MITS5003
Wireless Networks & Communication
Final AssessmentAssignment 1
June 2020
MITS5003 QUESTIONS TO BE ANSWERED
Enter your name and Student ID below
Student ID. ____46114_________________
Student Name: _______Sandeep Reddy Gade____________________
Assessment details:
 Final Assessment (Individual Work) – 50%.
 Time allowed 2 hours
 10 Theory and problems questions.
 Total marks =60
Submission details:
 The Final Assessment submission box will only be available for the time period
window beginning from when this document was made available for the length of
time specified above. This assessment must be completed and uploaded to the Final
Assessment submission box within this time period. The submission box will not
remain open beyond this time period. If you are late in downloading this document,
you will not have the same length of time to work on the assessment. The submission
box will still close at the same time for everyone.
 The submission box will be linked to Turnitin, and similarity scores WILL BE
CHECKED and assessments indicating high similarity will be penalized as per all
other assessments throughout the semester.
 You can submit this any time within the specified time frame (2 hours), DO NOT
leave this till the final minute as when the submission closes you won’t be able to
submit your work for marking.
 The assessment must be submited in a Word or PDF fle, with your student ID,
name, unit code and unit name.
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MITS5003 QUESTIONS TO BE ANSWERED
1.
a) Explain what multiplexing is in the context of wireless communication? List and define
the two main types of Multiplexing?( 6 Marks)
Multiplexing is a strategy in which different information signals channels are joined into a
single signal channel for communication transfer. The information which is to be transmitted
over a signal channel is kept apart from others to avoid the overlapping or interference.
The 2 main types of Multiplexing are :
1)Analog Multiplexing
Frequency Division Multiplexing (FDM) : FDM is an analog strategy. We use this method
mostly in TV and radios. FDM technique divides the bandwidth of a channel into several logical
sub channels. Each divided channel is assigned to different frequency Each user has its own sub
channels for its frequencies. Overlapping of frequencies is separated by Guard Band in this
technique.
Wavelength Division Multiplexing (WDM) : Wavelength division multiplexing is a
technology in which multiple optical signals of different wavelengths are joined into one
signal and is transmitted over the communication channel. So multiple signals are
transmitted simultaneously over a single communication channel.
2) Digital Multiplexing
Time Division Multiplexing (TDM): TDM is a technique in which different signals are
joined and processed one after the other on the same transmitting channel. It is a digital
multiplexing technique.
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MITS5003 QUESTIONS TO BE ANSWERED
b) Give three reasons for popular use of multiplexing?( 3 Marks)
1) Budget per kpbs is decreased with an increase of data rate
2) Equipment and transmission cost is decreased with an increase of data rate
3) Widely used interaction devices need latest data rate support
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MITS5003 QUESTIONS TO BE ANSWERED
2. Explain in your own words the three major advantages of digital transmission compared to analog
transmission? (5 Marks)
1) Information which is communicated is kept “private” and secure throughout the
encryption process. Details are not visible while the secure transmission is going on
2) Digital transmission is best suitable for multiplexing process. Multiplexing is easily done
through digital transmission compared to analog transmission. Transmission time is faster
and security is more in digital transmission.
3) Digital transmission is immune to noise. Signals can be easily calculated and measured in
digital transmission. Error detection and correction is easy in this transmission.
3. Differentiate between the two techniques used in packet switching networks. Please explain
them in your own words (4 Marks)
The 2 techniques in packet switching networks are circuit and packet switching.

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MITS5003 QUESTIONS TO BE ANSWERED
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MITS5003 QUESTIONS TO BE ANSWERED
4. A wireless communication channel separated by a distance of 12,500 meters is required to
operate at a center carrier frequency of 2400 MHz Calculate the attenuation loss in decibels?
(4 marks)
L=10 log(4(pi)d/lamda)2
Ldb=20log(2400)+20log(12500)-147.56
Ldb=121.98
5. Assume that two antennas are half-wave dipoles and each has a directive gain of 3dB. If the
transmitted power is 1W and the two antennas are separated by a distance of 10km, what is the
received power assuming that the antennas are aligned so that the directive gain numbers are
correct and the frequency used is 100 MHz?
[Hint: Use the Free space path loss equation]. (4 marks)
DIRECT GAIN= 3db
Transmitted gain at receiver end is 1w which is Pt
Distance = 10kms
Frequency= 100mhz
lamda=3mts
pt/pr=(4pi)2 d2/gtg0lamda2
pr=pt/(4lamda2)*d2*Gt*Gr*lamda2
Pr=1/(4lamda)2*(10000)*3*3*(3)2
Pr=81/394.7*106
Power at receiver end is Pr=0.205*10-6=0.000000205.
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MITS5003 QUESTIONS TO BE ANSWERED
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MITS5003 QUESTIONS TO BE ANSWERED
6.
a) The channel capacity analyses by Shannon and Nyquist places an upper limit on the bit
rate of a channel based on two different approaches. How are these two analysis related?
b) Calculate the channel capacity of an analog voice channel operating within the frequency
range from 1200Hz to 1500Hz and a signal-to-noise ratio (SNR) of 5 dB?
c) If the channel is required to operate at a capacity of 40Mbps and bandwidth of 4MHz,
find the SNR that will be required to achieve this channel capacity.
(4+4+4 = 12 marks)
a) Two various ideas. Despite the fact that Shannon limit needs Nyquist
rate to fnish the computation of limit with a given data transfer
capacity. Nyquist rate lets you know so as to remake a baseband
signal with data transfer capacity W from testing, you have to test the
sign at 2W rate. A decent instinct is to consider a sine wave. This
hypothesis is applying to a sign without commotion. Despite what
might be expected, Shannon’s Capacity hypothesis needs to
determine commotion conveyance, Under Gaussian clamor, Join with
Nyquist rate and ascertain the commotion power appropriately, you
get channel limit of transfer speed to be,
b) b)C= Blog2(1+SNR)
B=1500-1200
B=300
SNRDB=5
SNR=10LOG10(SNR)
SNR=3
C=600
C) 40 = 4 log2 (1+SNR)
SNR = 1023
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MITS5003 QUESTIONS TO BE ANSWERED
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MITS5003 QUESTIONS TO BE ANSWERED
7. List and describe the performance related three advantages and three disadvantages of Wireless
transmission media in Wireless communication networks (6 marks)

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MITS5003 QUESTIONS TO BE ANSWERED
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MITS5003 QUESTIONS TO BE ANSWERED
8. Consider an Orthogonal Frequency Division Multiplexing signal which is required to
operate at a capacity of 80 Mbps, if the cellular network operates at starting spectrum of 5
GHz with the Bandwidth of the channel being 80 MHz with center carrier frequency fc is
half of the Bandwidth, with each sub carrier has 5 MHz allocation.
a. Find the number of channels it can accommodate? And calculate the center carrier
frequency fc ?
b. Calculate the bit duration if the entire bandwidth is used to send the data stream,
and If OFDM will be implemented with the calculated channels, calculate the data
rate of each sub stream?
c. Calculate the each bit duration with OFDM if transmitted on a separate subcarrier,
with a spacing between adjacent subcarriers of fb ?
d. Explain how this will overcome multipath fading, If FDM was used instead of
OFDM, what will be the frequency spectrum required?
(6 Marks)
b)s1(t)*s2(t)=0
D) Orthogonal Frequency Division Multiplexing (OFDM) is a key wideband
advanced specialized technique utilized in remote transmission. Information is
part into a few streams and transmitted on different narrowband channels to
decrease obstruction and crosstalk.
Frequency selecting fading only affects some sub carriers and easily corrected
through error detection
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MITS5003 QUESTIONS TO BE ANSWERED
9. If the maximum distance for LOS communication between the transmitter and the
receiver is 20 Km, then calculate the heights of transmitting and receiving antennas such
that the height of receiving antenna is double of the transmitting antenna. (6 Marks)
Distance= 20km
K=4/3 by default
Distance=3.57(sqrt(kh1)+kh2)
20=3.57(sqrt(k)(sqrt(h1)+(h2)))
20=3.57(sqrt(4/3)sqrt(h1+h2))
20=4.122(sqrt(h1+h2))
4.852=sqrt(h1+sqrt(2h1))
4.852=sqrt(h1)*(2.414)
Sqrt(h1)=4.852/2.414
Sqrt(h1)=2.009
S0, h1=4.039
H2=8.079
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MITS5003 QUESTIONS TO BE ANSWERED
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MITS5003 QUESTIONS TO BE ANSWERED
10. In the emergence of 4th generation cellular networks and standardization, list and describe the two
contenders that successfully emerged? (4 marks)
LTE : It is a mobile communication broadband standard mostly used in mobile devices.
Both LTE & LTE Advances are wireless technologies
LTE Advanced: It is an enhancement version to LTE.It is Continual improvement to the LTE
radio technology and architecture.
Performance and flexibility is improved in LTE Advanced
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MITS5003 QUESTIONS TO BE ANSWERED
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