequency
¿ 2∗f 3= ( 2∗300 ) =600 Hz
Channel capacity
C=BW log2 ( 1+SNR )CITATION Tut 19 ¿ 1033(Tutorialspoint . dev , 2019)
BW =600 Hz∧ ( SNR )db=250 dB
( SNR )db=10 log10 (SNR)
250=10 log10 ¿)
SNR=1025
substituting the values into the equation gives
C=600∗log2 ( 1+ 1025 )=49828.9214 Hz
Nyquist theorem has a bit rate formular that is used in calculating the theoretical
maximum bit rate of any particular noiseless channel.
The formula is given by BitRate=2∗Bandwidth∗log2 (L) . In the equation, the bandwidth
is the channel bandwidth, the term L represent the number of signal levels representing
the data, BithRate is bit rate in bits per second. The bandwidth of the channel is a fixed
quantity that cannot be changed and hence data rate will be proportional to the number of
signal levels. (Tutorialspoint.dev, 2019)

Disadvantage of this approach: When the signal levels are increased, reliability of the
system reduces. (Tutorialspoint.dev, 2019)
Difference
The main difference is that circuit switching is mainly connection oriented while packet
switching virtual circuit is connectionless.
Advantages of Packet switching over Circuit Switching
1. Packet switching can perform data rate conversion.
2. Higher priority packet can be transmitted first since packet switching considers the
priorities.
3. Lower cost of communication.
4. During the same time different nodes of communications can be used.
Advantages of Circuit Switching over packet switching
1. Between the pairs of communication nodes dedicated and secured circuits can be
established. (KEARY, 2019)
2. Data communication process experience very few delays.
solution
Equation¿ be used
d= √2 h1 R+ √2h2 R
Where
d=distnace between transmeter ∧receiver=40 km=4∗104 m
h1∧h1=Heights of antennas
R=Earth radius=6400 km=64∗105 m

Taking the Assuotion that h1 is 3×h2
h1=3 h2
Solvingthe equation∈terms of h2
d= √ 2∗3 h2∗R+ √ 2 h2 R
Factoring out √2 h2 R
d= √ 2 h2 R ( 1+ √ 3 )
Dividing RHS∧LHSby ( 1+ √ 3 )
d
( 1+ √3 ) = √2h2 R
squaringboth sides∧making h2 subject of the formular
h2 = 1
2 R∗
( d
( 1+ √3 ) )2
Solving for h2
Making h 2the subject of the equation
h2 = 1
2 R∗
( d
( 1+ √3 ) )2
Solving for h 2 usingthe known value of R∧d
h2 = 1
2∗64∗105 ∗
( 4∗104
( 1+ √ 3 ) ) 2
=16.75 m
h1=3 h2= ( 3∗16.75 ) =50.25 m
References
Tutorialspoint.dev. (2019). Retrieved from Computer Network | Maximum data rate (channel capacity)
for noiseless and noisy channels: https://tutorialspoint.dev/computer-science/computer-

network-tutorials/computer-network-maximum-data-rate-channel-capacity-noiseless-noisy-
channels
Alencar, M. S. (2018). Modulation theory. River Publishers.
B.L. Theraja, R. S. (2006). Textbook of Electrical Technology Volume IV - Electronic Devices and Circuits. S
Chand & Co Ltd.
KEARY, T. (2019, March 19). Circuit Switching vs Packet Switching. Retrieved from
https://www.comparitech.com/net-admin/circuit-switching-vs-packet-switching/
University, T. O. (2019). Protocols in multi-service networks. Retrieved from Protocols in multi-service
networks: https://www.open.edu/openlearn/science-maths-technology/computing-ict/
protocols-multi-service-networks/content-section-4.3