Computation of Probability of Error for Integrate and Dump Filter Receiver block diagram and working , formula :-
As discussed earlier that a noise interference may lead to wrong decision at the receiver end. As a matter of fact, the probability of error denoted by Pe is a good measure of performance of the detector.
We know that the output of the integrator is expressed as,
y(t) = x0(t) + n0(t)
For the positive pulse of amplitude A, x0(t) is given as,
x0(t) = for x (t) A
Similarly, for the input pulse of amplitude – A, x0(t) is given by,
x0(t) = or x(t) = -A
Therefore, the output y(t) may be written as
x(t) = or x(t) = A …(6.53)
Similarly, y(t) = or x(t) = -A …(6.54)
Let us consider that x(t) = -A. Further, if noise n0(t) is greater than , then the output y(t) would be positive according to equation (6.54). After that the receiver will decide in favour of symbol + A, which is wrong decision. This means that an error is introduced.
Similarly, let us consider that x(t) = + A. If noise n0(t) > – then, the output y(t) will be negative according to equation (6.53). This leads to decision in favour of –A, which is erroneous. Based on the above discussion, we can make conclusions about probability of error in the form of a Table 6.4.
Table 6.4. Probability of error in integrate and dump filter receiver
|S. No.||Input x(t)||Value of n0(t) for error in the output||The probability of error Pe|
|1.||-A||An error will be introduced if n0(t)>||In this case, the probability of error may be obtained by calculating probability that n0(t) >|
|2.||+ A||An error will be introduced if n0(t) <||In this case, the probability of error may be obtained by calculating the probability that, n0(t) <|
As shown in Table 6.4, the error will be introduced depending upon probability that n0(t) takes a particular value. These probabilities may be obtained from PDF of n0(t). Recall that the probability density function (PDF) of the Gaussian distributed function is expressed by standard relation as under:
where, fX(x) = the PDF of random function x.
m = the mean value and
and = the standard deviation.
Here, because we have to evaluate PDF for white Gaussian noise, therefore, we have
x = n0(t)
Since this noise has zero mean value i.e., m = 0, therefore equation (6.55) may be written as,
The standard deviation is expressed as,
= [mean square value – square of mean value]1/2
This means that
Further, we have
mean square value =
Also, mean value mx = 0 for this noise.
Therefore, we have
Hence equation (16.56), can be written as,
At this stage, it may be noted that n0(t) is the function like ‘x’. It is a random variable and we are evaluating its PDF.
On simplifying above equation, we get
This equation describes PDF of white Gaussian noise.
Figure 6.21 shows the graphical representation of this PDF.
From the property of PDF, we know that,
This equation gives the probability that n0(t) takes values greater than .
The above integration gives the area under the curve from onwards. It has been illustrated by shaded region in figure. Similarly, the probability that n0(t) attains value less than is given by area under the curve from onwards on left side. This portion has also been shown shaded in figure 6.28.
FIGURE 6.28 Illustration of PDF of white Gaussian noise having zero mean.
Since the PDF curve is symmetric, therefore, we can write.
We know that the above probabilities represent error probability. Because, occurrence of – A or + A is mutually exclusive, therefore, the probability of error is given by either of the two in above equation i.e.,
Substituting value of P in above equation from equation (6.59), we get
Again, substituting value of fx[n0(t)] from equation (6.58), in above equation, we get
Now, let us put
Thus, d[n0(t)] =
when n0(t) then y
when n0(t) =
With all these substitutions, equation (6.61) takes the form
This equation can be rearranged as under:
The integration inside brackets may be evaluated with the help of complementary error function i.e.,
Recall that this is a standard result and generally evaluated using numerical methods.
Thus, with the help of this definition, equation (6.62) becomes,
Pe = …(6.63)
This equation describes the probability of error Pe of the integrate and dump filter receiver.
Now, since A2T = E, i.e., energy of the bit, therefore, we have
Pe = …(6.64)
NOTE : It may be noted that basically erfc’ is the monotonically decreasing function. Therefore Pe falls rapidly as the ratio increases. Hence, the maximum value of Pe is when is very very small. This means that even if the signal is lost entirely in the noise No even then, the probability of error Pe will be . Thus, the receiver would make incorrect decisions for half number of times.