Summary: SER formulation. Binary antipodal constellation. Generic binary constellation. Constellation gain. 2D constellations

Transcription

1 TUTORIAL ON DIGITAL MODULATIONS Part 8a: Error probability A [ ] 07] Roberto Garello, Politecnico di Torino Free download (for personal use only) at: 1

2 Part 8a: Error probability A Summary: SER formulation Binary antipodal constellation Generic binary constellation Constellation gain 2D constellations Hamming weight and distance BER formulation R. Garello. Tutorial on digital modulations - Part 8a [ ] 2

3 Transmission chain Transmitted bits Transmitted signal Received signal Received bits u T s( t ) r( t) source TX CHANNEL RX u R ( [ ] [ ]) P u i u i R T BIT ERROR RATE (BER) = P( u R [i] u T [i]) = P b (e) Probability that a received bit is different from the corresponding transmitted bit R. Garello. Tutorial on digital modulations - Part 8a [ ] 3

4 BER and SER Error probabilities: measure the goodness of a digital radio link Most important (the system exists to transmit bits): BIT ERROR RATE (BER) = P( u R [i] u T [i]) = P b (e) We will also consider (easier to compute but less important): SYMBOL ERROR RATE (SER) = P( s R [n] s T [n]) = P S (e) R. Garello. Tutorial on digital modulations - Part 8a [ ] 4

5 The transmission chain in the signal space Inverse grouping labeling AWGN Voronoi labeling degrouping u [ i] Z v [ n] H s [ n] M r[ n] R s [ n] M v [ n] H u [ i] Z d T 2 T k T R R k R 2 P ( e) = P( s [ n] s [ n]) S R T P ( e) = P( u [ i] u [ i]) b R T R. Garello. Tutorial on digital modulations - Part 8a [ ] 5

6 Received power The BER and SER performance should be evaluated as a function of the received power P RX (we will also denote it by S ). Important quantities: R b Bit Rate E b Energy per information bit T b = 1/ R b Bit time P RX =S = E b / T b = E b R b Received signal power R. Garello. Tutorial on digital modulations - Part 8a [ ] 6

7 Noise power N 0 / 2 AWGN channel noise spectral density B Signal Bandwidth P N =N = N 0 B Noise power R. Garello. Tutorial on digital modulations - Part 8a [ ] 7

8 E b /N 0 ratio We introduce the ratio E b /N 0 sort of Signal to Noise ratio referred to an information bit. Very important: this ratio is proportional to the received power Eb PRX = S = EbRb = Rb N N (Note: R b and N 0 are two numerical constants: the first one depends on the system, the second one on the channel. In case of thermal noise N 0 =kt.) 0 0 R. Garello. Tutorial on digital modulations - Part 8a [ ] 8

9 BER vs. E b /N 0 (P (P RX ) We will express the system error rate performance as a function of E b /N 0 it is equivalent to express them as a function of the received power P RX (the two quantities are proportional) Example of BER curve computed in the following: costellazione binaria antipodale BER E-3 1E-4 1E-5 1E-6 1E-7 1E-8 1E-9 1E-10 1E-11 1E-12 1E-13 1E Eb/N0 [db] Eb [ db ] (proportional to P RX [ dbm ]) N 0 R. Garello. Tutorial on digital modulations - Part 8a [ ] 9

10 Connection between E b /N 0 ratio and SNR S/N Signal to Noise ratio (SNR) E b /N 0 Signal to Noise ratio referred to an information bit Connection: P = S = E R = E RX b b b PN N N0 B N η 0 R b where η = spectral efficiency B R. Garello. Tutorial on digital modulations - Part 8a [ ] 10

11 SER formulation The SER is less important, but we start from its formulation since it is easier. Definition: P S (e) = P( s R s T ) The transmitted signal belongs to the constellation M { } i s M = s T All the signals have the same probability of being transmitted (here we use the hypothesis of ideal random binary sequence) P ( s = s ) = S T i R. Garello. Tutorial on digital modulations - Part 8a [ ] 11 1 m

12 Proof: ideal random binary sequence In fact, we model the binary information sequence as an ideal random sequence all the bits have same probability P( u [ i] = 0) = P( u [ i] = 1) = T T When we group them, each k-bit vector has the same probability The binary labeling is a one-to-one mapping, then each constellation signal has the same probability of being transmitted: R. Garello. Tutorial on digital modulations - Part 8a [ ] v H k P( vt[ n] = v) = = k 2 m s M P( s [ n] = s) = T 1 m

13 SER formulation We have: m m 1 P ( e) P ( e s s ) P( s s ) P ( e s s ) = = = = = S S T i T i S T i i= 1 m i= 1 For each constellation signal we must compute: P ( e s = s ) = P( s s s = s ) S T i R T T i By invoking the Voronoi region criterion, if the received vector belongs to the Voronoi region of the transmitted signal s i, the received signal s R is correct, otherwise it is wrong. Then we have ( ρ ( ) ) P ( e s = s ) = P( s s s = s ) = P V s s = s S T i R T T i i T i R. Garello. Tutorial on digital modulations - Part 8a [ ] 13

14 SER formulation First approach: ( ρ ( ) ) P ( e s = s ) = P V s s = s S T i i T i In this case we must compute the probability of transmitting s i and receiving a vector outside its Voronoi region. Note that we can compute it by ( ρ ( ) ) P ( ) ( ( ) ) S e st = si = P V si st = si = P ρ V s j st = si j i i.e. by summing all the probabilities of kind P( ρ V ( s ) s = s ) j T i which are the probabilities of transmitting s i and receiving a vector belonging to the Voronoi region of a different signal s j R. Garello. Tutorial on digital modulations - Part 8a [ ] 14

15 SER formulation First approach PS ( e st = si ) = P( ρ V ( si ) st = si ) = P( ρ V ( s j ) st = si ) j i R. Garello. Tutorial on digital modulations - Part 8a [ ] 15

16 SER formulation Second approach As an alternative we can also apply this approach: ( ρ ( ) ) P ( e s = s ) = P V s s = s = S T i i T i = 1 P( ρ V ( s ) s = s ) i T i This is equivalent to computing the probability of correct decision when s i is transmitted and then complement it. R. Garello. Tutorial on digital modulations - Part 8a [ ] 16

17 SER formulation Second approach P ( e s = s ) = 1 P( ρ V ( s ) s = s ) S T i i T i In this case we must compute P( ρ V ( s ) s = s ) i T i which is the probability of transmitting s i and receiving a vector within its Voronoi region R. Garello. Tutorial on digital modulations - Part 8a [ ] 17

18 SER and BER for binary constellations Note that if the constellation contains two signals (we will call it a binary constellation), each signal carries an information bit. At the receiver side: if the received signal is correct the received bit is correct if the received signal is wrong the received bit is wrong. Then, if and only if the constellation is binary (m=2), SER and BER are equal. R. Garello. Tutorial on digital modulations - Part 8a [ ] 18

19 SER/BER computation for binary antipodal signals A binary antipodal constellation is a one-dimensional constellation (d=1) composed by two signals (m=2), symmetrical with respect to the origin: M = { s = ( + A) s = ( A) } 1 2 The Voronoi regions are: V ( s ) = { ρ = ( ρ ), ρ 0 } V ( s ) = { ρ = ( ρ ), ρ 0 } R. Garello. Tutorial on digital modulations - Part 8a [ ] 19

20 SER/BER computation for binary antipodal signals We have: m 1 1 P ( e) = P ( e s = s ) = P ( e s = s ) + P ( e s = s ) m = S S T i S T 1 S T 2 i 1 2 Let us first compute P ( e s = s ) S T 1 and then P ( e s = s ) S T 2 R. Garello. Tutorial on digital modulations - Part 8a [ ] 20

21 SER/BER computation for binary antipodal signals P ( e s = s ) = P( ρ V ( s ) s = s ) = P( ρ < 0 s = s ) S T 1 2 T 1 1 T 1 We have: r = s + n r = ρ s = s T T 1 where ρ = ( ρ ) s = ( s ) = ( + A ) n = ( n ) It follows: ρ = A + n 1 1 R. Garello. Tutorial on digital modulations - Part 8a [ ] 21

22 SER/BER computation for binary antipodal signals P ( e s = s ) = P( ρ < 0 s = s ) = P( A + n < 0) = P( n < A) S T 1 1 T The random variable n 1 is Gaussian, with mean zero and variance N 0 /2 1 A PS ( e st = s1 ) = P( n1 < A) = P( n1 > A) = erfc 2 N 0 where we have used the erfc definition R. Garello. Tutorial on digital modulations - Part 8a [ ] 22

23 erfc Given a Gaussian random variable n with - mean µ - variance σ 2 - density We define f n 2 1 ( x µ ) ( x) = exp( ) 2 2 2πσ 2σ + 1 x µ P( n > x) = fn( x) dx = erfc 2 2σ x R. Garello. Tutorial on digital modulations - Part 8a [ ] 23

24 erfc where In fact 2 1 ( x µ ) P( n > x) = fn( x) dx = exp( ) dx = 2 2 2πσ 2σ π ( x µ ) 2σ 2 2 t erfc( x) = e dt π + + x x µ = 2 2σ t e dt = erfc x + x In case of zero mean and variance N 0 /2 1 x µ 1 x P( n > x) = erfc erfc 2 = 2σ 2 N 0 R. Garello. Tutorial on digital modulations - Part 8a [ ] 24

25 erfc Note that the erfc function is a decreasing function. In fact, it measures the probability of extracting a sample in a tail of a Gaussian pdf. The larger the argument x and the smaller the probability. The erfc function allows to express probabilities by closed-form formulas. There are a lot of functions and routines available for computing the erfc function. A not very good approximation, but very popular and very useful for heuristic considerations is erf x < e x R. Garello. Tutorial on digital modulations - Part 8a [ ] 25

26 Graphical computation The same result can be obtained by a graphical procedure. For computing P(e s T =s 1 ) we can draw the pdf of the received vector R. Garello. Tutorial on digital modulations - Part 8a [ ] 26

27 Graphical computation The error probability is the probability of receiving a vector in the Voronoi region of s 2. R. Garello. Tutorial on digital modulations - Part 8a [ ] 27

28 Graphical computation If we draw the pdf of the noise: R. Garello. Tutorial on digital modulations - Part 8a [ ] 28

29 Graphical computation Then the error probability is the probability of extracting a noise sample in the tail: The noise random variable is a Gaussian random variable with zero mean and variance N 0 /2. Then: 1 A PS ( e) = PS ( e st = s1 ) = P( n A) = erfc 2 N 0 R. Garello. Tutorial on digital modulations - Part 8a [ ] 29

30 SER/BER computation for binary antipodal signals Let us compute now the error probability for s T = s 2 P ( e s = s ) = P( ρ V ( s ) s = s ) = P( ρ > 0 s = s ) S T 2 1 T 2 1 T 2 We have: then r = s + n r = ρ s = s T ρ = ( ρ ) s = ( s ) = ( A) n = ( n ) T 2 ρ = A + n 1 1 R. Garello. Tutorial on digital modulations - Part 8a [ ] 30

31 SER/BER computation for binary antipodal signals P ( e s = s ) = P( A + n > 0) = P( n > A) S T A PS ( e st = s2) = erfc 2 N 0 R. Garello. Tutorial on digital modulations - Part 8a [ ] 31

32 GU constellations Note that the two error probabilities are equal P ( e s = s ) = P ( e s = s ) S T 1 S T 2 This is a consequence of the fact that the Voronoi regions of the two constellations are congruent We can generalize this result by introducing the concept of Geometrically Uniform constellations R. Garello. Tutorial on digital modulations - Part 8a [ ] 32

33 GU constellations A constellation M is geometrically uniform (GU) if the Voronoi regions of its signals are all congruent (they have the same shape, they can be overlapped by using elementary transformations: translations, reflexions, rotations). By using approach 1 we have: P ( e s = s ) = 1 P( ρ V ( s ) s = s ) = 1 f ( ρ s = s ) d ρ S T i i T i r T i V ( s ) The Gaussian pdfs have the same analytical expression for each transmitted signal. The integration areas have the same shape, then the probability is the same for each transmitted signal. i R. Garello. Tutorial on digital modulations - Part 8a [ ] 33

34 GU constellations For a GU constellation, the probabilities are the same for each transmitted signal. P ( e s = s ) S T i It follows m 1 P ( e) = P ( e s = s ) = P ( e s = s1 ) S S T i S T m i = 1 Then the error probability can be computed by considering a single signal (this highly simplifies error probability computation). R. Garello. Tutorial on digital modulations - Part 8a [ ] 34

35 SER/BER computation for binary antipodal signals We have It follows: then P ( e s = s ) = P ( e s = s ) S T 1 S T 2 1 PS ( e) = PS ( e st = s1 ) + PS ( e st = s2) = PS ( e st = s1 ) 2 1 A PS ( e) = PS ( e st = s1 ) = erfc 2 N 0 [ note that 1 d PS ( e) = PS ( e st = s1 ) = erfc 2 2 N 0 R. Garello. Tutorial on digital modulations - Part 8a [ ] 35

36 SER/BER computation for binary antipodal signals We have: 1 A PS ( e) = PS ( e st = s1 ) = erfc 2 N 0 We want to write it as a function of E b /N 0. E S E( s ) = E( s ) = A 1 2 E( s ) + E( s ) = = E S 2 Eb = = ES = A k A R. Garello. Tutorial on digital modulations - Part 8a [ ] 36

37 SER/BER computation for binary antipodal signals Fundamental result P ( e) S 1 E b = erfc 2 N 0 This is a binary constellation, then the BER and the SER are equal (if the received signal is wrong, the information bit is wrong, too. If the received signal is correct, the information bit is correct, too) : P ( e) b 1 E b = erfc 2 N 0 R. Garello. Tutorial on digital modulations - Part 8a [ ] 37

38 PLOTTING THE BER CURVES 1 E b Pb ( e) = erfc 2 N 0 The BER curves are plotted by posing - On the x axis the ratio E b /N 0 in db (linear scale in db) -On the y axis the P b (e) value in logarithmic scale Remember that the ratio E b /N 0 is proportional to the received power R. Garello. Tutorial on digital modulations - Part 8a [ ] 38

39 PLOTTING THE BER CURVES BER b costellazione binaria antipodale E-3 1E-4 1E-5 1E-6 1E-7 1E-8 1E-9 1E-10 1E-11 1E-12 1E-13 1E Eb/N0 [db] 1 E b P ( e) = erfc 2 N 0 R. Garello. Tutorial on digital modulations - Part 8a [ ] 39

40 THE ADVANTAGE OF THE SIGNAL SPACE APPROACH! Different signal constellations (different transmitted waveforms) with the same vector constellations achieve equal BER performance! As an example, the BER performance of a binary antipodal constellation does not depend on the basis signal: 1 b1 ( t) = PT ( t) T 2 b1 ( t) = PT ( t)cos(2 π f0t) T R. Garello. Tutorial on digital modulations - Part 8a [ ] 40

41 BER of a generic binary constellation Let us consider a generic binary constellation (for simplicity, we plot it in a two-dimensional space) M = { α, β } β α = ( α, α ) 1 2 β = ( β, β ) 1 2 α R. Garello. Tutorial on digital modulations - Part 8a [ ] 41

42 BER of a generic binary constellation For computing the BER we can think of CHANGING THE ORTHONORMAL BASIS. It is useful to choose a basis signal which is parallel to the direction joining αand β β α R. Garello. Tutorial on digital modulations - Part 8a [ ] 42

43 BER of a generic binary constellation This way, the BER computation is very easy. As an example, graphically: p β α 2 d 1 d P( e) = p = P n = erfc 2 2 4N R. Garello. Tutorial on digital modulations - Part 8a [ ] 43 0

44 BER of a generic binary constellation The BER of a binary constellation only depends on the Euclidean distance between the two points P( e) = 2 1 d erfc 2 4N 0 R. Garello. Tutorial on digital modulations - Part 8a [ ] 44

45 Binary orthogonal constellation: BER M = { s = ( A,0) s = (0, A) } 1 2 s s 1 = (0, 2 = (0, A) 0 ( A s1 = ( A,0) R. Garello. Tutorial on digital modulations - Part 8a [ ] 45

46 Binary orthogonal constellation: BER Voronoi regions: V ( s ) = { ρ = ( ρ, ρ ), ρ ρ } V ( s ) = { ρ = ( ρ, ρ ), ρ ρ } They are congruent, the constellation is GU P ( e) = P ( e s = s ) S S T 1 s s 1 = (0, 2 = (0, A) s s 0 A 1 = ( A,0) R. Garello. Tutorial on digital modulations - Part 8a [ ] 46

47 Binary orthogonal constellation: BER We have: d 1 A PS ( e) = PS ( e st = s1 ) = erfc = erfc 2 4N 2 2N 0 0 We must express it as a function of the ratio E b /N 0 E( s ) = E( s ) = A 1 2 E( s ) + E( s ) ES = = A 2 ES 2 Eb = = ES = A k R. Garello. Tutorial on digital modulations - Part 8a [ ] 47 2

48 Binary orthogonal constellation: BER Since the constellation is binary, SER=BER 1 1 E b Pb ( e) = Ps ( e) = erfc 2 2 N 0 R. Garello. Tutorial on digital modulations - Part 8a [ ] 48

49 Performance comparison If we compare the BER performance of a binary antipodal constellation and a binary orthogonal constellation: P ( e) b antipodal 1 E = erfc 2 P ( e) 1 1 E = erfc b b N b orthogonal N0 To achieve the same BER value, the orthogonal constellation requires a larger E b /N 0 (the erfc is a decreasing function). We say that, in terms of BER performance, the antipodal constellation is better than the orthogonal one: To achieve the same BER it requires a smaller E b /N 0 (then a smaller received power P RX ) or, dually At the parity of E b /N 0 (P RX ), it achieves lower BER. R. Garello. Tutorial on digital modulations - Part 8a [ ] 49

50 Performance comparison This is clear if we plot the two performance curves: P ( e) b antipodal 1 E = erfc 2 P ( e) 1 1 E = erfc b b b orthogonal N 2 2 N 0 0 BER E-3 1E-4 1E-5 1E-6 1E-7 1E-8 1E-9 1E-10 1E-11 1E-12 1E-13 1E Eb/N0 [db] ortogonale antipodale R. Garello. Tutorial on digital modulations - Part 8a [ ] 50

51 Constellation gain As an example, if we fix P b (e)=1e-6: The antipodal constellation requires E b /N 0 = 10.6 db; The orthogonal constelaltion requires E b /N 0 =13.6 db (then a higher received power). Dually, if we fix for example E b /N 0 =12 db: The antipodal constellation achieves P b (e) =1e-8, The orthogonal constellation achieves P b (e) =5e-5 (higher, then worse) To achieve the same BER, the orthogonal constellation requires 3 db more in terms of E b /N 0 (then a received power 3 db higher, i.e., double). This difference in terms of E b /N 0 (P RX ) at the parity of P b (e) is called gain R. Garello. Tutorial on digital modulations - Part 8a [ ] 51

52 Constellation gain BER E-3 1E-4 1E-5 1E-6 1E-7 1E-8 1E-9 1E-10 1E-11 1E-12 1E-13 1E Eb/N0 [db] E b N 0 A G E b N ortogonale antipodale 0 O R. Garello. Tutorial on digital modulations - Part 8a [ ] 52

53 Constellation gain For orthogonal/antipodal constellations we have G E b N0 O 1 = 10log10 = 10log10 = 10log10 2 = 3 [ db] E 1/ 2 b N 0 A R. Garello. Tutorial on digital modulations - Part 8a [ ] 53

54 Constellation gain Example: line of sight link P R = P T G G T R 4π d λ 2 The first constellation achieves P b (e)=1e-6 with a received power which is half of the received power required by the second constellation. At the parity of transmitted power, this means the distance can be increased by a factor 2 Dually, we can halve the transmitted power by keeping constant the distance (or use smaller antennas). R. Garello. Tutorial on digital modulations - Part 8a [ ] 54

55 Centre-of-mass The performance of a binary constellation depends on the distance between its two points. Why the antipodal constellation has better performance than the orthogonal one? The BER performance are given by Then they can be rewritten as 2 1 d P( e) erfc = 2 4N d E b P( e) = erfc 2 4 Eb N 0 The performance depends on the parameter d E 2 b Euclidean distance normalized by the energy R. Garello. Tutorial on digital modulations - Part 8a [ ] 55

56 Centre-of-mass At the parity of distance, the antipodal constellation minimizes the energy because its center-of-mass is in the origin. Given a constellation (of equiprobable signals) M = { s 1,, s i,, s m } Its centre-of-mass is s B 1 m m i = 1 = s i R. Garello. Tutorial on digital modulations - Part 8a [ ] 56

57 Centre-of-mass More in general, given a constellation M M = { s 1,, s i,, s m } with centre-of-mass s B not in the origin, let us consider a constellation M with centre-of-mass in the origin obtained by M = { s 1,, s i,, s m } s = s s ' i i B The two constellations have the same distance spectrum, but M has better performance, because its centre-of-mass is in the origin then its energy is minimized. R. Garello. Tutorial on digital modulations - Part 8a [ ] 57

58 Computing the SER of a 2D constellation We will do it both analytically and graphically R. Garello. Tutorial on digital modulations - Part 8a [ ] 58

59 4-PSK The 4-PSK constellation is a two-dimensional constellation (d=2) consisting of m=4 signals on a circle. M = { s = ( a, a) s = ( a, a) s = ( a, a) s = ( a, a) } s 2 s 1 a s 0 1 a s 2 3 s 3 4 R. Garello. Tutorial on digital modulations - Part 8a [ ] 59

60 4-PSK The Voronoi regions are: V ( s1 ) = { ρ = ( ρ1, ρ2), ( ρ1 0) ( ρ2 0)} V ( s2) = { ρ = ( ρ1, ρ2), ( ρ1 0) ( ρ2 0)} V ( s ) = { ρ = ( ρ, ρ ), ( ρ 0) ( ρ 0)} V ( s4) = { ρ = ( ρ1, ρ2), ( ρ1 0) ( ρ2 0)} They are congruent The constellation is GU s 2 s 1 a s 1 0 a s s s R. Garello. Tutorial on digital modulations - Part 8a [ ] 60

61 4-PSK: SER Since it is GU, we have: m 1 P ( e) = P ( e s = s ) = P ( e s = s1 ) S S T i S T m i = 1 Let us try to apply the approach 2: P ( e s = s ) = 1 P( ρ V ( s ) s = s ) S T 1 1 T 1 R. Garello. Tutorial on digital modulations - Part 8a [ ] 61

62 4-PSK: SER We have P( ρ V ( s ) s = s ) = P(( ρ 0) ( ρ 0) s = s ) 1 T T 1 where: r = st + n r = ρ = ( ρ, ρ ) s = s = ( a, a ) n = ( n, n ) 1 2 T Then, when s T =s 1 : ρ = a + n 1 1 ρ = a + n 2 2 R. Garello. Tutorial on digital modulations - Part 8a [ ] 62

63 4-PSK: SER P( ρ V ( s ) s = s ) = P(( a + n 0) ( a + n 0)) = P(( n a) ( n a)) 1 T n 1 and n 2 are statistically independent P( ρ V ( s ) s = s ) = P( n a) P( n a) 1 T R. Garello. Tutorial on digital modulations - Part 8a [ ] 63

64 4-PSK: SER Let us introduce We have 2 1 a p = P( n > a) = erfc 2 N 0 P( n a) = 1 P( n a) = 1 p 1 1 P( n a) = 1 P( n a) = 1 p 2 2 then P( ρ V ( s ) s = s ) = (1 p) 1 T 1 2 P ( e) = 1 P( ρ V ( s ) s = s ) = 1 (1 p) = 2 p p S 1 T R. Garello. Tutorial on digital modulations - Part 8a [ ] 64

65 4-PSK: SER To conclude, we must express the SER as a function of the ratio E b /N 0. We have: E( s ) = E( s ) = E( s ) = E( s ) = 2a ES = 2a 2 E b ES ES = = = k 2 a a 1 E ( ) b p = P n > a = erfc = erfc 2 N 0 2 N 0 R. Garello. Tutorial on digital modulations - Part 8a [ ] 65

66 4-PSK: SER For a 4-PSK constellation we have P ( e) = 2 p p S 2 where p 1 E b = erfc 2 N 0 R. Garello. Tutorial on digital modulations - Part 8a [ ] 66

67 Graphical computation Also in this case, it is possible to compute it graphically A 0 0 A R. Garello. Tutorial on digital modulations - Part 8a [ ] 67

68 Graphical computation Note that A 0 p (1 p ) Then, since the two noise random variables are s.i.: P ( e s = s ) = 1 P( ρ V ( s ) s = s ) = 1 (1 p) = 2 p p S T 1 1 T R. Garello. Tutorial on digital modulations - Part 8a [ ] 68

69 Computing the BER of a non-binary constellation Until now, binary constellations BER = SER How to compute the BER if the cardinality is m>2? R. Garello. Tutorial on digital modulations - Part 8a [ ] 69

70 Hamming weight For computing the BER, we shortly introduce these quantities k-bit vector v H k Hamming weight Examples w v = H ( ) number of components equal to 1 v = (101) w v = 2 H ( ) w = (001110) w w = 3 H ( ) R. Garello. Tutorial on digital modulations - Part 8a [ ] 70

71 Hamming distance two k-bit vectors v w, Hk Hamming distance d v, w = number of components which are different H ( ) R. Garello. Tutorial on digital modulations - Part 8a [ ] 71

72 EX-OR sum H = v = ( u,.., u,.. u ) { } k 1 i k By introducing the modulo-2 sum (EX-OR) We can define the modulo-2 sum between binary vectors: 1 v = ( u,.., u,..., u ) i 1k v = ( u,.., u,..., u ) i 2k v + v = ( u + u,.., u + u,.., u + u ) 72 R. Garello. Tutorial on digital modulations - Part 8a [ ] i 2i 1k 2k

73 Hamming distance This equivalent definition of Hamming distance holds: Examples: d v, w = w v + w H ( ) ( ) H v = (101) w = (110) d v, w = w v + w = w 011 = 2 ( ) ( ) ( ) H H H v = (00111) w = (11110) d v, w = w v + w = w = 3 ( ) ( ) ( ) H H H R. Garello. Tutorial on digital modulations - Part 8a [ ] 73

74 SER and BER The SER only depends on the constellation shape in the Euclidean space. The BER depends both on the constellation shape and the binary labeling R. Garello. Tutorial on digital modulations - Part 8a [ ] 74

75 BER computation When the received signal is correct (s R = s T ), also the information vector is correct (v R = v T ). When the received signal is wrong (s R s T ), the binary information vector is certainly wrong (v R v T ), but the average number of wrong information bits depends on the labeling and is given by dh ( vr, vt ) k R. Garello. Tutorial on digital modulations - Part 8a [ ] 75

76 BER computation Note that only the first approach can be applied to compute the BER The probability of receiving a vector belonging to the Voronoi region of a signal s j different from the transmitted signal s i must be weighted by the number of wrong bits dh ( v j, vi ) P ( ρ V ( s j ) s T = s i ) k R. Garello. Tutorial on digital modulations - Part 8a [ ] 76

77 BER computation We have where m 1 P ( e) = P ( e s = s ) b b T i m i = 1 Pb ( e st = si ) = Pb ( e, sr = s j st = si ) = j i dh ( v j, vi ) = P ( s R = s j s T = s i ) = k j i dh ( v j, vi ) = P( ρ V ( s j ) st = si ) k j i where ( ) and ( ) v = e s v = e s 1 1 i i j j R. Garello. Tutorial on digital modulations - Part 8a [ ] 77

78 BER computation m 1 P ( e) = P ( e s = s ) b b T i m i = 1 dh ( v j, vi ) Pb ( e st = si ) = P( ρ V ( s j ) st = si ) k j i R. Garello. Tutorial on digital modulations - Part 8a [ ] 78

79 4-PSK: Gray labeling Let us consider this labeling (also called Gray labeling in the following): e : H M v v v v 2 = (00) s 1 1 = (10) 2 2 = (11) 3 3 = (01) s s s /s 4 11/s 3 a a 00/s 1 10/s 2 R. Garello. Tutorial on digital modulations - Part 8a [ ] 79

80 4-PSK: BER m 1 P ( e) = P ( e s = s ) b b T i m i = 1 We must use approach 1: dh ( v j, v1 ) Pb ( e st = s1 ) = P( ρ V ( s j ) st = s1 ) = k j i = P( ρ V ( s2) st = s1 ) + P( ρ V ( s3) st = s1 ) + P( ρ V ( s4) st = s1 ) R. Garello. Tutorial on digital modulations - Part 8a [ ] 80

81 4-PSK: BER Pb ( e) = P( ρ V ( s2) st = s1 ) + P( ρ V ( s3) st = s1 ) + P( ρ V ( s4) st = s1 ) a 1 E ( ) b p = P n > a = erfc = erfc 2 N 2 N 0 0 P ( ρ V ( s ) s = s ) = P ( n < a ) P ( n > a ) = p (1 p ) 2 T P( ρ V ( s ) s = s ) = P( n < a) P( n < a) = p 3 T P( ρ V ( s ) s = s ) = P( n > a) P( n < a) = (1 p) p 4 T Pb ( e) = p(1 p) + p + (1 p) p = p 2 2 R. Garello. Tutorial on digital modulations - Part 8a [ ] 81

82 Graphical computation P( ρ V ( s ) s = s ) = P( n < a) P( n > a) = p(1 p) 2 T A 0 s 2 s 1 0 A s3 s4 R. Garello. Tutorial on digital modulations - Part 8a [ ] 82

83 Graphical computation P( ρ V ( s ) s = s ) = P( n < a) P( n < a) = p 3 T A 0 s 2 s 1 0 A s3 s4 R. Garello. Tutorial on digital modulations - Part 8a [ ] 83

84 Graphical computation P( ρ V ( s ) s = s ) = P( n > a) P( n < a) = (1 p) p 4 T A 0 s 2 s 1 0 A s3 s4 R. Garello. Tutorial on digital modulations - Part 8a [ ] 84

85 4-PSK: BER For a 4-PSK constellation with Gray labeling we have P ( e) b 1 E b = erfc 2 N 0 as for the binary antipodal constellation. R. Garello. Tutorial on digital modulations - Part 8a [ ] 85

86 The 4-PSK constellation as a Cartesian product In fact, when a Gray labeling is used, the 4-PSK constellation can be viewed as the Cartesian product of two binary antipodal constellation, one on the x axis (first bit) and the other on the y axis (second bit) The channel introduces two noise samples on the two axis, which are s.i. R. Garello. Tutorial on digital modulations - Part 8a [ ] 86

87 The 4-PSK constellation as a Cartesian product Given the received vector ρ[n]=(ρ 1 [n], ρ 2 [n]), when the Voronoi region decision is applied: The sign of the first component ρ 1 [n] (along the x axis) uniquely determines the first bit The sign of the second component ρ 2 [n] (along the y axis) uniquely determines the second bit The Gray-labelled La 4-PSK is equivalent to two binary antipodal constellations transmitted in parallel R. Garello. Tutorial on digital modulations - Part 8a [ ] 87