2 Signals. 2.1 Elementary algebra on signals

Transcription

1 2 Signals We usually use signals o represen quaniies ha vary wih ime. An example of a signal is he size of he sea swell a some locaion in False Bay: a any paricular ime he waves in he bay have an ampliude (size), and his ampliude varies from one momen o he nex. here are buoys in he ocean ha measure his quaniy, so you can find he informaion on he inerne (he websie repors i for a buoy jus off Milneron). Surfers and sailors find his informaion useful because by seeing a recen plo of wave ampliude versus ime hey know wheher hey should boher o go ou and enjoy he ocean. Even beer, by seeing he curren swell size and measuring a whole lo of oher signals i is possible o predic wha he waves will be like in he shor-erm fuure, so you can plan for omorrow his is he basis for weaher forecasing. he swell size can have wo values for an insan in ime. For example, righ now i doesn make sense o say ha he size is boh 2m and 3m. his is rue for all quaniies ha you can measure. A mahemaical funcion has he same properies: he funcion f() only provides a single value for any paricular value of he independen variable. hus i makes sense o consider a signal o be a funcion. Following he previous example, we could herefore consider he swell size of he waves in a paricular locaion o be a funcion of ime, represened by s(). If you know his funcion and I give you an insan of ime =, you can ell me exacly wha he value of he funcion is, s( ), and by implicaion I know he size of he swell. here are some aspecs of signals ha can be confusing. One is his: as far as he mahemaics is concerned a signal exiss for all values of ime, so as far as we re concerned he signal s() above is known for all possible values of i makes sense o hink of a signal s() as an objec, raher han as a collecion of values. he fac ha we migh no have observed he signal for a lo of is exisence is irrelevan. In principle (ignoring he fac ha due o coninenal drif False Bay didn always exis) here has always been a specific value of swell size for any insan in ime, even hough we weren measuring i. Anoher possibly confusing aspec is ha, for purposes of he heory, we really need o allow signals o ake on complex values for any given he value of s() is herefore a number, bu his value may have a real and an imaginary componen or a magniude and a phase. Admiedly, i s hard o hink of a real-worldsignal ha can be complex valued, bu as far as he mahemaical heory is concerned i is an essenial ingredien. We jus have o make sure ha our mahemaical represenaion evenually leads o a real-valued signal for any physical variable ha i produces. he upsho is ha signals can be represened as funcions of ime. I makes sense o represen a signal by a funcion s(). As far as signal processing is concerned, we could work wih s() as a mahemaical eniy, and i makes sense o ask quesions like Wha happens when we pu s() hrough a lowpass filer? 2. Elemenary algebra on signals A signal is simply a funcion, usually of ime. he basic algebra on funcions, and herefore of signals, requires an undersanding of wo operaions, addiion and muliplicaion. 2.. Addiion of signals here arewo casesha we can hink abou regardingaddiion ofsignals: we could add a consan value k o a signal, or we could add wo signals and y(). In some sense hese wo cases are equivalen. Consider a signal. A signal is simply a funcion. For each value of, he funcion reurns

2 a value. If we wan o plo a funcion, we draw an axis represening he independen variable. hen, for every value of (he domain of he funcion), we find he corresponding value of he funcion a his insan. he graph of hen indicaes his value for he chosen. If we repea his process for all possible values of, he graph races ou a curve ha we consider o be he graph of. he sum of wo signals is anoher signal. Suppose we know wo signals and y(). he sum of hese wo signals is anoher signal (or funcion) ha has he form z() = +y(). If you wan o draw z(), you choose a value of, say for example = 2, and you find he value of z a his poin: z(2). From he known relaion z(2) = x(2)+y(2) you hen find he values x(2) and y(2), add hem ogeher, and you have z(2). he graph of z() a = 2 mus pass hrough his poin. Procedurally, you repea his operaion for every value of, finding he corresponding values of z() from he relaion z() = +y(). Adding o signals (or funcions) in his conex is simple (and you know how o do i): for each value of, add he values of and y(), and he resul is he value of z() a ha insan. In his way you can plo z() as a funcion of ime: Signal value y() z() = + y() he signal z() is herefore obained by jus adding o y() poin by poin. Adding a consan value o a signal is simple. Suppose z() = +k for some fixed k. For each insan in ime, he value of z() is jus he value of a he same insan, wih he consan k added o i. he graph of z() looks he same as he graph of, bu i is shifed in he range direcion by a disance k. Signal value z() = Alernaively, we could hink of z() = + k as z() = + y() wih y() = k (a consan signal wih value k for all ). 2-2

3 2..2 Muliplicaion of signals Inhesamewayasforaddiion,wosignalscanalsobemulipliedpoinbypoin. Ifz() = y(), hen he value of z() a an insan = is jus he produc of he values of and y() a he same insan. hus z( ) = x( )y( ). his is rue for all possible values of : Signal value y() z() = y() Muliplicaion by a consan follows in he obvious way: Signal value 2 - z() = Basic signals o work wih signals we need a basic vocabulary. here are a number of signals ha we will deal wih rouinely he uni sep One really simple signal is he uni sep, which we denoe by u(). he definiion is simple: { u() = <. his is a signal (or funcion) ha has a value of zero for negaive values of ime, and a value of one for non-negaive values of ime. A plo of he uni sep follows: u() 2-3

4 he ineresing poin for he uni sep happens a =, where he signal changes value from o. his raises an ineresing quesion, hough wha is negaive ime? he answer is ha we can pu = anywhere we like as far as he mahemaics is concerned. I could say ha I consider = o be he insan when Mandela was inauguraed. In ha case negaive ime corresponds o pre-994 and oday corresponds o pos-994. he origin for a signal is imporan for pracical purposes, bu for mahemaics we can pu i wherever is convenien for our purpose. he uni sep signal changes insananeously from one value (zero) o anoher (one). I is herefore useful for represening changes in signals ha occur a specific values of ime. You will see laer ha i is useful for characerising sysems: if you pu a uni sep ino a sysem, hen he response of he sysem o his inpu is very informaive (and is called he sep response of he sysem). No real signal can look exacly like a uni sep, since i has a disconinuiy a =. Suppose for example ha u() represen he value of he curren hrough a paricular resisor in a circui hrough ime. his would mean ha he curren would have o change insananeously from a value of zero amps o a value of one amp. his could never occur in realiy: changing a curren insananeously would require infinie energy in pracice, and even he universe doesn conain infinie energy. Noneheless, i is valid mahemaically and in realiy we could make a signal ha almos looks like a uni sep and i s a paricularly useful idealisaion. In signals and sysems heory we are able o deal wih signals ha are undefined a some poins. Sricly speaking, he heory is consisen as long as he signal is defined everywhere excep a a counable (bu possibly infinie) se of poins. For he uni sep u() above we don really wan o concern ourselves wih wha he value is a = i could be one, as shown, bu i could also be zero or any oher value. For purposes of working ou wha happens when a signal goes hrough a sysem, he value a = urns ou o be irrelevan. We can express our ambivalence formally by jus leaving u() undefined a =, and represening he uni sep as below: u() As drawn, u() is no a funcion i akes on all possible values beween zero and one a =, while a funcion mus be single valued. he graph above would be unaccepable in formal mahemaics, bu for our applicaions i conveys he required informaion Sinusoidal signals An general canonical sinusoidal signal akes he form s() = Acos(ω+φ), where A is he ampliude (half of he peak-o-peak variaion), ω is he frequency (in radians/second), and φ is he phase, which basically corresponds o a specificaion of he posiion of he sinusoid along he ime axis. Why is his called a sinusoid while here s a cos in he expression? Well, since sin(x) = cos(x π/2), he only difference beween a sin and a cos is a phase shif, which can be accommodaed by an appropriae value of φ. hus when we alk abou he jusificaion for his lies in wha i means for wo signals o be equal o one anoher. An obvious way o define equaliy beween f() and g() is o require ha f() = g() for all. Under his definiion he signals f() and g() have o be defined everywhere. In signals heory his is no wha we mean by equaliy. Raher, we consider he wo signals o be equal if R (f() g())2 d = : here mus be zero energy in he difference beween f() and g(). his is a much less sric definiion of equaliy, since he wo signals can be differen from one anoher as long as he difference has zero (squared) area. his is also a more physically meaningful measure of similariy: any physical device ha measures a signal has o have some energy ransferred o i, so if wo signals are equivalen in erms of energy hen here is no pracical way of elling hem apar. 2-4

5 sinusoids we mean he class of signals ha varies in a way ha looks harmonic, wihou paying oo much aenion o he exac form of he funcion specified. A signal is periodic wih a period of if = x( +) for all. For he sinusoidal signal above o be periodic we herefore require ha Acos(ω+φ) = Acos(ω(+)+φ) = Acos(ω+ω +φ) for all, which is rue as long as ω = 2πk for some ineger k. he smalles value of for which his is rue is for k =, so he fundamenal period of he signal is = 2π ω. he phase φ of a sinusoid deermines he posiion of he signal along he ime axis. Suppose for now ha φ =. he signal s() above hen corresponds o a sinusoid ha akes on is peak value of A a =. However, for nonzero phase we can wrie he signal as = Acos(ω( + φ ω )), which is jus he signal Acos(ω) shifed o he lef by φ ω = φ 2π. I is eviden hen ha a specified phase shif shifs he signal in ime by an amoun ha is proporional o, he wavelengh of he sinusoid. he phase herefore deermines he shif of he signal measured in unis of he wavelengh, raher han in unis of ime he complex exponenial he complex exponenial is he mos imporan signal in he heory of signals and sysems. A complex exponenial of frequency ω radians per second can be expressed as s() = e jω, where j =. his is an example of a complex-valued signal (or funcion), since by Euler s formula we can wrie i in he recangular form s() = cos(ω)+jsin(ω). hus i has a real componen of cos(ω) and an imaginary componen of sin(ω), boh of which are sinusoidal. Alernaively, i has a consan magniude of and a phase of ω. he complex exponenial is probably he only signal ha you ll ever see ha conains boh an variable and a ω variable in is expression. his isn a coincidence: i s he signal ha links he wo domains of ime and frequency. When we alk abou he componen of a signal a frequency ω, we re really referring o he par of he signal ha looks like e jω. Because a complex exponenial signal akes on complex values we can only really visualise i using wo plos. Below are plos of he real par and he imaginary par of a complex exponenial signal wih frequency ω = 2π: 2-5

6 real(e j (π) ) imag(e j (π) ) Alernaively, he same complex exponenial could be represened in magniude and phase form, leading o he wo plos below: e j (π).5 (e j (π) ) Noe ha he phase is ω, which is a sraigh line, bu since we can ell he difference beween a phase of zero and a phase of 2π we normally plo i in he range π o π (which leads o he riangular phase plo above). An unconvenional bu oherwise informaive way of visualising i is shown below: 2-6

7 .5 imag(e j (π) ) real(e j (π) ).5 5 [phasor diagrams?] he complex exponenial signal of frequency ω is easily shown o be periodic. Assuming a period, he required condiion is ha e jω = e jω(+) = e jω e jω. his holds as long as e jω =, so we mus have ω = 2πk for some ineger k. As for he case of he sinusoid, he fundamenal period is herefore = 2π ω. Mos of Fourier analysis involves forming weighed sums (or linear combinaions) of complex exponenials. o his end i s really useful o have a clear picure of wha i means o muliply a complex exponenial by a consan, generally complex valued. Le = e jω be a complex exponenial and consider he signal y() = z, where z is complex and can herefore be wrien as z = ρe jφ. hen y() = z = ρe jφ e jω = ρe j(ω+φ) = ρe jω(+ φ ω ) = ρx(+ φ ω ). Exercise: he signal y() is herefore sill a complex exponenial a frequency ω. Is ampliude(or magniude) has been scaled by ρ = z and i has been shifed in ime by φ ω, bu i aside from ha i s he same signal. hus muliplying a complex exponenial by a complex value leaves he frequency consan, jus changing is magniude and is phase. Finally, noe ha here s nohing in he heory ha requires ω. Negaive frequencies exis and make sense: he jus correspond o a complex exponenial wih a negaive in he exponen. For example, he complex exponenial signal e jπ has a frequency of ω = π, and i s no he same signal as e jπ of frequency ω = π. Earlier in his secion you were shown he signal e jπ in real and imaginary form, and in magniude and phase form. Draw (or use a compuer o draw) equivalen plos for he signal e jπ, and convince yourself ha hese wo signals are differen (bu equally legiimae) he Dirac dela he Dirac dela funcion is no a funcion in he usual sense (acually i s an insance of somehing called a disribuion funcion, which is like a generalised funcion). However, i is also one of he mos imporan funcions in signals and sysems heory. 2-7

8 he Dirac dela δ() can be defined in erms of wo properies: ǫ δ() = for δ()d = for all ǫ >. he firs condiion says ha he Dirac dela is zero almos everywhere along he ime axis. he only insan in ime where i is nonzero is a he origin. Noe ha he value δ() is no defined. he second condiion says ha if we inegrae he dela funcion over any inerval ha includes he origin, hen we ge a value of. In oher words, he area underneah he dela funcion over any inerval ha includes he origin is uniy. I s a srange definiion: we couldn draw δ() since i s no really a funcion. here is however a way of visualising i as a limiing process. o formulae i in his manner, we firs need o define anoher signal, he uni pulse of oal widh : p () 2 2 Exercise: I is zero everywhere, excep in he range [ /2,/2] where i has a value of one. As for he case of he uni sep, we ve drawn his signal wih verical lines a /2 and /2, expressing he fac ha we are indifferen o is value a hese poins. Convince yourself ha p () = u(+/2) u( /2), where u() is he uni sep defined earlier. Now consider he signal p (), shown below: p () 2 2 I has a heigh of and a oal widh of /. I shares a lo of he properies of he Dirac dela as defined. I is zero everywhere, excep on he inerval [ 2, 2 ]. Also, since he signal is a recangle of heigh and widh / i has a oal area of one, so as long as we choose ǫ /2 we find ha ǫ p ()d =. o make his signal really behave like a dela funcion, we need o ake he limi as. he pulse hen becomes infiniely narrow and infiniely high, bu sill has an area of one. We could herefore reasonably define he dela funcion as δ() = lim p (). he inerval [ 2, 2 ] over which he pulse above is nonzero becomes infiniely small as, so he firs condiion in he Dirac dela definiion is me. Also, if you pick any value of ǫ > i defines he inerval [ ǫ,ǫ]. For he enire pulse, which always has uni area, is conained in his inerval. herefore lim p ǫ () =. he second condiion in he definiion is herefore also me. I s an advanced opic, bu he reason ha disribuion funcions work is because in principle we can do all our mahemaics wih he signal p () insead of wih δ(). All he resuls will hen 2-8

9 depend on. However, afer we re finished we can ake he limi as, so he disappears. he heory of disribuion funcions says ha for mos purposes i doesn maer when we ake his limi: before or afer is jus as good. here are whole exbooks wrien on he heory, bu since oher people have deermined ha i works we simply use i as a ool for our purposes. Since a Dirac dela is no really a funcion, we can draw i. he convenion is herefore adoped ha we represen δ() as shown below: () δ() he noion is suggesive: he funcion as shown is clearly zero everywhere excep a he origin, as required. A he origin, however, we don indicae a value, bu raher use an upward arrow. he () nex o he arrow indicaes ha i has an effecive area of. I also makes sense o hink abou a dela funcion muliplied by a consan value. For example, he signal 5δ() can be hough of as he funcion 5p (), in he limi as. Since he area under 5p () is always exacly 5, we can see ha ǫ 5p ()d = 5 as long as ǫ. aking he limi as of he above we see ha ǫ 5δ()d = 5 for all ǫ >, so he oal area under 5δ() is five. Using he same represenaion as before we would draw he signal as follows: (5) 5δ() he (5) nex o he arrow now indicaes ha he dela has a size or weigh of 5, in he sense ha he oal area underneah he arrow is 5 unis. Alernaively, he arrow as indicaed has 5 unis of area associaed wih i. From he definiion of he Dirac dela we can deermine ha i is an even funcion. Le δ () be he ime-reversed dela funcion, so δ () = δ( ). Since δ() = for i mus be rue ha δ () = for. Also, ǫ δ ()d = ǫ δ( )d = ǫ δ(τ)dτ = for all ǫ >, where he variable subsiuion τ = has been used. hus δ () is a funcion ha saisfies exacly he same properies as were used o define δ(), so i mus be rue ha δ() = δ () = δ( ) and δ() is even. Alernaively, since we consruced he dela funcion in a limi process from a recangular pulse, which is even, i seems eviden ha δ() mus iself be even. 2-9

10 2.3 Using he heory 2.3. Sifing propery of he dela funcion hink abou any signal, and consider an insan in ime. he sifing propery of he dela funcion saes ha δ( )d = x( ). Whenever you are faced wih a saemen ha you don undersand, i usually helps o draw somehing ha you do undersand. In his case, we should hink abou he quaniy inside he inegral, δ( ). I s clearly a funcion of ime, and is a fixed consan number. I s also he produc of wo signals, namely and δ( ). he signal δ( ) is a dela funcion, bu i has been shifed in ime so ha i occurs a ime =. he wo signals and heir produc are shown below: f() = x( ) x( ) () (x( )) δ( ) δ( ) Exercise: he signal δ( ) (shown a he boom) is a dela funcion a =, bu i now has an area (or size) of x( ) since δ( ) has been muliplied by he value of a he same ime insan. he sifing propery is claiming ha he oal area underneah his boom funcion is x( ), which is obviously rue he plo only conains a dela funcion of size x( ), which almos by definiion has an area of x( ). he sifing propery les you do some ineresing algebra. Insead of he signal above, suppose he signal was he consan signal f() shown as a doed red line. he produc f()δ( ) would look exacly he same as he boom plo: he dela funcion only samples he value of he op signal a =, and boh and f() have exacly he same value a his poin. hus δ( ) = f()δ( ) = x( )δ( ), since when you plo hese hey look exacly he same. Suppose = sin(π) and = /4. Plo δ( ) and x( )δ( ). hese wo signals should look exacly he same. Convince yourself ha his would be rue for any signal and any value. Since he inegrands are idenical, we mus have δ( )d = x( )δ( )d. However, x( ) is consan (since we chose ). We can herefore ake i ou of he inegral, and he sifing propery follows: δ( )d = x( )δ( )d = x( ) δ( )d = x( ). 2-

11 Noe ha his las inegral vanishes because a shifed dela funcion has uni area. he previous saemen leads o he following procedure. Since he inegrand δ( ) is only nonzero when =, i doesn maer wha he value is excep a =. Inside he inegral we can herefore replace all insances of (excep he one associaed wih he dela funcion) wih he value. his change doesn change he hing ha is being inegraed. hus δ( )d = x( )δ( )d. Finally, he sifing propery also les us wrie any signal in an ineresing and useful form: = x(τ)δ( τ)dτ. Suppose we choose a value of and wan o calculae. In he above expression is hen a fixed value, and τ is he inegraion variable. When considered as a funcion of τ, he inegrand is only nonzero when τ =, so inside he inegral we can replace x(τ) wih wihou changing he quaniy being inegraed. hus x(τ)δ( τ)dτ = δ( τ)dτ = δ( τ)dτ =. Exercise: You really need o be clear abou wha i means o be inegraing over a variable. In he expression x(τ)δ( τ)dτ, he variable τ is he inegraion variable, and in his expression is a consan. o evaluae he inegral you should herefore hink abou a fixed value of, and plo x(τ), δ( τ), and x(τ)δ( τ) as a funcion of τ. Repea he previous example in his conex, and convince yourself ha he expression (or decomposiion) of above is correc he generalised derivaive he sandard formulaion of calculus defines differeniaion in erms of gradiens in a limiing process. If a funcion f () is he derivaive of he funcion f(), hen a any ime insan = we mus have f f( +h) f( ) ( ) = lim. h h o use his expression o calculae he derivaive a, he value f( ) mus be defined. Also, for he limi o be meaningfully defined i mus no maer wheher h from he posiive side or from he negaive side: he same value mus be obained in he limi for boh cases. Consider he signals below: f() g() he signal f() exhibis he second problem jus oulined. While i is defined everywhere(including a = ), he slope depends on which side of he origin we are on: for negaive i is zero, while for posiive i is one. hus he slope as we approach he origin from he lef is, while i is if we approach from he righ. Since no meaningful value can be deermined for f (), he derivaive doesn exis a he origin. he signal g() exhibis boh problems. A he origin is value is no defined, as indicaed by he verical line a his poin on he graph. Since g() is no defined, we can even begin o use he 2-

12 formula o find he derivaive. he slope when approaching = from he lef and from he righ is also differen, so he second problem also sill persiss. In order o meaningfully define a derivaive for he signals above we herefore need o proceed in a differen manner. A soluion is o hink of he inverse process of differeniaion, namely indefinie inegraion. If y() = d d, hen elemenary calculus les us express in erms of y(): = y(τ)dτ. Noe he appearance of he in he inegraion limis, and he use of he dummy variable τ. Admiedly here could be an undeermined inegraion consan added o he righ-hand side, bu i is no imporan for our purposes. Since differeniaion and indefinie inegraion are inverse operaions, we proceed o define differeniaion in erms of inegraion: he generalised derivaive of is he funcion x () which, when i is inegraed in he indefinie sense, yields. Consider for example he indefinie inegral of he Dirac dela funcion: = δ(τ)dτ. he value of a = 5 is x( 5) = 5 δ(τ)dτ, which can be evaluaed and reurns a number. o find his number i is easies o draw he funcion being inegraed: () δ(τ) τ he value of x( 5) is he area underneah his funcion over he inerval (, 5], and since he funcion is always zero over his range we clearly have x( 5) =. On he oher hand, x(5) = 5 δ(τ)dτ is he area under he dela funcion drawn over he range (,5]. he signal is no zero over his range i includes a dela funcion of weigh, which has exacly one uni of area. hus x(5) =. For any value of, he signal has a value equal o he area underneah δ(τ) over he inerval (,]. For < we are no inegraing over he dela funcion a he origin, so =. For > he inegraion always includes he dela funcion a he origin, so =. We don know wha he value of for =, so we leave i undefined. In any case, we can draw he indefinie inegral as follows: hus we see ha he indefinie inegral of he dela funcion is he uni sep. According o he definiion of he generalised derivaive, he derivaive of he uni sep is herefore he dela funcion. Now consider again he funcion g() shown earlier in his secion. I isn defined a = and cerainly doesn have a derivaive a his poin. Noneheless, in he same way we can define he generalised derivaive g () o be he funcion which, when inegraed in he indefinie sense, yields g(). Specifically, we are looking for a signal g () such ha g() = 2-2 g (τ)dτ.

13 Once you see he answer you ll quickly work ou how o consruc generalised derivaives. Consider he signal shown below, also expressed as a funcion of τ in he form of x(τ): () () x(τ) τ We wanocalculaehe indefinie inegraly() = x(τ)dτ. For <, he areaunderneah his curve over he inerval (,] is zero, so y()) = in his range. For an infiniesimal disance o he righ of he origin, indicaed = +, he inegral is over he range (, + ]. his includes he dela funcion a he origin, which has area, so y( + ) =. For >, he inegral is over he range (,], which includes he impulse a he origin as well as unis of area conribued by he usual par of he inegral. hus y() = + over his range. he funcion y() is hus idenical o g(). Since he inegral of he funcion above is g(), i follows ha is he generalised derivaive of g(). hus he pair below form an inegral-derivaive pair: Indefinie inegral g () () slope = g() Derivaive Evidenly he derivaive of g() can be wrien in he analyical form g () = u()+δ(). he firs erm, u() is he ordinary derivaive (gradien of he funcion) ignoring any disconinuiies. he second erm, δ() is an impulse of size a he origin, and conribues o he disconinuiy of size a he origin in g(). Some hough should lead o he following conclusion. he generalised derivaive consiss of wo pars: one is he ordinary derivaive (or slope) of he signal, ignoring any disconinuiies. In he example above his is he par u(). However, here is an addiional componen comprised of dela funcions, which accouns for any disconinuiies in he signal being differeniaed. For he case of g(), if we walk along he signal from lef o righ here is a disconinuiy of size a =. his is accouned for in he derivaive by a Dirac dela of size a =, or in oher words by including a erm δ(). In general we need one impulse locaed a each poin of disconinuiy, and is weigh will be he size of he disconinuiy. Noe ha if he funcion akes a sudden disconinuous sep downwards a a poin, hen he generalised derivaive will conain a negaive impulse of he appropriae size a ha poin. 2-3