Cyclic Redundancy Check | Assignment

Solution 1
A) .
b) .

c) .
d) .

e) .
f) .

g) .
Solution 2

Solution 3
Solution 4

Solution 5

Solution 6
I. Handoff occurs at location A based on the average measurement of the received signal.
II. The effect of the threshold, the threshold TH1 is at a higher point than the value 2 because the
relative value compared depends on the signal strength of one of the base signal, handoff
position L4 occurs.
III. The effect of the threshold, the threshold TH2 is a large point than this value 2 because the
relative value compared depends on the signal strength of one of the base signal, handoff
position L4 occurs.
IV. For TH3, the mobile station moves too far to a new cell due to the very long delay. As a
result, the quality of the communication link from base station A is degraded, as well as the call
may be disconnected. In addition, this increases interference with co-channel users. Therefore,
this can create overlapping cell coverage areas.

V. The handoff occurs at point C because the handoff only occurs when the new base signal
is sufficiently stronger (due to the hysteresis margin h ) than the current base signal .
VI. When the T- mobile signal A is passed, his handoff is made by taking into account the relative
signal strength between the hysteresis as well as the threshold, as well as when the current
signal level falls below the threshold, the target-based signal is given by the given
hysteresis, Over margin on new base signal only if stronger than current one . This does not
happen, so a handoff occurs.
VII. When the T- mobile signal A is passed, his handoff is made by taking into account the relative
signal strength between the hysteresis as well as the threshold, as well as when the current
signal level falls below the threshold, the target-based signal is given by the given
hysteresis, Over margin on new base signal only if stronger than current one . This does not
happen, so a handoff occurs.
VIII. Handoff occurs at point L1 if the threshold is TH3.
Solution 7
Cyclic Redundancy Check (CRC): The term Cyclic Redundancy Check is the most dominant instrument for
mistake recognition. Dissimilar to equality checking dependent on expansion, Cyclic Redundancy Check
depends on paired division. Cyclic Redundancy Check Now, rather than understanding the equality
object by including bits, Cyclic Redundancy Check or Cyclic Redundancy Check arrangement of excess
bits called leftover portion is added as far as possible of the data unit, the consequence of the
information unit 2 precisely at the foreordained parallel A number that will be distinct. In the event that
there is no leftover portion in this progression, the information unit is viewed as unblemished as well as
is along these lines acknowledged. The rest show that the information unit was harmed in travel as well
as should be dismissed. The repetitive bits utilized in the Cyclic Redundancy Check are acquired by
isolating the information unit by a given divisor, the rest of the Cyclic Redundancy Check. Cyclic
Redundancy Check is in 2 requires one of value. It must be actually one piece not exactly the divisor, as
well as must be attached as far as possible of the information string with the goal that the subsequent
piece arrangement is actually distinguishable by the divisor [1].

Third, the n-bit Cyclic Redundancy Check acquired in stage 2 replaces the O included toward the finish of
the information unit. The information unit initially lands at the recipient information, trailed by the
Cyclic Redundancy Check. Beneficiary the whole string 1 is treated as a unit, Cyclic Redundancy Check
partition by a similar divisor that was utilized to discover the rest of. On the off chance that the string
lands without blunder, the Cyclic Redundancy Check checker produce the staying zeros as well as the
information unit passes. In the event that the string is changed during the move, the division delivers a
nonzero leftover portion as well as the information unit doesn't pass [2].
Solution 8
I. Total occupied time =
II. Total number of calls =
III. Time occupied per call for U1 =
Time occupied per call for U2 =
Time occupied per call for U3 =
Time occupied per call for U4 =
Time occupied per call for U5 =
Mean time occupied per call =
IV. Calls per min for U1 =
Calls per min for U2 =

Calls per min for U3 =
Calls per min for U4 =
Calls per min for U5 =
Mean time occupied per call =
V. Traffic intensity per user A:
A = average call request rate * average holding time
Total traffic intensity = traffic intensity per user * number of users
Total traffic intensity for U1 =
Total traffic intensity for U2 =
Total traffic intensity for U3 =
Total traffic intensity for U4 =
Total traffic intensity for U5 =
Total traffic intensity for U1, U2, U3, U4 as well as U5 is 51.99 Erlangs
Solution 9
Square: Does not make dark spots. Nonetheless, the good ways from the inside to the corner is longer
than the separation to any side. This makes an issue in giving a similar degree of sign at each purpose of
the circle. Since the good ways from the inside to any point in the circle is the equivalent, it is protected
to give a similar degree of sign at all focuses. Notwithstanding, putting the circles together makes a great
deal of zone that no circle covers. These territories are called dark spots just as can't get signals from
close by circles. Hexagon: A hexagon or honeycomb structure defeats the entirety of the above issues.
The good ways from the inside to any point is equivalent to well as can be set with the goal that no dark
dabs are made.

References
[1] P. E. Boudreau, W. C. Bergman, and D. R. Irvin. Performance of a cyclic redundancy check and its
interaction with a data scrambler. IBM Journal of Research Development, 38(6):651-658, 2014.
[2] F. Monteiro, A. Dandache, A. M'sir, and B. Lepley. A fast CRC implementation on FPGA using a
pipelined architecture for the polynomial division. In the 8th IEEE International Conference on
Electronics Circuits and Systems, volume 3, pages 1231-1234, 2016.