Electrical Engineering Homework: Signals and Systems

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Table of Contents
Q1................................................................................................................................................................2
Q2................................................................................................................................................................3
Q3................................................................................................................................................................4
Q4................................................................................................................................................................6
Q5................................................................................................................................................................8
Q6................................................................................................................................................................9
Q7..............................................................................................................................................................10
Q8..............................................................................................................................................................11
Q9..............................................................................................................................................................12
References.................................................................................................................................................13

Q1
The TCP/IP is used for transmitting the data packets between the destination and source nodes.
The Internet Protocol (IP) is used for transmitting the data packets from one node to another
using the addressing method called IP address. The IP address is responsible for the delivery of
the packets to the right person or machine (Chung, Lorenz and Somaya, 2019). The IP address
provides the uniqueness to each machine so that the other machine on the network can directly
interact with specific machine. Network Access Layer is bottom layer of the TCP/IP model
which provides access to the internet. This layer provides the physical connection of the device
with the network. Hardware devices present on this layer are responsible for converting the
signal into the format that can be understood by the computer. The mapping process takes place
over the network in which IP address converted into address which is understood by routers
called Router Address. Both methods are used for the transmission of the data packets between
the nodes.

Q2
The telephonic communication network is used for communicating between the Chinese PM and
French PM. The diagram is shown below which includes the layers included in the transmission
of the data between China and France. The translators are included which convert the data into
English.
Figure 1: Communication between China and France

Q3
a)
i) Maximum Amplitude can be calculated by the highest point of the graph which is 15
So Maximum Amplitude of the given wave = 15
ii) Frequency of the wave is given by the formula
F = (1/T) F shows the frequency of the wave and T shows time period
So, the F = (1/3) = 0.33 Hz
iii) Time Period for a given wave = 3 s
iv) formula for Phase angle is shown as X(t) = A*m*sin ( θ) = 0
 The Phase Angle of the given wave = 0∘
b)

i) Maximum Amplitude can be calculated by the highest point of the graph which is 4
So Maximum Amplitude of the given wave = 4
ii) Frequency of the wave is given by the formula
F = (1/T) F shows the frequency of the wave and T shows time period
So, the F = (1/6.5) = 0.1538 Hz
iii) Time Period for a given wave = 6.5 s
iv) formula for Phase angle is shown as X(t) = A*m*sin ( θ) = 0
 The Phase Angle of the given wave = 0∘
i) Maximum Amplitude can be calculated by the highest point of the graph which is 7.5
So Maximum Amplitude of the given wave = 7.5
ii) Frequency of the wave is given by the formula

F = (1/T) F shows the frequency of the wave and T shows time period
So, the F = (1/2.25) = 0.444 Hz
iii) Time Period for a given wave = 2.25 s
iv) formula for Phase angle is shown as X(t) = A*m*sin ( θ) = 0
 The Phase Angle of the give wave = 90∘

Q4
a. 3 s𝑖𝑛 (2 𝜋 (200) 𝑡)
Equation used for Y(t)= A sin (2πft+ φ)
Using this equation, we can find for a given equation:
Amplitude (A) = 3
Frequency (f) = 200 Hz
Time period (t) = 0.005 s
Phase angle (φ) = 0∘
Required Diagram:

b. 14 s𝑖𝑛 (2 𝜋 (50) 𝑡 + 90)
Equation used for Y(t)= A sin (2πft+ φ)
Using this equation, we can find for a given equation:
Amplitude (A) = 14
Frequency (f) = 50 Hz
Time period (t) = 0.02 s
Phase angle (φ) = 90∘

Required Diagram:
c. 4 s𝑖𝑛 (650 𝜋 𝑡 + 180)
Equation used for Y(t)= A sin (2πft+ φ)
Using this equation, we can find for a given equation:
Amplitude (A) = 4
Frequency (f) = 325 Hz
Time period (t) = 0.003 s
Phase angle (φ) = 180∘
Required Diagram:

d. 6 s𝑖𝑛 (700 𝜋 𝑡 + 270)
Equation used for Y(t)= A sin (2πft+ φ)
Using this equation, we can find for a given equation:
Amplitude (A) = 6
Frequency (f) = 350 Hz
Time period (t) = 0.002 s
Phase angle (φ) = 270∘
Required Diagram:

Q5
Shortest path can be selected using the attributes of the waves = 6 Hz = 6*109
Wavelength (W) = (3 * 108) / (6*109)
W = 0.495 m
LdB = 20 log (35.85 * 106) – 20 log (0.49) + 21.98
LdB = 99.68
44 dB and 48 dB are the values can be used for the antenna
The free space lost for a given wave can be shows as : LdB = 99.68 -(44+48)
 LdB = 7.68