Electrical Engineering: Communication and Signal Analysis Homework

This document presents a comprehensive solution to a communication and signal analysis homework assignment. The solution addresses various aspects of signal analysis, including calculating frequency, amplitude, and phase from given graphs and equations. It also explores the properties of sinusoidal waveforms, the Nyquist bit rate, and the concepts of packet switching and circuit switching in network communication. Furthermore, the document includes an analysis of a given signal, determining its fundamental frequency and bandwidth. The assignment also involves a practical application of a formula related to antenna propagation. The document is supported by multiple references to provide a complete and reliable solution.

Question 1:
The communication can be described as:
1. Guest ask for the menu from the host, and the host give him the menu to select his/her pizza
and place order.
2. The host look for the telephone number of the pizza store, dials the number. The telephone
lines connects at the pizza delivery center and host places the order to the order clerk.
3. Host acknowledge the order of pizza to the guest.
4. The clerk confirms payment details and the order as well, approves the order and sent it to the
pizza cook.
Question 2:
Hypothetical telephone conversation between French and Chinese PMs.
Both the PM talks as if they are talking to the one another. When the Chinese PM talks, he
conveys directly to the French PM his remarks, however the communication is passed through
the translators via the telephone. The translator at the Chinese PM’s end translate the message
into Chinese whereas the one at the French PM’s end translate into English.
Question 3:
For graph 1
Time period t= 5.2 - 2.2= 3 sec
Frequency f = 1/t = 1/3 = 0.33 hz
Maximum Amplitude A= 15
Phase P= phase is zero
For graph 2:

Time period t= 8.2 – 1.8 = 6.4 sec
Frequency f = 1/t = 1/6.4 = 0.156 hz
Maximum Amplitude A= 4
Phase P= phase is zero
For graph 3:
Time period t= 5.8 – 3.4 = 2.4 sec
Frequency f = 1/t = 1/2.4 = 0.416 hz
Maximum Amplitude A= 7.8
Phase P= phase is zero

Question 4:
a. 10 Sin (2π (100) t)
Comparing the equation with
We can see that Amplitude A= 10
Frequency ω = 2π 100 = 628 hertz
Time period T= 2π/ω = 0.01 sec
Frequency f= 1/T= 100
Phase = 0
b. 20 Sin (2π (30) t + 90)
Comparing the equation with
We can see that Amplitude A= 20
Frequency w= 2π (30) = 188.4

Time period T = 2π/ω = 0.033
Phase = 90
c. 5 Sin (500πt + 180)
Comparing the equation with
We can see that Amplitude A= 5
Frequency w= 500π = 1570 hz
Time period T = 2π/ω = 0.004 sec
Phase = π

d. 8 Sin (400πt + 270)
Comparing the equation with
We can see that Amplitude A= 8
Frequency w= 400π= 1256
Time period T = 2π/ω = 0.005
Phase = 3 π/ 2

Question 5:
Solution: b)
Question 6:
Solution: At is frequency (4 * 109) hz
And distance (35,863 * 106) m

Question 7:
S (t) = 5 sin (200 π t) + sin (600 π t)
Solution: We know for the period T, frequency f = 1/T, and the angular Frequency
For a given sinusoidal signal, is expressed in the following forms:
The fundamental frequency is expressed as the GCD of all frequency components that are in the
signal, similarly, the fundamental period T is the LCM Least Common Multiple of all period
components.
Noω for the given equation:
S (t) = 5 sin (200πt) + sin (600πt)
ω 1= 200π, f1=100, T1= 1/100, ω 2=600π, f2=300, T 2= 1/300
Fundamental frequency= f0= GCD {100, 300} = 100
Also we can give the bandwidth of the signal as 300 – 100= 200
Question 8:
The maximum bit for noiseless channels as per the can be given as Nyquist bit rate formula

Since bandwidth of a signal cannot be changed being a fixed quantity, it implies that the data r of
levels in signal is directly proportional to each other. That is to increase the data rate the number
of signal levels should be increased.
The disadvantage of this approach is that increase in the levels of signal makes the system less
reliable.
Question 9:
Packet Switching and Circuit Switching are the main models that are used for facilitating
connections within enterprise networks
Circuit switching: When a dedicated circuit or channel is required to be establish between users
speaking with one another on call circuit switching is used. Channel used for communication
between the two users is kept reserve in circuit switching and is used once for communication.
Packet switching: does not require to establish channels for communication. A channel is
always available throughout the network for the users. Messages are broken and send
individually to the networks in the form of packet. These packets are handled by virtual circuit or
datagram.
Packet switching is preferred than circuit switching because of the efficiency of communication.
Without the need of dedicated channel the packet can travel their own paths to reach their
destination. Whereas in the circuit switching other devices on the network are unable to use the
channel until the communication between users is terminated.
Question 10:
According to the condition given in question, h1= 2h2,
Putting value of d, K and h1 in formula
40= 3.57 ¿(8/3) h2 + √4/3 h2)
11.428= √h2 (1.632+ 1.154)
h2 = (4. 101)2
h2= 16.81 m
h1= 33.63 m

References:
Keary, T 2019, 'Circuit Switching vs Packet Switching', https://www.comparitech.com/net-
admin/circuit-switching-vs-packet-switching/ (viwewed on 24 April 2019)
Marshall, D. 2001, 'Nyquist's Sampling Theorem',
https://users.cs.cf.ac.uk/Dave.Marshall/Multimedia/node149.html (viewed on 24 April 2019)
Rajaraman, R 2010, 'Antennas & Propagation'.
http://www.ccs.neu.edu/home/rraj/Courses/6710/S10/Lectures/AntennasPropagation.pdf (viewed
on 24 April 2019)
Storr, W n.d, 'Electronics Tutorial,Sinusoidal Waveforms',
https://www.electronics-tutorials.ws/accircuits/sinusoidal-waveform.html (viewed on 24 April
2019)
White, BE n.d, 'Layered Communications Architecture for the Global Grid'.
https://www.mitre.org/sites/default/files/pdf/white_layered.pdf (viewed on 24 April 2019)