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. The lower limi is a infiniesimally small amoun less han zero. Now, suppose ha he widh w ges very small, indeed as small a +, an 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 1

2 3/2 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 δ(d = u(. More formally he dela funcion is defined in associaion wih any arbirary funcion f (, as The dela funcion... f (δ(d = f (. Picking ou values of a funcion in his way is called sifing of f ( by δ(. We can also see ha f (δ( d = f (, a resul ha we will reurn o. f( δ( f( δ( A δ( (a (b (c Figure 3.2: (a,b Sifing. (c Wih an ampliude A. 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 3/3 3.2 Properies of he δ-funcion Fourier ransform of he dela funcion: FT [δ(] = 1 Proof: Use he definiion of he δ-funcion and sif he funcion f ( = e iω : δ(e iω d = e iω = 1. Symmery: The δ-funcion has even symmery. δ( = δ( Parameer Scaling: δ(a = 1 a δ( Proof: To prove his reurn o he fundamenal definiion, If a, subsiue (a for (no swap in limis f (aδ(ad(a = a f (aδ(ad = f ( f (δ(d = f ( Bu f (aδ(d = f ( a δ(a = δ(. Now if a <, subsiue (a for (bu need o swap limis as a negaive f (aδ(ad(a = a f (aδ(ad = f ( Bu f (aδ(d = f ( a δ(a = δ(. So a δ(a = δ( covers boh cases, and he saed definiion follows immediaely.

4 3/4 3.3 Fourier Transforms ha involve he δ-funcion Fourier Transform of e iω FT [ e iω ] = e i(ω ω d = 2πδ(ω ω. Fourier Transform of 1 FT [1] = 2πδ(ω. 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. 2 Fourier Transform of sin ω FT [cos ω ] = FT [ 1 ( e iω e ] iω = iπ (δ(ω ω δ(ω + ω. 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. n= n= 3.4 Convoluion 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. The convoluion inegral. The value of f g a is (f g( = f (g( d

5 The process is commuaive, which means ha (f g( (g f ( or Example 1 f (g( d 3/5 f ( g(d 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(e a u( e b( d. Firs, make skeches of he funcions f ( and g( as varies. Funcion f ( looks jus like f ( of course. Bu g( 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( mus drop when =, ha is when =. f( g( g( g( (a (b (c (d Figure 3.3: 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 3.4 shows he resul. So now o he inegraion. For <, he funcion on he boom lef of Figure 3.4 is everywhere zero, and he resul is zero. For u(e a u( e b( d = e b ( e (b a d = e b e (b a 1 b a

6 3/6 g( f( g( f( f( g( f( g( < > Figure 3.4: So f g( = { ( e a e b /(b a for for < I is imporan o realize ha he funcion a he boom righ of Figure 3.4 is NOT he convoluion. Tha is he funcion you are abou o inegrae over for a paricular value of. Figure 3.5 shows he > par of he convoluion for b = 2 and a = (exp(-x-exp(-2*x Figure 3.5: f g( ploed for > when b = 2 and a = 1.

7 3.4.2 Example 2 [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 Au(. [A] Follow he rouine. Funcion f ( looks exacly like f (, bu g( is refleced and shifed. Muliply and inegrae over from o. Because g only has finie range, we can pinch in he limis of inegraion, and he convoluion becomes f g = a f (d + +a f (d 3/7 f( a 1 g( 1 a f( g( a 1 +a 1 Figure 3.6: When f ( = A, a consan, i is obvious by inspecion ha he convoluion is zero for all. When f ( = A + (B Au(, we have o be more careful because here is a disconinuiy in he funcion. From Figure 3.7(a: The convoluion is zero for all < a and all > a (Diagram posiions 1,2,5. The maximum value is when = (Posiion 4. By inspecion, or using he inegrals above, (f g( = = a(b A. For a < <, (Posiion 3 f g = a (Ad + Ad+ +a Bd = aa+ A+(+aB = (a+(b A 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 3.7(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.

8 3/8 g( B f( f( g( f * g a(b A A 2a a a a 1: < a 2: = a 3: a<< 4: = 5: >a a a (a (b Figure 3.7: 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 in he pas. The response now is y( = h(. Whack Now Succession of whacks x( Figure 3.8: Suppose you adminisered a succession of impulses of differen srenghs x. The porion dy( of he response due o impulse a ime earlier is dy( = x(h( d, so ha he oal response now is y( = where all. x(h( d,

9 Now suppose ha h is causal. If > hen h( =, and herefore, provided h( is causal, he ime response y( o an inpu x( is y( = x(h( d = x h. 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(ωH(ω 3/9 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] = = = = = [(x h(] e iω d x(h( de iω d

10 3/1 For absolue clariy, le s swich he order of inegraion, hen wrie = + p. In he inner inegral is a consan, so ha d = dp. FT [x h] = = = = = = = p= x(e iω d x(h( e iω dd x(h(pe iω e iωp dpd p= h(pe iωp dp = FT [x] FT [h] = X(ω H(ω So, he firs imporan hing we discover is The ime-convoluion/frequency-modulaion propery of Fourier ransform f ( g( F (ωg(ω or FT [f ( g(] = FT [f (] FT [g(] = F (ωg(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 or FT [h(] = H(ω

11 3/11 x( *h( y( X( ω H( ω Y( ω Figure 3.9: 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 2π 3.1 Example1 [Q1] Show ha δ( f ( = f (. [A1] One could do his he long way by wriing quicker way is o argue ha f (δ( d, ec, bu a FT [δ( f (] = FT [δ(] FT [f (] = 1. F (ω FT 1 [FT [δ( f (]] = FT 1 [F (ω] δ( f ( = f ( 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

12 3/12 Previously his was expressed as an inegral bu we now recognize ha inegral as he convoluion inegral!

13 3.11 More examples 3/13 [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 3.1: [A2] Ploed below as funcions of are f (, he reversed and shifed g( for = 3a/2, 3a/2 < < a/2 and = a/2, and he resuling f (g(. g( f( fg 1 a g( f( a fg 1 a g( f( a a +a/2 fg 1 a a a Figure 3.11: 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 ( 2d = 2 + 3a. 2 a

14 3/14 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 3.12: To find he Fourier ransform of his convoluion, we can use f g F (ωg(ω. Funcion f ( is a op ha of heigh 1, bu is widh is 2a. Parameer Scaling Propery from Lecure 2 ha says if If f ( F (ω hen FT [f (α] = 1 ( ω α F α We can use he 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. 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 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. 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..

15 3/15 [Q3] A signal f ( is ampliude modulaed by cos ω ino a signal y(. Find he resuling Fourier Transform Y (ω in erms of F (ω. [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 2π where our g( = cos ω. By wriing cos ω = 1 2 one can deermine ha ( e iω + e iω FT [cos ω ] = π (δ(ω + ω + δ(ω ω. δ(ω+ω π δ(ω ω π ω ω ω Hence Figure 3.13: Frequency specrum of cos ω FT [f ( cos ω ] = 1 2π F (ω π (δ(ω + ω + δ(ω ω = 1 2 (F (ω + ω + F (ω ω So suppose f ( had a ampliude specrum as skeched below. Afer ampliude modulaion, he ampliude specrum would spli ino wo pars. F( ω Cosine Modulaion F( ω ω ω ω ω Figure 3.14: An ampliude specrum spli ino wo by modulaion wih a cosine wave.

16 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.