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
)
dγ
i
=
∫
0
γ
1
Γ
exp
(
−
γ
i
Γ
)
dγ
i
P
(
γ
i
≤
γ
)
=
∫
0
γ
p
(
γ
i
)
dγ
i
=
∫
0
γ
1
Γ
exp
(
−
γ
i
Γ
)
dγ
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 dB−20 dB=−10 dB=0.1γ/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
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"} {}
