Telecommunication Engineering: Signals, Systems, and Networks

Desklib offers past papers and solved assignments for students. This project covers various aspects of telecommunication engineering.

ASSIGNMENT NUMBER 1
Solution 1:
STEP 1- Place Order
a) The guest will ask for the pizza menu, the host will provide the menu to the guest.
b) Guest will select the pizza from the menu and will place the order for pizza.
c) Guest will look for the contact number and dials the phone number. The telephone will
get connected to the telephone line and will send the tones to the number which is dialed.
d) The host will talk with the order clerk, it will allocate the order of the guest for the pizza,
an acknowledgment will be done with the guest and then the guest will wait for its order.
e) The clerk will receive the order and will ask for the information regarding the payments,
the approval of the payment information will be done and then the order will be placed to
the cook.
STEP 2- Pizza Delivery
a) The cook will prepare the pizza according to the given requirements of the guests and
will give the order to the clerk.
b) The pizza will be packaged by the clerk, a delivery receipt will be prepared and then it
will be placed into a delivery van.
c) By the transmission line which is the road, the delivery van will transfer the ordered pizza
to the guests.
d) The host will confirm the details of the payment and will return the clerk after that
confirmation of delivery takes place.
e) At last, the pizza will be provided to the guests by the host.
Solution 2:

Solution 3:
a)
i. Maximum Amplitude is 15.
ii. Frequency = (1/Time Period)
iii. Time Period = 3 sec, therefore, Frequency = 1/3 Hz = 0.3333 Hz
iv. Phase => net = Am Sin (∅ ), where ∅ , is the phase,
Here, wt = 0 , therefore, net = 0,
Therefore, Am Sin (wt +∅ ¿= 0, so it can be stated that ∅ = 0.
Hence, phase = 0o
b)

i. Maximum Amplitude is 4.
ii. Frequency = (1/Time Period)
iii. Time Period = 6.5 sec, therefore, Frequency = 1/6.5 Hz = 0.1538 Hz
iv. Phase => net = Am Sin (wt + ∅ ¿where ∅ , is the phase,
Here, wt = 0 , therefore, net = 0,
Therefore, Am Sin (wt + ∅) = 0, so it can be stated that ∅ = 0.
Hence, phase = 0o
c)
i. Maximum Amplitude is 7.5.
ii. Frequency = (1/Time Period)
iii. Time Period = 2.25 sec, therefore, Frequency = 1/2.25 Hz = 0.444 Hz
iv. Phase => net = Am Sin (∅ ), where ∅ , is the phase,
Here, wt = 0, but amplitude = max = Am
Therefore, Am Sin (wt + ∅) = Am, so by this it can stated that Sin ∅ = 1,
Hence, phase = 90o

Solution 4:
a) 10Sin(2 π(100)t)
Amplitude: 10
Time Period: 1/Frequency = 1/100 = 0.01 sec
Frequency: 100 Hertz
Phase: 00
Wave Form:
b) 20Sin(2 π(30)t + 90)
Amplitude: 20
Time Period: 1/Frequency = 1/30 = 0.33 sec
Frequency: 30 Hertz
Phase: 0o
Wave Form:

c) 5Sin(500 πt + 180)
Amplitude: 5
Time Period: 1/Frequency = 1/250 = 0.04 sec
Frequency: 500/2 = 250 Hertz
Phase: 1800
Wave Form:
d) 8Sin(400 πt +270)
Amplitude: 8
Time Period: 1/Frequency = 1/200 = 0.005 sec
Frequency: 400/2 = 200 Hertz
Phase: 2700
Wave Form:

Solution 5:
Given: 480x500 pixels and Intensity value: 35 per pixel
Assume, number of pictures sent per second are: 30
a) By given information it can be stated that pixel per second will be:
480 x 500 x 30 = 7200000.
As it can be seen that each pixel will take 32 values, therefore,
Log2 (32) = 5, therefore, there are 5 bits per pixel.
Now, the source rate can be calculated by the multiplication of the:
7200000 x 5 = 36 x 106 Mbps
b) Now bandwidth given is: 4.5 MHz and Signal-to-noise ratio is given as 35 DB
10Log10 (signal-to-noise ratio) = 31632
Capacity of the channel is: 4.5 x 106 (1 + 31632) log2
= 4.5 x 106 x log2 (3163) = 4.5 x 106 x11.6272 = 52.32 x 106 bps
Solution 6:
First, we’ll convert the given values: 4GHz = 4 x 109 Hz and 35863 Km = 35.863 x 106 m
Now by performing calculations, we’ll get:
20 Log10 (4 x 109) + 20 Log10 (35.863 x 106) – 147.56 = 195.574 dB
Solution 7:
Given equation is: S (t) = 5 Sin (200 πt) + Sin (600 πt)
Assume there are three frequencies those have signals f1, f2, f3 , and if these signals are periodic
in nature then each of the given frequencies are integer and are multiple of fo ,
F1 = n1 F0, F2 = n2 F0, F3 = n3 F0, therefore, F0 = GCD (F1, F2, F3)
Therefore, we’ll get: F0 = GCD (100, 300) = 100
And finally we’ll get the fundamental frequency as 100.
S (t) = 5 Sin (200 πt) + Sin (600 πt)
Now using Fourier transform Sin (2 πAt): 1s = 1/2i [δ (f - A) - δ (f + A)]

Therefore, S (t) = 5. 1/2i [δ (f - 100) - δ (f + 100)] + 1/2i [δ (f - 300) - δ (f + 300)]
Now spectrum is: S (t) = 1/2i [5 (δ (f - 100) - δ (f + 100)] + [δ (f - 300) - δ (f + 300)]
The formula for bandwidth is Fmax - Fmin = (f + 300) – (f - 100) = f + 300 – f + 100
On solving, bandwidth is 400Hz.
The maximum data rate is the noiseless signal of the channel capacity.
The bit rate capacity = 2 x Bandwidth x Log2 L, therefore, the channel capacity = 2 x Bandwidth
x Log2L where, L = number of levels.
Bandwidth = 400 Hz
Now for M = 2, Channel capacity = 2 x 400 x log22 = 800 bits / sec
M = 4, Channel Capacity = = 2 x 400 x log24 = 1600 bits / sec
M = 8, Channel Capacity = = 2 x 400 x log28 = 2400 bits / sec
Solution 8:
The noiseless channel, as the Nyquist bit rate formula state that the max bitrate is:
Bitrate = 2 x Bandwidth x Log2 (L)
The equation of the bandwidth provides the bandwidth channel, L is the signal level which is
used to represent data and the bit rate is in bit rate per second. And the bandwidth is a fixed
quantity that cannot be changed; therefore the data rate is directly proportional with the number
of the signal level.
Solution 9:
The switching techniques are used in order to provide the physical path so that the information
can be passed from one node to another. The Circuit switch is used in the physical layer and the
bandwidth in this is allocated in order to transfer the data. The Circuit switch includes two
different types that involve space and time division.
But in the case of the packet switch, there is a virtual circuit which use addresses to work and the
difference between both the circuits is that the circuit switch uses physical layer and packet
switch use data link layer.
The advantage of the virtual approach is that even if the resource allocation is demanded then
the source can check the resource availability without actually reserving.
Solution 10:
The Line of Sight Communication is the LOS,
H t: height of transmitter antenna
H r: height of the receiver antenna.
The formula used to calculate height is:
Dm = √ 2Rht + √ 2Rhr
Where R is the Radius of Earth
R ~ 64 x 105 m
Given is the height of one antenna is double than another: the equation is:
Dm =√ 2Rh + √2R (2h)

Dm = √ 2Rh + 2 √2Rh
= √ Rh (2+ √2)
Now, for further calculation,
√h= dm / Rh (2+ √2)
h = (dm) 2 / R (2+ √ 2) 2
Given value of dm is 40km which is converted as 40 x 103 m,
h = (dm) 2/ R (2+ √2) 2
= (40 x 103)2/ (2+ √2) 2 x 64 x 105
= 16 x 108 / (2+ √ 2) 2 x 64 x 105
h = 16 x 103/ (2+ √2) 2 x 64 = 1000/11.6568 x 4 = 1000/46.627 m
h = 21.446 = 21.45 m
Therefore,
Height of the First Antenna = 21.45m
Height of the Second Antenna = 42.90m