Electrical Engineering Assignment: Information Signal Analysis

This document presents a comprehensive solution to an electrical engineering assignment focusing on information signals. The assignment addresses several key areas, including modeling the pizza ordering and delivery process using layer models, and analyzing communication scenarios between the French and Chinese prime ministers using English translators. The solution then delves into signal analysis, computing amplitude, frequency, time period, and phase for various waveforms, both graphically and mathematically. It covers calculations for different sinusoidal equations, and the transmission of a digitized TV picture, including pixel rate and bit rate calculations. The solution further explores channel capacity, bandwidth, and the impact of signal levels on data rates. Switching techniques, including circuit switching and packet switching are compared and contrasted. Finally, the document applies the concepts to calculate the antenna height for line-of-sight communication, demonstrating a practical application of the principles discussed. References are provided at the end.

INFORMATION SIGNAL
[Author Name(s), First M. Last, Omit Titles and Degrees]
[Institutional Affiliation(s)]

Question-1
Placing an order for pizza
The diagram below shows how order for pizza is placed
Brief explanation
 Customer proposes the menu for pizza, customer receives the menu from the owner, a
choice is then made by the customer and puts pizza order.
 Customer gets the phone-number, collects his or phone and makes a call. A phone
connection to telephone-line is underway and rings after dialing
 The owner converses with pizza sales-clerk, presents the customer with-order-of the
pizza, customer acknowledges the placed-order and gets on hold.
 The pizza sales-clerk receives placed-order for pizza, request for further details
concerning bill settlement, and obtains the pizza-bill settlement details and presents
placed-order pizza-maker.
Bringing of ordered-pizza
Diagram shown below is an illustration of how pizza is brought using a suitable delivery means

This is necessary procedure that should be followed to get delivery:
 Pizza-Maker makes pizza as per the customer’s placed-order and assign it to the pizza
sales-clerk handling the placed-order (Nathan, et al., 2014).
 Pizza sales-clerk packs it, organizes for release-receipt and loads it to delivery-van
 The van transports total pizzas that were ordered from sales-clerk and drops it to the
owner. From this point main-road is used as a transmitter-line between placing-order and
delivery of made pizzas.
 The owner endorses bill-details and gets back to sales-clerk and accept pizza-delivery.
 In the end, the owner presents ordered-pizza to customers.
Question-2
The Chinese-prime minister and French-prime minister should have an English-translator.

The following system of numbering best describe the processes of the interaction at every level
Question-3
Consequently, the highest amplitude = 15

Periodic time (T) = 3-seconds
Therefore, frequency is given by 1
T = 1
3 =0.3333 Hz
Sin (wt+ϕ) = 0
Phase-angle = 00
Angle is 00
Consequently, Phase-angle = 00
Angle = 00
Amplitude = 4
Time = 6.5-seconds
Therefore, frequency 1
T = 1
6.5 =0.1538 Hz
Sin (wt+ϕ) = 0
Phase-angle = 00

Angle = 00
Amplitude = 7.5
Periodic time (t) = 2.25
Therefore, frequency = 1
T = 1
2.25 =0.444 Hz
0
Sin (wt+ϕ) = Amplitude
Phase-angle = 00
Sin ϕ = 1
Consequently, Phase-angle = 900
Question-4
a).
From, A sin (2 ( f ) t ;

Where, A = amplitude, f = frequency, t = Periodic time = 1
f
Hence, 10 sin (2 π (100 ) t
Consequently,
Amplitude = 10
Frequency = 100Hz
Periodic time = 1
100 =0.01 seconds
Phase-angle = 00
This is illustrated by the wave-form in the figure below
b).
20 sin (2 π ( 30 ) t
Amplitude = 20
Frequency = 30Hz

Time-period = 1/30 = 0.33 seconds
Phase angle = 0
This wave-form can be shown as indicated by the figure below;
c).
5 sin (500 π (180 ) t
Amplitude = 5
Frequency = 500
2 = 250 Hz
Time period = 1
250= 0.004 seconds
Phase-angle is 1800
Shift-plot to left by 1800 phase-shift
This wave-form can be shown as indicated by the figure below;

d).
8 sin (400 π ( 270 ) t
Amplitude = 8
Frequency = 400
2 = 200 Hz
Time period = 1
200= 0.005 seconds
The phase-angle = 2700
This wave-form can be shown as indicated by the figure below;
Question-5
a).

Pixel/ second = (480×500) × 30
Pixel/ second= 7,200,000
Each-pixel can take like 32 values
Therefore, Log2 (32) = 5
So, 5 bits / pixel.
Source rate R = (7,200,000×5)
Source rate R = 36,000,000 = 36×106 Mbps
Question-5
b).
B = 4.5×106
SNRdbB = 35= 10log10 (SNR)
=31,632
C = 4.5×106 (1+31632) log 2
C = 4.5×106 (log2(31632)
C= 4.5×106×11.6272
= 52.32×106 bps
Question-6

4 GHz = 4 x 109 Hz
35,863 Km = 35.863 x 106 m
=> (20log10 (4x10^9)) + 20log10 (35.863x10^6) - 147.56
= 195.574 dB
Question-7
(𝑡) = 5 sin(200𝜋𝑡) + sin(600𝜋𝑡)
Consider three frequencies expressed as signal f1, f2, f3
Where the signal is periodic, every-frequencies is a multiple of f0.
f1= n1f0
f2= n2f0
f3 = n3f0
f0 = gcd (f1, f2, f3)
Given signal as f0 = Gcd (100,300)
Fundamental frequency =100
(𝑡) = 5 sin(200𝜋𝑡) + sin(600𝜋𝑡)
Fourier transform of sin (22 πt) is 1
2i [ δ ( f − A )−δ ( f + A )
Therefore, S(f) = 5. 1
2i ( δ ( f −100 )−δ ( f +100 ) + 1
2i (δ ( f −300 )−δ ( f +300 )

Spectrum of S(t) is
S(f) =. 1
2i ( δ ( f −100 )−δ ( f +100 ) + δ ( f +300 )−δ ( f +300 )
Bandwidth = f max – f min
= (f+300) - (f-100)
= f+300-f+100
Bandwidth = 400Hz
Channel-capacity of a Noiseless-signal is the maximum-data-rate
For a noiseless-channel, the Nyguisist-bit-rate defines the theoretical-maximum bit rate.
Bit Rate (capacity)= 2× bandwidth × log2L
Therefore, Channel-Capacity = 2× Bandwidth × Log2 L
L = Number of Levels
Bandwidth = 400 Hz (calculated above)
For M=2,
Then substituting, Channel-capacity = 2×400× Log2 2
Channel-capacity = 800 bits per seconds.
For M=4