พระยาศรีสุนทรโวหาร น้อย อาจารยางกูร , Phraya Si Sunthon Wohan .

today is google doodle showing a doodle on it’s search engine bar to remember Phraya Si Sunthon Wohan on his 198th birthday , he was a very famous author, poet, and teacher . he was the master of his field and achieved many peaks where no one other can achieve.

wish him a very happy birthday on his birthday today.

**EXAMPLE 6.17. If the rectangular pulse is applied to a RC filter shown in figure 6.43 instead of a matched filter, evaluate the maximum signal to noise ratio ****which can be obtained by RC filter and compare it with that obtained by corresponding matched filter. **

*(U.P. Tech., Sem. Exam., 2003-04) *

**DIAGRAM**

**FIGURE 6.35 **An RC filter to receive rectan-gular pulse.

**Solution:** When rectangular pulse is applied as input to the RC filter, its output x_{0}(t) will be as shown below in figure 6.36(b).

The output x_{0}(t) of the RC filter can be expressed in two parts as under:

x_{0}(t) = A(1-e^{-t/RC}) for 0 __<__ t ≤ T

A(1-e^{-t/RC})_{e} ^{-(t-T)/RC} for t > T …(i)

** **

Here, the first part represents charging of the capacitor when pulse is applied for 0 ≤ t ≤ T. The second part represents a discharging of the capacitor when the pulse x(t) is removed after t > T. The maximum value of x_{0}(t) occurs at t = T as shown in figure 6.44(b). Therefore, substituting t = T in first part of equation (i), we get

x (T) = A (1-e-T/RC) …(ii)

**DIAGRAM**

**FIGURE 6.36 **(a) Rectangular pulse input to the RC filter (b) Output of the RC filter in response to the rectangular input pulse.

Now, let us calculate the transfer function of the RC filter. Figure 6.37 shows the values of different parameters of RC filter in frequency domain.

From figure 6.37, we can write H(f) as,

**EQUATION **

Let us multiply numerator and denominator of above equation by 1-j2fRC, i.e.,

**EQUATION**

**EQUATION **

**DIAGRAM**

**FIGURE 6.37 **Frequency domain representation of RC network

Magnitude of H(f) will be,

**EQUATION**

**EQUATION **…(iii)

We know that psds of input and output noise are related as,

S_{no}(f) = S_{ni}(f)

The psd of white Gaussian noise is, S_{ni}(f) = ; hence, above equation becomes,

S_{no}(f) =

The normalized noise power is obtained by integrating its power spectral density (psd), i.e.,

**EQUATION** …(iv)

Substituting from equation (iii) in above equation,

**equation **

* By property of psd.

Since *`f ‘* is squared in above equation, therefore, we can change integration limits from (-,) to (0, ) i.e.,

**equation**

Let 2*f*RC = y

so that 2RC *df* = dy

or dr =

Therefore, equation (v) becomes,

**equation **

Here, let us use the following standard result of integration:

**equation **

Now, we can rearrange equation (vi) as,

**equation **

Comparing this equation with equation (vii), we get

*m* = 1 and *n* = 2

Thus, **equation**

The signal to noise power ratio is given as

=

Substituting x_{0}^{2}*(t)* from equation (ii) and from equation (viii), we get

**equation**

Again, let v= , then above equation becomes,

**EQUATION**

For maximum value of should be equal to zero, i.e.,

**equation**

⸫ 2*v* e^{-v} (1-e^{-v}) = (1-e^{-v})^{2}

or 2v e* ^{-v}* = 1 – e

^{-v}

or 2v e^{-v }+ e^{-v} = 1

or e^{-v} (1 + 2v) = 1

or e^{v} = 1 + 2v

Solving, we get v = 1.26

Since v =, we can write,

^{ }

Maximum value of will be obtained substituting v = = 1.26 in equation (ix)

Thus, **equation**

In a last example we have obtained for a rectangular pulse input to the matched filter.

Thus, we have

(matched filter) =

With this result we can write equation (x) as under:

**EQUATION**

Thus, because of RC filter, the maximum signal to noise ratio obtained is reduced to 0.814 of that of matched filter. **Hence Proved **

**EXAMPLE 6.18. Find the impulse response of the matched filter for a Gaussian signal pulse given by, **

**EQUATION**

** The noise on the channel is a white noise with power density spectrum of **** **

** Calculate the maximum signal to noise ratio **** achieved by this filter. **

**Solution:** The given gaussian pulse is,

**EQUATION**

We know that the impulse response of the matched filter is given as,

h(t) = x (T –*t*)

Let = 1, then above equation will be,

h(t) == x (T-t = *…(ii)*

This equation gives the required impulse response of the matched filter for Gaussian signal pulse.

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

**equation**

Here, *E* is the energy of the signal x(t). It can be calculated as,

**equation**

Substituting value of x(t) from equation (i), we get

**equation**

Now, substituting

= y^{2}

so that = *y*

or dt = dy

however, the integration limits will remain same.

Now, substituting these values in equation (iv), we get

**equation**

Let us use the following standard result of integration:

**EQUATION**

Here, putting b = 0 and a = 1, we get

**equation**

Using the result of above equation, we can obtain equation (v) as,

**equation**

Substituting this value of *E* in equation *(iii)*, the maximum signal to noise ratio is given as,

**equation**

**EXAMPLE 6.19. Two messages are transmitted by mark and space using a simple binary pulse shown in figure 6.38. Find error probability of the optimum receiver assuming that the probability of x(t) being present is 0.5. Assume that the channel noise is a white noise of power spectral density **** , where N = 10 ^{-4}. **

**DIAGRAM**

**FIGURE 6.38 **Transmitted signal x(t).

**Solution:** The pulse of figure 6.38 can be represented mathematically as,

x(t) = for 0 < t ≤ T/2

2A – for T/2 < t ≤ T …(i)

The energy of signal x(t) can be calculated as under:

*E *= (t) dt

Substituting value of x(t), we get

**EQUATION**

or** EQUATION**

or** EQUATION**

or** EQUATION**

or** EQUATION**

or E = …(ii)

Since the channel noise is white noise, the optimum filter is same as matched filter. Hence, we can use the relations derived for matched filter. The probability of error of matched filter is given as,

P_{e} = …(iii)

Here, it may be noted that this probability of error, P_{e} is obtained for the transmission and reception of two symbols x_{1} (t) and x_{2}(t). Even though not stated earlier, it is assumed that probability of occurrence of x_{1} (t) and x_{2}(t) is same. Since there are only two symbols, probability of occurrence of x_{1} (t) and x_{2},(t) both will be 0.5. Hence, this example probability of x(t) [i.e., one of the symbols] is given 0.5. Thus, the probability of other symbol which is represented by space will also be 0.5. Thus, the error probability given by equation (iii) above can be used to calculate P_{e}, in this example.

Here N_{0} = *N*= 10^{-4} given

and

since T = 10^{3} given.

Substituting these values in equation (iii) we get

P_{e} =

or P_{e} =

This is the required probability of error. **Ans. **

**EXAMPLE 6.20. Figure 6.39 shows signal x(t) **

** (a) Determine the impulse response of the filter matched to this signal and sketch it. **

** (b) Plot the output of the matched filter. **

**Solution:** The impulse response of the matched filter is given x(t)

as

h(t) = x(T – t) assuming = 1

The impulse response is sketched in figure 6.40(a), it can be described as,

**EQUATION**

**diagram **

**FIGURE 6.39** *Input signal.*

The output of the matched filter is obtained by convolution of x(t) and h(t), i.e.,

y(t) = x(t) Ä h(t)

Figure 6.47(b) shows the output signal. The waveform has maximum value at t = T, it is .

**DIAGRAM**

**FIGURE 6.40** *(a) The signal pulse x(t), (b) Impulse response of the signal of (a), (c) Output waveform of the matched filter *

**EXAMPLE 6.21. A received binary NRZ signal assumes the voltages levels of 500 millivolts and 500 millivolts respectively for ‘1’ and ‘0’ transmission with a bit rate or r bits/second. The signal is corrupted by additive white Gaussian noise with a two-sided spectral density of 10^{-6} volts 2/Hz. The received signal is processed by an Integrate and Dump circuit in every bit interval and compared with a zero threshold to take a bit decision. **

** Assuming ‘1’ and ‘0’ transmission to be equally likely, the maximum value of r such that the bit error probability < 10^{-5}. **

**Given: **** at x = 4.27**

**EQUATION **

**Solution:**

**FIGURE 6.41.**

The transmitted signal is

**EQUATION**

then, energy of the difference signal at the filter input

**EQUATION**

Bit error probability,

**EQUATION
**

Give P_{s} ≤ 10^{-5}

then, **equation**

or = 4.27

Now as = 10^{-6} volts/Hz

⸫ **EQUATION**

**EQUATION**

We have, Pe_{0} =

Now, the probability of error, on condition that sending symbol is 0,

**equation**

Setting l = 0, and putting

= – z

we have, **EQUATION**

Now, the average probability of symbol error P_{e} in the receiver is given by,

P_{e} = P_{0} P_{e0} + P_{1}, P_{e1}

where P_{e} and P_{1} are the prior probabilities of binary symbols 0 and 1, respectively,

Since P_{eo} = P_{e}, and assuring the symbols 0 and 1 occur with equal probability i.e.,

p_{0} = p_{1} =

then, p_{e} =

Hence, average probability of symbol error in a binary coded PCM receiver depends solely on E_{b}/ N_{0}, the ratio of transmitted signal energy per bit to noise spectral density.

**EXAMPLE 6.22. Prove that signal to noise ratio **

** where E is the energy of the input signal s(t) and **

**h**

**/2 is the power spectral density of the input noise**

**h**

**(t).**

**Solution:** Suppose the frequency response of the linear filter is H().

Then the matched filter output,

**EQUATION**

where s() =

Output noise power,

**EQUATION **

**EQUATION** …(i)

**EQUATION**

Now, using Schwarz inequality, we have

**EQUATION** …(ii)

Using equations (ii) and (i), we get

**EQUATION**

where E = d, energy of input signal.

**EXAMPLE 6.23. Compute the matched filter output over (0, 7) to the pulse waveform**

**EQUATION**

**Solution:** For the given s(t), the impulse response of the matched filter is,

h(t) = s(T – t).

or h(t) = e^{-(T-t)}

Now, output z(t) = s(t) Ä h(t)

or **EQUATION**

or **EQUATION**