EUROPEAN UNIVERSITY CYPRUS | SCHOOL OF SCIENCE
DEPARTMENT OF COMPUTER SCIENCE AND ENGINEERING | FALL 2013 SEMESTER
CSC322 – DATA COMMUNICATIONS AND COMPUTER NETWORKS – ASSIGNMENT(S) 1 & 2
SET ON MONDAY 18-NOV-2013  DUE ON MONDAY 25-NOV-2013 (BY 23:59)
BY ANDREAS GRONDOUDIS
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Answer the following questions
Questions
1. What is a PDU (Protocol Data Unit)?
How are PDUs used?
2. Define the term wavelength
3. For any signal, which is greater (wider),
absolute or effective bandwidth?
4. What happens to the data rate if we
half the bandwidth?
5. Which type of signals is more
susceptible to noise, analog or digital?
6. A transmission system is using 8 signals
levels and has a bandwidth of 2GHz,
what (according to Nyquist) is the
channel capacity?
7. A signal with fundamental frequency of
200
MHz
is
(using
Fourier)
approximated using the first 7
components frequencies. What is the
absolute bandwidth of this signal?
Explain your answer
Answers
A PDU is the minimum size of information that is
exchanged between adjacent layers in order to
effect communication. For example when TCP at
host A wants to send some information to host’s B
TCP then it will create a TCP PDU (also known as
TCP segment) and that will be passed down to the
IP layer. The IP layer will use the TCP PDU as the
data (payload) for its own PDU (the IP datagram)
which will in turn be passed down to the network
layer and so on.
The distance that the signal will ‘travel’ in the
duration of 1 period (1 repetition)
Absolute bandwidth
It will be halved as well
analog
C = 2 * B * log2(8) = 2 * B * 3 = 6 * B = 6*2 =
12Gbps
Fundamental is 200MHz. the seven first
components are f, 3f, 5f, 7f, 9f, 11f, 13f this means
the 7th component is 13*200MHz = 2600MHz.
Now subtract the first from the last (13f-f) =
2400MHz. the absolute bandwidth is 2.4GHz
8. The same signal as above is (using
Fourier) approximated using the first 17
component frequencies. What is the
absolute bandwidth of this signal?
9. Given the signals in questions 7 and 8, if
they were used in a channel utilizing 2
signal levels; what will their respective
capacities be (according to Nyquist)?
10. An optical fiber link is right next to a
power cable. Is it affected by
interference and why?
11. What does microwave require that
broadcast radio does not?
12. NRZI encodes a zero as what?
13. What is the common feature between
the Manchester and Differential
Manchester encoding schemes?
14. How does Manchester encode a 1 bit?
15. In multilevel encoding techniques,
which is better Bipolar AMI or
Pseudoternary?
16. What is scrambling and why is it used
for?
17. Given 2 signals one with frequency f
and another with a frequency of 3f,
which of the two attenuates more and
why?
18. The signal to noise ratio of a line is very
high. Is this bad or good and why?
19. List (just list) wireless transmission
impairments?
20. How does binary frequency shift keying
work?
Fundamental is 200MHz this means that the 17th
component is 33f therefore 33*200MHz =
6600MHz. Now subtracting the first from the last
(33f-f) = 6400MHz. the absolute bandwidth is
6.4GHz
The first would produce = 2*2.4 = 4.8Gbps while
the second would produce = 2 * 6.4 = 12.8Gbps
No it is not. Fiber is electromagnetically isolated as
it does not carry power but rather light
Line-of-Sight
A zero in NRZI is encoded by the lack of transition,
therefore when going from the previous bit to the
next the voltage would stay at the same level
(irrespective or high or low)
They both have a midbit transition that can be
used to act as a clocking mechanism
It uses the midbit transition. A midbit transition
from low to high encodes a 1.
They are the same as one is the exactly inverse of
the other
Scrambling is the insertion of extra dummy bits in
known positions inside the bit stream in order to
introduce transitions that will help prevent loss of
synchronization between sender and receiver
The one with the 3f because we said that higher
frequencies attenuates faster
A good thing as this means that the power of the
signal is must larger than the power of the noise
which means that the noise can do little to
interfere with the signal.
Free space loss; refraction; atmospheric
absorption; multipath
The system would use 2 frequencies. One
frequency will be used to represent a binary 0 and
the other frequency would be used to represent a
binary 1.
21. What is quantizing noise?
The result of the approximation of pulse amplitude
samples to closest (existing) integer value which
when the streams are encoded means that some
information is getting lost and cannot be exactly
recreated at the receiving end.
Even-numbered errors (when 2 or 4 or 6 or any
even number of bits go bad)
For every character we require 1 bit for the start
element, 2 bits for the stop element, (no parity)
and 7 bits for the character a total of 10 bits. Given
that the word has 7 characters the total number of
bits required will be 10*7 = 70 bits
22. Parity is not able to detect which type
of errors?
23. A MODEM (using asynchronous
transmission) is transmitting the 7
character word Andreas. Assuming that
each character will be encoded to 7
bits, and that parity is not used, and
that the stop element is 2 bits long,
how many bits will be needed to
transmit the word? (Please explain your
answer)
24. What is the Hamming distance between
a. distance is 2
the following streams:
b. distance is 3
a. 01000 and 01011
c. distance is 5
b. 01111 and 01000
c. 10101 and 01010
For questions 25-28 below, the following codeword mapping has been used:
Row
1
2
3
4
Data
000
001
010
100
Codeword
11111
10101
00000
11000
25. What is the minimum distance of the
codeword set?
26. Given the answer to above question
what is the maximum number of error
bits that this set can detect?
27. If the following stream: 10010 was
received; what can you say about it?
(explain your answer)
28. If the following stream: 10001 was
received; what can you say about it?
(explain you answer)
1 and 2 distance is 2; --- 1 and 3 distance is 5; --- 1
and 4 distance is 3; --- 2 and 3 distance is 3; --- 2
and 4 distance is 3; --- finally 3 and 4 distance is 2;
---therefore the minimum distance of this
codeword set is 2. So Dmin = 2
The set can detect (Dmin-1)-bit errors so 1-bit
errors
Distance with row 1 is 3 --- distance with row 2 is 3
--- distance with row 3 is 2 --- distance with row 4
is 2. There is an error but we cannot correct it
Distance with row 1 is 3 --- distance with row 2 is 1
--- distance with row 3 is 2 --- distance with row 4
is 2. There is an error, 1 possibility with distance 1
so we probably have had a single bit error and got
10001 instead of 10101. Therefore the word must
have been 10101 therefore the original data must
have been 001
29. If we are encoding MFSK with 16
16 expressed in a 2 power is 2^4 so we can encode
frequencies, then how many bits can
4 bits for each frequency
we encode? (explain you answer)
30. Theoretically speaking, if we combined
8 frequencies. 4 given with 2 possible amplitudes
a shift keying scheme with 8
and the other 4 given with 2 possible phases (4*2
frequencies and 4 of those where using
+ 4*2 = 8 + 8 = 16). So we have 16 possible
two different amplitudes and the other
combinations, as a power of 2 it is expressed 2^4
4 were using two different phases then
so we can have each combination encoding 4 bits
how many bits would we be able to
encode? (explain your answer)
31. Consider
the
scenario
(again
8 frequencies multiplied by 2 possible amplitudes
theoretically) that we now have 8
(8*2 = 16). 16 combinations multiplied by 4
frequencies and all 8 can have 2
possible phases (16*4 = 64). So we have 64
different amplitudes as well as 4
possible combinations as a power of 2 it is
different phases. How many bits would
expressed as 2^6 so we can have each
we be able to encode now? (again,
combination encoding 6 bits
please explain your answer)
32. If Delta Modulation was used to encode the information, what would the resulting bit stream
be?
15
10
5
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19 20
Your answer (start from 0): 011111101110100000001
33. Using decimal (rather than binary)
First put 0 zeros after the number
arithmetic, given a dividing pattern of
564378000
256; calculate a 3 digit CRC number that
Then divide by 256
must be inserted after the number
2204601.xxxxxx
1
564378. (See footnote on this)
Get the next number and multiply by 256
2204602*256 = 564378112
112 is the CRC
1
Similar to the example seen in class and explained on the ‘additional’ slides for chapter 6