OPTIMUM FILTER , what is OPTIMUM RECEIVER , optimum filter transfer function bandwidth is typically in the range of

By   June 30, 2020
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OPTIMUM FILTER bandwidth is typically in the range of , OPTIMUM RECEIVER , optimum filter transfer function.
In the previous article, we discussed Integrate and Dump receiver. But, at this point, let us think whether Integrate and dump filter is an optimum* filter for the purpose of minimizing the probability of errors Pe. For this purpose, we shall discuss a generalized filter to receive binary coded signals. It is known as optimum filter. Therefore, let us consider the generalized Gaussian noise which is having zero mean. In the last section, we assumed ‘white’ Gaussian noise of zero mean.
Let the received signal be a binary waveform. Let us again assume that the polar NRZ signal is used to represent binary l’s and 0’s. i.e.,
for binary ‘1’,
x1 (t) = + A for one bit period T
and for binary ‘0’ ,
x2(t) = — A for one bit period T
Hence, the input signal x(t) will be either x1(t) or x2(t) depending upon the polarity of the NRZ signal. Figure 6.29 shows the block diagram of a receiver for such a binary coded signal.
As shown in this figure, the noise n(t) is added to the signal x(t) in the channel during the transmission.
Thus, input to the optimum filter is [x(t) + n(t)] i.e.,
Input to the receiver = x(t) + n(t)
* Here, the word optimum means to have maximum advantages.
FIGURE 6.29 Mock ditigrom of receiver for binary coded signal.
and Output from tin receiver = x01(t) + n0(T) or x02(T) +n0(T)
Also, in the absence of the noise n(t), the output of the receiver will be,
y(T) = x01(T)    if          x(t) = x1(t)
and                                  y(T) =  x02(T)        if                x(t) = x2 (t)
Hence, in the absence of noise, decisions are taken clearly. However, if noise is present then, we select x1(t) if y(T) is closer to x01(T) than x02(T) and we select x2(t) if y(T) is closer to x02(T) than x01(T).
Therefore, the decision boundary will be in the middle of x01(T) and x02(T).
It is expressed as,
Decision boundary  =                                                                …(6.65)
6.22.1. Calculation of Probability of Error (Pe) for Optimum Filter
The probability of error (Pe) may be evaluated on the same basis as in the last section. Here, we shall consider generalized Gaussian noise. Let us consider that x2(t) was transmitted, but x01(T) is greater than x02(T). If noise n0(T) is positive and larger in magnitude compared to the voltage difference  – x02(T), then, in this case, the incorrect decision will be taken. This means that the error will be generated if,
EQUATION                                      …(6.66)
We have obtained the probability density function (PDF) for n0(t) in the previous section. It is given as
EQUATION                                      …(6.67)
Here, n0(t) is the random function whose PDF is given by above equation. Also,  is its standard deviation and the function has zero mean value.
Hence, to evaluate the probability of error, we must integrate the area under the PDF curve from . This portion of the curve has been shown shaded in figure 6.30.
Therefore, we have
FIGURE 6.30 Evaluation of Pe for optimum filter receiver.
Using equation (6.67), the last equation becomes,
EQUATION                                      …(6.68)
Let us substitute                                     EQUATION
so that                                        N0(t) = Y
or                                                         d[n0(t)] = Y
when                                                   n0(t) = ,        then y =
and when                                            n0(t) =
then, we have                          y =
Substituting all these values in equation (6.30), we obtain,
Let us rearrange this equation as under:
EQUATION                                      …(6.69)
To solve this integration, let us use the following standard result.
Then, equation (6.69) becomes
EQUATION                                      …(6.70)
NOTE : This is the required expression for error probability Pe of optimum filter. It may be noted that the aerie’ function is the monotonically decreasing function. Hence, Pe decreases as the difference x01,(T) – x02(T) becomes greater and the rms noise voltage  becomes smaller. The optimum filter has to maximize the ratio  in such a manner that the probability of error Pe is maximum.
6.22.2. Evaluation of Transfer Function for the Optimum Filter
            Now, let us find the transfer function of the optimum filter in such a way that it will maximize the ratio.
For this, let us represent the difference signal x01(T) – x02(T) by x0(T).
Thus, we have
x0(T) = x01(T) – x02(T)                                      …(6.71)
This means that the optimum filter has to maximize the ratio
Now, let us derive the transfer function of the optimum filter such that the square of the ratio  is maximized.
This square of the ratio is,
where  is known as the signal to noise power ratio. Thus, we are maximizing square of  for mathematical convenience. In this equation,  is the normalized signal power in 1 W load.
Further,  =  = E[n02(T)] is normalized noise power because mean value of noise is zero.*
Thus, the optimum filter has to maximize the ratio
EQUATION                                      …(6.72)
Again, let X0(f) be Furrier transform of x0(t). If X(f) is the Furrier transform of input difference signal x(t) [i.e., x(t) = x1(t) – x2(t)], then we have,
X0(f) = H(f) X(f)                                             …(6.73)
Here, H(f) is the transfer function of optimum filter. Also, x0(T) is found by taking inverse Fourier transform of X0(f) i.e.,
x0(T) = IFT [X0 (f)] = EQUATION
*   – (mean value)2.
We know that
Also, the input noise to the optimum filter is n(t). Let, its power spectral density (psd) be Sni(f). The output noise of optimum filter is n0(t). Let its power spectral density be Sn0(f).
We know that the input and output power spectral densities of noise are related as,
Sn0(f) =  Sni(f)                                               …(6.75)
The normalized noise power can be obtained by integrating the power spectral density.
Thus, we have
EQUATION                                      …(6.76)
Substituting for x0(T) from equation (6.73) and  from equation (6.75) in equation (6.72), we
EQUATION                                      …(6.77)
Here, let us use the well known Schwarz’s inequality which states that
EQUATION                                      …(6.78)
Let                                                          EQUATION
With all these substitutions, the equation (6.76) becomes,
Applying Schwarz’s inequality of equation (6.77) to the numerator of above equation, we have
Now, substituting value of (f) in the above equation, we have
Thus, we have
EQUATION                                      …(6.78a)
The signal to noise power ratio expressed by equation (6.78) will be maximum when we consider equality.
In other words we must have
This is possible only when equality applies in Schwarz ‘s inequality i.e.. in equation (6.77),
This equality is possible only if we have  = k* 2
Substituting the values of  and  , we obtain,
Therefore, when this equation is satisfied, the signal to noise ratio  will be maximum.
In this case. the transfer function H(f) can be obtained from above equation as under:
It may be noted that at the beginning of this derivation, we stated that
Now we can have the following conclusion:
The optimum filter which minimizes the probability of error (Pe) has to maximize the ratio . To maximize this ratio, the filter has the transfer function given by following expression,
EQUATION                                      …(6.79)
and the maximized ratio is expressed as,
* By property of pad.