Communication Systems Homework: Solutions for Signal Analysis Problems

This document presents solutions to a Communication Systems assignment, addressing a range of topics. The solutions begin with a system-level example of a pizza ordering process and then progresses to message exchange between two PMs using translators. Detailed calculations and explanations for signal analysis are provided, including amplitude, time period, frequency, and phase. The document also covers data rate calculations, channel capacity based on Shannon's theorem, and path loss calculations in a wireless communication scenario. Further, the solutions delve into the analysis of waveforms, fundamental frequencies, and bandwidth. The assignment also includes a comparison between circuit switching and packet switching, along with the calculation of antenna heights based on communication range. Finally, the solutions are supported by the bibliography of a standard textbook on the topic.

Q1. The guest passes the desire for a Pizza to the host. On receiving the
request, host starts looking up for address and number of the Pizza shop. On
finding the number , the hosts makes a telephone call to the listed number.
The telephone on other end notifies the Pizza Clerk to check Telephone. This
connects the Host with the clerk. Host places order for Pizza and mentions
his address. Clerk forwards the request to the cook so he can prepare the
dish. Once dish is prepared, the order is packed and parceled to the host
address as informed to the clerk earlier. The Pizza reaches hosts address by
delivery van.
.
Q2.
This process shows the sequence of message exchange between Two PM’s
who cannot understand each other’s language but Translators help them
communicate with each other using a third common language. Any
French PM
sends
Message
Translator
translates
message
Telephone
relays the
message
Chinese PM
receives
Message
Translator
translates
message
Telephone
receives the
message

conversation from French PM is translated to English and forwarded to
Chinese translator over Telephone. The Chinese translator receives the
message and translates it from English to Chinese and apprises the PM with
the message in Chinese. Vice versa happens when Chinese PM wants to send
a reply or a message to the French PM.
Q3 .
a. Amplitude : 15 ; Time period = 3 ; Frequency = 1/3 ; Phase =0
b. Amplitude : 4; Time period = 7 ; Frequency = 1/7 ; Phase =0
c. Amplitude : 8; Time period = 3 ; Frequency = 1/3 ; Phase =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)

A
mplitude = 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.
Bits for 32 intensity levels = 5 bits
Frames in 1 second = 30
Bits transferred in 1 sec = 480 x 500 x 5 x 30 = 36 * 106 bits
That is the source data rate. 36 Mbps
Channel Bandwidth in Hz = 4.5 MHz = 45 x 105
Signal To Noise Ratio (SNR) = 35 decibels

Then Channel Capacity in bits = B * log2 *(1+SNR)
= 45 x 105 x log2(1 + 103.5)
= 45 x 105 x log2 (1 + 3162)
= 45 x 105 x log2(3163)
= 45 x 105 x 11.6272 = 52.32 x45 x 106 Bps
Q6. Path Loss for given frequency 4 GHz and distance of lower earth Orbit is
20Log10 (4 x 109) + 20log10 (35.863x106) – 147.56dB
192.04 + 151.092 – 147.56dB = 195.57dB
Q7.
a. Fundamental Frequency = 100Hz.
b. Spectrum
= 5 ( ∆ ( f −100 )−∆ ( f +100 ) )
+1 ¿
c. Bandwidth = (f + 300) – (f -100) = 400Hz.
d. Channel Capacity using Nyquist Theorem =
Channel Capacity = 2 * 400 * Log22 = 800 bits per second (M = 2
)
Channel Capacity = 2 * 400 * Log24 = 1600 bits per second (M =
4)
Channel Capacity = 2 * 400 * Log28 = 2400 bits per second (M =
8)

Q8. Channel capacity of the medium is dependent upon Bandwidth and SNR.
If Bandwidth is fixed then to increase the data rate, we must increase out
SNR high enough so that more level of signals can be accommodated in the
channel. This will allow each bit time to transfer more information. For
example if number of levels is 2 then only 1 bit information is transferred at
a time. If number of levels is increased to 4, then 2 bits of information is
transferred at a given time. Thus increasing the data rate without increase in
bandwidth.
Q9.Circuit switching is an old technique from times of telephony where
sender and receiver were connected together by switches so they could talk
to each other. In this type of switching the route is electrically connected
from one point to another. This allows communication without any explicit
headers being exchanged or transmitted about destination. This actually is
also the limitation that a circuit switch can only connect two points in a given
time. Packet switching is a modern concept, the ownership of the channel
remains with network managers but on request of transmitting station, a set
of paths is set aside to allow data only from designated source channel to
flow through it. This makes a connection between sender nad receiver that is
similar to circuit switching but is comprised of several smaller paths. This
type of communication has some overheads with each packet as packet now
must identify paths allocated to it while travelling. But it still saves lot of time
and increase network resource sharing helps reduce costs of network.

Q10. Antenna has height h1 and h2, such that h2 = 2 x h1.
Distance between them is then calculated as dm =√2 Rh 1+ √2 Rh2
where R is Radius of Earth.
When Distance is known, the same formula can be used for calculation of
height of Antenna. Substituting the values we get,
40 x 103 =√2 Rh 1+ √2 R 2 h1
40 x 103 = √ h 1( √ 2 R+2 √ R)
√h 1 = 40 x 103 / ( √2 √R+2 √ R)
h1 = 1600 x 106 / (3.414)2 x 64 x 105
= 1600 x 106 / 11.6568 x 64 x 105
= 21.44 m
Therefore, antenna 1 height = 21.44m and antenna 2 height = 42.88
m
Bibliography
Haykin, S. (2007). Communication Systems. Wiley.