Explains the repetition code for error correction.
Perhaps the simplest error correcting code is the
repetition code .
Repetition code
The upper portion depicts the result of directly modulating
the bit stream
into a transmitted signal
using a baseband BPSK signal set.
is the datarate produced by the source coder. If that bitstream passes through a (3,1) channel coder to yield the bit
stream
, the resulting transmitted signal requires a bit
interval
three times smaller
than the uncoded version. This reduction in the bit intervalmeans that the transmitted energy/bit decreases by a factor of
three, which results in an increased error probability in thereceiver. Here, the transmitter sends the data bit several times, an odd
number of times in fact. Because the error probability
is always less than
, we know that more of the bits should be correct
rather than in error.Simple majority voting of the received bits (hence the reason
for the odd number) determines the transmitted bit moreaccurately than sending it alone.
For example, let's consider the three-fold repetition code:for every bit
emerging from the source coder, the channel coder
produces three. Thus, the bit stream emerging from the channelcoder
has a data rate three times higher than that of the
original bit stream
.
The coding table illustrates when errors can becorrected and when they can't by the majority-vote decoder.
Coding table
Code
Probability
Bit
000
0
001
0
010
0
011
1
100
0
101
1
110
1
111
1
In this example, the transmitter encodes
as
.
The channel creates an error (changing a
into a
)
that with probability
.
The first column lists all possible received datawords and thesecond the probability of each dataword being received. The
last column shows the results of the majority-vote decoder.When the decoder produces
, it
successfully corrected the errors introduced by the channel(if there were any; the top row corresponds to the case in
which no errors occurred). The error probability of thedecoders is the sum of the probabilities when the decoder
produces
.
Thus, if one bit of the three bits is received in
error, the receiver can correct the error; if more than oneerror occurs, the channel decoder announces the bit is
1 instead of transmitted value of
0 . Using this repetition code, the
probability of
equals
.
This probability of a decoding error is always less than
, the uncoded value, so long as
.
Demonstrate mathematically that this claim is
indeed true. Is
?
This question is equivalent to
or
.
Because this is an upward-going parabola, we need only checkwhere its roots are. Using the quadratic formula, we find
that they are located at
and
. Consequently
in the range
the error rate produced by coding is smaller.
Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
Gautam
A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate
velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.
Mehmet
A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat
which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer
I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.
Someone
Scratch that
Someone
temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
Someone
about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle
pratica A on solution of hydro chloric acid,B is a solution containing 0.5000 mole ofsodium chlorid per dm³,put A in the burret and titrate 20.00 or 25.00cm³ portion of B using melting orange as the indicator. record the deside of your burret tabulate the burret reading and calculate the average volume of acid used?
No. According to Isac Newtons law. this two bodies maybe you and the wall beside you.
Attracting depends on the mass och each body and distance between them.
Dlovan
Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?
Specific heat capacity is a measure of the amount of energy required to raise the temperature of a substance by one degree Celsius (or Kelvin). It is measured in Joules per kilogram per degree Celsius (J/kg°C).
AI-Robot
specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin
ROKEEB
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