MITS5003 Wireless Networks & Communication Assignment 1

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MITS5003 Wireless Networks &
Communication - 2019S2
Assignment No 1

Contents
Ans.1..........................................................................................................................................3
Ans. 2.........................................................................................................................................3
Ans. 3.........................................................................................................................................4
(a)...........................................................................................................................................4
(b)...........................................................................................................................................6
Ans. 4.........................................................................................................................................8
(a)...........................................................................................................................................8
(b)...........................................................................................................................................9
(c).........................................................................................................................................10
(a).........................................................................................................................................11
Ans. 5.......................................................................................................................................12
Ans. 6.......................................................................................................................................13
Ans. 7.......................................................................................................................................14
Ans. 8.......................................................................................................................................15
Ans. 9.......................................................................................................................................16
List of Figures
Figure 1: Communication plan...................................................................................................4
Figure 2: Given 1st figure..........................................................................................................4
Figure 3: Given 2nd figure...........................................................................................................6
Figure 4: Obtained graph for (a)................................................................................................9
Figure 5: Obtained graph for (b)..............................................................................................10
Figure 6: Obtained graph for (c)..............................................................................................11
Figure 7: Obtained graph for (d)..............................................................................................12

Ans.1
The Network Access layer
The smallest level in the TCP / IP communication structure is the Network Access Layer. The
protocols in this section provide the device with the means of transmitting information to
other computers via a straight connected network. It describes how the network can be used
to send an IP datagram. Network Access Layer protocols must, unlike higher-level protocols,
be aware of information of the fundamental network such as packet structures, targeting, etc.
used to type the transmitted data properly to meet network limitations.
Users often ignore the Network Access Layer. The TCP / IP layout masks the functionality of
the reduced levels, and higher-level protocols are all recognized protocols such as IP, TCP,
UDP, etc.
While
The Internet Protocol
the Internet layer or the IP layer defined as internetworking group protocols or methods for
the internet protocol suite and using these
The internet layer or IP layer is a group of internetworking techniques, protocols and
specifications used for carrying packets from the host to the destination host, if necessary,
specified by an IP address, defined by Internet Protocol or IP, from the originating host
across the network. The title of the internet layer comes from the task of building the internet
or enabling internet, which is the notion of linking various networks via gateways.
The Protocols using the IP-based messages are for internet layers. The internet layer does not
contain protocols that specify the communication between local on-line network nodes to
maintain the links between local nodes, such as the local network topology, and that normally
use protocols depending on the framing of link-type signals. The link-layer includes such
protocols.
Ans. 2
Following is the communication diagram of the given problem

Figure 1: Communication plan
According to the mentioned Numbering on the lines, the communication will be established.
Following is a small description of communication way on the basis of numberings.
1. Chinese Prime Minister => Telephone
2. Telephone attended by => Translator (English)
3. Translator (English) => French Prime Minister
4. French Prime Minister => Translator (English)
5. Translator (English) => Telephone in to Chinese Prime Minister
6. Telephone => Translator (English)
7. Translator (English) => Chinese Prime Minister
8. Chinese Prime Minister => Translator (English)
Ans. 3
(a)
Figure 2: Given 1st figure

Max. Amplitude (Am) is = 15 from figure.
Time Period (T) is = 3 sec
Frequency:
f = 1
T = 1
3 =0.333 Hz
Phase: Phase can be calculated by the following formula
Am sin(ωt+ ϕ ¿)¿
Here, ϕ is described as Phase
Therefore,
When ωt is = 0
Then
Am sin( ϕ ¿)=0 ¿
Hence,
ϕ =0 or phase will be zero.

(b)
Figure 3: Given 2nd figure
Max. Amplitude (Am) is = 4 from the figure.
Time Period (T) is = 6.5 sec
Frequency:
f = 1
T = 1
6.5 =0.153 Hz
Phase: Phase can be calculated by the following formula
Am sin(ωt+ϕ ¿)¿
Here, ϕ is described as Phase
Therefore,
When ωt is = 0
Then
Am sin( ϕ ¿)=0 ¿
Hence,
ϕ =0 Or phase will be zero.

Max. Amplitude (Am) is = 7.75 from the figure.
Time Period (T) is = 2.25 sec
Frequency:
f = 1
T = 1
2.25 =0.444 Hz
Phase: Phase can be calculated by following formula
Am sin(ωt+ ϕ ¿)¿
Here, ϕ is described as Phase
Therefore,
When ωt is = 0; Amplitude is maximum
Then
Am sin(ϕ ¿)=Am ¿
or
sin(ϕ ¿)=1¿

And,
ϕ =90o Or phase will be 90o.
Ans. 4
It is known that, for the equation
A sin (2 π ( f ) t )
Here,
A=Amplitude
f = frequency
Time period is given by 1
f
(a)
Given eq. is
3 sin(2 π (200)t )
Amplitude = 3
frequency= 200 Hz
Time period=
T = 1
f = 1
200 =0.005 sec
Phase = 0

Figure 4: Obtained graph for (a)
(b)
Given eq. is
14 sin (2 π (50) t+ 90)
Amplitude = 14
frequency= 50 Hz
Time period=
T = 1
f = 1
50 =0.02 sec
Phase = 90o

Figure 5: Obtained graph for (b)
(c)
Given eq. is
4s𝑖𝑛 (650𝜋𝑡 + 180)
Amplitude = 4
frequency=
f = 650
2 =325 Hz
Time period=
T = 1
f = 1
325 =0.003 sec
Phase = 180o

Figure 6: Obtained graph for (c)
(a)
Given eq. is
6s𝑖𝑛 (700𝜋𝑡 + 270)
Amplitude = 6
frequency=
f = 700
2 =350 Hz
Time period=
T = 1
f = 1
350 =0.0028 sec
Phase = 270o

Figure 7: Obtained graph for (d)
Ans. 5
Given is,
Free space loss is at 6 GHz
To a synchronous satellite from earth distance 35,863km
Path or space loss is given by (Semeghini 2018)
( loss ) dB=20 log10 ( 4 πR
λ )
Given is f =6 GHz and distance R = 35863
Therefore,
λ= c
8 = 3× 108
6× 109 = 1
20 m
Now, Loss will be
( loss ) dB=20 log10 ( 4 × 3.14 ×35863 ×20
1 )
( loss ) dB=20 log10 ( 8963569.6 )
( loss ) dB=20 ×6.952