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 )
It 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 with the 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 it ∈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 it ∈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 it ∈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