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Error rate due to noise In this section, an expression for the probability of error will be derived The analysis technique, will be demonstrated on a binary PCM based on polar NRZ signaling The channel noise is modeled as additive white Gaussian noise π€(π‘) of zero mean π0 and power spectral density 2 1 Bit error rate The receiver of the communication system can be modeled as shown below x(t) y(t) Y The signaling interval will be 0 β€ π‘ β€ ππ , where ππ is the bit duration 2 Bit error rate The received signal can be given by +π΄ + π€ π‘ , π π¦ππππ 1 π€ππ π πππ‘ π₯ π‘ = βπ΄ + π€ π‘ , π π¦ππππ 0 π€ππ π πππ‘ It is assumed that the receiver has acquired knowledge of the starting and ending times of each transmitted pulse The receiver has a prior knowledge of the pulse shape, but not the pulse polarity 3 Receiver model The structure of the receiver used to perform this decision making process is shown below 4 Possible errors When the receiver detects a bit there are two possible kinds of error 1. 2. Symbol 1 is chosen when a 0 was actually transmitted; we refer to this error as an error of the first kind Symbol 0 is chosen when a 1 was actually transmitted; we refer to this error as an error of the second kind 5 Average probability of error calculations To determine the average probability of error, each of the above two situations will be considered separately If symbol 0 was sent then, the received signal is π₯ π‘ = βπ΄ + π€ π‘ , 0 β€ π‘ β€ ππ 6 Average probability of error calculations The matched filter output will be given by 1 ππ π¦= π₯ π‘ ππ‘ ππ 0 1 ππ π¦ = βπ΄ + π₯ π‘ ππ‘ = βA ππ 0 The matched filter output represents a random variable π 7 Average probability of error calculations This random variable is a Gaussian distributed variable with a mean of βπ΄ The variance of π is π2 π = π0 2ππ The conditional probability density function the random variable π, given that symbol 0 was sent, is therefore ππ π¦ 0 = 1 ππ0 /ππ π π¦+π΄ 2 β π0 ππ 8 Average probability of error calculations The probability density function is plotted as shown below To compute the conditional probability, π10 , of error that symbol 0 was sent, we need to find the area between π and β 9 In mathematical notations π10 = π π¦ > π π π¦ππππ 0 π€ππ π πππ‘ β π10 = ππ¦ π¦ 0 ππ¦ π π10 = 1 ππ0 /ππ β π π¦+π΄ 2 β π0 /ππ ππ¦ π 10 The integral 2 erfc(π§) = π β π βπ§2 ππ§ π Is known as the complementary error function which can be solved by numerical integration methods 11 By letting π§ = π¦+π΄ π0 /ππ , ππ¦ = became π10 π0 /ππ ππ§, π10 β 1 = π π βπ§2 ππ§ π΄+π π0 /ππ π10 1 = ππππ 2 π΄+π π0 /ππ 12 By using the same procedure, if symbol 1 is transmitted, then the conditional probability density function of Y, given that symbol 1 was sent is given ππ π¦ 1 = 1 ππ0 /ππ π π¦βπ΄ 2 β π0 ππ Which is plotted in the next slide 13 The probability density function π(π|1) is plotted as shown below 14 The probability of error that symbol 1 is sent and received bit is mistakenly read as 0 is given by β π01 = ππ¦ π¦ 1 ππ¦ π π10 = 1 ππ0 /ππ β π π¦βπ΄ 2 β π0 /ππ ππ¦ π 15 Expression for the average probability of error π10 1 = ππππ 2 π΄βπ π0 /ππ The average probability of error ππ is given by ππ = π0 π10 + π1 π01 π0 π΄+π π1 π΄βπ ππ = ππππ + ππππ 2 2 π0 /ππ π0 /ππ 16 In order to find the value of π which minimizes the average probability of error we need to derive ππ with respect to π and then equate to zero as πππ =0 ππ From math's point of view (Leibnizβs rule) πerfc(π’) 1 βπ’2 =β π ππ’ π 17 Optimum value of the threshold π value If we apply the previous rule to ππ , then we may have the following equations 2 (π΄+π) π0 β π π π 0 π πππ 1 =β ππ π π0 ππ 2 + 1 2 (π΄βπ) π1 β π π π 0 π π π0 ππ 2 =0 By solving the previous equation we may have π0 = π1 (π΄βπ)2 β π π0 ππ (π΄+π)2 β π π0 ππ ππ¦ (π¦|1) = ππ¦ (π¦|0) 18 From the previous equation, we may have the following expression for the optimum threshold point π0 π0 ππππ‘ = ππ 4ππ π1 If both symbols 0 and 1 occurs equally in 1 the bit stream, then π0 = π1 = , then 2 1. ππππ‘ = 0 19 1 2. ππ = 2 ππππ π΄ π0 /ππ 1 = ππππ 2 π΄2 π0 /ππ 1 = ππππ 2 π΄2 ππ π0 1 = ππππ 2 πΈπ π0 20 Average probability of error vs πΈπ π0 A plot of the average probability of error as a function of the signal to noise energy is shown below 21 As it can be seen from the previous figure, the average probability of error decreases exponential as the signal to noise energy is increased For large values of the signal to noise energy the complementary error function can be approximated as 22 Example1 Solution 23 Example1 solution 24 Example 2 25 Example 2 solution a) The average probability of error is given by ππ = 1 2 erfc πΈπ π0 where πΈπ = π΄2 ππ , 1 2 We can rewrite ππ = erfc 1 2 erfc 2π΄ π , where π = π΄2 ππ π0 1 2 = erfc 2π΄2 π 2π 0 π = π0 2ππ 26 Example 2 solution 27 Example 2 Solution 28 Example 2 solution 29 Example 2 solution 30 Example 2 solution 31 Example 3 32 Example 3 solution Signaling with NRZ pulses represents an example of antipodal signaling, we can use 1 × 10 1 ππ = ππππ 2 πΈπ π0 1 = ππππ 2 π΄2 ( β3 1 ) 56000 10β6 33 Example 3 solution Using the erfc(π₯) table we find 1 2 π΄ 56000 = 2.18 10β6 π΄2 = 0.266 With 3-dB power loss the power at the transmitting end of the cable would be twice the power at the receiving terminal, this means that ππ‘π₯ = 2πππ₯ = 0.532 π€ππ‘π‘ 34