Telecommunication Systems Assignment: Signals, Channels, and Networks

Desklib provides past papers and solved assignments for students. This assignment covers telecommunication systems.

Assignment No 1
1. Using the layer models in Figure, describe the ordering and delivery of a pizza, indicating the
interactions at each level.
Answer.
In this model, the ordering and the delivering of the pizza happen in the below-mentioned steps.
Step 1: The Guest selects a pizza from the menu, and places the order with the host. Here, the guest is
trying to place an order with the pizza cook.
Step 2: The Host will then communicate with the clerk, regarding the pizza request. Host allots the order
to the clerk.
Step 3: Here telephone lines act as physical means of transport from host to clerk.
Step 4: Once the order is received by the clerk, the order is passed to cook after the payment
confirmation.
Step 5: The pizza is prepared by the cook and via the clerk it is delivered to the host and the guest.
2. The French and Chinese prime ministers need to come to an agreement by telephone, but neither
speaks the other’s language. Further, neither has on hand a translator that can translate to the
language of the other. However, both prime ministers have English translators on their staffs. Draw a
diagram similar to Figure 1 to depict the situation, and describe the interaction and each level.

Answer.
In this model, the French and Chinese Prime Minister will communicate through their translators. The
telephone line will act as a transport medium between the two translators. The Chinese Prime Minister
will speak and the translator will communicate through the telephone line to the other translator, which
will then translate the English words to the French. This process will happen vice versa.
3. From the following figures, compute the maximum amplitude, frequency, time period and phase
for each of the wave. The x-axis represents the time in sec and y-axis represents the amplitude.
Amplitude=A= 15
Frequency= f = 0.33 Hz
Time Period= 3 sec

Phase = 0
Amplitude=A= 4
Frequency= f = 0.15 Hz
Time Period= 6.5 sec
Phase = 0

Amplitude= A = 7.5
Frequency= f = 0.43 Hz
Time Period= 2.3 sec
Phase = -180
4. Compute the amplitude, frequency, time period and phase for each of the following equations and
also draw their respective waveforms.
a. 10𝑆𝑖(2𝜋(100)𝑡)
b. 20𝑆𝑖(2𝜋(30)𝑡+90)
c. 5𝑆𝑖(500𝜋𝑡+180)
d. 8𝑆𝑖(400𝜋𝑡+270)
Answer.
a. 10𝑆𝑖(2𝜋(100)𝑡)
Amplitude=A= 10
Frequency= f = 100 Hz
Time Period= 1/100= 0.01 sec
Phase = 0

b. 20𝑆𝑖(2𝜋(30)𝑡+90)
Amplitude=A= 20
Frequency= f = 30 Hz
Time Period= 1/30= 0.33 sec
Phase = 0
c. 5𝑆𝑖(500𝜋𝑡+180)
Amplitude=A= 5
Frequency= f = 500/2 = 250 Hz

Time Period= 1/250= 0.004 sec
Phase = 180
d. 8𝑆𝑖𝑛(400𝜋𝑡+270)
Amplitude=A= 8
Frequency= f = 400/2 = 200 Hz
Time Period= 1/200= 0.005 sec
Phase = 270
5. Suppose that a digitized TV picture is to be transmitted from a source that uses a matrix of
480×500 picture elements (pixels), where each pixel can take on one of 32 intensity values. Assume
that 30 pictures are sent per second. (This digital source is roughly equivalent to broadcast TV
standards that have been adopted.). a. Find the source rate R (bps).

b. Assume that the TV picture is to be transmitted over a channel with 4.5-MHz bandwidth
and a 35-dB signal-to-noise ratio. Find the capacity of the channel (bps).
a.
Size: 480*500
No. Of pictures per sec: 30
Sample size= Size * No. Of pictures
Said that each pixels can take one of 32 intensity values: 32
Log232 = 5 * Log2 2= 5
Source rate = Sample Size * 5 = 36 Mpbs
b.
Bandwidth = 4.5 MHz
Signal to noise ratio = 35 dB
10 log 10 (SNR) = 35 dB
log 10 (SNR) = 35/10
SNR = 103.5
Channel capacity = B * log (1 + SNR) = 4.5 * 106 log ( 1 +103.5) = 52.335 * 106 bps
6. Determine the isotropic free space loss at 4 GHz for the shortest path to a synchronous satellite
from earth (35,863 km).
Answer.
Free space loss = 20 log 10 (4*pi*r/ λ)
f = 4 GHz = 4 * 106 Hz
λ = C/ f = 3 * 108 * 4 * 106 Hz = 3/40 m
loss = 20 log 10 (3.14 * 4 * 35863 * 1000/ 0.075 ) = 3911.42 dB

7. Given a signal as follows, compute the fundamental frequency, spectrum and bandwidth. Also
calculate the channel capacity using Nyquest criteria using M= 2, 4, 8, where M is the number of
levels.
(𝑡)=5sin(200𝜋𝑡)+sin(600𝜋𝑡)
Answer.
2π(f1)t = 200 π t
f 1=100
2π(f1)t = 600𝜋𝑡
f 2 = 300
Fourier Transform: 1/2i [ƍ(f-100)- f+(100)] + 1/2i [ƍ(f-300)- f+(300)]
Bandwidth = fmax – fmin = (f + 300) – (f-100)
BW = 400 Hz
Channel capacity = 2 * BW * log2(L)
For M = 2, Channel capacity = 800bps
For M = 4, Channel capacity = 1600bps
For M = 8, Channel capacity = 2400bps
8. Explain how the data rate over a channel can be increased, without increasing the bandwidth?
What is the disadvantage of this approach?
Hint: Nyquist Theorem
Channel capacity = 2 * BW * log2(L)
According to this formula, the data rate over the channel is directly proportional to the number of levels.
It is having a bad effect as the levels may reduce the reliability of the system.

9. What is the main difference between Packet switching Virtual Circuit and Circuit Switching? Also
discuss the advantages of Packet switching Virtual Circuit over Circuit Switching.
The difference between the Packet switching and the Circuit switching are as follows:
The Circuit switching is connection oriented. It was designed for Voice communication, a bit inflexible
because transmission follows the same path. Messages are sent from the source in order. It is
implemented at the physical level. It has two parts, Space division and Time division Switching.
The packet switching is connectionless and earlier designed for data Transmission. It is flexible because
there is a route for every packet. Packets are assembled at the destination. Packet switching used two
technologies, datagram approach and Virtual circuit approach. It is implemented at the network layer.
10. In a LOS communication, consider d = 40km, the requirement is to make two antennas
(transmitter and receiver) such that the height of one antenna should be twice of the other.
Considering this, find the appropriate heights of these two antennas.
Answer .
The maximum LOS distance is:
D = √(2Rh1) + √(2Rh2) = 40 Km
R = 64 * 105 m , h2 = 2*h1 (height of one antenna should be twice of the other)
D = √(2Rh1) + √(2R2h1)
√h1 = D/ (2 + √2) √R , h1 = 21.44 m and h2 = 42.88 m