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Course by: Tuan Do-Hong. E-mail the author

# Diversity Techniques

Module by: Sinh Nguyen-Le, Tuan Do-Hong. E-mail the authorsEdited By: Sinh Nguyen-Le, Tuan Do-HongTranslated By: Sinh Nguyen-Le, Tuan Do-Hong

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

The bit-error-probability, PB¯PB¯ 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 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 PB(x)PB(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=xSNR=x size 12{ ital "SNR"=x} {}, where x=α2Eb/N0x=α2Eb/N0 size 12{x=α rSup { size 8{2} } {E rSub { size 8{b} } } slash {N rSub { size 8{0} } } } {}, and p(x)p(x) size 12{p $$x$$ } {} is the pdfpdf size 12{ ital "pdf"} {} of xx size 12{x} {} due to the fading conditions. With EbEb size 12{E rSub { size 8{b} } } {} and N0N0 size 12{N rSub { size 8{0} } } {} constant, αα size 12{α} {} is used to represent the amplitude variations due to fading.

For Rayleigh fading, αα size 12{α} {} has a Rayleigh distribution so that α2α2 size 12{α rSup { size 8{2} } } {}, and consequently xx size 12{x} {}, have a chi-squared distribution:

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

where Γ=α2¯Eb/N0Γ=α2¯Eb/N0 size 12{Γ= {overline {α rSup { size 8{2} } }} {E rSub { size 8{b} } } slash {N rSub { size 8{0} } } } {} is the SNRSNR size 12{ ital "SNR"} {} averaged through the “ups and downs” of fading. If each diversity (signal) branch, i=1, 2, ..., Mi=1, 2, ..., M size 12{i=1," 2, " "." "." "." ", "M} {}, has an instantaneous SNR=γiSNR=γi size 12{ ital "SNR"=γ rSub { size 8{i} } } {}, and we assume that each branch has the same average SNRSNR size 12{ ital "SNR"} {} given by ΓΓ size 12{Γ} {}, then

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

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

P ( γ i γ ) = 0 γ p ( γ i ) i = 0 γ 1 Γ exp ( γ i Γ ) i 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 ) = 1 exp ( γ Γ ) size 12{ {}=1 - "exp" $$- { {γ} over {Γ} }$$ } {}

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

P ( γ 1 , . . . , γ M γ ) = 1 exp ( γ Γ ) M 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>γSNR>γ size 12{ ital "SNR">γ} {} is:

P ( γ i > γ ) = 1 1 exp ( γ Γ ) M 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 SNRSNR size 12{ ital "SNR"} {} is Γ=20 dBΓ=20 dB size 12{Γ="20"" dB"} {}, determine the probability that all four branches are received simultaneously with an SNRSNR size 12{ ital "SNR"} {} less than 10 dB10 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 dBγ=10 dB size 12{γ="10"" dB"} {}, and γ/Γ=10 dB20 dB=10 dB=0.1γ/Γ=10 dB20 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

SNRSNR size 12{ ital "SNR"} {} will drop below 10 dB10 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 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 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 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 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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