Electrical Engineering Homework: Signals and Systems

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Contents
Answer 1....................................................................................................................................2
Answer 2....................................................................................................................................2
Answer 3....................................................................................................................................3
Answer 4....................................................................................................................................4
Answer 5....................................................................................................................................5
Solution 6...................................................................................................................................6
Answer 7....................................................................................................................................6
Answer 8....................................................................................................................................6
Answer 9....................................................................................................................................7
References..................................................................................................................................8

Answer 1
IP is the Internet Protocol that helps a system in building a connection with different
computer systems. The main role of the IP is in order to assign the high aspects of network
that are different and can be used in any of the application (Kozierok, 2005).
The TCP/IP contain the layer that is also termed as the Network access layer and is used for
the transfer of data as well as receive the data from the main network which is used for the
physical network. This is considered as the main suite of protocol which helps in establishing
the communication that can connect diverse devices that are available on internet.
Answer 2
Figure 1: Answer 1

Figure 2: Answer 2
Answer 3
The Max Amplitude of the wave is: 15
The time period is: 3 sec
Frequency is: One divided by time period = one divided by 3 equals to 0.33-hertz frequency.
The phase is zero by calculating the equation x (t) = Am sin (θ)

The Max Amplitude of the wave is: 4
The time period is: 6.5 sec
Frequency is: One divided by time period = one divided by 6.5 equals to 0.1538-hertz
frequency.
The phase is zero by calculating the equation x (t) = Am sin (θ)
The Max Amplitude of the wave is: 7.5
The time period is: 2.25 sec
Frequency is: One divided by time period = one divided by 2.25 equals to 0.444-hertz
frequency.
The phase is ninety 900 by calculating the equation x (t) = Am sin (θ)
Answer 4
Given: 3𝑆𝑖𝑛 (2𝜋 (200) 𝑡)
The Max Amplitude of the wave is: 3
Frequency is: 200 hertz
The time period is: 0.005 sec

Phase is: zero
Given: 14𝑆𝑖𝑛 (2𝜋 (50) 𝑡 + 90)
The Max Amplitude of the wave is: 14
Frequency is: 50 hertz
The time period is: 0.02 sec
The phase is: ninety (90)
Given: 4𝑆𝑖𝑛 (650𝜋𝑡 + 180)
The Max Amplitude of the wave is: 4
Frequency is: 325 hertz
The time period is: 0.003 sec
The phase is: 180

Given: 6𝑆𝑖𝑛 (700𝜋𝑡 + 270)
The Max Amplitude of the wave is: 6
Frequency is: 350 hertz
The time period is: 0.002 sec
The phase is: 270
Answer 5
The smallest path given as 6 hertz, convert it: 6 * 109 is 654
Calculation of the wavelength: (3 * 108) divided by 654 will give as 0.495m
Now, LdB is 99.684
The calculation is given as:
20 log (35.853 * 106) – 20 Log (0.495) + 21.98
Antennas Calculation: 99.68452 subtracted with 44 subtracted with 48 is equal to 7.68452

Solution 6
S (𝑡) = 5 sin (100𝜋𝑡) + sin (300𝜋𝑡) + sin (600𝜋𝑡)
Fifty 50, Hundred fifty 150 and three hundred 300 are the fundamental frequencies.
Fourier transform: S(f)= (1/2i) {(d(f-150)-d(f+150))+5(d(f-50)-d(f+50))+d(f-300)-d(f+300))}
Formula: Fmax = Bandwidth + Fmin
Bandwidth is 50 plus 150 plus 300 is 500 MHz
Nyquist Bitrate:
Bitrate = 2 Bandwidth
If M is 2, then Channel capacity is calculated as 2 x 500 Log 2 2 = 1000 bits/sec
If M is 4, then Channel capacity is calculated as 4 x 500 Log 2 2 = 2000 bits/sec
If M is 8, then Channel capacity is calculated as 8 x 500 Log 2 2 = 4000 bits/sec
Answer 7
The rate of data is understood as the rate by which the data can be transfer on the channel
from two diverse ends. The use of Nyquist bit rate formula is used if the noiseless channel is
there:
Bit Rate equals to 2 multiplies Log2 (L) multiplies Bandwidth (Tan et al., 2014).
This equation is understanding using the explanation of the bandwidth of the channel and the
letter L used helps in signifying the signal numbers whereas it helps in expressing and
representation of the data and the data rate that is used in the definition of the bit per second.
The disadvantage is related to the reliability of the system which minimizes the signals.
Answer 8
Switching technique is used in order to provide the path that is appropriate and can assist the
transmission information between the nodes that are connected. A circuit switching that is

used in the physical layer as well as for the information passing and the different allocated
bandwidths. Time-division and the space divisions are two different divisions that are
categorized from circuit division.
The packet switching is also a type of switching which differs from the circuit switch as the
virtual work is present on the data link layer in case of packet switching but in the circuit
switching it is on the physical layer (Muñoz and Makinwa, 2013).
The advantage of using packet switching is that the availability of the resources that are
allocated can be checked.
Answer 9
D is given as 60 km so to convert it: 60 multiplied by 1000 convert it into meters.
H1 is equaled to 2 x and H2 is equals to x
LOS formula is: d equals to 3.57√KH equals to (d/3.57)2 equals to KH, and K equals to 4 / 3
Replace values: 60/3.57)2 give 4/3(2x + x) gives (60/3.57)2 equals 4x
So x is 70.58
Therefore, H1 = 141.18m and H2 = 70.58m

References
 Kozierok, C.M., 2005. The TCP/IP guide: a comprehensive, illustrated Internet
protocols reference. No Starch Press.
 Tan, H.N., Inoue, T., Tanizawa, K., Kurosu, T. and Namiki, S., 2014, September. All-
optical Nyquist filtering for elastic OTDM signals and their spectral defragmentation
for inter-datacenter networks. In 2014 The European Conference on Optical
Communication (ECOC) (pp. 1-3). IEEE.
 Muñoz, J.R.A. and Makinwa, K.A., 2013. Analysis and Design of Decision Feedback
Equalizers for bitrates of 10Gbps and Beyond in Submicron CMOS.