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**MATCHED FILTER (U.P. Tech-Semester Exam. 2002-03, 2004-05)**

In previous section, we have discussed optimum filter. For this filter, we have considered generalized Gaussian noise. When this noise is white Gaussian noise, then the optimum filter is known as

**matched filter.**Recall that for the white Gaussian noise, the power spectral density is given as,

S

_{ni}(f) = …(6.81)

**6.23.1. Calculation of Impulse Response for the Matched Filter**

**The transfer function of the optimum filter is expressed as,**

**EQUATION**…(6.82)

In this equation, if we substitute S

_{ni}(f) = i.e., psd of white noise, then H(f) becomes transfer function of matched filter.

This means that

Transfer function of matched filter,

**EQUATION**

or X

^{*}(f) …(6.83)

From the property of Fourier transform, we know that

X

^{*}(f) = X(-f) …(6.84)

Using this property, we can write equation (6.83) as under:

The impulse response of a matched filter can be evaluated by taking inverse Fourier transform of above equation.

Thus, we have

h(t) = IFT [H(f)] = IFT …(6.85)

The inverse Fourier transform of X(-f) is x(-t) and represents time shift of

*T*seconds.

Hence, we have

FT[x(-t)] = X(-f)

and FT [x(T-t)] = X(-f)

With the help of all these properties of Fourier transform, the equation (6.86) takes the form

h(t) = …(6.86)

Note that we have considered x(t) = x

_{1}(t) – x

_{2}(t), therefore, equation (6.49) will become,

h(t) = [x

_{1}(T-t) -x

_{2}(T-t)] …(6.87)

Thus, two equation (6.86) and (6.87) give the required impulse response of the matched filter.

**EXAMPLE 6.12. A polar NRZ waveform has to be received with the help of a matched filter. Here, binary 1 is represented by a rectangular positive pulse. Also, binary zero is represented by a rectangular negative pulse. Determine the impulse response of the matched filter. Also sketch it.**

**Solution**: Let x

_{1}(t) represent the positive rectangular pulse with duration T as shown in figure 6.31(a).

**DIAGRAM**

**FIGURE 6.31.**

Let x

_{2}(t) represent a negative rectangular pulse with duration

*T*as shown in figure 6.31(b)

Thus, we have

x

_{1}(t) = + A for 0

__<__t

__<__T

and x

_{2}(t) = + A for 0

__<__t

__<__T …(i)

Let us calculate the difference signal x(t) i.e.,

x(t) = x

_{1}(t) — x

_{2}(t) for 0

__<__t

__<__T

or x(t) = A – (-A)

or x(t) = 2A for 0

__<__t

__<__T …(ii)

This difference signal has been shown in figure 6.31(c). The time reversed version of x(t) will be,

x(-t) = 2A for -T

__<__t

__<__0

*…(iii)*

It may be noted that that in the above equation, the time reference is inverted. This signal has been shown in figure 6.31(d).

Now let us delay x(- t) by T seconds.

i.e., x(T – t) = 2A for 0

__<__t

__<__T …(iv)

The signal has been shown in figure 6.31(e)

Impulse response of a matched filter is expressed as,

h(t) =

Substituting x(T –

*t*) from equation (iv), we get

h(t).2A for 0

__<__t

__<__T

or h(t).2A for 0

__<__t

__<__T

Figure 6.32 shows the sketch of this impulse response.

It may be noted that with = 1, this figure will be same as figure 6.31(e) and figure 6.31(c). This proves that the shape of the impulse response of the matched filter is similar (or matched) to the shape of the input signal x(t).

**Hence, it is known as matched filter.**

DIAGRAM

**FIGURE 6.32**

*Impulse response of a matched filter for a rectangular pulse input.*

**6.23.2. Calculation of Probability of Error (P**

_{e}) for the Matched FilterTo evaluate the probability of error for matched filter, let us again start with optimum filter and we shall consider the special case of white Gaussian noise.

We know that error probability of optimum filter is expressed as,

**EQUATION**…(6.88)

In this equating, we have

**EQUATION**

In this equation, let us substitute S

_{ni}(f) for white Gaussain noise.

Thus, we have

psd of white noise = Sni(f) =

Hence,

**EQUATION**…(6.89)

Also, Parseval’s power theorem states that,

**EQUATION**…(6.90)

In the last integral we have taken limits from 0 to T because x(t) exists from 0 to T only. We know that x(t) = x

_{1}(t) – x

_{2}(t).

Hence, above equation becomes,

**EQUATION**

or

**EQUATION**

**EQUATION**…(6.91)

where, dt = E

_{1}i.e., energy of x

_{1}(t) by standard relations.

and dt = E

_{2}i.e., energy of x

_{2}(t) by standard relations.

and = E

_{12}represents energy due to autocorrelation between x

_{1}(t) and x

_{2}(t).

Now, if we choose x1(t) = -x2(t), then these energies will be equal, i.e.,

E

_{1}= E

_{2}= -E

_{12}= E …(6.92)

Substituting all these values in equation (6.92), we get

**EQUATION**…(6.93)

Substituting this value of

**EQUATION**df in equation (6.89), we get

**EQUATION**…(6.94)

Therefore,

**EQUATION**

Substituting this value of in equation (6.88), we get probability of error of a. matched filter as under:

P

_{e}= erfc …(6.95)

**NOTE**Here, erfc is the monotonically decreasing function. Therefore, when we substitute as the argument of erfc, we obtain minimum value of probability of error P

_{e}. This means that the above equation really provides minimum probability of error of matched filter.

Thus,

**Minimum error probability of matched filter:**

P

_{e}= erfc

…(6.96)

**6.23.3. Few Points about Error Probability of Error, P**

_{e}of Matched FilterFrom the equation (6.96), we can draw the following important conclusions:

(i) It is obvious from equation (6.96) that the error probability depends only upon the signal energy

*E*. It does not depend on the shape of the signal.

(ii) We have obtained equation (6.96) by considering that x

_{1}(t) = x

_{2}(t). Also, in previous sections we have derived error probability for Intergrate and Dump filter where we have taken x

_{1}(t) = + A and x

_{2}(t) = – A.

In other words, x

_{1}(t) = -x

_{2}(t) = A and we obtained error probability of integrator as,

Pe = erfc …(6.97)

This result is same as error probability of matched filter given be equation (6.97).

Hence, Integrator is equivalent to matched filter when x

_{1}(t) = – x

_{2}(t) = A or in other words, we can say that for a rectangular bipolar pulse input, the integrate and dump filter is same as a matched filter.

**EXAMPLE 6.13. Prove that the maximum signal to noise ratio for the matched filter is found to be,**

**Solution:**To derive this equation, let us consider the signal to noise power ratio of optimum filter which is given as,

**EQUATION**

We know that when white noise is present, then optimum filter is known as matched filter.

psd of white noise is S

_{ni}(f) = . Substituting this value in equation (i), we get

**EQUATION**

Also, Rayleigh’s energy theorem states that,

**EQUATION**

With this result, equation (ii) becomes

Hence, maximum signal to noise power ratio of matched filter will be

Now, this equation can be rearranged as under:

Here,

*E*is energy of signal x(t) and is power spectral density (psd) of white noise.

Thus, we have

**Hence Proved**

**EXAMPLE 6.14. Prove that for a matched filter, the maximum signal component occurs at t = T (i.e., sampling instant) and has magnitude equal to E, i.e., energy of the signal x(t).**

**Solution:**We know that the Fourier transform of the output signal x

_{0}(t) is expresse as,

X

_{0}(f) = X(f) H(f) …(i)

where X(f) = Fourier transform of input signal x(t)

and H(f) = is transfer function of matched filter

Substituting value of H(f) from equation (6.83), we get

**EQUATION**…(ii)

Now, since X(f) X*(f) = , therefore, equation (ii) becomes,

**EQUATION**…(iii)

From this expression, we can obtain x

_{0}(t) by taking inverse Fourier transform.

Thus, we have

**EQUATION**

or

**EQUATION**

At t = T, x

_{0}(t) will become,

**EQUATION**

or

**EQUATION**

By Rayleigh’s energy theorem, we can write

**EQUATION**

Hence, equation (iv) becomes

x

_{0}(T) =

Also, maximum value of x

_{0}(T) will result when = 1

Thus, x

_{0}(T) = E when t = T

**Hence Proved**

**EXAMPLE 6.15. Prove that the output signal of a matched filter is proportional to a shifted version of the autocorrelation function of the input signal to which the filter is matched.**

**Solution:**For a matched filter, the output signal x

_{0}(t) may be obtained from its spectrum X

_{0}(f) by taking inverse Fourier transform.

This means that

**EQUATION**

Substituting X

_{0}(f) from equation (iii) [last example], we get

**EQUATION**

or

**EQUATION**

Recall that the energy spectral density (esd) is given as

Y(f) =

* By definition of IFT.

** According to the Standard definition of esd.

Hence, equation (i) becomes,

**EQUATION**

We know that one of the property of esd states that the autocorrelation function R() and energy spectral density Y(f) form a Fourier transform pair.

This means that

R() Y (f) …(iii)

Thus, we have

**EQUATION**

and

**EQUATION**

Applying equation (iv) to RHS of equation (ii), we get

x

_{0}(t) = R (t-T)

This proves that x

_{0}(t) is proportional to autocorrelation function of x(t) shifted to right by

*T*.

Here, is a constant term and it depends upon psd of noise.

**Hence Proved**

**EXAMPLE 6.16. In a binary transmission, one of the messages is represented by a rectangular pulse x(t). An other message is transmitted by the absence of the pulse. Evaluate the signal to noise ratio at t = T. Assuming white noise with psd equal to**

**. Also sketch the impulse response of the matched filter and output of the matched filter.**

**Solution:**Evaluation of Impulse response:

The rectangular pulse can be represented as

**EQUATION**…(i)

Further, we know that impulse response of the matched filter is given by,

h(t) = x (T-t) …(ii)

From equation (i), we can write,

**EQUATION**…(iii)

Let us delay this signal by

*T*to get

**EQUATION**…(iv)

Substituting this value in equation (ii), we obtain

**EQUATION**…(v)

This equation gives impulse response of the matched filter. Figure 6.33 illustrates the sketch of this response.

**DIAGRAM**

**FIGURE 6.33.**

**Calculation 0f**

We know that the maximum signal to noise ratio of the matched filter is given by,

= …(vi)

Here, is given as psd of white noise.

Further, we can evaluate energy of signal x(t) by,

**EQUATION**

Substituting x(t) = A and changing he limits from 0 to T as described in equation (i), the last equation becomes,

**EQUATION**

Substituting this value in equation (vi), we obtain,

which is the required value of

**Output of the Matched Filter:**

We know that the matched filter is an LTI system. Thus, the output of the matched filter will be the convolution of input and impulse response.

Thus, we have y(t) = x(t) Ä h(t)

where y(t) = output of matched filter,

h(t) = impulse response of matched filter,

and x(t) = input signal.

Without going into mathematical details, we can state that the convolution of two similar rectangular pulses is a triangular pulse. Here, x(t) and h(t) are two rectangular pulses. Therefore, y(t) will be a triangular pulse. This has been shown in figure 6.34.

As figure 6.34 shows the matched filter maximizes its output at t = T. Hence, y(t) has a value ofA

^{2}T at = T.

**DIAGRAM**

**FIGURE 6.34.**