# MATCHED FILTER in digital communication | matched filter is used for coherent detection

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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,
Sni(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 Sni(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) = x1(t) – x2(t), therefore, equation (6.49) will become,
h(t) =  [x1(T-t) -x2(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 x1(t) represent the positive rectangular pulse with duration T as shown in figure 6.31(a).
DIAGRAM
FIGURE 6.31.
Let x2(t) represent a negative rectangular pulse with duration T as shown in figure 6.31(b)
Thus, we have
x1(t) = + A                   for       0 < t < T
and                                          x2(t) = + A                   for       0 < t < T                             …(i)
Let us calculate the difference signal x(t) i.e.,
x(t) = x1(t) — x2(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 (Pe) for the Matched Filter
To 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 Sni(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) = x1(t) – x2(t).
Hence, above equation becomes,
EQUATION
or                                               EQUATION
EQUATION                                      …(6.91)
where,  dt = E1 i.e., energy of x1 (t) by standard relations.
and                   dt = E2 i.e., energy of x2 (t) by standard relations.
and                   = E12 represents energy due to autocorrelation between x1(t) and x2(t).
Now, if we choose x1(t)  = -x2(t), then these energies will be equal, i.e.,
E1 = E2 = -E12 = 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:
Pe =  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 Pe. This means that the above equation really provides minimum probability of error of matched filter.
Thus, Minimum error probability of matched filter:
Pe =  erfc
…(6.96)
6.23.3. Few Points about Error Probability of Error, Pe of Matched Filter
From 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 x1(t) = x2(t). Also, in previous sections we have derived error probability for Intergrate and Dump filter where we have taken x1(t) = + A and x2(t) = – A.
In other words, x1(t) = -x2(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 x1(t) = – x2(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 Sni(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 x0(t) is expresse as,
X0(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 x0(t) by taking inverse Fourier transform.
Thus, we have
EQUATION

or                                                          EQUATION
At t = T, x0(t) will become,
EQUATION
or                                                          EQUATION
By Rayleigh’s energy theorem, we can write
EQUATION
Hence, equation (iv) becomes
x0(T) =
Also, maximum value of x0(T) will result when   = 1
Thus,                           x0(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 x0(t) may be obtained from its spectrum X0(f) by taking inverse Fourier transform.
This means that
EQUATION
Substituting X0(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
x0(t) =  R (t-T)
This proves that x0(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 ofA2T at = T.
DIAGRAM
FIGURE 6.34.