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

**THE OPTIMUM FILTER (i.e. OPTIMUM RECEIVER**

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 P_{e}. 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’,

x_{1} *(t)* = + A for one bit period *T*

and for binary ‘0’ ,

x_{2}(t) = — A for one bit period T

Hence, the input signal x(t) will be either x_{1}(t) or x_{2}(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.

**DIAGRAM**

**FIGURE 6.29** *Mock ditigrom of receiver for binary coded signal.*

and Output from tin receiver = x_{01}(t) + n_{0}(T) or x_{02}(T) +n_{0}(T)

Also, in the absence of the noise n(t), the output of the receiver will be,

y(T) = x_{01}(T) if x(t) = x_{1}(t)

and y(T) = x_{02}(T) if x(t) = x_{2} (t)

Hence, in the absence of noise, decisions are taken clearly. However, if noise is present then, we select x_{1}(t) if y(T) is closer to x_{01}(T) than x_{02}(T) and we select x_{2}(t) if y(T) is closer to x_{02}(T) than x_{01}(T).

Therefore, the decision boundary will be in the middle of x_{01}(T) and x_{02}(T).

It is expressed as,

Decision boundary = …(6.65)

**6.22.1. Calculation of Probability of Error (P _{e}) for Optimum Filter**

The probability of error (P

_{e}) may be evaluated on the same basis as in the last section. Here, we shall consider generalized Gaussian noise. Let us consider that x

_{2}(t) was transmitted, but x

_{01}(T) is greater than x

_{02}(T). If noise n

_{0}(T) is positive and larger in magnitude compared to the voltage difference – x

_{02}(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 n

_{0}(t) in the previous section. It is given as

**EQUATION**…(6.67)

Here, n

_{0}(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

**EQUATION**

**diagram**

**FIGURE 6.30**Evaluation of P

_{e}for optimum filter receiver.

Using equation (6.67), the last equation becomes,

**EQUATION**…(6.68)

Let us substitute

**EQUATION**

so that N

_{0}(t) = Y

or d[n

_{0}(t)] = Y

when n

_{0}(t) = , then y =

and when n

_{0}(t) =

then, we have y =

Substituting all these values in equation (6.30), we obtain,

**EQUATION**

Let us rearrange this equation as under:

**EQUATION**…(6.69)

To solve this integration, let us use the following standard result.

**EQUATION**

Then, equation (6.69) becomes

**EQUATION**…(6.70)

**NOTE**: This is the required expression for error probability P

_{e }of optimum filter. It may be noted that the aerie’ function is the monotonically decreasing function. Hence, P

_{e}decreases as the difference x

_{01},(T) – x

_{02}(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 P

_{e}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 x

_{01}(T) – x

_{02}(T) by x

_{0}(T).

Thus, we have

x

_{0}(T) = x

_{01}(T) – x

_{02}(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[n

_{0}

^{2}(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 X

_{0}(f) be Furrier transform of x

_{0}(t). If X(f) is the Furrier transform of input difference signal x(t) [i.e., x(t) = x

_{1}(t) – x

_{2}(t)], then we have,

X

_{0}(f) = H(f) X(f) …(6.73)

Here,

*H(f)*is the transfer function of optimum filter. Also, x

_{0}(T) is found by taking inverse Fourier transform of X

_{0}(f) i.e.,

x

_{0}(T) =

*IFT*[X

_{0}(f)] =

**EQUATION**

* – (mean value)

^{2}.

We know that

**Equation**

Also, the input noise to the optimum filter is n(t). Let, its power spectral density (psd) be S

_{ni}(f). The output noise of optimum filter is n

_{0}(t). Let its power spectral density be S

_{n0}(f).

We know that the input and output power spectral densities of noise are related as,

S

_{n0}(f) = S

_{ni}(f) …(6.75)

The normalized noise power can be obtained by integrating the power spectral density.

Thus, we have

**EQUATION**…(6.76)

Substituting for x

_{0}(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,

**EQUATION**

Applying Schwarz’s inequality of equation (6.77) to the numerator of above equation, we have

**EQUATION**

**EQUATION**

Now, substituting value of

*(f)*in the above equation, we have

**EQUATION**

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

**EQUATION**

This is possible only when equality applies in Schwarz ‘s inequality i.e.. in equation (6.77),

**EQUATION**

This equality is possible only if we have = k

^{*}

_{2}

Substituting the values of and , we obtain,

**EQUATION**

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:

**EQUATION**

It may be noted that at the beginning of this derivation, we stated that

**EQUATION**

Now we can have the following conclusion:

The optimum filter which minimizes the probability of error (P

_{e}) 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,

**EQUATION**

* By property of pad.

**EQUATION**