Computer Networks Fundamentals: Homework Assignment

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Table of Contents
Q 1..............................................................................................................................................2
Q 2..............................................................................................................................................3
Q 3..............................................................................................................................................4
Q 4..............................................................................................................................................6
Q 5..............................................................................................................................................9
Q 6............................................................................................................................................10
Q 7............................................................................................................................................11
Q 8............................................................................................................................................12
Q 9............................................................................................................................................13
References:...............................................................................................................................14

Q 1
IP is stand for Internet protocol and it is communication protocol which used in same
computer network and internet . It is mainly known as TCP/IP . Foundational protocol are the
TCP that’s why its called commonly TCP/IP. Internet protocol is used in end to end
communication of data . And its specifies how data would be
addressed ,packetized ,transmitted , received and routed . This model is known as DOD
model mean Department of defence .In TCP/IP there are four layer and First Layer is
Network Layer .In network Layer It transmit data segment in the form of packets between
networks . It manage the hardware protocol and addressing available in layers .
Network layer assign data segment from source to destination Internet protocol address or IP
address .Network layer is determine the best way to data delivery form source to destination
for example your friend’s IP address is destination and your IP is source .

Q 2
PM of china No conversation No conversation PM of frame
Translator conversation is done Conversation Traslator
PM of china No conversation telephone No conversation PM of france
Translator conversation is done Conversation Translator
Translator between Japanese and German
Proper
Information
Proper
Information

Q 3
1.
a). As shown in diagram amplitude is 15 because its highest Y axis value is 15
b). For calculation of Frequency of given diagram ,it can be calculated using 1 / T .here T is
Time period .
so time period T= 3
so Frequency F = 1/T = 0.33Hz
c). Phase angle can be calculated using this formula
Z(t) = Am sin(x) = 0
So finally face angle is Zero
2.
a). As shown in diagram amplitude is 4 because its highest Y axis value is 4 .
b). For calculation of Frequency of given diagram ,it can be calculated using 1 / T .here T is
Time period .

so time period T= 6.5 seconds
so Frequency F = 1/T = 1/6.5 = 0.1535 Hz
c). Phase angle can be calculated using this formula
Z(t) = Am sin(x) = 0
So finally face angle is Zero
3.
a). As shown in diagram amplitude is 7.5 because its highest Y axis value is 7.5 .
b). For calculation of Frequency of given diagram ,it can be calculated using 1 / T .here T is
Time period .
so time period T= 2.25
so Frequency F = 1/T =1/2.25 = 0.444Hz
c). Phase angle can be calculated using this formula
Z(t) = Am sin(x) = Am
So face angle has value is 900.

Q 4
a). 3sin(2 π(200)t)
This is wave equation and it contain 4 data which defined below
Amplitude of given wave equation is = 3
Frequency of given wave equation is = 200Hz
Time period of given wave equation is = 0.005sec
Phase angle of given wave equations is = 0
Wave diagram can be show like below - -
b) 14sin(2π(50)t +90)
This is wave equation and it contain 4 data which defined below
Amplitude of given wave equation is = 14
Frequency of given wave equation is = 50Hz
Time period of given wave equation is = 0.02sec
Phase angle of given wave equations is = 90
Wave diagram can be show like below - -

c). 4sin (650πt + 180)
This is wave equation and it contain 4 data which defined below
Amplitude of given wave equation is = 4
Frequency of given wave equation is = 325Hz
Time period of given wave equation is = 0.003 sec
Phase angle of given wave equations is = 180
Wave diagram can be show like below - -

d). 6sin (700πt + 270)
This is wave equation and it contain 4 data which defined below
Amplitude of given wave equation is = 6
Frequency of given wave equation is = 350Hz
Time period of given wave equation is = 0.002 sec
Phase angle of given wave equations is = 270
Wave diagram can be show like below - -
.

Q 5
In this problem shortest path can be given like 6Hz= 6*109
So wavelength can be given like W= ( 3*108 ) / ( 6*109 ) = 0.495 m
So LbB = 20 log (35.85 * 106) – 20 log (0.495) + 21.98
LbB = 99.68452
Now for calculation of antennas,
44dB and 48dB.these are two values of antennas .
So free space lost is --
LdB = 99.68452-44-48
LdB= 7.68452

Q 6
Now question is defined S(t)
S(t) = 5sin(100πt) + sin(300πt) + sin(600πt)
Some frequency 50 ,150 and 300 has been calculated as fundamental frequencies .
So we have to applies a Fourier transform on given equations so we get
S(F)= (1/2i) {(d(F-150)-d(F+150))+5(d(F-50)-d(F+50))+d(F-300)-d(F+300))}
Now, f max can be calculated using addition of f min into bandwidth like
f max = f min + bandwidth
bandwidth can be calculated using addition of all frequency like
500MHz = 300Hz + 150Hz + 50Hz
Using Formula of Nyquist Bitrate
twice of Bandwidth is equal to Bitrate
Calculation for M=2
So Channel Capacity of given wave is 2*500 log2 2 = 1000 bits/second
Same calculation For M=4
So Channel Capacity of given wave is 4*500 log2 2= 2000 bits/second
Same calculation for M=8
So Channel Capacity of given wave is 8*500 log2 2= 4000 bits/second

Q 7
Data rate is consumed speed of transmission of data . Over a channel how fast data we can
send in network in bits per second this is consideration for data communication .Its mainly
depends on 3 factors First is level of digital signal ,second is quality of channel and third is
bandwidth .For calculation of data rate on is given by Nyquist for Noiseless channel and
second is Shannon for channel of noisy .
Formula is given like
Bit rate = 2 * bandwidth * log2 (L)
Here L is number of signal level for representation of data .
Disadvantage :
1. reduction of system’s reliability
2. Signal level is increased .

Q 8
Both technique ,circuit switching and packet switching is used to connect multiple
communicational device with another one .
Circuit switching is connectional oriented and has purpose to initially design for
communication of voice .Flexibility of circuit switching is inflexible . Order of message
received like sender from the source .there are two technology used in circuit switching first
is space division switching and second is time division switching .this layer is implemented
in physical layer .There is a physical path here which connect source to destination nut this
functionality of physical path is not available in packet switching It has advance version
which reverse the bandwidth but not same for packet switching. In this switching there is a
wastage of bandwidth, has occurred .
Packet switching is connectionless orientation . packet switching has main purpose to
designed data transmission initially . Flexibility of This switching is flexible because router is
available and created for packet to travel to destination .It assembled to the destination so it
received from out of the order .packet switching is assigned at network layer .Physical path
is not available for source to destination .It has no advance feature like reverse of bandwidth
but this feature is available in circuit switching. There is no wastage of bandwidth .
Advantages :
Packet switching is more use full as compare to circuit switching .It has advanced way for
data communication over circuit switching . Packet switching has been converting in
different packets .It can solved the problem of on-demand of resource allocation . In packet
switching network is used efficient manner and easily get packets or bits .In this switching it
high data transmission process is done very easily . Main advantage of circuit switching is
there is a dedicated path allocated between sender and receiver transmission mode . Once
circuit has established ,no delay is occurred .