Communication Systems Homework: Signal Analysis and Transmission

This document presents solutions to a Communication Systems homework assignment, addressing various concepts within the field. The solutions cover a range of topics, including signal analysis, frequency and time domain characteristics of signals, and calculations of parameters like amplitude, time period, and frequency. The assignment also explores concepts in data transmission, such as source rate, channel capacity, and the application of Shannon's theorem. Furthermore, the solutions delve into communication protocols like packet switching and circuit switching, comparing their advantages and disadvantages. The assignment also includes problems related to antenna height and the calculation of free space path loss. Finally, the document concludes with the analysis of composite signals and the determination of bandwidth and channel capacity based on Nyquist criteria.

Question and answers (1 to 9)

TABLE OF CONTENTS
QUESTION 1...................................................................................................................................1
QUESTION 2...................................................................................................................................1
QUESTION 3 ..................................................................................................................................2
QUESTION 4...................................................................................................................................4
QUESTION 5...................................................................................................................................7
QUESTION 6...................................................................................................................................8
QUESTION 7...................................................................................................................................8
QUESTION 8...................................................................................................................................9
QUESTION 9...................................................................................................................................9
QUESTION 10.................................................................................................................................9

QUESTION 1
The menu of pizza is given to guests by the host so that they can place the order of their favourite
pizza. When guest requires to place an order then, the information is transmitted by the host.
From the guest end the order is placed to cook. The information delivered by the host is sent to
clerk who then give the order with cook. For communicating the pizza order between host and
clerk the physical layer or system is provided by the telephone. The pizza is given to clerk by
cook along with the order form which acts as the header to pizza. Further the delivery van
encloses all respective orders consisting delivery address along with the pizza boxes into van so
that pizza can be delivered. The physical connection for the delivery action is provided by the
road.
QUESTION 2
Both French and Chinese prime ministers are interacting as if they are communicating directly
with each other. For instance when French PM addresses the message directly to the Chinese
prime minister. However, the information is actually transmitted through translators via phones.
The message by the French PM is first translated into English and is then delivered to the
translator of Chinese PM who then convert the message into Chinese language.
1

Now as per the given scenario an intermediated node is used to translate the messages before it is
delivered to the destination. The link between Germany and France transmits the information in
Japanese while that between China and Germany delivers message in German language. The
intermediate node performs the translation on each link.
QUESTION 3
2

Maximum amplitude: 15
Time period (T): 3 seconds
Frequency (f): 1/ T = 1/3 = 0.333 Hz
Phase: 0˚
Maximum amplitude: 4
3

Time period (T): 6.5 seconds
Frequency: 1/T = 1/6.5 = 0.153 Hz
Phase: 0˚
Maximum amplitude: 7.5
Time period (T): 2.4
Frequency: 1 / 2.4 = 0.416 Hz
Phase: 90˚
QUESTION 4
A. 10 Sin (2 π (100) t)
Amplitude: 10
Time period (T): 0.01 seconds
Frequency: 100 Hz
Phase: Zero
4

B. 20 Sin (2 π (30) t + 90)
Amplitude: 20
Time period (T): 1/30 = 0.03 seconds
Frequency: 30 Hz
Phase: 90˚
5
Illustration 1: 10 Sin (2 π (100) t)
Illustration 2: 20 Sin (2 π (30) t + 90)

C. 5 Sin (500 π t + 180)
Amplitude: 5
Time period (T): 1/250 = 0.004 seconds
Frequency: 250 Hz
Phase: 180˚
D. 8 Sin (400 π t + 270)
Amplitude: 8
Time period (T): 1/200 = 0.005 seconds
Frequency: 200 Hz
Phase: 270˚
6
Illustration 3: 5 Sin (500 π t + 180)

QUESTION 5
The number of pictures send per second = 30
Pixel size = 480 * 500
Intensity value which can be taken by each pixel = 32
Solution
A.
Sample size = Pixel size * pictures transmitted per second
= 480 * 500 * 30
Sample size = 7.2 * 10^6
Number of bits required for representing each pixel = 5
(2^5 = 32)
Source Rate (R)= Sample size * Number of bits required for representing each pixel
= 7.2 * 10^6 * 5
Source Rate (R)= 36 * 10^6 bps
B.
Channel bandwidth = 4.5 MHz
Signal-to-noise ratio (SNR) = 35 dB
7
Illustration 4: 8 Sin (400 π t + 270)

According to the Shannon's theorem the achievable channel capacity is given by the following
formula:
Channel capacity (C)= B * log (1 + S/N)
Where
B = Line bandwidth
S/N = signal-to-noise ratio
On substituting the given values we get
C = 4.5 * 10^6 * log (1 + 35)
Channel capacity C = 23.26 Mbps = 23.26 * 10^6 bps
QUESTION 6
Frequency = 4 GHz
Distance between earth and satellite = 35,863 km
Solution
Isotropic free space path loss is calculated as:
20 log (4 * 10) + 20 log (35.863 * 10) - 147.56 = 195.6 dB
QUESTION 7
S (t) = 5 Sin (200 π t) + Sin (600 π t)
Solution
W1 = 200 π f1 = 100 Hz
W2 = 600 π f2= 300 Hz
The fundamental frequency = GCF (100, 300) = 100 Hz
Bandwidth = 300-100 = 200 Hz
According to the Nyquist criteria the maximum channel capacity C is given by formula:
C = 2B log M where
B is the bandwidth in Hz
M = number of signal levels
M = 2 C = 2 * 200 * log (2) = 400 bps
M = 4 C = 2 * 200 * log (4) = 800 bps
M = 8 C = 2 * 200 * log (8) = 1200 bps
8

Thus, channel capacity is maximum when number of levels is 8.
QUESTION 8
The speed at which data is transmitted is regulated and demonstrated by the data rate. The
data rate depends upon bandwidth, number of signal levels and noise level of the channel. As per
the Nyquist formula for the bit rate is = C = 2B log M
Bandwidth is constant and thus in order to increase the data rate number of signal levels can be
increased. However, it suffers from a disadvantage that increased number of signal levels can
lower the reliability of the digital system.
QUESTION 9
Difference between packet switching virtual circuit and circuit switching
Both of the switching techniques are preferred for providing connection between multiple
number of communicating devices. The key difference is that the circuit switching technique is
connection oriented while the packet switching or virtual circuit is connectionless. Circuit
switching is use for the voice communication and is less flexible. It also receives messages in
systematic order. On the other hand the packet switching is preferable technique for the
transmission of data and thus it also allows receiving message packets in random order. This
switching is implemented at network layer while the circuit switching is implemented at physical
layer.
Advantages of packet switching over circuit switching
The biggest advantage of packet or virtual switching over the circuit switching is that it
has greater efficiency. Since the data packets have ability to determine own path and to reach the
desired destination they are more efficient. However, in circuit switching use of channel by
network devices is restricted until voice communication does not terminate. Further packet
switching is more easy and affordable. As compare to circuit switching the packet switching also
allows to have fair utilisation of the resources and also allows the pipelining.
QUESTION 10
The relation between transmitting and receiving antenna: h1 =2* h2
d = 40 km
9

Solution
The maximum distance between antennas for the LOS communication = 3.57* (√kh1 + √kh2)
k = adjustment factor = 4/3
h1= 2h2
So max distance d = 3.57* (√(k2*h2) + √kh2)
40 = 3.57 * √kh2 * [1 + √2]
40 = 8.618 * √kh2
√kh2 = 4.641
kh2 = 21.53
k = 4/3
h2 = 16.14 km
h1= 2 * 16.14 = 32.28 km
Height of both the antennas = 16.14 km, 32.28 km
10