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This section shows the error-performance improvements that can be obtained with the use of diversity techniques.

The bit-error-probability, P B ¯ size 12{ {overline {P rSub { size 8{B} } }} } {} ­­­­, averaged through all the “ups and downs” of the fading experience in a slow-fading channel is as follows:

P B ¯ = P B ( x ) p ( x ) dx size 12{ {overline {P rSub { size 8{B} } }} = Int {P rSub { size 8{B} } \( x \) p \( x \) ital "dx"} } {}

where P B ( x ) size 12{P rSub { size 8{B} } \( x \) } {} is the bit-error probability for a given modulation scheme at a specific value of SNR = x size 12{ ital "SNR"=x} {} , where x = α 2 E b / N 0 size 12{x=α rSup { size 8{2} } {E rSub { size 8{b} } } slash {N rSub { size 8{0} } } } {} , and p ( x ) size 12{p \( x \) } {} is the pdf size 12{ ital "pdf"} {} of x size 12{x} {} due to the fading conditions. With E b size 12{E rSub { size 8{b} } } {} and N 0 size 12{N rSub { size 8{0} } } {} constant, α size 12{α} {} is used to represent the amplitude variations due to fading.

For Rayleigh fading , α size 12{α} {} has a Rayleigh distribution so that α 2 size 12{α rSup { size 8{2} } } {} , and consequently x size 12{x} {} , have a chi-squared distribution :

p ( x ) = 1 Γ exp ( x Γ ) size 12{p \( x \) = { {1} over {Γ} } "exp" \( - { {x} over {Γ} } \) } {} x 0 size 12{x>= 0} {}

where Γ = α 2 ¯ E b / N 0 size 12{Γ= {overline {α rSup { size 8{2} } }} {E rSub { size 8{b} } } slash {N rSub { size 8{0} } } } {} is the SNR size 12{ ital "SNR"} {} averaged through the “ups and downs” of fading. If each diversity (signal) branch, i = 1, 2, . . . , M size 12{i=1," 2, " "." "." "." ", "M} {} , has an instantaneous SNR = γ i size 12{ ital "SNR"=γ rSub { size 8{i} } } {} , and we assume that each branch has the same average SNR size 12{ ital "SNR"} {} given by Γ size 12{Γ} {} , then

p ( γ i ) = 1 Γ exp ( γ i Γ ) size 12{p \( γ rSub { size 8{i} } \) = { {1} over {Γ} } "exp" \( - { {γ rSub { size 8{i} } } over {Γ} } \) } {} γ i 0 size 12{γ rSub { size 8{i} }>= 0} {}

The probability that a single branch has SNR size 12{ ital "SNR"} {} less than some threshold γ size 12{γ} {} is:

P ( γ i γ ) = 0 γ p ( γ i ) i = 0 γ 1 Γ exp ( γ i Γ ) i size 12{P \( γ rSub { size 8{i} }<= γ \) = Int rSub { size 8{0} } rSup { size 8{γ} } {p \( γ rSub { size 8{i} } \) dγ rSub { size 8{i} } = Int rSub { size 8{0} } rSup { size 8{γ} } { { {1} over {Γ} } "exp" \( - { {γ rSub { size 8{i} } } over {Γ} } \) dγ rSub { size 8{i} } } } } {}

= 1 exp ( γ Γ ) size 12{ {}=1 - "exp" \( - { {γ} over {Γ} } \) } {}

The probability that all M size 12{M} {} independent signal diversity branches are received simultaneously with an SNR size 12{ ital "SNR"} {} less than some threshold value γ size 12{γ} {} is:

P ( γ 1 , . . . , γ M γ ) = 1 exp ( γ Γ ) M size 12{P \( γ rSub { size 8{1} } , "." "." "." ,γ rSub { size 8{M} }<= γ \) = left [1 - "exp" \( - { {γ} over {Γ} } \) right ] rSup { size 8{M} } } {}

The probability that any single branch achieves SNR > γ size 12{ ital "SNR">γ} {} is:

P ( γ i > γ ) = 1 1 exp ( γ Γ ) M size 12{P \( γ rSub { size 8{i} }>γ \) =1 - left [1 - "exp" \( - { {γ} over {Γ} } \) right ] rSup { size 8{M} } } {}

This is the probability of exceeding a threshold when selection diversity is used.

Example: Benefits of Diversity

Assume that four-branch diversity is used, and that each branch receives an independently Rayleigh-fading signal. If the average SNR size 12{ ital "SNR"} {} is Γ = 20 dB size 12{Γ="20"" dB"} {} , determine the probability that all four branches are received simultaneously with an SNR size 12{ ital "SNR"} {} less than 10 dB size 12{"10"" dB"} {} (and also, the probability that this threshold will be exceeded).

Compare the results to the case when no diversity is used.

Solution

With γ = 10 dB size 12{γ="10"" dB"} {} , and γ / Γ = 10 dB 20 dB = 10 dB = 0 . 1 size 12{ {γ} slash {Γ} ="10"" dB" - "20"" dB"= - "10"" dB"=0 "." 1} {} , we solve for the probability that the

SNR size 12{ ital "SNR"} {} will drop below 10 dB size 12{"10"" dB"} {} , as follows:

P ( γ 1 , γ 2 , γ 3 , γ 4 10 dB ) = 1 exp ( 0 . 1 ) 4 = 8 . 2 × 10 5 size 12{P \( γ rSub { size 8{1} } ,γ rSub { size 8{2} } ,γ rSub { size 8{3} } ,γ rSub { size 8{4} }<= "10"" dB" \) = left [1 - "exp" \( - 0 "." 1 \) right ] rSup { size 8{4} } =8 "." 2 times "10" rSup { size 8{ - 5} } } {}

or, using selection diversity, we can say that

P ( γ i > 10 dB ) = 1 8 . 2 × 10 5 = 0 . 9999 size 12{P \( γ rSub { size 8{i} }>"10"" dB" \) =1 - 8 "." 2 times "10" rSup { size 8{ - 5} } =0 "." "9999"} {}

Without diversity,

P ( γ 1 10 dB ) = 1 exp ( 0 . 1 ) 1 = 0 . 095 size 12{P \( γ rSub { size 8{1} }<= "10"" dB" \) = left [1 - "exp" \( - 0 "." 1 \) right ] rSup { size 8{1} } =0 "." "095"} {}

P ( γ 1 > 10 dB ) = 1 0 . 095 = 0 . 905 size 12{P \( γ rSub { size 8{1} }>"10"" dB" \) =1 - 0 "." "095"=0 "." "905"} {}

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Source:  OpenStax, Principles of digital communications. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10805/1.1
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