Communication Systems Assignment: Signal Processing and Network Design

This document presents the solutions to a communication systems assignment, covering a range of topics from signal analysis to network design. The first question illustrates a communication scenario using a pizza order, followed by an analogy between the communication of two prime ministers using translators and a common medium. Questions 3 and 4 involve analyzing signal characteristics like amplitude, frequency, and phase, and calculating equations for waveforms. Question 5 delves into pixel counts, intensity levels, and data rates in image transmission, including calculations for channel capacity and SNR. Question 6 calculates signal loss due to frequency and distance. Question 7 analyzes signal frequencies, fundamental frequencies, and bandwidth. Question 8 discusses methods to increase channel data rates. Question 9 explains the differences between circuit and packet switching, including virtual circuits. Finally, Question 10 involves calculations for antenna heights and the distance between them based on the Earth's radius, demonstrating practical applications of communication systems principles. All solutions are designed to help students understand and solve the assignment.

Q1.
Guest Cook Cook
Guest asks Host for a Pizza Clerk asks Cook to bake Pizza Cook informs clerk when
Pizza done
Host Pizza Clerk Pizza Clerk
Host places call using
Telephone
Pizza Clerk receives signal
from Telephone
Clerk sends Delivery van
to Host Address
Telephone Telephone Delivery Van
Telephone transmits signal on
medium
Telephone receives signal from
medium
Medium
Q2.
French Prime Minister Chinese Prime Minister
Sends / Receives message in French Sends / Receives message in Chinese
Translator (English ↔French) Translator ( English ↔Chinese)
Sends / Receives message in English Sends / Receives message in English
Telephone Telephone
Sends / Receives message using medium
characteristics
Sends / Receives message using medium
characteristics
Medium
The process of conversion of messages from one language to another is shown in the table. This
translation allows Two Prime Ministers to communicate with each other using a common
language of translators and of underlying medium

Q3.
Amplitude Frequency Time Period Phase
A 15 0.33 Hz 3 0
B 4 0.155Hz 6.5 0
C 7.5 0.40 Hz 2.5 90
Q4. Equation for waveform A Sin (2ℿ (f)t + θ)
10 Sin(2 ℿ(100) t)
Amplitude = 10. ; Frequency = 100 Hz ; Time Period = 1/100 =10msec Phase = 0◦

20 Sin(2 ℿ(30) t + 90)
Amplitude = 20. ; Frequency = 30 Hz ; Time Period = 1/30 = 33.33msec; Phase = 90◦

5 Sin(500 ℿ t + 180)
Amplitude = 5; Frequency = 250 Hz ; Time Period = 1/250 = 4msec; Phase = 180◦

8 Sin(400 ℿ t + 270)
Amplitude = 8; Frequency = 200 Hz; Time Period = 1/200 = 5msec ; Phase = 270◦
Q5.
Pixel Count 480 * 500 240,000
Intensity level Bits Log232 5
Frame Per Second 30 30
Total Bits Transferred / Source Data Rate 240000 * 5 * 30
36,000,000

Channel Bandwidth (B) 4.5 MHz 4.5 * 106
SN Ratio 35dB 10log1035 = log2(1+103.5)
Capacity B * SNR 4.5*106 * log2(1+3162)
4.5*106 *11.6272
52.33 *106 bps
Q6.
Loss Due to Frequency 20Log10 (4 x 109) 192.04 dB
Loss Due to Distance 20log10 (35.863x106) 151.092 dB
Total Loss 20Log10 (4 x 109) + 20log10
(35.863x106) – 147.56dB
195.57dB
Q7. Frequency in signals = 100 Hz, 300Hz
Therefore, Fundamental Frequency = 100Hz.
Spectrum = ∆ ( f ±1 00 ) + ∆ ( f ±300 )
Bandwidth = (f + 300) – (f -300) = 600Hz.
M= 2 Channel Capacity 2 * 600 * Log22 1200bps
M= 4 Channel Capacity 2 * 400 * Log24 2400bps
M= 8 Channel Capacity 2 * 400 * Log28 3600bps
Q8.Channel data rate can be increased by increasing levels. From table above it is evident
without increase in bandwidth, the capacity is increased by improving signal levels. Basically it
helps transfer more bits in same time hence increasing the data rate.

Q9. Data is transferred between nodes once connection is established. This connection can be
direct connection without any intermediate nodes or channeled through other nodes in between.
When connection is established directly between two sources with no other device being able to
listen to the communication, it is called circuit switching. But when same connection is
established by means of other intermediate devices, it is called Packet switching. Data from one
node travels to other in form of packets that are channeled using header address values. When
this shared channel is reserved for connection between two points, it becomes a virtual circuit.
Such a virtual circuit provides all benefits of circuit switching with small amount of overhead but
vastly improving system resource utilization.
Q10.
Antenna 1 Height t1 meters
Antenna 2 height t2 meters 2 t1 meters
Radius of Earth R 64 * 105 meters
Distance d 40*103meters
Relationship d √ 2 R t1+ √ 2 R 2t1
40*103
√2t1 ( √ R+ √2 R)
40 x 103 / (( √2+1) √R) √ 2t1
1600 x 106 / 2 * (2.414)2 x 64
x 105
t1
21.44 m t1
Antenna 1 height 21.44meter
Antenna 2 height 42.88meter