Earlier, this chapter, we have discussed various digital modulation techniques. When a signal is transmitted over a baseband or bandpass* channel, the noise interferes it. As a result of this, the baseband signal loses its shape (which was digital) and it becomes quite difficult to detect a particular value of digit. In this chapter, we have discussed some techniques for detection of baseband signal. These techniques are Integrator, optimum filter, matched filter and correlator. All these techniques maximize the signal to noise ratio (S/N ratio) of the received signal. Generally, the signal to noise ratio is maximum at the end of symbol period i.e., T. The output is then sampled at t = T and a decision is taken. The probability of error Pe, in the output is minimum when signal to noise ratio is maximum. In bandpass transmission, the noise interference depends upon the type of modulation technique used. Thus, the probability of error Pe, in the output also depends upon the particular type of digital modulation technique.
In this article, we shall discuss this effect of noise on detection of signals for various binary modulation techniques.
8.18.1. Probability of Error for PCM System
            Let us consider the binary pulse code modulated signal as shown in figure (8.41). This is a baseband signal. Let us evaluate its probability of error Pe, at the receiver in presence of white gaussian noise. Let this signal be received by matched filter receiver. The signal shown in this figure is clean, but actually it’s shape will be distorted due to presence of noise. As shown in this figure, let us consider that binary ‘1’ is represented by a pulse a amplitude A and duration T, and call this signal x1(t).
FIGURE 8.41 Illustration of a Binary coded PCM signal.
*          In bandpass transmission, the baseband signal modulates some carrier to give PSK, FSK, QPSK etc. At the receiver end, the baseband is separated from the carrier. This baseband signal is then detected using filter, correlator etc techniques to get binary seqence
Thus, we have
For binary ‘1’               x1(t) = A          for       0 £ t £ T
and for binary ‘0’,                    x2(t) = 0           for       0 £ t £ T                      …(8.103)
Note that binary ‘0’ is represented by zero amplitude i.e., no pulse. This is represented by x2(t) as in equation (8.103).
We know that error probability, Pe, of the optimum filter is expressed as
equation                                      …(8.104)
Here                                         equation
These equations are applied to matched filter when we consider white gaussian noise. This means that in the above equation, let us substitute Sni(f) for white gaussian noise which is Sni(f) =  (psd of white noise),
In this case, above equation becomes
or                                               equation                                                …(8.105)
Recall the Parseval’s power theorem which states that
Therefore, equation (8.105) becomes
In this equation x(t) = x1(t) — x2(t). But it is clear from equation (8.103) that x2(t) = 0 always. Thus, for this Binary PCM,
we have x(t) = x1(t).
Therefore, we can write above equation (8.106) as under :
x1(t) = A for 0 £ t £ T from equation (8.103). Thus, we can change limits to 0 to T in last equation i.e.,
Therefore, we have
Substituting this value of   in equation (8.104), we get,
Simplifying, we get
Pe =                                          …(8.108)
This is the required expression for probability of error of matched filter diction of Binary PCM signal. The normalized energy of one bit of x1(t) will be,
Since x1(t) = A from 0 to T as shown in figure 8.41. With the result of above equation, we can write equation 8.108.
Pe =                                          …(8.109)
Which is an alternate expression for probability of error of binary PCM signal.
In sixth chapter, we have discussed Amplitude Shift Keying (ASK). Some number of carrier cycles are transmitted to send ‘1’ and no signal is transmitted for binary ‘0’. Thus, we have
For Binary ‘1’                          x1(t) =  cos (2πfct)
For Binary ‘0’                          x2(t) = 0 (i.e., no signal)                      …(8.110)
Here Ps is the normalized power of the signal in 1Ω load. i.e., power Ps, = . Hence, A =  . Therefore, in above equation for x1(t) amplitude ‘A’ is replaced by .
We know that the probability of error of the optimum filter is given as,
Pe =                                    …(8.111)
Here                                        equation
These equations can be applied to matched filter when we consider white gaussian noise. The power spectral density (psd) of white gaussian noise is given as,
Sni(f) =
Substituting this value of Sni(f) in above equations, we get,
EQUATION                                      …(8.112)
Recall Parseval’s power theorem which states that,
Hence, equation (8.112) becomes,
We know that x(t) is present from 0 to T. Hence, limits in above equation can be changed as under :
Further x(t) = x1(t) — x2(t).
For ASK, x2(t) is zero, hence x(t) = x1(t).
Therefore, last equation becomes,
Substituting expression for x1(t) from equation (8.110) in above equation, we get,
We know that cos2 q = . Here applying this formula to cos2 (2πfct), we get,
We know that T is the bit period and in this one bit period, the carrier has integer number of cycles. Thus the product f0T is an integer. This has been illustrated in figure 8.42. From figure, we can write,
T =
or                                             fc T = 2
FIGURE 8.42 In one bit period T, the carrier completes its two cycles. The carrier has frequency fc.
As shown in above figure, the carrier completes two cycles in one bit duration. Hence,
fc T = 2
Therefore, in general if carrier completes ‘k‘ number of cycles, then,
fc T = k (Here k is an integer)
Therefore the sinc term in equation (8.114) becomes, sin 4pk and k is integer.
For all integer values of k, sin 4pk = 0. Hence, equation (8.114) becomes,
            ⸫                                                 equation
Putting this value in equation (8.111) we get error probability of ASK matched filter detection as,
Here Ps T = E, i.e., energy of one bit hence above expression will become
Error probability of ASK : Pe, =                                           …(8.117)
NOTE : Here, the complementary error function ‘erfc (x)’ is a monotonically decreasing function of x. This means that as x increases, the value of erfcx decreases. Therefore, the probability of error, Pe, decreases with increase in the ratio  . Hence, the probability of error depends only upon the signal energy and not its shape or any other parameter.
But, we know that
average energy per bit = Eb =
Therefore, the modified expression for error probability for ASK system is given by
Pe =  erfc
Also, an elaternate expression for ASK system is given by
Pe = Q
This is the expression for error probability of ASK using matched filter detection.*
EXAMPLE 8.1. An ON/OFF system uses pulse waveforms as described below
Additive white gaussian noise (AWGN) with a power spectral density (psd) of 10-15 W/Hz is added to the signal. Evaluate the probability of bit error when P(s1) = P(s2) = 1/2. Take A = 0.2µV and T= 2µs.
Solution. Given that A = 0.2 mV, T = 2 µs; P(s1) = P(s2) = 1/2
We have
Eb =T.P (s1) + (0)2. T.P.(s2) =  x                 (i)
For ASK, we know that probability of error of given by
Pe = Q
In above expression, substituting value of Eb from equation (i), we get
Pe = Q
Substituting all the values, we obtain
Solving we get
= Q 8 x 10-4                                   Ans.
Example 8.2. A binary ASK system for equally probable massages uses 100μ sec bits and channel has N0 = 1.338 x 10-5 W/Hz. Determine the peak transmitted pulse amplitude to maintain Pe ≤ 2.055 x 10-5.
Solution: We know that for ASK, the probability of error is given by
Pe =
*          This is also known as the error probability for a coherent ASK system.
or                                               ≥ 2.9     or    ≥ 8.46
or                                             Eb ≥ 8.46 x 2 x 1.338 x 10-5
But                                          Eb =
This means that  ≥ 8.46 x 2 x 1.338 x 10-5
Also,                            T =  = 100μ sec
Hence, we have           A2
Solving, we get           A = 4.53 volts.                        Ans.
EXAMPLE 8.3. In a binary PCM system using ON-OFF signalling, 1 is represented by the pulse
            and elsewhere and symbol zero is represented by switching OFF the pulse. The receiver uses a matched filter. Assuming that noise is Gaussian and power spectral density (psd) = N0/2. Calculate average probability of error when symbols 1 and 0 occur with equal probability.
Solution: Let us assume that during some bit interval, the input signal is held at + 1V, then, at the sample time, the output signal s0(T) =
Then, the noise component is n0(f). If N0(T) is negative and larger than , it will result in an error in the decoding process. The probability of such a misinterpretation is the probability that n0(t) < –   is given by
FIGURE 8.43.
The probability density of noise n0(t) is given by

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