Communication Systems Analysis and Design Homework Solution

This document presents a comprehensive solution to a communication systems homework assignment, addressing a range of topics within electrical engineering. The solution begins with a practical example of order placement, detailing the communication process from guest request to order fulfillment. It then delves into signal analysis, including amplitude, frequency, and phase calculations for various waveforms, and provides equations for waveform representation. The assignment continues with calculations of data transmission rates, bandwidth, and channel capacity using the Shannon theorem. It also covers isotropic path loss calculations, fundamental frequency determination, and the application of the Nyquist theorem. Furthermore, the solution includes a comparison of circuit and packet switching techniques, and concludes with calculations related to antenna height and distance based on given parameters. The document provides detailed explanations and calculations for each problem, offering a thorough understanding of the concepts involved.

Q1. Guests makes a request to have a Pizza. The hosts receives the request and initiates procedure to
inform the order clerk about this request. The host makes use of Telephone as physical medium to
convert this request into signal that can be transferred to Pizza shop over a network. The Telephone
makes all necessary signals to ring a bell on other end. Once Clerk hears the bell he reponds to the
request by picking up the phone. The Host and Clerk are now connected and can exchange messages.
The message is exchanges and clerk decoded it as order for Pizza. The clerk conveys the request to the
cook for a Pizza as per specifications required by the guests. This completes the process of order
placement by Guest to the cook.
Next when Pizza is baked at the shop, the parcel is handed over to delivery boy and van is rushed
towards the house of the host. The delivery person hands over the pizza to the client.
Q2.
Q3 .
a. Amplitude : 15 units Frequency = 1/3 units Time period = 3 units Phase =0
b. Amplitude : 4 units Frequency = 1/6.5 units Time period = 6.5 units Phase =0
c. Amplitude : 7.5 units Frequency = 1/2.5 units Time period = 2.5 units Phase =90
French PM
Speaks to translator in French
French to English
Translator
Translates PM's message from
French to English
Translates incoming message
from English to French
Telephone
Conveys MEssage from one
transalator to another in English
Chinese PM
Speaks to translator in Chinese
Chinese to English
Translator
Translates PM's message from
Chinese to English
Translates incoming message from
English to Chinese
Telephone
Conveys Message from one
transalator to another in English

Q4. Equation for waveform A Sin (2ℿ (f)t + θ)
a.
10 Sin(2 ℿ(100) t)
Amplitude = 10.
Frequency = 100 Hz
Time Period = 1/f
= 1/100 =10msec
Phase = 0◦
b.
20 Sin(2 ℿ(30) t + 90)
Amplitude = 20.
Frequency = 30 Hz
Time Period = 1/f
= 1/30 = 33.33msec
Phase = 90◦
c.
5 Sin(500 ℿ t + 180)
Amplitude = 5.
Frequency = 500/2
= 250 Hz
Time Period = 1/f = 1/250 =
4msec
Phase = 180◦
d.

8 Sin(400 ℿ t + 270)
Amplitude = 8.
Frequency = 400/2 = 200 Hz
Time Period = 1/f = 1/200 = 5msec
Phase = 270◦
Q5.
a. Given Pixel count = 480 * 500.
Each pixel has 32 intensity levels. It needs Log232 bits for transmission of information I,e, 5 bits
Information transferred for one frame = 480*500*5 bits = 1200000 bits
Total information transferred in 1 sec = 12 x 10^5 * 30 = 36 * 10 ^ 6 bits
This is the source rate in Bits per second 36 * 10 ^ 6
b. Bandwidth of Channel = 4.5 MHz.
SNR = 35dB
Using Shannon Theorem
Channel Capacity = B * log2 *(1+SNR)
= 4.5*10^ 6* log2*(1 + 1035/10)
= 4.5 * 10 ^ 6 * log2 * (1 + 3162)
= 4.5 *10 ^ 6 * log2(3163)
= 4.5 *10 ^ 6 *11.6272 = 52.32 * 10 ^ 6 Bits per second
Q6. Isotropic Path Loss is given by
20Log10(Frequency) + 20log10(Distance) – 147.56dB
20Log10(4 x 109) + 20log10(35.863x106) – 147.56dB
192.04 + 151.092 – 147.56dB = 195.57dB
Q7.
a. Fundamental Frequency is maximum common frequency among given two signals
i.e. G.c.d. of (100, 300) = 100Hz.
b. Spectrum

= 1
2i ¿
c. Bandwidth = fmax – fmin
= (f + 300) – (f -100) = 400Hz.
d. Channel Capacity using Nyquist Theorem = 2 * Bandwidth * Log2L ( L number of levels)
For M = 2 Channel Capacity = 2 * 400 * Log22 = 800 bits per second
For M = 4 Channel Capacity = 2 * 400 * Log24 = 1600 bits per second
For M = 8 Channel Capacity = 2 * 400 * Log28 = 2400 bits per second
Q8. Channel Capacity for a channel is given using Nyquist Theorem
2 * Bandwidth * Log2L ( L number of levels).
When we cannot increase bandwidth we must increase number of levels the signal can achieve to
indicate a set of bits. This allows us to increase data rate by sending more information in same amount
of time slot. Since more number of level means each signal represents increased number of bits.
Q9. Circuit switching is physically a medium is connected between source and destination. In Packet
switched Virtual circuit, the path between Source and destination is not physically allocated. It is
reserved for the usage to allow packets to flow between two ends at the pre-determined rate. Circuit
switching the circuit is on or connected even when no data flow takes place. In Packet switching the
bandwidth is allocated to other users in case of no data flow. At the same time, Packet switching
provides all good features of circuit switching like in order packet receiving, fixed time delay across
different packets, connection reliability and minimum overheads while providing advantage of optimal
usage of network resources by sharing bandwidth on demand basis.
Q10. Given two Antenna of height h1 and h2, the distance between them is then calculated using
formula
dm =√2 Rh 1+ √2 Rh2 where R is Radius of Earth = 64 * 10 ^5 m
we are given h1 = 2 x h2 and dm = 40Km
40 * 10^3 =√2 Rh 1+ √2 R 2 h1
= √h 1( √2 √ R+2 √R)
 √h 1 = 40 x 103 / ( √ 2 √ R+2 √ R)
H1 = 1600 x 106 / (2 + 1.414)2 x 64 x 105
= 1600 x 106 / 11.6568 x 64 x 105

= 1000 / 11.6568 x 4
= 21.446 m
Thus we have antenna 1 height = 21.446m and antenna 2 height = 42.892 m
Bibliography
Stallings, W. (2013).
Data and Computer Communications. Pearson.