The δ-function & convolution. Impulse response & Transfer function

Transcription

1 Topic 3 The δ-funcion & convoluion. Impulse response & Transfer funcion In his lecure we will described he mahemaic operaion of he convoluion of wo coninuous funcions. As he name suggess, wo funcions are blended or folded ogeher. We will hen discuss he impulse response of a sysem, and show how i is relaed o he ransfer funcion of he sysem. Firs hough we will define a special funcion called he δ-funcion or uni impulse. I is, like he Heaviside sep funcion u(), a generalized funcion or disribuion and is bes defined by considering anoher funcion in conjuncion wih i. 3.1 The δ-funcion Consider a funcion { 1/w < < w g() = oherwise One hing of noe abou g() is ha w g()d = 1. (3.1) (3.2) The lower limi is a infiniesimally small amoun less han zero. Now, suppose ha he widh w ges very small, indeed as small a +, a number an infiniesimal amoun bigger han zero. A ha poin, g() has become like he δ funcion, a very hin, very high spike a zero, such ha δ()d = + δ()d = 1 (3.3) 1

2 3/2TOPIC 3. THE δ-function & CONVOLUTION.IMPULSE RESPONSE & TRANSFER FUNCTION g() 1/w δ() (a) w (b) Figure 3.1: As w becomes very small he funcion g() urns ino a δ-funcion δ() indicaed by he arrowed spike. In some sense i is akin o he derivaive of he Heaviside uni sep u() Zero δ()d = u(). (3.4) Zero Infinie More formally he dela funcion is defined in associaion wih any arbirary funcion f (), as The dela funcion... f ()δ()d = f (). Figure 3.2: Heaviside (3.5) Picking ou values of a funcion in his way is called sifing of f () by δ(). We can also see ha f ()δ( τ)d = f (τ), (3.6) a resul ha we will reurn o. f() δ() f() δ( τ) A δ( τ) τ (a) (b) (c) Figure 3.3: (a,b) Sifing. (c) Dela funcion wih ampliude A. τ

3 3.2. PROPERTIES OF THE δ-function 3/3 Alhough he δ-funcion is infiniely high, very ofen you will see a described as he uni δ-funcion, or see a δ-funcion spike wih an ampliude A by i. This is o denoe a dela-funcion where δ()d = 1 or Aδ()d = A. 3.2 Properies of he δ-funcion Fourier ransform of he dela funcion: FT [δ()] = 1 (3.7) Proof: Use he definiion of he δ-funcion and sif he funcion f () = e iω : δ()e iω d = e iω = 1. (3.8) Symmery: The δ-funcion has even symmery. δ() = δ( ) (3.9) Parameer Scaling: δ(a) = 1 a δ() Proof: To prove his reurn o he fundamenal definiion, a, subsiue (a) for (no swap in limis) Bu f (a)δ(a)d(a) = a (3.1) f ()δ()d = f () If f (a)δ(a)d = f () (3.11) f (a)δ()d = f () a δ(a) = δ(). (3.12) Now if a <, subsiue (a) for (bu need o swap limis as a negaive) Bu f (a)δ(a)d(a) = a f (a)δ(a)d = f () (3.13) f (a)δ()d = f () a δ(a) = δ(). (3.14) So a δ(a) = δ() covers boh cases, and he saed definiion follows immediaely.

4 3/4TOPIC 3. THE δ-function & CONVOLUTION.IMPULSE RESPONSE & TRANSFER FUNCTION 3.3 Fourier Transforms ha involve he δ-funcion Fourier Transform of e iω FT [ e iω ] = Proof For you o fill ou... e i(ω ω) d = 2πδ(ω ω ). (3.15) Fourier Transform of 1 FT [1] = 2πδ(ω). (3.16) Proof. You could obain his eiher by puing ω = jus above, or by using he dual propery, FT [1] = 2πδ( ω), hen he even symmery propery δ( ω) = δ(ω). Fourier Transform of cos ω FT [cos ω ] = FT [ 1 ( e iω + e )] iω = π (δ(ω ω ) + δ(ω + ω )). (3.17) 2 Fourier Transform of sin ω FT [cos ω ] = FT [ 1 ( e iω e )] iω = iπ (δ(ω ω ) δ(ω + ω )). (3.18) 2i Fourier Transform of Complex Fourier Series yes, his can be useful! [ n= ] n= FT C n e inω = 2π C n δ(ω nω ). (3.19) n= 3.4 Convoluion n= We urn now o a very imporan echnique is signal analysis and processing. The convoluion of wo funcions f () and g() is denoed by f g. The convoluion is defined by an inegral over he dummy variable τ.

5 3.4. CONVOLUTION 3/5 The convoluion inegral. The value of f g a is (f g)() = f (τ)g( τ)dτ (3.2) The process is commuaive, which means ha (f g)() (g f )() or Example 1 f (τ)g( τ)dτ f ( τ)g(τ)dτ (3.21) I is easier o see wha is going on when convolving a signal f wih a funcion g of even or odd symmery. However, o ge ino a sric rouine, i is bes o sar wih an example wih no symmery. [Q] Find and skech he convoluion of f () = u()e a wih g() = u()e b, where boh a and b are posiive. [A] Using he firs form of he convoluion inegral, he shor answer mus be he uninelligible f g = u(τ)e aτ u( τ)e b( τ) dτ. (3.22) Firs, make skeches of he funcions f (τ) and g( τ) as τ varies. Funcion f (τ) looks jus like f () of course. Bu g( τ) is a refleced ( ime reversed ) and shifed version of g(). (The reflecion is easy enough. To check ha he shif is correc, ask yourself where does he funcion g(p) drop? The answer is a p =. So g( τ) mus drop when τ =, ha is when τ =.) f(τ ) g(τ ) g( τ ) g( τ ) τ τ τ (a) (b) (c) (d) τ Figure 3.4: We now muliply he wo funcions, BUT we mus worry abou he fac ha is a variable. In his case here are wo differen regimes, one when < and he oher when. Figure 3.5 shows he resul.

6 3/6TOPIC 3. THE δ-function & CONVOLUTION.IMPULSE RESPONSE & TRANSFER FUNCTION g( τ) f( τ ) g( τ ) f( τ ) τ τ f(τ )g( τ) f(τ )g( τ ) < > τ τ Figure 3.5: So now o he inegraion. For <, he funcion on he boom lef of Figure 3.5 is everywhere zero, and he resul is zero. For u(τ)e aτ u( τ)e b( τ) dτ = e b ( ) e (b a)τ dτ = e b e (b a) 1 b a (3.23) So f g() = { ( e a e b) /(b a) for for < (3.24) I is imporan o realize ha he funcion a he boom righ of Figure 3.5 is NOT he convoluion. Tha is he funcion you are abou o inegrae over for a paricular value of. Figure 3.6 shows he > par of he convoluion for b = 2 and a = 1.

7 3.4. CONVOLUTION 3/7.25 (exp(-x)-exp(-2*x)) Example 2 Figure 3.6: f g() ploed for > when b = 2 and a = 1. [Q] Derive an expression for he convoluion of an arbirary signal f () wih he funcion g() shown in he figure. Deermine he convoluion when f () = A, a consan, and when f () = A + (B A)u(). [A] Follow he rouine. Funcion f (τ) looks exacly like f (), bu g( τ) is refleced and shifed. Muliply and inegrae over τ from o. Because g only has finie range, we can pinch in he limis of inegraion, and he convoluion becomes f g = a f (τ)dτ + +a f (τ)dτ (3.25) f() a 1 g() 1 a f() g( τ) a 1 +a τ 1 Figure 3.7: When f () = A, a consan, i is obvious by inspecion ha he convoluion is zero for all. When f () = A + (B A)u(), we have o be more careful because here is a disconinuiy in he funcion. From Figure 3.8(a): The convoluion is zero for all < a and all > a (Diagram posiions 1,2,5).

8 3/8TOPIC 3. THE δ-function & CONVOLUTION.IMPULSE RESPONSE & TRANSFER FUNCTION The maximum value is when = (Posiion 4). inegrals above, (f g)( = ) = a(b A). For a < <, (Posiion 3) By inspecion, or using he f g = a (A)dτ + Adτ + +a Bdτ = aa+ A+( +a)b = (a+)(b A) (3.26) showing ha he increase in correlaion value is linear. Symmery ells us ha he decrease for < < a will also be linear. One can see from Figure 3.8(b) ha his convoluion provides a rudimenary deecor of seps in he signal. The maximum in he correlaion is proporion o he sep size, and would go negaive if he sep was downwards. g( τ) B f(τ) f(τ) g( τ) f * g a(b A) A τ τ τ 2a a a a 1: < a 2: = a 3: a<< 4: = 5: >a a a (a) (b) Figure 3.8: 3.5 The Impulse Response Funcion You are very used o describing sysems using he ransfer funcion in he frequency domain. So used o i in fac, ha you may have forgoen o wonder why can we describe a sysem s response in he ime domain? The answer is You can, bu you didn know abou convoluion unil now. In he ime domain, a sysem is described by is Impulse Response Funcion h(). This funcion lierally describes he response of sysem a ime o an uni impulse or δ-funcion inpu adminisered a ime =. Suppose ha now is ime, and you adminisered an impulse o he sysem a ime τ in he pas. The response now is y() = h( τ). Suppose you adminisered a succession of impulses of differen srenghs x. The porion dy() of he response due o impulse a ime τ earlier is dy() = x(τ)h( τ)dτ, (3.27)

9 3.6. HOW DO WE CONNECT THIS UP... 3/9 Whack Now Succession of whacks x() τ Figure 3.9: so ha he oal response now is y() = where all τ. x(τ)h( τ)dτ, (3.28) Now suppose ha h is causal. If τ > hen h( τ) =, and herefore, provided h() is causal, he ime response y() o an inpu x() is y() = x(τ)h( τ)dτ = x h. (3.29) The oupu is he inpu CONVOLVED wih he Impulse Response Funcion. 3.6 How do we connec his up... Can we reconcile he following hings you now know abou sysems and signals? 1. The emporal oupu is he emporal inpu CONVOLVED wih he Impulse Response Funcion. 2. The frequency domain oupu is he frequency domain inpu MULTIPLIED by he Transfer Funcion. 3. The frequency domain signal is he Fourier Transform of he emporal signal Mahemaically, i mus be ha he FT of a convoluion is a produc. y() = x() h() FT FT?? Y (ω) = X(ω)H(ω) (3.3) (3.31)

10 3/1TOPIC 3. THE δ-function & CONVOLUTION.IMPULSE RESPONSE & TRANSFER FUNCTION 3.7 The Fourier Transform of a Convoluion To prove his we need o develop he inegraion for he Fourier Transform of a convoluion. Now x h is a perfecly respecable funcion of, so FT [x h] = = = = τ= [(x h)()] e iω d (3.32) x(τ)h( τ)dτe iω d (3.33) For absolue clariy, le s swich he order of inegraion, hen wrie = τ + p. In he inner inegral τ is a consan, so ha d = dp. FT [x h] = = = τ= = τ= τ= p= x(τ)e iωτ dτ x(τ)h( τ)e iω ddτ (3.34) x(τ)h(p)e iωτ e iωp dpdτ (3.35) p= h(p)e iωp dp (3.36) = FT [x] FT [h] (3.37) = X(ω) H(ω) (3.38) So, he firs imporan hing we discover is The ime-convoluion/frequency-modulaion propery of Fourier ransform f () g() F (ω)g(ω) (3.39) or FT [f () g()] = FT [f ()] FT [g()] = F (ω)g(ω) 3.8 Impulse response and Transfer Funcion The second imporan connecion is ha The Fourier Transform of he Impulse Response Funcion is he Transfer Funcion h() H(ω) (3.4) or FT [h()] = H(ω)

11 3.9. THE TIME-MODULATION/FREQUENCY-CONVOLUTION PROPERTY 3/11 x() *h() y() X( ω) H( ω) Y( ω) Figure 3.1: 3.9 The ime-modulaion/frequency-convoluion propery There is a furher resul involving convoluion ha we sae now. The modulaion/convoluion propery of Fourier ransform f ()g() 1 F (ω) G(ω) (3.41) 2π 3.1 Example1 [Q1] Show ha δ() f () = f (). [A1] One could do his he long way by wriing way is o argue ha f (τ)δ( τ)dτ, ec, bu a quicker FT [δ() f ()] = FT [δ()] FT [f ()] = 1. F (ω) (3.42) FT 1 [FT [δ() f ()]] = FT 1 [F (ω)] (3.43) δ() f () = f () (3.44) We could also show ha f () δ( ± α) = f ( ± α). Boh of hese are manifesaions of he Sifing propery of δ-funcions f () δ( ± α) = f ( ± α) (3.45) Previously his was expressed as an inegral bu we now recognize ha inegral as he convoluion inegral!

12 3/12TOPIC 3. THE δ-function & CONVOLUTION.IMPULSE RESPONSE & TRANSFER FUNCTION 3.11 More examples [Q2] Convolve he wo op ha funcions f () and g() shown in below, and plo he resuling convolved signal r(). Find is Fourier Transform R(ω). f() g() 2 1 a a a/2 a/2 Figure 3.11: [A2] Ploed below as funcions of τ are f (τ), he reversed and shifed g( τ) for = 3a/2, 3a/2 < < a/2 and = a/2, and he resuling f (τ)g( τ). g( τ) f(τ) fg 1 τ a g( τ) f(τ) a fg 1 τ a g( τ) f(τ) a a +a/2 fg 1 τ a a a Figure 3.12: From he plos i is obvious ha for all < 3a/2 f g is zero, so he convoluion inegral is zero. For 3a/2 < < a/2 he inegral is +a/2 ( 2dτ = 2 + 3a ). (3.46) 2 a

13 3.11. MORE EXAMPLES 3/13 A = a/2 here is complee overlap, a which poin he inegral is 2a. I will coninue consan a 2a unil =. The convoluion has even symmery. The resul is given in Figure f() * g() 2a 3a/2 a/2 a/2 3a/2 Figure 3.13: To find he Fourier ransform of his convoluion, we can use f g F (ω)g(ω). Funcion f () is a op ha of heigh 1, bu is widh is 2a. We can use he Parameer Scaling Propery from Lecure 2 ha says if If f () F (ω) hen FT [f (α)] = 1 ( ω ) α F (3.47) α We have o be VERY careful. I is emping o say ha because our funcion is 2 imes wider han he sandard op ha, he scaling mus be 2. WRONG! The scaling is α = 1/2. For he uni op-ha of half-widh a/2 Π(ω) = 2 ( ωa ) ( ωa ) ω sin = a sinc. (3.48) 2 2π So for our parameer scaled version ( ) 1 ω F (ω) = (1/2) Π = 2 2 ( ) 2ωa (1/2) 2ω sin = 2 ( ωa ) 2 ω sin (ωa) = 2a sinc. (3.49) π Funcion g() is a op ha of heigh 2 and half-widh a/2. From Lecure 2, Secion 2.4, G(ω) = 2Π(ω) = 4 ( ωa ) ( ωa ) ω sin = 2a sinc. (3.5) 2 2π So he Fourier Transform of he convoluion is FT [f g] = 8 ( ωa ) ω 2 sin (ωa) sin = 4a 2 sinc 2 ( ωa ) 2π sinc ( ωa ) π. (3.51)

14 3/14TOPIC 3. THE δ-function & CONVOLUTION.IMPULSE RESPONSE & TRANSFER FUNCTION [Q3] A signal f () is ampliude modulaed by cos ω ino a signal y(). resuling Fourier Transform Y (ω) in erms of F (ω). Find he [A3] We could use he mehod given in Lecure 2 Ampliude modulaion by a cosine. Insead, le s use he propery a Secion 3.6 of his Lecure. f ()g() 1 F (ω) G(ω) (3.52) 2π where our g() = cos ω. By wriing cos ω = 1 2 one can deermine ha ( e iω + e iω ) (3.53) FT [cos ω ] = π (δ(ω + ω ) + δ(ω ω )). (3.54) δ(ω+ω ) π δ(ω ω ) π ω ω ω Hence Figure 3.14: Frequency specrum of cos ω FT [f () cos ω ] = 1 2π F (ω) π (δ(ω + ω ) + δ(ω ω )) (3.55) = 1 2 (F (ω + ω ) + F (ω ω )) (3.56) So suppose f () had a ampliude specrum as skeched below. Afer ampliude modulaion, he ampliude specrum would spli ino wo pars. FT[ f() ] Cosine Modulaion FT[ f()cos(ω ) ] ω ω ω ω Figure 3.15: An ampliude specrum spli ino wo by modulaion wih a cosine wave.

15 3.12. SUMMARY 3/ Summary We sared by developing he definiion of he uni dela funcion δ(). Alhough his is an infiniely high spike, i is also infiniely narrow, so ha he inegral over i is uniy. The dela funcion was more formally defined via is sifing properies. We looked a properies of he dela funcion and a Fourier ransforms ha involve he dela funcion We hen defined he convoluion inegral, and looked a how o lay ou he inegral graphically. We defined he Impulse Response funcion and showed ha he emporal oupu of a sysem is he emporal inpu convolved wih he inpu. Finally, by showing ha he FT of a convoluion of wo emporal funcion is he produc of heir individual FTs, we found ha our old friend he Transfer Funcion is he Fourier Transform of he Impulse Response.

16 3/16TOPIC 3. THE δ-function & CONVOLUTION.IMPULSE RESPONSE & TRANSFER FUNCTION

17 Topic 4 From Modulaion o Sampling, Aliasing & Reconsrucion 4.1 Inroducion A he end of he las lecure, we saw how muliplying a signal f () by a cosine g() = cos ω caused he frequency specrum F (ω) of f () o spli ino wo pars. The spli arose because he Fourier Transform G(ω) of he cosine is a pair of dela funcions G(ω) = π (δ(ω + ω ) + δ(ω ω )) (4.1) and he convoluion in he frequency domain urns ino sifing by he wo dela funcions: FT [f () cos ω ] = 1 F (ω) G(ω) 2π (4.2) = 1 2π F (ω) π (δ(ω + ω ) + δ(ω ω )) (4.3) Ampliude Modulaion = 1 2 (F (ω + ω ) + F (ω ω )). (4.4) We called ha process ampliude modulaion. To modulae simply means o vary, and in communicaions engineering one can consider insering informaion ono a carrier signal by modulaing he ampliude, frequency or phase of he signal. In our previous example, we hough of muliplying he signal f () by he carrier cos ω, bu in communicaions (i) one hinks he oher way round, and (ii) he signal is considered o be a perurbaion o he carrier. Tha is, if he carrier is A cos ω, hen he ampliude modulaed signal is v() = A(1 + m f ()) cos ω. (4.5) This is easily decomposed ino v() = A cos ω + m f ()A cos ω, (4.6) 1

18 4/2 TOPIC 4. FROM MODULATION TO SAMPLING, ALIASING & RECONSTRUCTION and our knowledge of linear superposiion ells us ha he specrum is he sum of ha of he original carrier (which is jus wo dela funcions) and he ineresing modulaed par. The consan m is called he modulaion index or, more descripively, he modulaion deph. A.M. in comms (Double Side Band, Carrier no suppressed) F( ω) ω ω ω ω Figure 4.1: Sampling In his lecure we look a how sampling a signal affecs is frequency specrum. Ampliude modulaion involves muliplicaion of signals in he ime domain, and so does sampling. They are really he same process bu one ends o sample wih a disconinuous funcion and modulae wih a coninuous one. Eiher way, we will again be concerned wih convoluions of he Fourier ransforms of he original signals. Mos ofen, sampling is a precursor o digiisaion bu in his course we consider only analogue signals. 4.2 Insananeous regular sampling Recall from Lecure 1 our discussion of discree-ime analogue signals. Wha sor of sampling signal p() would generae he sampled signal y()? The sampling signal needed is he infinie impulse rain p() = δ( kt s ) k= (4.7) where T s is he sampling inerval or sampling period.

19 4.3. THE FOURIER TRANSFORM OF THE SAMPLING AND SAMPLED SIGNAL 4/3 f() y() Sampling Inerval Ts p() Figure 4.2: 4.3 The Fourier ransform of he sampling and sampled signal Using he modulaion propery from he las lecure, he frequency specrum of he sampled signal mus be he convoluion Y (ω) = 1 F (ω) P (ω), (4.8) 2π for which, prey obviously, we need he Fourier ransform of he sampling signal. Fourier ransform of a pulse rain: If p() = k= δ( kt s) and ω s = 2π/T s, P (ω) = ω s k= δ(ω kω s ) (4.9) Proof: The nea way of doing his is o find he complex Fourier series firs! Trea he pulse rain a periodic funcion conaining jus one dela funcion per period. The fundamenal frequency is ω = ω s, so ha, using sifing in he inegral, C k = 1 T s Ts /2 T s /2 δ()e ikω s d = 1 T s = ω s 2π. (4.1) U sing he earlier definiion of he Fourier ransform of a complex Fourier Series [ n= ] n= FT C n e inω = 2π C n δ(ω nω ). (4.11) we find n= P (ω) = 2π k= k= n= C k δ(ω kω s ) = ω s k= δ(ω kω s ) (4.12)

20 4/4 TOPIC 4. FROM MODULATION TO SAMPLING, ALIASING & RECONSTRUCTION A he end of las lecure we saw wha happened o F (ω) when convolved wih a pair of dela funcions δ(ω ± ω s ). Now in addiion we have convoluion wih δ(ω), δ(ω ± 2ω s ) and so on. Hence he sampled specrum becoms Y (ω) = ω s 2π k= F (ω kω s ) = 1 T s k= F (ω kω s ) (4.13) An example of he oupu appears in Figure 4.3. Noe ha he ampliude of each copy is a facor of 1/T s imes he original magniude. A F(ω) ω P( ω) * A/Ts Y( ω) 2ω ω ω 2ω s s s s ω Figure 4.3: Pulse rain modulaion a T = 2π/ω s ampliude by 1/T s. gives duplicae specra every ω s, bu reduced in 4.4 Wha does he specrum represen? Firs le us alk in Hz, raher han rad s 1. As skeched in Figure 4.4, suppose he original signal was from some (dull) music which conained frequencies only up o 1 khz. Wha would you hear if you sampled his a 2.5 khz? The resuling specrum would conain componens from dc-1. khz, and khz and khz and all (2.5n 1.) (2.5n+1.) khz. Now human hearing goes up o some 2 khz alhough he sensiiviy is no uniform. So you would hear he music, bu wih higher frequency inny and whisly addiions. (Musical appreciaion 11: whereas a low frequencies 1 khz spans he seven ocaves from he lowes organ pedal noe upo a soprano s op C, a 18 khz a 1 khz deviaion spans jus one semione; hence he whisle...)

21 4.5. RECOVERING THE ORIGINAL SIGNAL 4/5 A khz Y( ω) A/Ts khz 1. Original Afer sampling a 2.5kHz Figure 4.4: 4.5 Recovering he original signal I seems exraordinary ha he process of sampling does no damaged he frequency specrum in he original base band. Even more remarkable is ha i can be recovered uncorruped by filering he oupu wih an ideal low-pass filer in our example wih wih a band-limi of jus greaer han 1. khz, as shown in Figure 4.5. Y( ω) T s Original Sampled Afer filering by ideal LP filer Figure 4.5: This leads us o a major and fundamenal resul Nyquis-Shannon Sampling Theorem: Precise reconsrucion of a coninuous-ime signal from is samples is possible 1. if he signal is bandlimied; and 2. he sampling frequency is greaer han wice he signal bandwidh. We have acually assumed a furher iem: 3. Ideal low-pass filers exis.

22 4/6 TOPIC 4. FROM MODULATION TO SAMPLING, ALIASING & RECONSTRUCTION 4.6 Aliasing Le us carry on assuming (1) and (3) ha he signal f () is band-limied such ha f () = for all ω > ω max, and ha ideal filers exis. This means ha if here remain gaps beween he secions of specrum we can always inser a filer o recover he original. However, if hey become enangled his is impossible. This condiion occurs if ω s < 2ω m, as Figure 4.6 shows. This enanglemen is called aliasing. When a signal is aliased, i is eviden from Figure 4.6 ha is higher frequency componens sar appearing a lower frequencies. If he sampling frequency is reduced as low as ω m, hen he componens a ω m are acually aliased o zero frequency. F( ω) Y( ω) ω s ω m No aliased ω m ω ω m ω s 2ω s Y( ω) ω s ω m Aliased ω m ω s 2ω s 3ω s Figure 4.6: Aliasing in he frequency domain So o reinforce he message: To avoid aliasing, one mus sample a ω s > 2ω m (4.14) The minimum sampling frequency, 2ω m, is ofen called he Nyquis limi.

23 4.6. ALIASING 4/7 An example of aliasing ha can be followed in he ime and frequency domains is ha of sampled cosine wave shown in Figures 4.7 and 4.8. As he sampling frequency ω s is reduced owards ω m he frequency of he apparen cosine in he ime domain reduceds o zero. The word apparen is insered o sress ha he ime domain drawing conains only he lowes frequency inerpreaion of he daa. The frequency specrum sill conains componens a higher frequencies. When ω s = ω m, he acual value obained depends on he phase. Figure 4.7 shows he maximum value, bu shifing he Figure 4.7: Aliasing in he ime domain phase by 1/4 period in eiher direcion would produce a zero oupu. ω s ω s 2ω s ω ω m +ω m ω s +ω ω m s ω m 2ω s ω m ω s +ω m ω s ω s 2ω s ω ω m +ω m ω s +ω ω m s ω m 2ω s ω m ω s +ω m ω s ω s 2ω s ω ω m +ω m ω s +ω ω m s ω m 2ω s ω m ω s +ω m Figure 4.8: Aliasing in he frequency domain

24 4/8 TOPIC 4. FROM MODULATION TO SAMPLING, ALIASING & RECONSTRUCTION 4.7 Spaial aliasing in sampled images You are likely o have come across aliasing in he spaial domain when viewing pixelaed images. Each pixel provides a spaial sampling of he underlying image, and if he spaial frequency in he image is oo high, he resuling image is corruped. In he image below he line spacing decreases, and he effecs of aliasing differ a each image raser causing paerns o appear. Figure 4.9: Aliasing in he spaial domain: he paern as i appears on (a) a high resoluion priner and (b) a low resoluion screen. You may have seen similar effecs on elevision when viewing a sriped shir or similar. There aliasing causes colours o appear. The reason is ha he red, green and blue pixels on he camera are spaially separaed in a Bayer paern, and he so he effec of spaial aliasing is differen for each channel resuling in spurious colours being generaed. Figure 4.1: Colour aliasing: for full colour, drink vodka or visi websie.

25 4.8. AVOIDING ALIASING 4/9 4.8 Avoiding aliasing There is jus only one recipe for avoiding aliasing in sampling signals, and ha is 1. o deal only wih signals ha are sricly band-limied a some ω m ; 2. o sample a ω s > 2ω m ; and 3. o use an ideal filer o recover he original. However, in mos cases (1) and (3) are impracical. Insead signal processing engineers 1. Decide on he useful upper limi o he bandwidh, ω m. The signal is hen passed hrough an ani-aliasing filer, a low-pass filer which cus off signals above ω m. (The descripion ani-aliasing filer is misleading i helps avoid aliasing a he expense of compromising he original signal!) 2. Then sample a ω s some way above 2ω m. This will provide bands of zero beween he blocks in he frequency specrum 3. Apply a reconsrucion filer, again a low-pass filer wih a cu-off some way above ω m, as illusraed in Figure 4.11 Pulse rain Original Ani Aliassing Reconsrucion Oupu L.P. Filer L.P. Filer Figure 4.11: Ani-Aliasing, Sampling, and Reconsrucion. Mos remarkably, hey can also reconsruc he signal in he emporal domain, bu le us sick wih he idea of reconsrucion filering for he momen. As illusraed in Figure 4.12, if he low-pass filer is close o he ideal brick-wall filer, he zero bands menioned in (2) above don have o be wide, and so he sampling frequency ω s can be reduced closer o 2ω m. Conversely, if he filer has only a low-order roll-off, he zero bands will need o be wider so ha he nex block of filled specrum is sufficienly aenuaed. Sampling a ω s well above 2ω m will achieve his.

26 4/1 TOPIC 4. FROM MODULATION TO SAMPLING, ALIASING & RECONSTRUCTION High order roll off Can have smaller gap Low order roll off Needs bigger gap Figure 4.12: The qualiy of he reconsrucion filer deermines he gap required, and hence how much higher han ω m he sampling frequency ω s should lie. 4.9 Reconsrucion in he ime domain The argumen based on a low-pass filer will have convinced you ha precise reconsrucion is possible if he filer is an ideal brick-wall filer. Bu a brick-wall low-pass filer is nohing oher han a op-ha in he frequency domain. Le us assume ha he inpu signal f () is band-limied o ω m. Our reconsrucion filer, because i is ideal, can have (minimum) half-widh ω m oo, so is ransfer funcion is { Ts ω < ω T s Π m (ω) = m (4.15) ω > ω m The facor of T s resores he original ampliude. The FT of he sampled inpu is 1 2πF (ω) P (ω), so he FT of he oupu of he reconsrucion filer is O(ω) = T s Π m (ω) 1 (F (ω) P (ω)) (4.16) 2π Now recall from Lecure 3 he convoluion/modulaion properies ha f () g() F (ω)g(ω) and f ()g() 1 2πF (ω) G(ω) Hence he emporal oupu is [ ] 1 o() = T s FT 1 [Π m ] FT 1 F (ω) P (ω) (4.17) 2π = T s FT 1 [Π m ] (f ()p()) (4.18) (4.19) We know ha f ()p() = f () k= δ( kt s ) = k= f (kt s )δ( kt s ). (4.2)

27 4.9. RECONSTRUCTION IN THE TIME DOMAIN 4/11 Now le us work on FT 1 [Π m ], which provides good revision of wha we can do. In Lecure 2 we worked ou ha a uni pulse of half-widh a/2 had as i Fourier Transform Π a/2 () a sin(ωa/2) ωa/2 = sin(ωa/2) ω/2 We also proved he Dual Propery ha Hence. (4.21) f () F (ω) implies F () 2πf ( ω) (4.22) sin(a/2) /2 sin(ω m ) π 2πΠ a/2 ( ω) (4.23) Π m ( ω) = Π m (+ω) (4.24) (4.25) where flipping he sign of ω explois Π s even symmery. Puing his all ogeher gives o() = T s sin(ω m ) π = T s k= = T s k= f (kt s ) k= f (kt s )δ( kt s ) (4.26) f (kt s ) sin(ω m( kt s )) π( kt s ) sin(ω m ( τ)) δ(τ kt s )dτ (4.27) π( τ) Remember ha, for proper reconsrucion, ω s 2ω m or 2π/T s 2ω m. (4.28) This expression is quie sraighforward, bu is quie amazing! I says ha he emporal reconsrucion ino f () of a sampled band-limied signal f (kt s ) can be achieved by inerpolaing he samples wih sinc funcions. One draws a sinc funcion scaled by he sample s size a he posiion of each sample, and hen adds hem up.

28 4/12 TOPIC 4. FROM MODULATION TO SAMPLING, ALIASING & RECONSTRUCTION 4.1 Example We need a bandlimied funcion as a es, so le us use f () = exp( 2 /2) cos(), which is bandlimied somewha above ω m = 1. Now we require ω s 2, bu for convenience le s make ω s = 2π which makes T s = 1. The exponenial envelope here allows us o approximae he funcions o zero for > 1, so our samples will be, using T s = 1 { exp( k 2 /2) cos(k) k = 1, 9,, 9, 1 f (kt s ) = f (k) = oherwise (4.29) So using he reconsrucion inerpolaion wih ω m = 2, T s = 1, and using he approximaion above o() = T s k= f (kt s ) sin(ω m( kt s )) π( kt s ) 1 k= 1 sin(2( k)) f (k) π( k) (4.3) Figure 4.13: The orignal signal f (), and samples aken a T s = 1.

29 4.1. EXAMPLE 4/13 (a) (b) Figure 4.14: (a) Sinc funcions associaed wih each sample. (b) Their summaion is idenical o he original signal!

30 4/14 TOPIC 4. FROM MODULATION TO SAMPLING, ALIASING & RECONSTRUCTION 4.11 A noe on oher inerpolaion funcions We have shown ha scaled sinc funcions are he correc funcions wih which o reconsruc a sampled signal he raionale is ha he FT of a sinc funcion is a op-ha, providing an ideal brick-wall filer in he frequency domain. This allows you o hink how oher inerpolaion funcions would behave as approximaions o he sinc funcion. For example, Suppose we made a Gaussian approximaion o he cenral hump of he sinc funcion. The Gaussian inerpolaion funcion would provide a Gaussian shaped filer. A riangular approximaion o he hump of he sinc funcion would give a sinc 2 shaped filer in he frequency domain. Figure 4.15: Possible approximaions o he scaled-sinc inerpolaion funcion. The filer in he frequency domain would no longer be a brick wall Summary We discussed aspecs of ampliude modulaion, and noed ha sampling was a form of modulaion. We esablished he mahemaical form for regularly spaced emporal sampling, and developed is Fourier ransform. We hen convolved his wih he signal, showing ha he baseband was replicaed a higher frequencies. We considered he problem of aliasing, and how o avoid i, and hen discussed reconsrucion of a sampled signal.