Faculty Of Computer Studies
Course code: T215 A
Course Title: Communication and information Technologies
Final Exam- Key solution
First semester 2014/2015
T215A_Final_Exam_Key Solution_Fall 2014
Page 1 of 8
Part 1
This part carries 15% of the total examination marks. You should attempt all questions in this
part by choosing the correct answer for each question and write it on the external answer
sheet. Each question carries 1.5 marks:
1) In TCP/IP model, frame is used to designate for a piece of data at the__________
a. Transport layer.
b. Internet layer.
c. Application layer.
d. Network interface layer.
2) A mobile terminal offering___________GPRS would enable both voice and data to be
used simultaneously.
a. Class B.
b. Class A.
c. Class D.
d. Class C.
3) When mobile network operators try to restrict its customers to a closed set of services,
this is known as_________________.
a. Service support.
b. Denial service.
c. Walled garden.
d. Consumer satisfaction.
4) The mobile phone can pick up the incoming radio signals that carry speech or text
through the use of________________
a.
b.
c.
d.
Loudspeaker.
Microphone.
Battery.
Aerial.
5) Unlike GSM, EDGE uses a very different sort of digital modulation called ‘phase
modulation’, specifically a version called:
a. 64 PSK
b. 32 PSK
c. 16 PSK
d. 8 PSK
T215A_Final_Exam_Key Solution_Fall 2014
Page 2 of 8
6) Newer displays are based on organic light-emitting diode (OLED) polymers that emit light
directly and in doing so, they use_____________________________
a.
b.
c.
d.
Less power than a separate backlight.
More power than a separate backlight.
Same power as a separate backlight.
Identical power to a separate backlight.
7) A ____________ doesn’t consume its electrodes; instead, it uses a separate supply of
chemical energy.
a.
b.
c.
d.
Fiber cell.
Fuel cell.
Florescent cell
Ferromagnetic cell.
8) When the camera and the GPS module in the same mobile phone are linked together in
order to label images with the location where they were captured, then this
called_______________
a.
b.
c.
d.
Geointerpolation.
Geomapping.
Geotagging.
Geolabelling.
9) The small cells used in urban environments are sometimes called _____________
a.
b.
c.
d.
Picocells.
Microcells.
Millicells.
megacells
10) Metadata can be used to facilitate searching for files on a hard disk. It can also be used
in other sorts of searches. One example relates to photographs and other images,
which are increasingly saved with metadata known as:
a.
b.
c.
d.
EXIF
JPEG
MPEG
ZIP
T215A_Final_Exam_Key Solution_Fall 2014
Page 3 of 8
Part 2
This part carries 35% of the total examination marks. You should attempt all questions in this
part. Each question carries 5 marks:
1- SIMs which stands for subscriber identity modules is a small card in the mobile terminal on
which electronic data is stored. State five examples of such stored data inside SIMs.
-
The SIM Serial Number (SSN), also known as the Integrated Circuit Card ID (ICC-ID).
The International Mobile Subscriber Identity (IMSI).
An authentication key (that is, a special code used for security purposes) assigned by the
mobile service provider.
The identity of the MSC in whose area the mobile terminal is currently located.
(For most network operators) the user’s personal contacts and other similar data.
2- There are at least two processers inside the mobile phone. What are they? Explain briefly why
they are used?


Digital signal processor or DSP: dedicated to processing the signals that has been
received or is to be transmitted.
A general-purpose processor: controls the activities the phone carries out, including
detecting the user’s key presses, putting messages on the screen and ensuring an
orderly and timely flow of data around the phone.
3- To combat for the fading problem, the GSM has introduced some techniques. Explain briefly two
of these techniques.
•
The use of a training sequence: (1 Mark)
A sequence of bits called training sequence is transmitted as the central portion of each time slot
is used. It allows the digital circuitry to be ‘trained’ to its best behavior. What happens is that
the precise characteristics of the compensating circuit are automatically adjusted until the best
reception of the known training sequence is obtained.
(1.5 marks)
•
frequency hopping: (1 Mark)
The fading phenomenon is frequency dependent, and transmission on one frequency can suffer
degradation at a particular location while another frequency can be comparatively free from
problems. So GSM allows for the rapid switching from one carrier frequency to another:
frequency hopping.
(1.5 Marks)
T215A_Final_Exam_Key Solution_Fall 2014
Page 4 of 8
4- What is the essential difference between a circuit-switched network and a packet-switched
network? Give an example for each type of network.
•
In a circuit-switched network, such as a telephone network, a circuit is set up
between the sender and the recipient and is maintained for the duration of the
phone call, even if no one is speaking. The circuit cannot be used for any other call
during this time.
(2.5 marks)
•
In a packet-switched network, such as the internet, a message is split up into
packets which are sent one after the other from the sender to the recipient.
Although the path taken by the packets can be the same for all packets, it does not
have to be. In addition, packets from other messages can use the path whenever
the original message has no packets to send.
(2.5 marks)
Part 3
This part carries 50% of the total examination marks. You should attempt all the problems in
this part.
Problem 1: (10 marks)
a) Assuming that radio waves propagate at a speed of 3×108 m/s, estimate the propagation
time for a path length of 6 km between a mobile and a base station. Express your answer in
seconds and micro seconds?
(5 marks)
6 km = 6000 m
(2 marks)
Time = distance/speed
(1 mark)
Propagation time = 6000/ (3×108) = 20 × 10-6 s = 20 μs
(2 marks)
Problem 2: (20 marks)
Part A: (8 marks)
i.
A battery has a capacity of 2000 mAh and a mass of 50 grams. Find its gravimetric
capacity? (4 marks)
Gravimetric capacity = capacity / mass (2 marks for formula)
Gravimetric capacity = 2000/50 = 40 mAh/g. (2 marks for correct answer)
ii.
If another battery has the same gravimetric capacity of battery in part (i) and having a
capacity of 2400 mAh, what is its mass? (4 marks)
Mass = capacity / gravimetric capacity = 2400 / 40 = 60 g
T215A_Final_Exam_Key Solution_Fall 2014
Page 5 of 8
Part B: (12 marks)
A blue ray disk can store a 30 GB of raw video file.
i.
How long would it take to store a 30 GB of raw video file on this blue ray disk if we use a
writing speed of 500 Mbit/s?
(4 marks)
Time = File size/Writing speed
(1 Mark)
So Time = 30 GB/500 Mbits/s = (30 x 230 x 8 bits)/ (500 x 106 bits/s) ≈ 515.4 s
(3 Marks: 1 for bits, 1 for bits/s and 1 for the final answer)
ii.
How small would the 30GB raw video file become if it is compressed with a compression
ratio of 20?
(4 Marks)
Compression ratio = uncompressed file size/compressed file size
Compressed file size = uncompressed file size/Compression ratio
Compressed file size = 30GB/20 = 1.5GB.
iii.
(2 Marks)
(2 Marks)
Express the size of the video file after compression in bits.
(4 Marks)
Size of file after compression= 1.5 GB
1.5 GB = 1.5 x 230 bytes (1Mark) = 1.5 x 230 x 8 bits (1Mark) = 1.5 x 230 x 23= 1.5 x 233 bits
So 1.5 GB = 12 884 901 888 bits
(2 Marks)
In GSM, there are 8 separate time slots that makeup what is known as a “TDMA frame’’. Each
GSM time slot lasts for 577 µs (microseconds) and contains more than simply a sample of each
individual signal.
i.
How long does the TDMA frame last, express your answer in µs and ms? (4 marks)
Time = number of slots x period of a single slot = 8 x 577 µs = 4616 µs (3 marks)
= 4.616 ms (1 mark)
ii.
Each time slot comprises a total of 148 bits. Of these 148 bits, only 114 represent voice
or other data. The rest are used for a variety of control purposes. How many bits of data
are in one TDMA frame?
(4 marks)
Number of bits = number of data bits per slot x number of slots = 114 x 8 = 912 bits
iii.
How many bits for control purposes are in one TDMA frame?
(4 marks)
Number of control bits = 148 – 114 = 34 bits
Number of bits = number of control bits per slot x number of slots = 34 x 8 = 272 bits
iv.
What is the total number of bits, including data and control, in a TDMA frame? (2 marks)
Total = number bits per slot x number of slots =148 x 8 = 1184 bits
T215A_Final_Exam_Key Solution_Fall 2014
Page 6 of 8