CALCULATION OF ERROR PROBABILITY IN ASK , Probability of Error for PCM System ?

**DETECTION OF BINARY MODULATION TECHNIQUES IN PRESENCE OF NOISE **

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 P* _{e}*, 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 P

_{e}, 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 P**

*, 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 x*

_{e}_{1}(t).

**diagram**

**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’ x

_{1}(t) = A for 0 £ t £ T

and for binary ‘0’, x

_{2}(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 x

_{2}(t) as in equation (8.103).

We know that error probability, P

*, of the optimum filter is expressed as*

_{e}**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 S

_{ni}(f) for white gaussian noise which is S

_{ni}(f) = (psd of white noise),

In this case, above equation becomes

**equation**

or

**equation**…(8.105)

Recall the Parseval’s power theorem which states that

Therefore, equation (8.105) becomes

…(8.106)

In this equation x(t) = x

_{1}(t) — x

_{2}(t). But it is clear from equation (8.103) that x

_{2}(t) = 0 always. Thus, for this Binary PCM,

we have x(t) = x

_{1}(t).

Therefore, we can write above equation (8.106) as under :

x

_{1}(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

…(8.107)

Substituting this value of in equation (8.104), we get,

**equation**

Simplifying, we get

**P**

_{e}=**…(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 x

_{1}(t) will be,

Since x

_{1}(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.

**8.19 CALCULATION OF ERROR PROBABILITY IN ASK**

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’ x

_{1}(t) = cos (2πf

*t)*

_{c}For Binary ‘0’ x

_{2}(t) = 0 (i.e., no signal) …(8.110)

Here P

*is the normalized power of the signal in 1Ω load. i.e., power P*

_{s}*, = . Hence, A = . Therefore, in above equation for x*

_{s}_{1}(t) amplitude ‘A’ is replaced by .

We know that the probability of error of the optimum filter is given as,

P

*= …(8.111)*

_{e}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,

S

_{ni}(f) =

Substituting this value of S

_{ni}(f) in above equations, we get,

**EQUATION**

**EQUATION**…(8.112)

Recall Parseval’s power theorem which states that,

Hence, equation (8.112) becomes,

…(8.113)

We know that x(t) is present from 0 to T. Hence, limits in above equation can be changed as under :

Further x(t) = x

_{1}(t) — x

_{2}(t).

For ASK, x

_{2}(t) is zero, hence x(t) = x

_{1}(t).

Therefore, last equation becomes,

**equation**

Substituting expression for x

_{1}(t) from equation (8.110) in above equation, we get,

**equation**

We know that cos

^{2}q = . Here applying this formula to cos

^{2}(2πf

*t), we get,*

_{c}**equation**

**equation**

…(8.114)

We know that

*T*is the bit period and in this one bit period, the carrier has integer number of cycles. Thus the product f

*is an integer. This has been illustrated in figure 8.42. From figure, we can write,*

_{0}TT =

or f

_{c}T = 2

**diagram**

**FIGURE 8.42**

*In one bit period T, the carrier completes its two cycles. The carrier has frequency f*

_{c}.As shown in above figure, the carrier completes two cycles in one bit duration. Hence,

f

_{c}

*T*= 2

Therefore, in general if carrier completes ‘

*k*‘ number of cycles, then,

f

_{c}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**

**⸫ equation**

Putting this value in equation (8.111) we get error probability of ASK matched filter detection as,

**equation**

Here P

_{s}T = E, i.e., energy of one bit hence above expression will become

**Error probability of ASK : P**

_{e}, =**…(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, P

*, 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.*

_{e}But, we know that

average energy per bit = E

_{b}=

Therefore, the modified expression for error probability for ASK system is given by

P

_{e}= erfc

Also, an elaternate expression for ASK system is given by

P

*= Q*

_{e}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**

**equation**

**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(s_{1}) = P(s_{2}) = 1/2. Take A = 0.2µV and T= 2µs.**Solution.**Given that A = 0.2 mV, T = 2 µs; P(s

_{1}) = P(s

_{2}) = 1/2

We have

E

_{b}=T.P (s

_{1}) + (0)

^{2}. T.P.(s2) = x (i)

For ASK, we know that probability of error of given by

P

*= Q*

_{e}In above expression, substituting value of E

*from equation (i), we get*

_{b}P

*= Q*

_{e}Substituting all the values, we obtain

**Equation**

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 N**

_{0}= 1.338 x 10^{-5}W/Hz. Determine the peak transmitted pulse amplitude to maintain P*≤ 2.055 x 10*_{e}^{-5}.**Solution:**We know that for ASK, the probability of error is given by

P

*=*

_{e}or

* This is also known as the error probability for a coherent ASK system.

or ≥ 2.9 or ≥ 8.46

or E

_{b}≥ 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 A

^{2}≥

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**

**equation**

**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) = N**

_{0}/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 s

_{0}(T) =

Then, the noise component is n

_{0}(f). If N

_{0}(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 n

_{0}(t) < – is given by

**equation**

FIGURE 8.43.

The probability density of noise n

_{0}(t) is given by

**equation**