Fundamentals of Telecommunications: Homework Set 1

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Table of Contents
Q1 Answer..................................................................................................................................................2
Q2 Answer..................................................................................................................................................3
Q3 Answer..................................................................................................................................................4
Q4 Answer..................................................................................................................................................5
Q5 Answer..................................................................................................................................................8
Q6 Answer..................................................................................................................................................9
Q7 Answer................................................................................................................................................10
Q8 Answer................................................................................................................................................11
Q9 Answer................................................................................................................................................12
References................................................................................................................................................13

Q1 Answer
The first layer of the TCP/IP model is called the network access layer. This layer consists of the
data link layer and physical address layer. This layer is responsible for determining the details of
the data. In this layer, the medium through which the data is transferred is determined. In this
layer, the transfer to data takes place. It is the lowest layer in the TCP/IP model.
The second layer of the TCP/IP model is called the internet layer or the IP layer. This layer
comes between the transport and network access layer. The data is packed in packets. These
packets are called IP datagrams. These packets contain the address of destination and the source
which helps the data to move across the networks. The protocols that are included in the Internet
layer are ICMP, IGMP, RARP, IP, and ARP. This layer is used by all the layers in the TCP/IP
model.

Q2 Answer
1. The French prime minister talks to his/her translator using the French language.
2. The Chinese prime minister talks to his/her translator using the Chinese language.
3. The French translator converts the French message to the English language.
4. Similarly, the Chinese translator converts the Chinese message to the English language.
5. Both translators talk to each other via telephone.
6. The conversation is then translated to the French language for the French prime minister
and similarly to the Chinese language for the Chinese prime minister.
7. This is the whole process through which two prime ministers of two different countries
can converse directly via telephone.
8. For reference, the figure is attached above.

Q3 Answer
First Graph
The amplitude of the wave is 15
The time period of the wave is 3 sec
The frequency of the wave is 1
3 =0.33 Hz
The phase angle of the wave is 0˚
Second Graph
The amplitude of the wave is 4
The time period of the wave is 6.5 sec
The frequency of the wave is 1
6.5 =0.154 Hz
The phase angle of the wave is 0˚
Third Graph
The amplitude of the wave is 7.5
The time period of the wave is 2.3 sec
The frequency of the wave is 1
2.3 =0.435 Hz
The phase angle of the wave is 90˚

Q4 Answer
a) 3 𝑆𝑖𝑛 (2𝜋 (200) 𝑡)
the amplitude of the wave is 3,
the frequency of the wave is 200 Hz,
the time period of the wave is 1
200 =0.005 sec
the phase angle of the wave is 0˚
b) 14 𝑆𝑖𝑛 (2𝜋 (50) 𝑡 + 90)
the amplitude of the wave is 14,
the frequency of the wave is 50 Hz,
the time period of the wave is 1
50 =0.02 sec

the phase angle of the wave is 90˚
c) 4 𝑆𝑖𝑛 (650𝜋𝑡 + 180)
the amplitude of the wave is 4,
the frequency of the wave is 325 Hz,
the time period of the wave is 1
325 =0.003 sec
the phase angle of the wave is 180˚

d) 6 𝑆𝑖𝑛 (700𝜋𝑡 + 270)
the amplitude of the wave is 6,
the frequency of the wave is 350 Hz,
the time period of the wave is 1
350 =0.002 sec
the phase angle of the wave is 270˚

Q5 Answer
Distance from the earth= 35863 Km
Frequency= 6GHz
Isotropic free loss= 20 log 10(6 x 109)+20 log10(35863 x 103) – 147.56 dB
Isotropic free loss= 199.09dB

Q6 Answer
Given equation
𝑠 (𝑡) = 5 sin (100𝜋𝑡) + sin (300𝜋𝑡) + sin (600𝜋𝑡)
There are 3 frequencies in this equation,
First frequency, f1= 50 Hz
Second frequency, f2= 150 Hz and,
Third frequency, f3= 300 Hz.
The fundamental frequency is calculated by determining the GCD of f1, f2, and f3.
The GCD of f1, f2, and f3 is 50. Therefore, the fundamental frequency is 50 Hz.
The spectrum equation is written below,
s(t )= i
2 {(d (f −150)−d ( f +150))+ d(f −300)−d ( f +300 ) +5 ( d ( f −50 ) −d ( f +50 ) ) ¿ }
The bandwidth is calculated by adding all the frequencies.
Therefore, Bandwidth (H) = 50 + 150 + 300 = 500 Hz
The formula to find the channel capacity is,
Channel Capacity = 2*(Bandwidth)*(log2 (Signal level))
For Signal level = 2
Channel Capacity = 2*(500)* (log2 (2)) = 1000
For Signal level = 4
Channel Capacity = 2*(500)* (log2 (4)) = 2000
For Signal level = 8
Channel Capacity = 2*(500)* (log2 (8)) = 3000

Q7 Answer
There are various reasons which are responsible for the variation of the speed of a channel. The
channel is affected by bandwidth allocation. The channel is affected by the number of signal
levels and it is also affected by the quality of the channel.
To increase the data rate, the number of signal levels should be increased while making
bandwidth a fixed and constant quantity.
To achieve a high data rate, Nyquist formulas are used.
1. Noiseless Channel
The bandwidth is fixed and the number of signal levels is varied to increase or decrease
the data rate of the channel.
Channel capacity =2( Bandwidth)( log2 Signal Level)
2. Noisy Channel
The bandwidth is fixed and the ratio of signal/noise is increased to increase or decrease
the date rate of the channel.
Channel capacity =(Bandwidth)¿

Q8 Answer
The difference between circuit switching and packet switching are mentioned below.
Circuit Switching Packet Switching
In this type of switching, the information
about the path is not provided to the packet
and hence first the route has to be established
for the transmission of the data.
In this type of switching, the packet is
provided with the route information. The
dynamic route is followed by the data packet.
Ideal for communication via voice. Ideal for data transmission
Uniform delay No uniform delay
It is processed at the physical layer It is processed at the network layer
High consumption of resources Low consumption of resources
If at any part the connection is lost, the whole
connection goes down
If at any place the connection is lost, only the
packets routing through that part is restarted
and not the whole connection
This type of switching is preferable for big
amount of data
This type of switching is not preferable for
big amount of data
It is not connectionless It is connectionless
It has a physical path It does not have a physical path
The data packets travel using the established
physical path
The data packets travel independently.
It reserves the bandwidth It does not reserve the bandwidth
The bandwidth is wastage There is no wastage of bandwidth

Q9 Answer
According to the question,
The distance (Z) is given, Z = 60 Km
The first antenna’s height, A = 4p
The second antenna’s height, B = p
Applying the line of sight formula,
Z=(3.57) √ KH
Squaring both sides
Z2 =(3.57)(K∗H )
602=(3.57)( 4 ( 4 p+ p )
3 )
3600=(12.745)( 20 p
3 )
3600∗3
12.745∗20 = p
p=42.37
A = 169.48
B = 42.37