Signal space Review on vector space Linear independence Metric space and norm Inner product

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1 Sgnal space.... Revew on vector space.... Lnear ndependence Metrc space and norm Inner product Orthonormal bass Waveform communcaton system Some examples... 6

2 Sgnal space he noton of sgnal space s fundamental n communcaton. It allows a smple geometrc approach to complex communcaton problems.. Revew on vector space Vector spaces form an algebrac structure based on Felds. A vector space s composed of a set V of elements called vectors. hs set possesses a composton law called "addton of vector". he set of vectors along wth the addton of vectors forms an abelan group. In addton, there exsts a multplcaton operaton between elements (called "scalars") from a feld F and vectors from gvng vectors. In ths set of notes, we are gong to use bold letters to desgnate vectors and normal letters for scalars. o recaptulate, we must have: Gven V = {x, x, }, the elements must satsfy: x, y, z V ( x + y) + z = x + ( y + z ) {assocatvty} unque V x + = + x = x {addtve dentty} x V ; x V x + ( x) = ( x) + x = {addtve nverse} x, y V x + y = y + x {commutatvty} and α F and x V αx F such that α, β F and x V ( α + β ) x = αx + βx α F and x, y V α( x + y) = αx + αy We say that we have a vector space V over the feld F. We can defne many dfferent types of vector spaces. Example: n he set of n-tuples from R s a vector space: x = ( x, x,, x n ) R. he feld s the set of real numbers R. A set that s very mportant s the set of tme lmted functons that have a fnte energy. he feld s the set of complex numbers C. We denote ths set L (). he vector space s the set L () over the set of complex numbers C. x( t) = t, such that x ( t ) <. When we consder t as a tme [ ] functon, we denote t x(t) and when we consder t as a vector we desgnate t x.

3 . Lnear ndependence When we wor wth vectors, t s convenent to be able to express vectors as lnear combnaton of other vectors. A vector y s a lnear combnaton of the followng n vectors x, x,, x n f we can express t as: y n = = α x where the scalars α, α,,α n are not all zero. Now, a set {x, x,, x n } of vectors from V forms a lnearly ndependent set f no one of the vectors can be expressed as a lnear combnaton of the others. hs means that n αx = f and only f α = α = = α n =. = Dmenson of the space In a vector space, the maxmum number of lnearly ndependent vectors s called the dmenson of the space. Some spaces are fnte dmensonal. he vectors that represent felds n electromagnetc theory are three dmensonal vectors. However, functon spaces are usually nfnte dmensonal spaces. Example: Consder the set of complex functons of the real varable t defned as follows: exp( jt) t π ϕ ( t) = elsewhere + We can show that αϕ ( t) = f and only f α = for all Z. = Bass (plur. Bases) When we possess a set B = {u, u,, u n } of vectors from V where n = dmv, we can express any vector from V as a lnear combnaton of the vectors composng B. We say that the set of vectors B forms a "bass". So, for x V, we can wrte x n = = α u. We say that the set B spans the space V and we can wrte also V = span{u, u,, u n }. At that tme, the vectors can be represented by the sequence of coeffcents x = ( α, α,, αn ). Example: Consder the nfnte set of functons defned prevously. Any functon that s nonzero only n the nterval [, π] and that satsfes the Drchlet condtons (see EE3 course) can be represented as the followng lnear combnaton: 3

4 + x( t) = α exp( jt) (.) = Equaton (.) s nothng but the Fourer seres representaton of the sgnal x(t). hs sgnal s represented usng the above bass and ts coordnates are the Fourer coeffcents α. π α = x( t)exp( jt ) π (.) he vector x can then be represented by the nfnte sequence: x = (, α, α, α,, α, ) Subspace If we are gven a vector space V over the feld F, we can defne a subspace as follows: W s a subspace of V f t s a subset of V closed under the addton of vectors and multplcaton by scalars from F. In fnte spaces, we always have dmw < dmv..3 Metrc space and norm Dstance If we want to compare elements from a gven space, we have to defne the noton of metrc or dstance. he dstance between two elements a and b of a set S s the postve real number d(a, b) satsfyng:. a, b S d( a, b), wth d( a, b ) = f and only f a = b. a, b, c S d( a, c) d( a, b) + d( b, c) (trangular nequalty) he set S s called a metrc space. Norm If now the set S possesses the structure of a vector space over C, we can defne the length of the vectors usng the concept of norm. he norm of a vector x s the postve real number x satsfyng:. x V x, wth x = f and f x=. x, y V x + y x + y 3. x V and α C αx = α x When a norm s defned n a vector space, we can deduce from t a dstance between ponts n the space. If we consder that a vector s consttuted by ponts: Its orgn (at the orgn of the bass) and ts tp, we can use t to 4

5 represent ponts n the space spanned by the bass. So, the vector x represents also the pont n the above space havng as coordnates the coordnates of x. So, gven two ponts represented by x and y, the dstance between x and y can be expressed by: d ( x, y) = x y (.3) Now f a metrc space has the structure of a vector space over C, a norm can be defned usng the dstance. Snce a vector x s represented by two ponts (the orgn and ts tp) and the orgn s represented by zero, then we can wrte: x = d(, x ) (.4).4 Inner product he concept of nner product can be defned only for vectors spaces defned over the feld of complex (real) numbers C (R ). Gven a vectors space V defned over C, the nner product of two vectors s defned as follows: Let, V x, y satsfyng: x y, the nner product of x and y s the complex number. x V ( x, x ), wth x, x = f and only f x =. x, y V ( x, y) = ( y, x ) * 3. x, y, z V and α, β C ( αx + βy, z) = α ( x, z) + β ( y, z) Example: Consder the two dmensonal real vectors represented by ther coordnates: x = ( x, x) where x, x R. he nner product of the two vectors x. y = x y + x y = x y cos x, y. It s also called x and y s usually wrtten as "dot product". If we consder the space L () defned prevously, the nner product s defned as: * (, ) x y = x t y t (.5) Eucldan Norm he propertes of the nner product can be used to defne a norm n the vector space: the Eucldan norm (, ) x = x x (.6) In the two dmensonal space shown above, the length of a vector x havng as coordnates x and x s x = x + x 5

6 In the space L (), the norm of a functon s: x = x ( t ) and ths s the energy of the sgnal x(t). Propertes of the nner product Cauchy-Schwartz nequalty he Cauchy-Schwartz nequalty s an mportant property of the nner product. It s commonly used to solve some optmzaton problems that occur n communcaton theory. It s also used to defne the concept of an angle between two vectors. Gven two non zero vectors x and y from the vector space V defned over C, the followng nequalty apples: x, y x y wth equalty f x = αy, α beng a non zero scalar. Proof: Let x and y be two nonzero vectors from the vector space V defned over C and the arbtrary nonzero scalar λ, consder the postve number: Developng the product, we obtan: λ beng arbtrary, let λ = x y y D( λ) = x λy = x λy, x λy * D( λ) = x λ y, x λ x, y + λ y (, ) D( λ ) = x x y y * * (, ) (, ) (, ) = x λ x y λ x y + λ y. Usng ths value, D(λ) becomes:. Snce D(λ), the nequalty s proved. We wll have equalty f D(λ) =. In ths case, x = λy. Usng the Cauchy-Schwartz nequalty, we can defne the angle between the vector x and y as: (, ) cosθ = x y x y (.7) If x = λy, the angle between the two vectors wll be or 8 (cosθ = ±). hs means that the two vectors are collnear. If the two vectors are nonzero, 6

7 they wll be orthogonal f (x, y) = (θ =± π ). We use the notaton x y to say x orthogonal wth y. In the L () space, the Cauchy-Schwartz nequalty can be expressed as: * x( t) y ( t) x( t) y( t) Pythagorean heorem If two vectors x and y are orthogonal, they satsfy the Pythagorean theorem: x + y = x + y Proof: (, ) (, ) (, ) x + y = x + y x + y = x + x y + y x + y. However, x y, so (x, y) = (y, x) =..5 Orthonormal bass In sgnal spaces (and even n general vector spaces) t s convenent to represent vectors usng an "orthonormal bass". Consder a space V such that dmv = n. (n can be nfnte). he set of vectors { ϕ, ϕ,, ϕn} forms an orthonormal bass f the vectors satsfy: ( ϕ, ϕ ) = = hese vectors are lnearly ndependent and ther number s equal to the dmenson of the space. So, we can wrte V = span{ ϕ, ϕ,, ϕ n}. hs means that any vector n V can be expressed as: n = = x α ϕ (.8) One advantage of an orthonormal bass s the smple expresson of the coeffcents. If we compute the nner product of x wth one bass vector, we obtan: n n ( x, ϕ m ) = α, m = α (, m ) ϕ ϕ = ϕ ϕ = So: 7

8 m (, ) α = x ϕ (.9) In the case of the L () space, the expresson of the coeffcents s then α = x( t) ϕ ( t) (.) m In the sgnal space L (), the expresson (.8) s called Fourer decomposton of the sgnal x(t) and s wrtten as: n x( t) = α ϕ ( t) (.) = he coeffcents α are called the Fourer coeffcents. m m Orthogonal spaces Consder a vector space V of dmenson n. It can be decomposed nto the "sum" of two "orthogonal" subspaces V and V such that: x V and x V x x x V x V x V x = x + x Usng orthonormal bases, we can easly decompose any vector nto the two above components. Assume that we are gven an orthonormal bass such that V = span{ ϕ, ϕ,, ϕn} and let us defne two subspaces V = span{ ϕ,, ϕm} and V = span { ϕ,, m+ ϕ n}. It s clear that V and V are orthogonal. If we consder any vector of V, ts frst m coordnates defne an element of V whle the last n m coordnates defne an element of V. So: m x V x = α ϕ V and x = α ϕ V. he coeffcents α = = m+ are computed usng equaton (.9): α = ( x, ϕ ) n. he vector x s the orthogonal projecton of x on V whle the vector x s the orthogonal projecton of x on V. he Gram-Schm procedure he Gram-Schm procedure s a procedure for dervng an orthonormal bass from a set of M vectors spannng a space of dmenson K M. So, we are gven the set { x, x,, xm } of vectors, not necessarly lnearly ndependent. he procedure follows the subsequent steps: 8

9 . ϕ = x x ; ths s an ntalzaton step.. For = to M let: = (, ) g x x ϕ ϕ ; We subtract from x ts projecton on the = dmensonal space spanned by the already found bass vectors. hs mples that g belongs to a space that s orthogonal to the space spanned by the bass vectors ϕ. If g s zero, the vector x must be dscarded. It belongs to the prevous space and t s not lnearly ndependent. he bass vector s then: ϕ = g. g At the end of step, we wll have a set of K orthonormal bass vectors. If x, x,, xm s composed of lnearly ndependent vectors, we wll have the set { } K = M. Otherwse, the dmenson of the space wll be smaller than the number of gven vectors. Expresson of the nner product n an orthonormal space Let us consder two vectors x and y wth coordnates (x, x,, x K ) and (y, y,, y K ) n an orthonormal bass. It s easy to show that: K K K * ( x, y ) = x ϕ, y ϕ j j = x y (.) = j= =.6 Waveform communcaton system Let us consder the followng communcaton system Source of nformaton m s (t) y(t) ransmtter Recever n(t) We are gven a source of nformaton that can produce M dfferent symbols. he transmtter assgns a sgnal s (t) to the message m. he sgnals belong to the space L (). hs means that the duraton of a symbol s seconds and the rate of symbols (t s called the "baud rate") s r =. he total number of sgnals produced by the transmtter s M. hese sgnals belong to a fnte 9

10 dmensonal space. In fact, we can use these M sgnals to derve an orthonormal bass usng the Gram-Schm procedure. he derved bass s composed of K M orthonormal sgnals ϕ (t). Let us call ths space S K, S {,, K = span ϕ ϕ K }. he sgnals generated by the transmtter can be expressed n ths space as: (,,, ) K K s ( t) = α ϕ ( t) ; =,, M, = So, we have a correspondence between the sgnal s ( t) and the vector s = α, α,, α. ϕ ( t) m, s t ROM α, α α, K ϕ ( t) ϕ ( t ) K Fg.3- Possble transmtter structure Fg.3- shows a possble mplementaton of a transmtter. he symbol m selects a set of K coeffcents. he waveform produced by the transmtter s a sequence of fnte tme sgnals, each one beng a weghted sum of bass functons. he nose sgnal n(t) s assumed to be a whte Gaussan nose wth a power spectrum S n (f) = N. he receved sgnal y(t) s the sum of the transmtted sgnal and ths nose sgnal. Despte the fact that the sgnal at the output of the transmtter belongs to a fnte dmensonal space, the sgnal at the nput of the recever s a sgnal whch belongs to an nfnte dmensonal space. We can see ths by developng the nose process. Consder the followng orthonormal bass: {ϕ, ϕ, ϕ K, ϕ K+, } where the frst K bass vectors are the vector representaton of the bass functons used to represent the sgnals generated by the transmtter. We complete ths bass by

11 an arbtrary set of orthonormal functons to span an nfnte dmensonal space. In ths bass, we can show that the nose process s expressed as: K (.3) n( t) = N ϕ ( t) = N ϕ ( t) + N ϕ ( t) = = = K + Equaton (.3) shows that the nose process s the sum of two orthogonal processes. he frst one belongs to the same space as the sgnals generated by the transmtter (S K ). he second one s orthogonal to all these sgnals. he coordnates N are random varables and they are the projectons of the nose vector on the bass functons. N n, ϕ n( t) ϕ ( t) (.4) = = Snce the nose process s Gaussan, these varables are Gaussan random varables. We can derve ther statstcs as follows: E[ N ] = E n( t) ϕ [ ] t = E n t ϕ t = So, they are zero mean. her covarance s then: E N N j E n( t) ϕ [ ] t n u ϕ j u du = = E n t n u ϕ t ϕ j u du he nose process s whte. Its autocorrelaton functon s: N Rn ( t u) = E[ n( t) n( u) ] = δ ( t u) he covarance s then: N E N N j = ϕ t ϕ j t Snce the bass functons are orthonormal, the above covarance s: N = j E N N j = j So, the components of the nose process are ndependent Gaussan random varables wth zero mean and varances equal to N. hese components wll be the same n any orthonormal bass. A whte Gaussan nose s always represented by an nfnte dmensonal vector wth coordnates that are ndependent zero mean Gaussan vector wth varance equal to N. he sgnal observed by the recever s the sgnal y(t). It conssts of the sum of a fnte dmensonal vector (s (t)) and the nfnte dmensonal vector

12 representng the whte nose process. However, we can express ths receved sgnal as a sum of two orthogonal sgnals. y( t) = s ( t) + n ( t) + n ( t) = z( t) + n ( t) K where n ( t) = Nϕ ( t) SK and n( t) = N ϕ ( t) S = = K + K. he sgnal z(t) s z( t) = s ( t) + n ( t). It s clear that z( t) SK and that z( t) n ( t). We can show that the observaton of the vector representng the sgnal n (t) s not needed f we want to detect whch message has been transmtted. So, at the recever, we are gong to observe only the sgnal z(t) and not the sgnal y(t). he sgnal z(t) s the orthogonal projecton of y(t) on the space S K and t can be represented by a fnte dmensonal vector z = (z, z,, z K ). he sgnal y(t) on the other hand must be represented by an nfnte dmensonal vector y = (y, y,, y K, y K+, ) and the frst K coordnates correspond to the vector z. We can now wrte the expresson of the condtonal pdf of the vector z gven that the symbol m s transmtted. he coordnates of z are: z = α + N for =,, K. he numbers α, are the coordnates of the, sgnal s (t). So, the coordnates z are ndependent Gaussan random varables wth mean α, and varance N. he condtonal pdf of z s: K z α, f m ( m ) exp z z = = N N π or: K f ( z m ) exp ( z α ) = (.5) z m K, π N N = We can use equaton (.5) to determne the structure of an MAP recever. We have seen that the MAP rule s: P m f ( z m ) > P m f ( z m ) for all j ; j =,, M" "Decde m f z m j z m j Replacng and tang logarthms, we obtan: K "Decde m f ( z α, ) N ln P( m ) = mn " (.6) he frst term n the above expresson s just the square of the dstance between the vector z and the vector s. So we can re-express the rule (.6) as: j

13 { N P( m )} z s ln mn (.7) If the symbols are equprobable (ML recever), the second term of (.7) becomes rrelevant and the decson rule smplfes to: Decde m such that z s mn (.8) he recever s called a "mnmum dstance classfer". In order to buld the recever, we must determne the coordnates of the vector z. We have seen that they are the frst K coordnates of the vector y. From (.9), the coordnates of y are gven by: y = z = ( y, ϕ ) = y( t) ϕ t for =,, K. So, the recever s composed of three cascaded sectons. he frst one computes the coordnates of the vector z. he second one s a dstance computer and the thrd one apples a bas to tae nto account the unequal a pror probabltes. y(t) ϕ (t) ϕ (t) ϕ K (t) z z z K Dstance computer z s z s z s M N ln P m + + N ln P m + N ln P( m M ) Select mnmum Fg.3- Recever structure usng dstance computer he above structure uses K "correlators" that transform the random process y(t) to the random vector z. he dstance computer has K nputs (coordnates of z) and computes M dstances (between the receved vector z and 3

14 the "prototypes" s ). If the symbols are equprobable (ths s often the case), the bas N ln P m s not requred. A dfferent structure can be obtaned f we develop the square of the dstance. (, ) (, ) (, ) (, ) z s = z s z s = z z + s s z s (.9) In the above expresson, the squared norm of z does not depend on the message. he same value z wll appear for all symbols. It can be elmnated. he nner product of s wth tself s the energy of the sgnal s (t)., = = s t = E s s s So, the expresson (.8) smplfes to: z s N ln P m = cste + E z, s N ln P m he decson rule becomes: E N Decde m f ( z, s ) + ln P( m ) max. z s can be computed usng the sgnal y(t) present at he nner product, the nput of the recever because y(t) = z(t) + n (t) and n (t) z(t). So: ( z, s ) = ( y n, s ) = ( y, s ) = y( t) s t Fnally, the decson rule s: Decde m f E N y( t) s ln max t + P m (.) he decson rule (.) can be mplemented usng the structure shown n Fg.3-3. hs structure uses M correlators and does not requre a dstance computer. If the symbols are equprobable, the bas becomes only E. he correlaton recever has a smpler structure than the mnmum dstance one. However, t uses M correlators. In many cases, (PAM, MPSK and QAM) the dmenson of the sgnal space s less or equal to two whle the number of sgnals can be very hgh. In ths case, the mnmum dstance recever s better. However, when the dmensonalty of the sgnal space s hgh and M s equal to K, t s better to use the correlaton recever structure. 4

15 y( t ) s ( t ) E N ln P( m ) s ( t ) E N ln P( m ) + + Select maxmum s M t Fg.3-3 Correlaton recever + EM N ln P m M he dfferent correlators can be mplemented by matched flters. We have already studed ths equvalence n the matched flter course. he correlator output s t y( t) s t = y u s u du = s u y t u du he last ntegral s the evaluaton of the convoluton of a flter wth mpulse response h( t) = s ( t) wth the sgnal y(t) at the tme t =. So, we obtan the followng equvalence: y( t ) s ( t ) h( t) = s ( t) t= Sample at 5

16 he decson rule dvdes the observaton space nto decson regons I such that z L decde m. If the symbols are equprobable, the decson I s the set of ponts that are closer to the pont s than to the other ponts..7 Some examples For all examples, we assume equprobable symbols. Antpodal sgnalng hs communcaton system s a one-dmensonal bnary system. We use two sgnals: s ( t) = Eϕ( t) and s ( t) Eϕ( t) =. he functon ϕ(t) has an energy equal to one. Usng ϕ as a bass, the sgnal space s represented below. E E z We can use a mnmum dstance classfer. It uses only one correlator. Let the output of the correlator be called z. In ths partcular case, we don't have to compute the dstance between the observed value z and the ponts E and E. he decson regons are separated by the orgn. he optmum recever s: ϕ ( t) + Fg.3-4 Antpodal sgnallng recever Gven that the dstance between the mean of each dstrbuton and the N threshold s A = E and the varance s, the probablty of error s: A E P( E) = erfc = erfc σ N (.) hs s the same result as the one deduced usng the matched flter derved n the prevous chapter. 6

17 Pulse ampltude modulaton (PAM) he PAM system s also a one dmensonal system. However, n ths system, we use M dfferent ampltudes. he ampltudes are equally spaced and are such that the average value s zero. Furthermore, we assume that M = N n order to be able to encode the dfferent levels n bnary. So, M levels correspond to N = log M bnary dgts. he level separaton s A. s s s 3 s 4 3A A A 3A z Fg.3-5 PAM wth M = 4 he decson regons are ntervals separated by thresholds occurrng mdway between the sgnal levels. For example, the regon I correspondng to the sgnal s s the nterval [A, ]. he regon I correspondng to s s the nterval ], A]. For the general case of M dfferent sgnal levels, we are gong to have M thresholds. ϕ ( t) Analog to dgtal converter Fg.3-6 PAM recever he above fgure shows a typcal PAM recever. he analog to dgtal converter encodes drectly the sgnal levels nto bnary. It can be mplemented usng M comparators followed by a prorty encoder or t can be mplemented usng a successve approxmaton analog to dgtal converter. he 7

18 dfferent levels are usually encoded n Gray code, so that an error n one level wll cause only one bt to be n error. o compute the probablty of error, we have to tae nto consderaton whether the level s an ntermedate level (between two thresholds) or an extreme one (bounded by one threshold only). We have two extreme levels: s and s M. he condtonal probablty of error s the area of a tal of a Gaussan A N wth a threshold away from the mean by and a varance. So: A A P( E m ) = P( E mm ) = erfc = erfc N N For the ntermedate levels, we have two thresholds, so for the levels s up to s M, the condtonal probablty of error s the probablty of beng away from the mean by A on both sdes. So: A A = = = = P( E m ) P E m M erfc erfc N N Fnally, the probablty of error s the average: M M A P( E) = P( m ) P( E m ) = erfc (.) = M N We can express the ampltude A as a functon of the average energy of the sgnals: M M E = s = α M M = = where the coeffcents α are the ampltudes of the sgnals. For the PAM sgnalng where the spacng between the ponts has a value A M of A, the dfferent ampltudes are: ± ( ) for = to. he average energy s then: M A M E = ( ) = A M = 4 8

19 Replacng n (.), we obtan: M 3 E P( E) = erfc M M N (.3) Orthogonal sgnalng hs s a two dmensonal sgnalng scheme. he two sgnals are: s( t) = Eϕ( t) and s( t) = Eϕ( t). he two bass functons are orthonormal. E he coordnates of the two vectors are ( E,) and (, E ). We can show that the probablty of error s ndependent on translatons and rotatons n the sgnal space. So, under proper translaton and rotaton, the above sgnal space s equvalent to the followng one: E E E he above fgure s the same as the antpodal case. he probablty of error s: A P( E) = erfc σ E E N where A = = and σ = 9

20 he probablty of error s then: E P( E) = erfc N (.4) An example of orthogonal sgnalng s the orthogonal FSK system. he recever can be mplemented as follows: ϕ ( t) + + ϕ ( t) Fg.3-7 orthogonal recever structure Non coherent FSK In ths communcaton system, the phase of the receved waveform s unnown. It wll be modeled as a unformly dstrbuted random varable over [,π ]. We assume also that the frequences are selected so that the two carrers are orthogonal. he recever observes one of the followng two sgnals: s ( t) Acos ω t s ( t) = Acos ω t + Θ = ( + Θ ) and Θ s a random varable unformly dstrbuted over [, π]. he above representaton corresponds to an nfnte number of sgnals. Consder the followng four orthonormal functons: ϕ( t) = cosωt, ϕ( t) = snωt, ϕ3( t) = cosωt, ϕ4( t) = snωt and let θ be a realzaton of the random varable Θ. We can represent the above sgnals n the four dmensonal space spanned by ϕ, ϕ, ϕ 3, ϕ 4. he correspondng vectors are: s = A cos θ, A sn θ,,

21 s =,, A cos θ, A snθ he vector observed by the recever s one of the above sgnals plus nose. We have seen that the components of nose are ndependent Gaussan random N varables wth zero mean and a varance equal to. So, the four dmensonal z = z, z, z, z. observed vector s the followng Gaussan random vector: 3 4 When ω s transmtted and Θ = θ s observed, z has the followng pdf: f ( ω, θ ) exp z A cos z A sn z z ω θ ( π N ) N θ θ = z z , 3 4 A Usng E =, the above expresson becomes: f ω θ ( ω, θ ) exp z z = z E cosθ z E snθ z z ( π N ) N, 3 4 hs expresson becomes: f, ( ω, θ ) exp z z z3 z4 E E z cos z sn ω θ ( π N ) N θ θ z z = When ω s transmtted and Θ = θ s observed, z has the followng pdf: f, ( ω, θ ) exp z z z3 z4 E E z3 cos z4 sn ω θ ( π N ) N θ θ z z = A change of varable (from rectangular to polar) produces an expresson of the pdf that s more nformatve. So, let z = r cosψ, z = r snψ, z3 = r cosψ and z4 = r snψ. he above pdf's become: r r f ( r, ψ, r, ψ ω, θ ) exp r r E Er cos ( π N ) N θ ψ = r r f r, ψ, r, ψ ω, θ = exp r + r + E Er cos θ + ψ ( π N ) N If we can measure the phase of the carrer, we can use the above expressons for an MAP recever desgn. However, we don't have that

22 nformaton. he soluton s to tae a decson after averagng the above pdf's over θ. π f ( r, ψ, r, ψ ω ) = f ( r, ψ, r, ψ ω, θ ) dθ π r r E r r E r ( θ ψ ) dθ = exp exp cos ( ) N π N π N We use the fact that: π exp( z cos λ ) dλ = I ( z) π, I s the modfed Bessel functon of the frst nd of order zero. he pdf s: r r E f ( r, ψ, r, ψ ω) = exp ( r + r + E) I r (.5) ( π N ) N N When ω s transmtted, the pdf becomes: r r E f ( r, ψ, r, ψ ω) = exp ( r + r + E) I r ( π N ) N N (.6) he maxmum lelhood decson rule s then: E E "Decde ω f I r > I r, decde ω otherwse." N N he functon I s a monotone ncreasng functon of ts argument. he decson rule can be smplfed: "Decde ω f r > r, decde ω otherwse." he recever should compute r and r and compare them.

23 cos ω t r sn ω t cos ω t r sn ω t Fg.3-8 Non coherent FSK recever We can show that the above structure can be mplemented usng two bandpass matched flters tuned respectvely to ω and ω followed by envelop detectors. he mpulse response of the bandpass flter matched to ω s h( t) = p( t)cos( ω t + θ ) where p( t ) = for t and zero elsewhere and θ s arbtrary. BPF ω Envelop detector r BPF ω Envelop detector r Fg.3-9 Non coherent FSK recever he decson depends only on the ampltude of the output of the bandpass flters. he condtonal probablty of error can be computed as follows: Gven that ω s transmtted, we have an error f r s larger than r. So: P E ω f ( r, r ω ) dr dr = D 3

24 r D We need of course to compute the jont pdf of the two envelops r and r. π π (, ) = (,,, ) f r r ω f r ψ r ψ dψ dψ π π r r r + r + E E = ( π N ) exp I r dψ dψ N N Fnally: 4r r r + r + E E f ( r, r ω ) = exp I r (.7) N N N From the above expresson, we can see that the two varables (r and r ) are ndependent and ther pdf's are: r r + E E f ( r ω ) = exp I r (.8) N N N hs means that r s Rcan. r r f ( r ω ) = exp (.9) N N hs means that r s Raylegh. he condtonal probablty of error s: + + P( E ω ) = f ( r ω) f ( r ω) dr r dr After some smple manpulatons, we obtan: E P( E ω ) = exp N he other probablty of error has the same expresson. Snce the two symbols are equprobable, the probablty of error has the same expresson also. E P( E) = exp (.3) N r 4