Author: R. Garello. Tutorial on digital modulations - Part 3 2

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1 TUTORIAL ON DIGITAL MODULATIONS Part 3: The signal space. [ ] Roberto Garello, Politecnico di Torino Free download (for personal use only) at:

2 Author: Roberto GARELLO, Ph.D. Associate Professor in Communication Engineering Dipartimento di Elettronica Politecnico di Torino, Italy web: R. Garello. Tutorial on digital modulations - Part 3 2

3 Part 3: The signal space. Summary: Problem introduction. Definition of orthonormal basis. Gram Schmidt The signal space. Vectorial representation R. Garello. Tutorial on digital modulations - Part 3 3

4 Detection problem at the receiver side PROBLEM: given r(t) recover s(t) instead of working with real waveforms easier to solve working with VECTORS R. Garello. Tutorial on digital modulations - Part 3 4

5 Solving the detection problem by using vectors Given the signal constellation M = { s (t),, s i (t),, s m (t) } Build an orthonormal basis B Work in the signal space S generated by B Each signal of S can be expressed as a linear combination of the base elements each signal of S corresponds to a real vector (= coefficients of the linear combination) R. Garello. Tutorial on digital modulations - Part 3 5

6 Short introduction to needed vector theory tools Real vectors v = ( v v ) Scalar product (also called projection),..., k v w k = i= v w E v = v v = v 2 energy ( ) i Orthogonal vectors i i k i= k = viwi = i= v w Parallel vectors v = aw a R 0 R. Garello. Tutorial on digital modulations - Part 3 6

7 Short introduction to needed vector theory tools Linear combination of vectors = weighted sum with real coefficients d i= a b = a b a b a b a R i i i i d d i R. Garello. Tutorial on digital modulations - Part 3 7

8 Short introduction to needed signal theory tools Real signals (Any domain time v( t) t + For most of our applications 0 t < T Scalar product v( t) w( t) = v( t) w( t) dt E v = v( t) v( t) = v ( t) dt energy ( ) 2 Orthogonal signals v( t) w( t) = v( t) w( t) dt = 0 Parallel signals v( t) = aw( t) a R (they have the same shape) R. Garello. Tutorial on digital modulations - Part 3 8

9 Short introduction to needed signal theory tools Linear combination of signals = weighted sum with real coefficients d i== a b ( t) = a b ( t) a b ( t) a b ( t) a R i i i i d d i R. Garello. Tutorial on digital modulations - Part 3 9

10 The basis B Given the signal constellation M = { s (t),, s i (t),, s m (t) } we look for an orthonormal basis B = { b (t),, b j (t),, b d (t) } (d m) B = set of signals. orthogonal T 0 b ( t) b ( t) dt = 0 when j i j i R. Garello. Tutorial on digital modulations - Part 3 0

11 B = set of signals. orthogonal T 0 b ( t) b ( t) dt = 0 when j i j i 2. with unitary energy T 2 b t dt = j 0 ( ) 3. their number d is minimal and sufficient for writing each signal of M as a linear combination d s ( t) = s b ( t) s R i ij j ij j= R. Garello. Tutorial on digital modulations - Part 3

12 Construction of the basis B Given M, how to build B? For simple constellations, it is often not difficult to directly write a basis B Anyway, remember that there exists an algorithm which always provides a basis: Gram-Schmidt algorithm R. Garello. Tutorial on digital modulations - Part 3 2

13 Gram-Schmidt algorithm for a set of vectors For simplicity, we introduce the idea for a set of vectors (for signals is essentially identical). We have a set of vectors and we want to build an orthonormal basis for it. R. Garello. Tutorial on digital modulations - Part 3 3

14 The first vector identifies the first direction (we only have to normalize it, i.e., obtain a vector with unitary energy) R. Garello. Tutorial on digital modulations - Part 3 4

15 To build a direction orthogonal to the first one, we can project the second vector on the first direction, and subtract it. R. Garello. Tutorial on digital modulations - Part 3 5

16 This way, we have obtained two directions which are orthogonal (we only have to normalize them) We can now present the same algorithm for signals R. Garello. Tutorial on digital modulations - Part 3 6

17 Gram-Schmidt algorithm for signals M = { s (t),, s i (t),, s m (t) } STEP Given s (t) compute the first base element Define the first candidate c ( t) = s ( t) compute b ( t) = c ( t) E( c ) ( If c ( t ) = 0 b ( t ) = 0 At the end, all the will be deleted b ( t ) = 0 Meaning: the first signal determines the first direction R. Garello. Tutorial on digital modulations - Part 3 7

18 Gram-Schmidt algorithm Given s 2 (t), look for the second base element. STEP 2 Compute the projection on the first element define s T = s ( t) b ( t) dt c ( t) = s ( t) s b ( t) compute b ( t) 2 = c 2 ( t) E( c ) 2 ( If c ( t ) = 0 2 b ( t ) = 0 2 ) R. Garello. Tutorial on digital modulations - Part 3 8

19 Gram-Schmidt algorithm s Note: T = s ( t) b ( t) dt c ( t) = s ( t) s b ( t) if c ( t ) = 0 2 (s 2 (t) is proportional to b (t) ) b 2 (t)=0 and no new direction is found if c ( t) 0 2 (s 2 (t) is not proportional to b (t) ) b 2 (t) 0 and a new direction is found R. Garello. Tutorial on digital modulations - Part 3 9

20 Gram-Schmidt algorithm Given s i (t) 3 i m STEP i Compute the projection on the previous versors define compute T sij = si ( t) bj ( t) dt j i b ( t) i o i c ( t) s ( t) s b ( t) = i i ij j j= ci ( t) = ( If c ) i ( t ) = 0 bi ( t ) = 0 E( c ) i R. Garello. Tutorial on digital modulations - Part 3 20

21 Gram-Schmidt algorithm s Note: T = s ( t) b ( t) dt ij i j o i c ( t) s ( t) s b ( t) = i i ij j j= ci ( t ) = 0 versors ) b i (t)=0 and no new direction is found if (s i (t) is a linear combination of the previous if ci ( t) 0 (s i (t) is not a linear combination) b i (t) 0 and a new direction is found R. Garello. Tutorial on digital modulations - Part 3 2

22 Gram-Schmidt algorithm FINAL STEP Delete all b ( t ) = 0 i Renumber the survived non-zero b i (t) We have got the basis B = { b (t),, b j (t),, b d (t) } (d m) R. Garello. Tutorial on digital modulations - Part 3 22

23 Important We always have d m If we permute the starting signal constellation we may obtain a different basis the basis is not unique but all of them have the same dimension d R. Garello. Tutorial on digital modulations - Part 3 23

24 Exercize Given the signal constellation M = { s ( t) = + P ( t), s ( t) = P ( t)} T 2 T B Build an orthonormal basis B. B = b ( t) = + PT ( t) T R. Garello. Tutorial on digital modulations - Part 3 24

25 Exercize Given the signal constellation ( f 0 integer multiple of R = /T) M = { s ( t) = + P ( t)cos(2 π f t), s ( t) = P ( t)cos(2 π f t)} T 0 2 T 0 Build an orthonormal basis B. By using 2 B = b ( t) = + PT ( t)cos(2 π f0t) T 2 x sin 2x cos x dx = R. Garello. Tutorial on digital modulations - Part 3 25

26 Exercize Given the signal constellation ( f 0 integer multiple of R = /T) M = { s ( t) = + AP ( t)cos(2 π f t), s ( t) = + AP ( t)sin(2 π f t)} T 0 2 T 0 Build an orthonormal basis B. 2 2 B = b ( t) = PT ( t)cos(2 π f0t), b2 ( t) = PT ( t)sin(2 π f0t) T T R. Garello. Tutorial on digital modulations - Part 3 26

27 Basis construction As mentioned before, for simple constellations it is often possible to directly find a basis B without applying Gram Schmidt. It is sufficient to look for d signals satisfying the definition of orthonormal basis. orthogonal 2. with unitary energy 3. their number d is minimal and sufficient for writing each signal of M as a linear combination Remember that the basis B is NOT UNIQUE, but all the basis B for M have the same dimension d R. Garello. Tutorial on digital modulations - Part 3 27

28 Exercize Given the signal constellation M = { s ( t) = 0, s ( t) = + P ( t)} 2 T B Build an orthonormal basis B. R. Garello. Tutorial on digital modulations - Part 3 28

29 Exercize Given the signal constellation M = { s ( t) = + AP ( t)cos(2 π f t), s ( t) = + AP ( t)sin(2 π f t), T 0 2 T 0 s ( t) = AP ( t)cos(2 π f t), s ( t) = AP ( t)sin(2 π f t)} 3 T 0 4 T 0 B Build an orthonormal basis B. R. Garello. Tutorial on digital modulations - Part 3 29

30 The signal space S Given the basis B B = { b (t),, b j (t),, b d (t) } The signal space S generated by B is defined as d S = a( t) = a b ( t) a R j= j j j (set of all signals which can be expressed as linear combination of the basis signals) R. Garello. Tutorial on digital modulations - Part 3 30

31 d S = a( t) = a b ( t) a R j= j j j Note that we certainly have M S S contains the m starting signals of M, but it is much larger!!! R. Garello. Tutorial on digital modulations - Part 3 3

32 Exercize Given the basis B B = b ( t) = + PT ( t) T What is the signal space S? R. Garello. Tutorial on digital modulations - Part 3 32

33 Exercize Given the basis B What is the signal space S? 2 B = b ( t) = + PT ( t)cos 2 f0t T ( π ) R. Garello. Tutorial on digital modulations - Part 3 33

34 Exercize Given the basis B 2 2 B = b ( t) = PT ( t)cos(2 π f0t), b2 ( t) = PT ( t)sin(2 π f0t) T T What is the signal space S? ( ) ( ) Acos(2 π f t ϑ) = Acosϑ cos(2 π f t) + Asinϑ sin(2 π f t) R. Garello. Tutorial on digital modulations - Part 3 34

35 Linear combination Fixed B, each signal a(t) S can be expressed as linear combination of B signals: d = a( t) a b ( t) j= j j R. Garello. Tutorial on digital modulations - Part 3 35

36 Projection d = a( t) a b ( t) j= Note that the coefficient a j is the projection of a(t) on the basis signal b j (t) j j a j T = 0 a( t) b ( t) dt j If not convinced: since all the basis signals are orthogonal T T T a( t) b ( t) dt = a b ( t) b ( t) dt = a b ( t) b ( t) dt = a j i i j i i j j 0 0 i i 0 R. Garello. Tutorial on digital modulations - Part 3 36

37 Vector representation Fixed B, for each signal a(t) S we have d = a( t) a b ( t) j= j j There is a one-to-correspondence a( t) a = ( a,..., a,..., a ) j d a j T = a( t) b ( t) dt Projection on the versor b j (t) 0 j R. Garello. Tutorial on digital modulations - Part 3 37

38 Vector representation d = a( t) a b ( t) j= j j The signal a(t) corresponds to a real vector a with d real components, and viceversa a( t) a = ( a,..., a,..., a ) j d R. Garello. Tutorial on digital modulations - Part 3 38

39 Constellation vector representation We certainly have M S Each constellation signal s i (t) S Each constellation signal corresponds to a real vector with d components s ( t) s = ( s,..., s,... s ) i i i ij i d R. Garello. Tutorial on digital modulations - Part 3 39

40 Constellation vector representation From signal s i (t) to vector s i s ( t) i s T = 0 s ( t) b ( t) dt ij i j Projection on the versor b j (t) s = ( s,..., s,..., s ) i i ij id R. Garello. Tutorial on digital modulations - Part 3 40

41 Constellation vector representation Note that, for simple constellations, the vector components can often be obtained without computing the projections. We write s ( t) = s b ( t) s b ( t) +... s b ( t) i i ij j id d The basis signals b i (t) are known. We look for a set of coefficients s ij able to satisfy the equation. The solution is unique. R. Garello. Tutorial on digital modulations - Part 3 4

42 Constellation vector representation Constellation M as a signal set Constellation M as a vector set M = { s (t),, s i (t),, s m (t) } M = { s,, s i,, s m } R. Garello. Tutorial on digital modulations - Part 3 42

43 Constellation vector representation The signal space S is isomorphic to the Euclidean space R d (set of all vectors with d real components) We can draw it as a Cartesian space If d=, S Rand can be drawn as a -D line If d=2, S R 2 and can be drawn as the 2-D plane If d=3, S R 3 and can be drawn as the 3-D space We will draw M R d ( a vector constellation is a set of m points in the Euclidean space R d ) R. Garello. Tutorial on digital modulations - Part 3 43

44 Example Example of -D constellation 0 R. Garello. Tutorial on digital modulations - Part 3 44

45 Example Example of 2-D constellations R. Garello. Tutorial on digital modulations - Part 3 45

46 Signal energy Given a signal a(t) S Its energy is given by T 2 ( ) ( ) = E a a t dt 0 Given its vector representation a( t) ( a,..., a,... a ) j d It is easy to show that (Parseval identity) E( a) d = j= a 2 j R. Garello. Tutorial on digital modulations - Part 3 46

47 proof In fact, since d = a( t) a b ( t) j= j j T T d d T d j j j j j 0 0 j= 0 j= 0 0 j= 0 E ( a ) = a ( t ) dt = [ a b ( t )] dt = a b ( t ) dt = a Where we have used the orthogonality property T 0 b ( t) b ( t) dt = 0 se i j j i R. Garello. Tutorial on digital modulations - Part 3 47

48 Constellation energy Given a constellation with {,...,,..., } d M = s si sd R s = ( s,..., s,..., s ) i i ij id We have: E ( s ) d = The (average) constellation energy is equal to: j= where P(s i ) is the probability of transmitting s i E i m = s 2 ij P( s ) E( s ) s i i i= R. Garello. Tutorial on digital modulations - Part 3 48

49 Constellation energy Binary information sequencies: ideal random The binary vectors v H k are equiprobable The labeling is a one-to-one mapping e : H k M The constellation signals are equiprobable ( ) P s i = m The signal constellation is simply: s i M E s m = m i = E( s ) i R. Garello. Tutorial on digital modulations - Part 3 49

50 Energy per information bit Average energy necessary to transmit an information bit via M E b = E k S R. Garello. Tutorial on digital modulations - Part 3 50