Electrical Engineering: Quiz on Waveforms and Signals

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Contents
Q 1..............................................................................................................................................2
Q 2..............................................................................................................................................3
Q 3..............................................................................................................................................4
Q 4..............................................................................................................................................6
Q 5..............................................................................................................................................7
Q 6..............................................................................................................................................8
Q 7..............................................................................................................................................9
Q 8............................................................................................................................................10
Q 9............................................................................................................................................11

Q 1
IP is mainly described as the protocol used only for the internet. IP addresses are mainly
being used for connecting the various computer systems. TCP/IP mainly helps in assigning
all the higher aspects of the use of the network for the applications. IP that is Internet
Protocol assumes the proper existence of the network access layer but it does not at all define
it. Network Access Layer whereas encompasses the proper functions of different lower
layers. The main design of the TCP/IP helps in hiding the proper functions of different lower
layers.
Network Access Layer is one of the layers in the TCP/IP model which helps in transmitting
the data as well as receiving the entire data in the physical network. TCP/IP is considered as
the suite of the protocol for the communication that helps in interconnecting the various
network devices on the internet.

Q 2

Q 3
a). By the help of the diagram above, maximum amplitude could be seen as 15.
b). The time period is given or seen as 3 seconds. ‘T’ could be considered as the Time Period.
Now, the value of the Frequency is to be calculated and the solution is provided below as:
F = (1/T) = (1/3) = 0.33Hz
c). At next part, a formula is given as:
Calculation of the phase angle is to be done by-
X(t)= Am sin(θ) = 0
Phase angle is equals to zero.

a). By the help of the diagram above, maximum amplitude could be seen as 4.
b). The time period is given or seen as 6.5 seconds. ‘T’ could be considered as the Time
Period.
Now, the value of the Frequency is to be calculated and the solution is provided below as:
F = (1/T) = (1/6.5) = 0.1538 Hz
C). Since, we could easily see that wt value is provided as zero therefore, here also phase
angle would be zero.
a). By the help of the diagram above, maximum amplitude could be seen as 7.5.
b). The time period is given or seen as 2.25 seconds. ‘T’ could be considered as the Time
Period.
Now, the value of the Frequency is to be calculated and the solution is provided below as:
F = (1/T) = (1/2.25) = 0.444 Hz
c). At next part, a formula is given as:
Calculation of the phase angle is to be done by-
X(t)= Am sin(θ) = Am
The phase angle equals to 900.

Q 4
a). 3sin(2π(200)t)
Amplitude is calculated here = 3
Frequency is calculated here = 200 Hz
Time is calculated here = 0.005 sec
Phase angle would be 0.
b) 14sin (2π(50)t + 90)
Amplitude is calculated here = 14
Frequency is calculated here = 50 Hz
Time is calculated here = 0.02 sec
Phase angle would be 90.

c). 4sin (650πt + 180)
Amplitude is calculated here = 4
Frequency is calculated here = 325 Hz
Time is calculated here = 0.003 sec
Phase angle would be 180.

d). 6sin (700πt + 270)
Amplitude is calculated here = 6
Frequency is calculated here = 350 Hz
Time is calculated here = 0.002 sec
Phase angle would be 270.

Q 5
The shortest path could be calculated as 6Hz= 6*109
Then, wavelength would be calculated as W= (3x108)/(6x109) = 0.495 m
Now, LdB = 20log (35.853x106) – 20log (0.495) + 21.98
LbB = 99.68452
Now for the antennas,
Two values of the antenna are being taken as 44dB and 48dB.
Now, the free space lost would be:
LdB = 99.68452 -44-48
LdB= 7.68452

Q 6
Given as, S(t) = 5sin(100πt) + sin(300πt) + sin(600πt)
Fundamental frequencies here are being calculated as 50, 150 as well as 300.
By using the Fourier transform, we get the-
S(f)= (1/2i) {(d(f-150)-d(f+150))+5(d(f-50)-d(f+50))+d(f-300)-d(f+300))}
Now, Fmax = Bandwidth + Fmin
So, Bandwidth would be calculated as 50+150+300= 500 MHz
Using Nyquist bitrate formula-
Bitrate is equals to twice of Bandwidth
For M=2
Channel Capacity is 2*500 log2 2= 1000 bits per second
For M=4
Channel Capacity is 4*500 log2 2= 2000 bits per second
For M=8
Channel Capacity is 8*500 log2 2= 4000 bits per second