MITS5003 Wireless Networks & Communication

MITS5003 Wireless Networks & Communication - 2019SS
Assignment No 1
We have two hosts, host A and host B.
A want to communicate with B; first it establishes a virtual circuit.
When virtual circuit is established transport layer and the ATM layer needs to agree on the
services used, the contact has three parts first traffic to be offered, secondly service agreed upon
and compliance requirements.
If both sides cannot agree on a contact the virtual circuit won’t be setup.
- Computer A establishes Quality of Service (QoS) parameter with the network server residing in
node of computer A, and setup on route like: - A→ X→ B.
- Node A then reserve a switch connection say A1.
-A sends A1 to X requesting for reserve of connection A2.
- X then sends A2 to B requesting for reserve of connection A3.

-B then find computer A send its A3 and acknowledges send back to finish the connection and
computer A gets back A1.
Final summary for the connection:-
A → A 1( A) → A 2( X)→ A 3 (B)→ B.
B→ A 3( B)→ A 2(X )→ A 1( A)→ A.
Reserved connection along the way allows QoS .
When the communication is done, connection torn down, virtual circuits put back for reuse.
ATM -A dressing.
Permanent virtual circuit (PVC ) A → X → B
Switched virtual circuit ( SVC) A → X → B∨ A → Y → X → B .
Routes and switches on the actual path.
solution
a) 8 sin ( 2 π ( 500 ) t )
¿ is the form of a modulating signal of the form

M ( t )= Am sin ( 2 π ( f m ) t+∅ ) [1]
Where
Am =Amplitude of Modulating Signal
f m=Frequency of modulating signal
∅ = phase of modulating signal
Comparing withthe equation modulation
Amplitude(A ¿¿ m)=8 ¿
Frequency( f m )=500 Hz
Time period ( T )= 1
f m
= 1
500 Hz =0.002 s
Phase ( ∅ ) =0o
b) 3 sin ( 600 t+ 45 )
Rewriting¿ the form of mod ulating signal
¿ Am sin (2 π ( f m ) t +∅ )
600 t
2 π =95.493 t
600 t=2 π (95.493 t)
Substituting new value of 600 t into the equation
3 sin ( 2 π ( 95.493 t)+45 )
Amplitude(A ¿¿ m)=3 ¿
Frequency(f m )=95.493 Hz
Time period ( T )= 1
f m
= 1
95.493 Hz =0.0105 sec
Phase ( ∅ ) =45o
c) 6 sin ( 400 t +135 )
Rewriting¿ the form ofa modulating signal
¿ Am sin (2 π ( f m ) t +∅ )

400 t
2 π =63.662t
400 t=2 π (63.662 t)
Substituting new value of 400 t into theequation
6 sin ( 2 π (63.662tt )+135 )
Amplitude(A ¿¿ m)=6 ¿
Frequency(f m )=63.662 Hz
Time period ( T )= 1
f m
= 1
63.662 Hz =0.0157 sec
Phase ( ∅ )=135o
d) 2 sin ( 1000t +180 )
Rewriting¿ the form ofa modulating signal
¿ Am sin (2 π ( f m ) t +∅ )
1000t
2 π =159.1549 t
1000 t=2 π (159.1549 t)
Substituting new value of 1000 t into the equation
2 sin ( 2 π (159.1549 t)+180 )
Amplitude(A ¿¿ m)=2 ¿
Frequency(f m )=159.1549 Hz
Time period ( T )= 1
f m
= 1
159.1549 Hz =0.0063 sec
Phase ( ∅ )=180o

i. Maximum amplitude( A¿ ¿ m)=15 ¿
ii. Time period(T )=3 sec
iii. Frequency(f ¿¿ m)= 1
Time period (T ) = 1
3 sec =0.3333 Hz ¿
iv. Phase ∅
¿ the modulation equation m ( t ) =Am sin ( 2 π ( f m ) t+∅ )
2 π ( f m ) t=w (t)=0∧m ( t ) =0
sin ( ∅ ) =0o
Phase ∅ =00

i. Maximum amplitude( A¿ ¿ m)=4 ¿
ii. Time period(T )=6.5 sec
iii. Frequency(f ¿¿ m)= 1
Time period (T ) = 1
6.5 sec =0.1538 Hz ¿
iv. Phase ∅
¿ the modulation equation m ( t ) =Am sin ( 2 π ( f m ) t+∅ )
2 π ( f m ) t=w (t)=0∧m ( t ) =0
sin ( ∅ ) =0o
Phase ∅ =00

i. Maximum amplitude( A¿ ¿ m)=7.5 ¿
ii. Time period(T )=2.25 sec
iii. Frequency(f ¿¿ m)= 1
Time period (T ) = 1
2.25 sec =0.4444 Hz ¿
iv. Phase ∅
¿ the modulation equation m ( t ) =Am sin ( 2 π ( f m ) t+∅ )
2 π ( f m ) t=w (t)=0∧m ( t ) =Am
Am sin ( ∅ ) = Am
sin ( ∅ )=1o
∅ =Sin−¿ (1 )=90o ¿
Phase ∅ =900

Solution
i. Fundamental Frequency
Calculating different Frequncies= wt
2 π
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 is thelowest frequency f 0=50 Hz
ii. Spectrum
spectrumof a sin t= e jωt −e− jωt
2
Taking ℱ Transform( FT )
→ 2 π ( δ ( w−wo ) −δ ( w+wo ) )
2 j → πj ¿
The Overal spectrum of the sine signal

iii. Bandwidth BW =2∗f Max
f Max =f 3=300 Hz
BW =(2∗300)=600 Hz
iv. Channel capacity (CC )
CC =BW log2 (1+ S
N )
( S
N )
db
=250 dB
But
( S
N )db
=10 log10 ( S
N )
∴ 250 dB=10 log10 ( S
N )
Evalu ating for ( S
N )
1025= S
N
( S
N )=1025
substituting the calculated value into the original equation
CC =600∗log2 ( 1+1025 ) =49.829 kHz

Nyquist bit rate formula is used to define the theoretical maximum bitrate for a particular
noiseless channel. It is given by the equation Bitrate=2∗Bandwidth∗log2 ( L) where the channel
bandwidth is the bandwidth, the number of signals levels used to represent the data is denoted by
letter L and Bitrate is the bit rate in bits per second.
The bandwidth as seen from the Nyquist bit rate formula is fixed quantity that cannot be changed
and therefore the data rate will be directly proportional to the number of signal levels.
Disadvantage of this approach
The system reliability can be reduced by increasing the signal levels.
Differences between circuit switching and packet switching virtual circuit
Circuit switching Packet switching virtual circuit
1 It is connection oriented It is connectionless
2 Originally designed for voice
communication
Originally designed for data communication
3 It is inflexible since a single path is
established for transmission
It is flexible since individual packet can
travel through different routes to the final
destination.
4 It is implemented at a physical layer It is implemented on network layer [2]
Advantages of Packet switching over Circuit Switching
1. The cost of communication is lower.
2. It can perform data rate conversion.

3. Priorities can be used and therefore higher priority packet can be transmitted first.
4. Different pairs of nodes can be used for communication during the same time. [3]
Advantages of Circuit Switching over packet switching
1. The data communication process experiences fewer delays.
2. Dedicated and secured circuits are established between pairs of communicating nodes. [3]
solution
LOS comunication equatio n
d= √ 2 h1 R+ √ 2h2 R Equation 1
Where
d=distance apart=40 km
h1∧h2 are antennas(Transmitter∧Receiver )heights
R=Radius of the Earth=6400 km
Taking that h1 ¿ be three×h2
∴ h1=3 h2 Equation2
Substituting Equation2 into equation1
d= √ 2∗3 h2∗R+ √ 2 h2 R
d= √2 h2 R ( 1+ √3 )
D ivi di ng both sides by ( 1+ √ 3 ) ∧Squaring

( d
( 1+ √3 ) )2
=2 h2 R
Solving for h2
h2 = 1
2 R∗
( d
( 1+ √3 ) )2
= 1
2∗6400∗
( d
( 1+ √3 ) )2
=0.01675 km
h1=3 h2= ( 3∗0.01675 km ) =0.05025 km
Converting h1∧h2 into meters by multiplying with 1000
h1=50.25m
h2 =16.75m
Reference
[1] R. S. S. B.L. Theraja, Textbook of Electrical Technology Volume IV - Electronic Devices
and Circuits, S Chand & Co Ltd, 2006.
[2] T. Differences, "Difference Between Circuit Switching and Packet Switching," 22 August
2016. [Online]. Available: https://techdifferences.com/difference-between-circuit-switching-
and-packet-switching.html.
[3] T. KEARY, "Circuit Switching vs Packet Switching," 19 March 2019. [Online]. Available:
https://www.comparitech.com/net-admin/circuit-switching-vs-packet-switching/.