115A_1_Tutorial on Convolution

115A_1_Tutorial on Convolution - rm .1 5(3) (it) Mr) I 0...

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Unformatted text preview: rm .1 5(3) (it) Mr) I 0 WE; M Mr — nAT) ' i} HAT I A” (d) {tin/mans e nit?) “3"” man/2o — mama _ . A n u [—9. i 0 m3? f'w (ct {W {) nAT (f) 2.4 System Response to External Input: Zero—State Response 371 approaches an impulse x(ntfir)5(t m not) as Ar _> 0 (Fig. 2.3e). Therefore Air—>0 x(t) = lim Emerita writer) a: (2.28) To find the . sponse for this input Mr}, we consider the input and the correspond g output pairs, as shown in F1 2.3cm2.3f and also shown by directed arrow notation as fol '4 3: input 22> output 5(1) we htt) Mt —nAt) 22> Mt ——nA [x(nAt)At]§(. nAr) => EXOQ )At]it(t -~ nAt) iim x(nAr)6(t artist“) 1': rm"? tn thnArManAflAt AraU r Ar~>0 XE!) [sec Eq. (2.28)] yU) Therefore? yo) 3 Hngnnmmn mm: = / new ~ mt: (2.29) This is the result we 3 elk. We have obtained the system response y(t) to .- arbitrary input x{t} in terms of the un' impulse response Mr). Knowing ha“), we can determin the response y(t) to any input. 0 erve once again the all-pervasive nature of the system ’8 char . eristic modes. The system r sponse to any input is determined by the impulse response, which, in . rn, is made up of ch actert‘stic modes ofthe system. It " important to keep in mind the assumptions used in deriving Eq. (2.29). We ass ‘1 ed a lin r, titne- invariant (LTD system. Linearity allowed us to use the principle of superposition, . s d 'rne invariance made it possible to express the system’s response to 5 (t mm mm") as Mt m not). 2.4:]; The Convolution Integral The zeronstate response y(t) obtained in Eq. (2.29} is given by an integral that occurs frequently in the physical sciences, engineering, and mathematics. For this reason this integral is given a special name: the convolution integral. The convolution integral of two functions x} (r) and xfit) iln deriving this result we have assumed n time—invariant system. If the system is time varying, then the system response to the input 5(t i not) cannot he expressed as h(t m mix) but instead has the form he, nAr). Use of this form modifies Eq. (2.29) to 3’0} 2 x(r)h(t,t)dr rx. 172 CHAPTER 2 TIME—DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS is denoted symbolically by x10) a< xa (I) and is defined as ‘00 mo “an E / xttr'mtr m rice (2.30) .706 Some important properties of the convolution integral follow. THE COMMUTATIVE PROPERTY Convolution operation is commutative; that is, 261(1) 2: x2(z) = x2(r) >:< x; (r). This property can be proved by a change of variable. In Er; (2.30), if we let 2: = t w r so that r = I - z and dr : ~dz, we obtain mama) = — f X2CZ}X1(Iez)dz DC) / ixztexttr — ziaz 3X20) *Xiif) (2.31) rTHE DISTRIBUTIVE PROPERTY According to the distributive property, X1(I)*ix2(?) + MUN = 9510) *szl "’r" X10) * 13(3) {232) THE ASSOCIATIVE PROPERTY According to the associative property, xi {3) * [3620) * 1639)] ={X1(I)*X2(r}]* 353“) {2-33) The proofs of Eqs. (2.32) and (2.33) follow directly from the definition of the convolution integral. They are left as an exercise for the reader. THE SHIFT PROPERTY If x10) *x2(r) x C(i) then x16} *x2(t m T): x1(t — T) $X30) m C(t m— T) (2.34a) and xiv m T1) *xm‘ w T; : C(z‘ —- T1— T2) (23412)) Proof. We are given ‘00 x1(t)>t=x3(t) w / x1(r)x3(r w r)dr = C(r) u’CX. 2.4 System Response to External Input: Zero—State Response 173 Therefore 00 x10) *xzo‘ m ’1"); f x1(r)x2(i m T m I) dr v 700 2 CU — T) Equation (2.34b) follows from Eq. (2.34a). CONVOLUTION WiTl—I AN IMPULSE COHVOIBUOII of a function x (r) with a unit impulse results in the function x (I) itself. By definition of convolution 00 x(t)a<8(t‘) m / x(r)8(twr)dr (2.35) v. “"89 Because 5 (r — r) is an impulse located at r = it, according to the sampling property of the impulse {Er}. ($24)}, the integral in Eq. (2.35) is the value ofx(r) at r m t, that is, x(t). Therefore x0?) *50) = 1(3) (2.36) Actually this result was derived earlier [Er]. (2.28)]. THE WIDTH PROPERTY if the durations (Widths) of x; (r) and X20?) are finite, given by T; and T2, respectively, then the duration (width) ofxt (t) a: x2(r) is T; + T; (Fig. 2.4). The proof of this property follows readily from the graphical considerations discussed later in Section 2.40.. ZERO~STME RESE’ONSE AND CAUSALttY The (zerowstate) response y(t) of an LTIC system is 00 y(1)$ x0?) vth) 2/ x(r)h(t — 1:) d1” (2.37) w ea In deriving Eq. (2.37), we assumed the system to be linear and time invariant. There were no other restrictions either on the system or on the input signal x0). In practice, most systems are Figure 2.4 Width property of convolution. 174 CHAPTER 2 TIME—DOMAIN ANALYSIS OF CONT‘INUOUSnTlME SYSTEMS 2.4 System Response to External input: Zero-State Response 175 x”) == 0 h(r 7 T) 3 0 K O I T H’ r _ For an LTIC system with the unit impulse response MI) 2 e’z’uU), determine the response (a) y(t) for the input 160‘) m eerie) (2.39) i an : r hrr-e=0 “<0 '- ‘3: . i .I I Here both x(r) and Mr) are causal (Fig. 2.6). Hence, from Eq. (2.38a), we obtain 3 Figure 2.5 Limits ot‘ the convolution I (a) integral. ' 31(1) m/O x(r)h{i‘ m r)dr t 3 O r Because x(t) m e""fu(t) and Mr) 2 e’quU) cause}, so that their response cannot begin before the input. Furthermore, most inputs are also x(r) = e"""‘u(r) and Mr —— r) m ewzi"""”n(t — r) causal, which means they start at r : O. Causality restriction on both signals and systems further simplifies the limits of integration in Eq. (2.37). By definition, the response of a causal system cannot begin before its input begins. Consequently, the causal system’s response to a unit impulse 5(1‘) (which is located at t = 0) cannot begin before I 2 it Therefore, a causal system’s unit impulse response i1(t) is a causal ‘ _, . y(r) 2 6 Te ’0 “(it t> 0 Signal. I A w it is important to remember that the integration in Eq. (2.37) is performed with respect to ' r (not I). if the input x(t) is causal, x(r) = O for r < 0. Therefore, x(r) = 0 for r < O, as illustrated in Fig. 2.5a. Similarly, ifh(r) is causal, Mr W r) = 0 for r — r < 0; that is, for r > r, as depicted in Fig. 2.53. Therefore, the product x(r}h(t w r) = O everywhere except over the nonshaded interval 0 g r g 1” shown in Fig. 2.52 (assumingt a 0). Observe that if r is negative, x(r)h(r m r) m 0 for all r as shown in Fig. 2.5%). Therefore, Eq. (2.37) reduces to Remember that the integration is performed with respect to r (not I), and the region of integration is 0 S r g t. Hence, I 3 0 andr— 1: 3 0. Therefore, Mr) 3 l and Li(rwr) = 'l; consequently ii(I) ‘r y(t) m Mr) $110) z/ x(r)h(r «— 1:)d1: I2 0 ~() 3 0 t < 0 (2.383) The lower limit of integration in Eq. (2.38s) is taken as 0‘ to avoid the difficulty in integration that can arise if x (r) contains an impulse at the origin. This result shows that if x (r) and Mr) are both causal, the response _v(t) is also causal. Because of the convolution’s commutative property [Eq. (2.31)], we can also express Eq. (2.3821) as [assuming causal x0) and 110)} y(t) = h(r)x(t w r) dr t 3 0 - G" t = 0 r < 0 (2.38b) (c) Hereafter, the lower limit of 0’ will be implied even when we write it as 0. As in Eq. (2.383), Figure 2,6 Cenmiution 015x”) and [my this result assumes that both the input and the system are causal. 3.76 CHAEWER 2 TIMEvDOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS 2.4 System Response to External Input: Zero-State Response 177 TABLE 2.1 Convolution Table Because this integration is with respect to r, we can poll 6‘” outside the integral, giving r - -- - - .. .. . us NO- 77777 W x10) mt) 261(1) * xztt) = 9:28) at xitt) - i i ctr) at; _ I) My , 1) """""" ii y(r) : 64’] 6I d1 = e’ZKe’ —— l') = e” — 6—” r3 0 0 V 1 m elt 2 6”” tr) utt) ~-————— um Moreover, y(r) = 0 when t < 0 {see Eq. (23821)]. Therefore —/\ 5% i i E ,I _ i Mi?) Mi) m r t no=o —e”ma) t (i i r ‘ er I m elgf '3 The response is depicted in Fig. 2.60. - 4 e ‘i “(1} WW0) Mimi—«Em so) A1 # A2 5 e M it (I) e’" u (I) lie)" M (f) ., , 1 , 6 MM?) GMMU‘ ~13 “ z exescsssezs } ”” For an LTEC system with the impulse. response 1: t t} m 69'" 1-! (r ). detetmine the system response M N M . . ' 7 IN (1. A, to the input: - -' Lt ) e at!) NH “(2‘) ~ 2 Mt) . . [2:320 {a} Butt} . . . . . . so fiestas} ' u = -' -.' ' . -. . - "=[';{ - _ s rMuo) IN I eg:EEXLli M+Ne ‘ _ . . .- . -. - -- _ . ._. _ wt) (M+N+1)!r um AN SWE‘RS lfifiw-ng‘} « . Arm kg; -.E___ A: .g i I ‘ _ 9 remiufl} GWMU) e 2 e + (A._ Aflte t “(1‘) (be 9(6 _‘ w e MI) (A; m M? _ . _ . t t 10 IMEMMS) riveltufi) tittiei’t'uluéeklua) (N+M+1)! . __-. M J 7 V r . . _ : . 11 {Melting} twelgru(r) a“) - - ' kttM m k)!().1w A2)N+’<*1 Repeat Exercise £2.13 for the input x0) :2 e""'t4t'r}. ' N —1“NTM k ! M W +Zt ) t + )r e ANSWER _ '. . _ ' I -. _' ;_ 3 . )t 5H2 I atom ' ' ' '.aej a ' kgmwwnmwwonw”” __ l! __ 7m . __ 12 e"°“ cos (,3; + mug} emu“) W‘flfljfi .... “(3) i W THE CONVOLUTION TABLE $5 : {Emmi-WWW + m The task of eonvoiution is considerabiy simplified by a ready—made convolution table (Table 2.1). This tabte, which lists several pairs of signais and their convoiution, can conveniently determine 13 emu“) chm—E) 61mm + eagrutmr) R A R " y(t), a system response to an input ta), without performing the tedious job of integration. For JET)” e 2 > e M instance, we could have readily found the convolution in Example 2.5 by using pair 4 (with ‘3 M M A} m ml and A2 r: ‘2) to be (6* — e‘Z’MU). The following example demonstrates the utility 5 14 emf/Kit} ei'l’tMu—t) e—l¥_—:— u(mr) . 2 W l of this table. ; 7— .... .......... W .................................................... .. 2.4 System Response to External input: Zero-State Response 179 178 CHAPTER 2 TIME—DOMAEN ANALYSIS OF CONTENUOUS-TIME SYSTEMS EXERCISE E23 For an LTlC system with the unit impulse response Mt} {3"‘3‘ntn. determine the zero~stare : Find lhfi 100p current 310‘ l 013?“? RLCCiFCUii in Example 2.2 for the infill/TU) m 106 #3!” (f )1 3% - response vii) lithe input Mt) sin 3! an}. {Hint Use pair i2 from Table 2.1;] when all the initial conditions are zero. 4 AN SW ER ' %-l3e‘ *" + Ei3cos-{3: m 'l46.32"")]u.ttl . . 3 w The loop equation for this circuit {see Example l.i0 or Eq. (1.55)] is g {p2 + 31) + 2)y(t) a: we) The impulse response M!) for this system, as obtained in Example 2.4, is are = (262! — (one) O f qr flag-"'— — mean; + Hessian) The input is x(t) = 106‘3’a(t), and the response y(t) is i RESPONSE TO COMPLEX INPUTS yo) e no * he) 2s; The LTIC system response discussed so far applies to general input signals, real or compiex. m: jag—31 14(1) 2}: [25m — e"]u(r) However, if the system is real, that is, if Mr) is real, then we shall show that the real part of . . . w . ' the input generates the real part of the output, and a similar conclusion applies to the imaginary Using the distributive property oi the convolution {hq (2.32)], we obtain part- ym = 10673!” (I) * ge-ZIMU) W mafia“) * (tug) ‘ if the input is x(z‘) = Jolt) + jxi (I), where x,.(t) and x,(r) are the real and imaginary parts ofx(t), then for real Mr) m 20[e_3'a(t) a (Eaton — too—tan) =1: arson t=hr}>i<x,.r+'xj(t=hr>err+hra<xir=linear : Nowtheuseofpair4in'Fable2.1yields y” l l O J J] U H J U 0 3’0 J”) 10 5% where y,-(I) and y,- (t) are the reai and the imaginary parts of y(t). Using the right-directed arrow ya) = {6’3 __ [Mum _ Wit?“ _ affirm notation to indicate a pair of the input and the corresponding output, the foregoing resuit can be "’3 “ (‘2) ‘3 m (*7 l) expressed as follows. if 3 “209—3” * WW) + 5V” "“ WW) M?) m are + flair) we ya) a are a Dim i = (—5194 + 208—2! m lSe’S’fitU) 2' 5 than wwwm_mw‘m6wwwmggwmpgm,MW{mammalma;sag/immmwmxummthmvMWWWWMWMWfiNfiWfiMWWW‘N‘““fiMymw‘mmwwmwmwflmWManwgflmmjgwemfiwmmmmwww ? x}. ) 2% yr ( ) xiii} ==> y; (1‘) (2.40) EXERCISE E237 MULTIPLE INPUTS Multiple inputs to LTI systems can be treated By applying the superposition principle. Each input is considered separately, with all other inputs assumed to be zero. The sum of all these individual system responses constitutes the total system output when all the inputs are applied simultaneously. Rework Exercises E25 and E16 using the convolution table. EXERCISE 132.8 Use 316 mnmlumfl “5131‘? “5’ deter“an 2.4-2 Graphical Understanding of Convolution Operation 5’ #34“ l H "_ ‘” 331W} Convolution operation can be grasped readily by examining the graphics? interpretation of the AN SWE R convolution integral. Such an understanding is heipfu'l in evaluating the convolution integral of i s“ p, Mgr,” €- 72: i My} more complex signals. In addition, graphical convolution allows us to grasp visually or mentally Hus nnnunlnfinn 1‘n1'anrnl’rx rmrw'tli‘ “rid/sh nah Inn n-F nwnnf inaln 3n nnw\nl;nn 'Iqli‘nr-inm rum-l mnnvu 180 CHAPTER 2 TEMEDOMAIN ANALYSIS OF CONTINUOUS—TIME SYSTEMS other problems. Finally, many signals have no exact mathematical description, so they can be described only graphically. If two such signals are to be convolved, we have no choice but to perform their convolution graphically. We shall now explain the convolution operation by convolving the signals x(z‘) and got), illustrated in Fig. 2.7a and 2.7b, respectively. If C(t) is the convolution of .x(r) with gm, then C(r) = foo x(r)g(t we 1') dr « (2.4.l) DIG One of the crucial points to remember here is that this integration is performed with respect to r, so that t is just a parameter (iike a constant). This consideration is especially important when we sketch the graphical representations of the functions x(r) and g(t —- r) appearing in the integrand of Eq. (2.41). Both these functions should be sketched as functions of r, not of t. The function x(r} is .identicai to x(r), with r replacing t (Fig. 2.7c). Therefore, x(r) and x(r) will have the same graphical representations. Similar remarks apply to g(f) and g(r) (Fig. 2.7d). To appreciate what g(t — r) iooirs like, let us start with the function g(r) (Fig. 2.7d). Tirne reversai of this function (reflection about the vertical axis 7: x 0) yields g(-r) (Fig. 2.7e). Let us denote this function by (Mr) (to) x 8%?) Now (Mr) shifted by 2‘ seconds is (Mr M t), given by 9M? — 1} fl glwfi e“ U] = 5:0 - 7) Therefore, we first time-reverse g(r} to obtain g(——r) and then time—shift g(~r) by t to obtain g(r — I). For positive I, the shift is to the right (Fig. 2.7f); for negative I, the shift is to the left (Fig. 2.7g, 2.7h). The preceding discussion gives us a graphical interpretation of the functions .r(t) and g(t - r). The convolution C(l‘) is the area under the product of these two functions. Thus, to compute C(t‘) at some positive instant I = n, we first obtain g(e~r) by inverting g(r) about the vertical axis. Next, we right-shift or delay g(—r) by :1 to obtain gm m r) (Fig. 2.7f), and then we multiply this function by x(r), giving us the product x(r)g(t1 M r) (shaded portion in Fig. 2.71”). The area A, under this product is C(n), the value of C(I) at t = t1. We can therefore plot c-(tl) = A1 on a curve describing C(t}. as shown in Fig. 2.7i. The area under the product x(r)g(v~r) in Fig. 2.70 is C(O), the mine of the convolution fort m 0 (at the origin}. A similar procedure is followed in computing the value of C(r) at t m t; , where 12 is negative (Fig. 2.7g). In this case, the function g(——r) is shifted by a negative amount (that is. left—shifted) to obtain gm w r). Multiplication of this function with Mr} yicids the prodnct x(r}g(r2 e r). The area under this product is C(tg) w A2, giving us another point on the curve C(I) at r = I; (Fig. 2.7i). T his procedure can be repeated for all values of r. from —oo to 00. The result will be a curve describing C(t) for all time t. Note that when 1‘ 5 “3, x(r} and g(t — 1:} do not overlap (see Fig. 2.7h); therefore. C(r) = 0 fort 5 —3. m no L —i O 7—— —2 0 14—» (C) {d} (e) (f) (i) i E E l- 33 e3 :9 0; t 3 Figure 2.7 Graphicai explanation of the convolution operation. 182 CHAPTER 2 TIME—DOMAIN ANALYSES OF CONTINUOUS-TlME SYSTEMS SUMMARY OF THE GRAPHECAL PROCEDURE The procedure for graphical convolution can be summarized as follows 1. Keep the function x(r) fixed. 2. Visualize the function g(I) as a rigid wire frame, and rotate (or invert) this frame about the vertical axis (I m U) to obtain g(m—I). 3. Shift the inverted frame along the I axis by In seconds. The shifted frame now represents 8% m T)- 4. The area under the product of x(I) and gag —- I) (the shifted frame) is C(ro), the value of the convolution at: = to. 5. Repeat this procedure, shifting the frame by different values (positive and negative) to obtain ((1‘) for all values of t. The graphical procedure discussed here appears very complicated and discouraging at first reading. Indeed, some people claim that convolution has driven many electrical engineering undergraduates to contemplate theology either for salvation or as an alternative career (IEEE Spectrum, March 1991, p. 60).l Actuaily, the bark of convolution is worse than its bite. In graphical convolution, we need to determine the area under the product x(I)g(t -~ I) for all values of r from ~00 to 00. However, a mathematical description of x(r)g(t w I) is generally valid over a range of t. Therefore, repeating the procedure for every value of I amounts to repeating it only a few times for different ranges of I. Convolution: its bark is worse than its hitei l'Strange that religious establishments are not agitating for compulsory “convolution education” in schoo'is 2.4 System Response to External input: Zero-State Response 183 We can also use the commutative property of convolution to our advantage by computing x(t) >l< g(r) or g(r) * x0), whichever is simpler. As a rule of thumb, convolution computations are simplified ifwc choose to invert (rimareversc) the simpler of the two functions. For example, if the mathematical description of g0) is simpler than that of x(t), then x (t) >l< g(r) will be easier to compute than g(r) * x(r). In contrast, if the mathematical description of x(t) is simpler, the reverse will be true. We shall demonstrate graphical convolution with the following examples. Let us start by using this graphical method to rework Exampie 2.5. Determine graphically y(r) = x(r) >l< Mt) forx(t) = e“‘u(r) and h(t) = aware). In Fig. 2.821 and 2.8b we have x(r) and Mt), respectively; and Fig. 2.8c shows x(I) and h(—I) as functions of I. The function h(t — I) is now obtained by shifting h(-I) by r. If r is positive, the shift is to the right (delay); if r is negative, the shift is to the left (advance). Figure 2.8d shows that for negative 1‘, M: w I) [obtained by left-shifting h(v~«I)] does not overlap x(I), and the product x(I)h(t w I) = 0, so that y(t)m0 t<O Figure 2.8e shows the situation for r 3 0. Here x (I) and Mt — I) do overlap, but the product is nonzero only over the interval 0 g 1,’ g t (shaded interval). Therefore 3’“) = / x(I)FI(t —- I)dI :3 0 .0 All we need to do now is substitute correct expressions for x (I) and Mt —— I) in this integral. From Fig. 2.8a and 2.8b it is clear that the segments of x(t) and g(r) to be used in this convolution (Fig. 2.8e) are described by t x(t) m e‘ and Mr) 2 64’ Therefore 1' x(I) : e’ and 110‘ — I) = c’m’“ Consequently -i y(t) : / ew”rc""2{’"""”dI . o .l' 2672/6361? .0 Moreover, y(r) : 0 for r < 0, SD that W) m (6” — e ““ “'Z‘JW} Figure 2.8 Convolution of x{t) and Mr). .: 2 r; .2 .4 .1 "Q '4 '3 -s '4 s 4 :§ '§ :5 .3 .i ,4 'i ,2 l .l .2 a a 2.4 System Response to External Input: Zero-State Response Find (:0) = x0) »:< g(r) for the signals depicted in Fig. 2.9a and 2.9'b, Since x(t) is simpler than 3(1), it is easier to evaluate 3(1) * x(r) than x(r) 4: g(t). However, we shall intentionally take the more difficult route and evaluate x(.t) * g(t). From x(r) and g{r) (Fig. 2.921 and 2.9b, respectively), observe that g(:) is composed of two segments. As a result, it can be described as 2.9” segment A 30) m MZeZ’ segment B Therefore 29”“ "1) segment A 30‘ - r) = W.) —26' ’ segment B The segment of x0) that is used in convolution is x(t) m 1, so that Mr) 2 1. Figure 2.9c shows x(r) and g(_r). To compute 60‘) for I 3 0, we right»shift g(~r) to obtain 30‘ w t), as illustrated in Fig. 2.9d. Clearly, g0 —- r) overlaps with x(r) over the shaded interval, that is, over the range 1' _>_ 0; segment A overlaps with 260:) over the interval (0, I), while segment B overlaps with x{t) over (I, 00). Remembering that x(r) = l, we have (2(1) : /mx(r)g(r w t)dr o t ~o<z =f 2e-s-I>dz+/ MZQZU‘TMt () : =2(lwe_’)ml m'l~2e” I30 Figure 2.9e shows the situation for t < 0. Here the overlap is over the shaded interval, that is, over the range 1: 3 0, where only the segment B of g(t) is involved. Therefore c{2‘) = foox{t)g(r m “()6175 o m g(r—I)dt .0 ‘OC / WZEZU "" "v dr 0 ~e21 r < 0 H 185 a 2.4 System Response to External Input: Zero-State Response 287 gm 2 Am a Therefore Segment oft; i 1 w Ze’” r a 0 c t m g ( ) mer t g 0 o I w 0 H» 1 Segment ® 72le Figure 29f shows a plot of chi). (b) i? Find x(r) as gU) for the functions x(t) and g(t) shown in Fig. 2.10s and 2.10b. Here, x0) has a simpler mathematical description than that of gm, so it is preferable to Lime-reverse x0). Hence, we shall determine g0) a< x0) rather than x0) * gm. Thus cm = gm >u< xv) 00 m f g(r)x(t — I) dr First, we determine the expressions for the segments of x(t) and gm used in finding C(I). According to Fig. 2.10a and 2.1013, 1these segments can be expressed as x(t) m 1 and g(t) m %t so that " x(2‘ — I) = 1 and g(t} = %t Figure 2.10.: shows g(r) and x04}, whereas Fig. 2.10d shows g(t) and x0 — I“), which is x (—r) shifted by t. Because the edges ofx(m— t) are at 1' m m1 and 1, the edges of x (t — r) are at m1 +t and l. + I. The two functions overlap over the interval (0, 1 + t) (shaded interval), so that ctr) = f gmxtt m 2') dr = f it dz I Q mam—1)? —1 5:51 {2.423) This situation, depicted in Fig. 2.10d, is valid only for —1 5 t g I. For t 3 1 but 5 2, the situation is as illustrated in Fig. 2.10e. The two functions overlap only over the range m1 ml” t to 1 + t (shaded interval}. Note that the expressions for g(t) and x(t _ I) do not w. an m 1 1.1... "Ci./‘\ «"4 n.f«\ I? 2.4 System Response to External Input: Zero~Staie Response 189 change; only the range of integration changes. Therefore ~1—l—i‘ / érdt .eluw 7 t 1 g r g 2 (2.4%) C(t) il (AIM Also note that the expressions in Eqs. (2.423} and (2.42b) both apply at t = l, the transition point between their respective ranges. We can readily verify that both expressions yield a value of 2/3 at t = 1, so that C(l) m 2/3. The continuity of cm at transition points indicates .3 a high probability of a right answer. Continuity of C(t) at transition points is assured as long as there are no impulses at the edges of Mr} and g0). Fort 3 2 out 5 4 the situation is as shown in Fig. 2.10? The functions g(r) and x(r — r) overlap over the interval from ——1 + t to 3 (shaded interval}, so that 3 'l f i17le —l—I-! =w%(12w2rm8) 2Et54 (2.42s) C(t) Again, both Eqs. (2.4%) and (2.42:3) apply at the transition point r 2 2. We can readily verify that (2(2) m 4/3 when either of these expressions is used. For t 3 4, x(r — t) has been shifted so far to the right that it no longer overlaps with g(r) as depicted in Fig. 2.l0g. Consequently co) m 0 r a 4 (2.42d) We now turn our attention to negative values of I. We have already determined C(f) up to r = —E. For t < m] there is no overlap between the two functions, as iiiustrated in Fig. 2. ion. so that C(I) : 0 r g ~1 (2.42e) Figure 2. lOi plots C(t) according to Eqs. (2.42s) through (2.42e). 3&4 «rm- swmeamfimmm ammmwmmmaswmmsMwemamasmisewswmwwfixwowummmm n.me wmfiesmwsemsmwsw xwmnw was-Masses Kmmmwmwmmd WIDTH OF THE CONVOLVED FUNCTION The widths (durations) of 26(1), g(t), and C(t) in Example 2.9 (Fig. 2.10) are 2, 3, and 5, respectively. Note that the width of C(f) in this case is the sum of the widths of x(t) and 30). This observation is not a coincidence. Using the concept of graphical convolution, we can readily see that if x(r) and g(t) have the finite widths of T; and T2 respectiveiy, then the width of C(t) is equal to T; + T2. The reason is that the time it takes for a signal of width I .2 (duration) 7"; to corripletely pass another signal of width (duration) T2 so that they become m nonoverlapping is Ti + T3. When the two signals become nonoveriapping, the convolution goes 5 "Fiaan 2.10 Convolution ol‘xm and alt}. 1 a n fl n _ .n 190 CHAPTER 2 TIMEDOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS EXERCiSE €2.18 Rework Exampie 2.8'by evaluating g{_t) * .r(r) EXERCISE E211 Use graphical convolution to Show that .r(r‘; * gm m gm salt} ":2 cl!) in Fig. 2i 1. i W? itng ii Cm 1"“ r" ’ i ._..4_.._...MW.777,774HJ..M....._._.. o r w u " I w a Figure 2.1} Convolution of xi!) and grit). EXERCiSE E2512 Repeat Exercise 132.11% for the functions in Fig. 2. t2. Fit} 1: to) ._. _ _ = . . _ . Y ; I - . ' 1 ii In” - i} 2"“, Figure 3.12 Convointion {Ff-x3} and git). EXERCISE £2.13 Repeat Exercise E'Zi‘i for the functions in Fig. 1i}. Figure 2139 Convolution oft-(ii and gift. 2.4 System Response to External Input: Zero»8tate Response 191 THE PHANTOM OF THE SIGNALS AND SYSTEMS OPERA In the study of signals and systems we often come across some signals such as an impulse, which cannot be generated in practice and have never been sighted by anyone.“ One wonders why we even consider such idealized signals. The answer should be clear from our discussion so far in this chapter. Even if the impulse function has no physical existence, we can compute the system response Mr) to this phantom input according to the procedure in Section 2.3, and knowing Mr), we can compute the system response to any arbitrary input. The concept of impulse response, therefore, provides an effective intermediary for computing system response to an arbitrary input. in addition, the impulse response Mr) itself provides a great deal of information and insight about the system behavior. In Section 2.7 we show that the knowledge of impuise response provides much valuabie information, such as the response time, puise dispersion, and littering properties of the system. Many other usefui insights about the system behavior can be obtained by inspection of h(r). Similarly, in frequency-domain anaiysis (discussed in Eater chapters), we use an everlasting exponential (or sinusoid ) to determine system response. An everlasting exponential (or sinu— soid) too is a phantom, which no-body has ever seen and which has no physical existence. But it provides another effective intermediary for computing the system response to an arbitrary input. Moreover, the system response to everlasting exponentiai (or sinusoid) provides valuable infor— mation and insight regarding the system’s behavior. Clearly, ideaiized impuise and everlasting sinusoids are friendly and helpful spirits. Interestingly, the unit impulse and the everlasting exponential (or sinusoid) are the dual of each other in the time-frequency duality, to be studied in Chapter 7. Actually, the time-domain and the frequency-domain methods of analysis are the dual of each other. WHY CONVOLUTION? AN iN'IUlTlVE EXPLANATION OF SYSTEM RESPONSE On the surface, it appears rather strange that the response of linear systems (those gentlest of the gentie systems) should be given by such a tortuous operation of convolution, where one signal is fixed and the other is inverted and shifted. To understand this odd behavior, consider a hypothetical impulse response h(r) that decays linearly with time (Fig. 2.14a). This response is strongest at r = 0, the moment the impulse is applied, and it decays iinearly at future instants, so that one second later (at r w l and beyond), it ceases to exist. This means that the closer the impulse input is to an instant I, the stronger is its response at 1‘. Now consider the input x(r) shown in Fig. 2.14b. To compute the system response, we break the input into rectangular putses and approximate these pulses with impulses. Generally the response of a causal system at some instant: will be determined by all the impulse components of the input before 1‘. Each of these impulse components will have different weight in determining the response at the instant 1‘, depending on its proximity to I. As seen earlier, the closer the i'i‘he late Prof. S. 3. Mason, the inventor of signal flow graph techniques, used to tell a story of a student frustrated with the impulse function. The student said, “The unit impuise is a thing that is so small you can’t see it, except at one place (the origin), where it is so big you can’t see it. In other words, you can’t see it at all; 192 CHAPTER 2 TIME-DOMAEN ANALYSIS OF CONTINUOUS-TIME SYSTEMS hm 1 second (a) (b) Figure 2.14 intuitive explanation of convolution. impulse is to t, the stronger is its influence at t. The impuise at t has the greatest weight (unity) in determining the response at I. The weight decreases linearly for all impulses before 1' until the instant I — i. The input before I m i has no influence (zero weight). Thus, to determine the system response at I, we must assign a linearly decreasing weight to impulses occurring before I, as shown in Fig. 2.14b. This weighting function is precisely the function Mt - r). The system response at t is then determined not by the input x(r) but by the weighted input x(r)h(r m r), and the summation of all these weighted inputs is the convolution integral. ~3 Interconnected Systems A iar more complex system can often be viewed as the interconnection of several sma'l'ir subsystem each of which is easier to characterize. Knowing the characterizations of ese subsystems, it : comes simpler to analyze such large systems. We shall consider her . ' 0 basic interconnections, .. cade and parailei. Figure 2.15a shows 8; and 82, two LT ' subsystems connected in paraliel, d Fig. 2.15%) shows the same two systems connecte 'n cascade. In Fig. 2.15a, the devr : depicted by the symboi 2 inside a circle re p '- sents an adder, which adds signals at its inputs. Als he junction from which two (or a e) branches radiate out is caiied the pickofi'node. Every bra that radiates out from r pickoff node carries the same signal (the signal at the junction). in ‘ . 2.15a, for insta ’ , the junction at which the input is applied is a pickoff node from which ltwo 'a nches r a am out, each of which carries the input signal at the node. Let the impulse response of 8, and 83 b ' ;(r nd mm, respectively. Further assume that interconnecting these systems, as show ’ Fig. 2.i5, - es not load them. This means that the impulse response of either of these : stems remains unch i ed whether observed when these systems are unconnected or w :. they are interconnected. To find hptt), the itn se response of the paraiicl system ,, ' Fig. 2.153, we appiy an impulse at the input of 1 p. This resuits in the signal 8 (r) at the inputs of , . nd 82, heading to their outputs is I (t) and a _ I), respectively. These signais are added by the adder t , ieid J13 (r) + ham as the output . ’ 8,,. Consequently hilt) E 1110‘) + h2(1‘) (2.432!) 0 find 11,. (t), the impulse response of the cascade system Sc. in Fig. 2.15b, we appiy the in t 2.4 System Response to External Input: Zero~State Response 1 hptr) : hltr) + an; 80} 11(1) fi 112(1) a hEU) (D Figure 2.15 nterconnected systems. acts as the input to 8; he response of 82 to input 1110) is h, (I) >s< fight). erefore Piotr} = 11:0} * 1120) (24%) Because of co v utative property of the convolution, it follows that interchangi ‘ ; the systems Si and 82, as . own in Fig. 2.150, results in the same impuise response is, (t) a #120). This means that when , veral LTiC systems are cascaded, the order of systems does not affect t - impulse response f the composite system. in other words, linear operations, performed in scaée, comm c. The order in which they are performed is not important, at least theoretically“ angeof order, however, could affect performance because of physical limitations and sensitivities to ...
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This note was uploaded on 01/16/2011 for the course EL ENGR 115A taught by Professor Abidi during the Fall '10 term at UCLA.

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115A_1_Tutorial on Convolution - rm .1 5(3) (it) Mr) I 0...

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