2010-10-06
A One-way Telecommunication System
TSKS01 Digital Communication Lecture 7
Source
Digital Modulation – Detection
Packing
Error control
Digital to analog
Source encoder
Channel encoder
Modulator
Source coding
Channel coding
Digital modulation
Source decoder
Channel decoder
Demodulator
Medium
Channel
Mikael Olofsson Department of EE (ISY) Destination
Div. of Communication Systems
Unpacking
2010-10-01
Situation
Error correction Analog to digital
TSKS01 Digital Communication - Lecture 7
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Last Time – The Vector Model Noise
si (t ), i = 0,1,K , M − 1 0≤t 0, 0 1,0 | A = a0 ⎬ 2 ⎩ ⎭ s −s ⎫ s +s ⎧ ⎧ ⎫ = Pr ⎨ W0 > 0,0 1,0 − s0, 0 | A = a0 ⎬ = Pr ⎨ W0 > 1, 0 0, 0 ⎬ This works in all 2 2 ⎩ ⎩ ⎭ ⎭ directions. ⎛ d (s0 , s1 ) ⎞ ⎛ d (s0 , s1 ) 2 ⎞ d (s0 , s1 ) ⎫ s0 ⎧ ⎟ ⎟ = Q⎜ = Pr ⎨ W0 > ⎬ = Q⎜⎜ ⎜ 2N ⎟ ⎟ 2 ⎭ N 2 ⎩ 0 0 B1 ⎝ ⎠ B0 ⎝ ⎠ s 1 ⇒ Pe = Q(L) Similarily for Pr{ X ∉ B1 | A = a1 }.
i =0
Pe =
s0 + s1 2
s0
B0
= ∑ Pr{ A = ai } ⋅ Pr{ X ∉ Bi | A = ai } 1 ⎛ ML detection again ⎜ Pr{ A = ai } = M ⎝
fX|A(x|a1)
1 M
Distances:
TSKS01 Digital Communication - Lecture 7
M −1
⎛ di, j ⎞ ⎟ ⎟ ⎝ 2N0 ⎠
∑∑ Q⎜⎜ i =0 j ≠i
d i , j = d (si , s j ) 20
5
2010-10-06
The Nearest Neighbour Approximation We had the union bound:
φ1
s0 d0,1 = d1,0 = dmin
d0,2 = d2,0
⎛ di, j ⎞ ⎟ Q⎜ ∑∑ ⎜ 2N ⎟ i =0 j ≠i 0 ⎠ ⎝ Dominated by the smallest distance. 1 Pe ≤ M
3
1 d1,2 = d2,1 3
d min = min di , j ni = # j : di , j = dmin
φ0
Interprete x as the nearest signal. 2010-10-01
1 M
1 = M
M −1
⎞ ⎛ d Q⎜ min ⎟ ⎟ ⎜ 2 N j:d i , j = d min 0 ⎠ ⎝
∑ ∑ i =0
Alternative upper bound: As the union bound, but only consider pairs of points whose decision regions share a common border.
Very simple approximation: ⎞ ⎛ d Pe ≈ Q⎜ min ⎟ ⎜ 2N ⎟ 0 ⎠ ⎝
⎞ ⎛ d ni Q⎜ min ⎟ ∑ ⎜ 2N ⎟ i =0 0 ⎠ ⎝
M −1
TSKS01 Digital Communication - Lecture 7
At high SNR Eavg/N0: Both the union bound on and the nearest neighbour of the error probability are close to the real error probability.
Alternative approximation: As the nearest neighbour approximation, but consider the two or three smallest distances.
Nearest neighbour approximation:
Pe ≈
s2
M −1
i≠ j
s1
1
–1
Comments
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2010-10-01
Maximum á-Posteriori (MAP) Detection 1(2)
Very simple upper bound: ⎞ ⎛ d Pe ≤ (M − 1) ⋅ Q⎜ min ⎟ ⎜ 2N ⎟ 0 ⎠ ⎝
TSKS01 Digital Communication - Lecture 7
Maximum á-Posteriori (MAP) Detection 2(2) φ1
Back to this rule (MAP): Set aˆ = ai if f X | A ( x | ak ) ⋅ Pr{ A = ak } is maximized for k = i.
B0
The borders are still: Straight lines (hyperplanes) Orthogonal to straight lines between signal points.
s0 3
B1
Pr{A=a1}· fX|A(x|a1) Pr{A=a0}· fX|A(x|a0) B0
s1
φ0
B1
1
B2
–1
Equivalent reasoning leads to this rule: Set aˆ = ai if d 2 ( x , sk ) − N 0 ⋅ ln(Pr{ A = ak }) is maximized for k = i. 2010-10-01
Linköpings universitet
TSKS01 Digital Communication - Lecture 7
s1
1
φ0
s0
23
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2010-10-01
s2
3
But: Borders only cut the lines midway if the two probabilities are equal. Borders depend on N0. Note: A signal point may in extreme cases not even be in its own detection region.
TSKS01 Digital Communication - Lecture 7
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2010-10-06
Special Case: Orthogonal Decision Borders
Cost of Non-Optimal Detection
d0,1
B0 B1
s1
Define notation:
s0
d0,2
d1,2
s2
B2 B3
s3
Pythagoras:
NN:
Pe ≈ q1,2
d 02, 2 = d 02,1 + d12, 2
Alternative bound:
Pe ≤ q0,1 + q1,2
Alternative approx.:
Pe ≈ q0,1 + q1,2
Pe > q1,2 Lower bound: Exact: Pe ≤ q0,1 + q1,2 − q0,1q1,2 (orthogonal noise components are independent) 2010-10-01
Signals:
⎞ ⎟ ⎟ ⎠
θ0
d′
s0
α
s1
TSKS01 Digital Communication - Lecture 7
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Detected signal points:
φ0
Wrong basis function in receiver:
θ 0 (t )
(± s0 ,θ 0 ) = ± a ⋅ (φ0 ,θ 0 ) = ± a ⋅ φ0 ⋅ θ 0
Effective distance:
d ′ = d ⋅ cos(α )
Resulting error probability:
⎛ d′ Pe = Q⎜ ⎜ 2N 0 ⎝
2010-10-01
s0 (t ) = a ⋅ φ0 (t ) s1 (t ) = −a ⋅ φ0 (t )
What is the cost of this non-ideal situation?
d = 2a
Pe ≤ q0,1 + q1,2 + q0,2
UB:
⎛ di, j qi, j = Q⎜ ⎜ 2N 0 ⎝
⎞ ⎟ ⎟ ⎠
TSKS01 Digital Communication - Lecture 7
⋅ cos(α ) = ± a ⋅ cos(α )
Signals are projected on basis functions 26
Example of Non-Optimal Detection SNR:
θ0
d′
α
s1
s0
φ0
a2 =5 N0
Signal points: ± a Distances:
d = 2a
d = 2a
d ′ = d ⋅ cos(α )
Ideal detection:
⎛ d Pe = Q⎜ ⎜ 2N 0 ⎝
⎞ ⎟ = Q 10 ≈ Q(3.16) ≈ 7.9 ⋅10 − 4 ⎟ ⎠
At α = π /4:
⎛ d′ Pe = Q⎜ ⎜ 2N 0 ⎝
⎞ ⎟ = Q 5 ≈ Q(2.24) ≈ 1.25 ⋅10 − 2 ⎟ ⎠
2010-10-01
Linköpings universitet
( ) ( )
TSKS01 Digital Communication - Lecture 7
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www.liu.se
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