Fourier Series - CAU
Fourier SeriesMathematicians of the eighteenth century, including Daniel Bernoulli and Leonard Euler, expressedthe problem of the vibratory motion of a stretched string through partial differential equations that hadno solutions in terms of ‘‘elementary functions.’’ Their resolution of this difficulty was to introduceinfinite series of sine and cosine functions that satisfied the equations. In the early nineteenth century,Joseph Fourier, while studying the problem of heat flow, developed a cohesive theory of such series.Consequently, they were named after him. Fourier series and Fourier integrals are investigated in thisand the next chapter. As you explore the ideas, notice the similarities and differences with the chapterson infinite series and improper integrals.PERIODIC FUNCTIONSA function f ðxÞ is said to have a period T or to be periodic with period T if for all x, f ðx þ TÞ ¼ f ðxÞ,where T is a positive constant. The least value of T 0 is called the least period or simply the period off ðxÞ.EXAMPLE 1. The function sin x has periods 2!; 4!; 6!; . . . ; since sin ðx þ 2!Þ; sin ðx þ 4!Þ; sin ðx þ 6!Þ; . . . allequal sin x. However, 2! is the least period or the period of sin x.EXAMPLE 2. The period of sin nx or cos nx, where n is a positive integer, is 2! n.EXAMPLE 3. The period of tan x is !.EXAMPLE 4. A constant has any positive number as period.f (x)xxx(a)Periodf (x)Periodf (x)PeriodOther examples of periodic functions are shown in the graphs of Figures 13-1(a), (b), and (c) below.(b)(c)Fig. 13-1336Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
CHAP. 13]337FOURIER SERIESFOURIER SERIESLet f ðxÞ be defined in the interval ð#L; LÞ and outside of this interval by f ðx þ 2LÞ ¼ f ðxÞ, i.e., f ðxÞis 2L-periodic. It is through this avenue that a new function on an infinite set of real numbers is createdfrom the image on ð#L; LÞ. The Fourier series or Fourier expansion corresponding to f ðxÞ is given by1 !a0 Xn!xn!x"þ bn sinþan cosð1ÞLL2n¼1where the Fourier coefficients an and bn are8ð1 Ln!x dx¼f ðxÞ cosa n LL#Lð 1 Ln!x : bn ¼dxf ðxÞ sinL #LLn ¼ 0; 1; 2; . . .ð2ÞORTHOGONALITY CONDITIONS FOR THE SINE AND COSINE FUNCTIONSNotice that the Fourier coefficients are integrals. These are obtained by starting with the series, (1),and employing the following properties called orthogonality conditions:(a)ðLcosm!xn!xcosdx ¼ 0 if m 6¼ n and L if m ¼ nLLsinm!xn!xsindx ¼ 0 if m 6¼ n and L if m ¼ nLLsinm!xn!xcosdx ¼ 0. Where m and n can assume any positive integer values.LL#L(b)ðL#L(c)ðL#L(3)An explanation for calling these orthogonality conditions is given on Page 342. Their application indetermining the Fourier coefficients is illustrated in the following pair of examples and then demonstrated in detail in Problem 13.4.EXAMPLE 1. To determine the Fourier coefficient a0 , integrate both sides of the Fourier series (1), i.e.,ðLðLðL X1 na0n!xn!xof ðxÞ dx ¼an cosdx þdxþ bn sinLL#L#L 2#L n¼1NowðLa0dx ¼ a0 L;#L 2ðL#lsinn!xdx ¼ 0;LðL#Lcosn!x1dx ¼ 0, therefore, a0 ¼LLðL#Lf ðxÞ dx!xEXAMPLE 2. To determine a1 , multiply bothð L sides of (1) by cos L and then integrate. Using the orthogonality1!xf ðxÞ cosdx. Now see Problem 13.4.conditions (3)a and (3)c , we obtain a1 ¼L #LLIf L ¼ !, the series (1) and the coefficients (2) or (3) are particularly simple.case has the period 2!.DIRICHLET CONDITIONSSuppose that(1)(2)f ðxÞ is defined except possibly at a finite number of points in ð#L; LÞf ðxÞ is periodic outside ð#L; LÞ with period 2LThe function in this
338FOURIER SERIES(3)[CHAP. 13f ðxÞ and f 0 ðxÞ are piecewise continuous in ð#L; LÞ.Then the series (1) with Fourier coefficients converges toðaÞðbÞf ðxÞ if x is a point of continuityf ðx þ 0Þ þ f ðx # 0Þif x is a point of discontinuity2Here f ðx þ 0Þ and f ðx # 0Þ are the right- and left-hand limits of f ðxÞ at x and represent lim f ðx þ !Þ and!!0þlim f ðx # !Þ, respectively. For a proof see Problems 13.18 through 13.23.!!0þThe conditions (1), (2), and (3) imposed on f ðxÞ are sufficient but not necessary, and are generallysatisfied in practice. There are at present no known necessary and sufficient conditions for convergenceof Fourier series. It is of interest that continuity of f ðxÞ does not alone ensure convergence of a Fourierseries.ODD AND EVEN FUNCTIONSA function f ðxÞ is called odd if f ð#xÞ ¼ #f ðxÞ. Thus, x3 ; x5 # 3x3 þ 2x; sin x; tan 3x are oddfunctions.A function f ðxÞ is called even if f ð#xÞ ¼ f ðxÞ. Thus, x4 ; 2x6 # 4x2 þ 5; cos x; ex þ e#x are evenfunctions.The functions portrayed graphically in Figures 13-1(a) and 13-1ðbÞ are odd and even respectively,but that of Fig. 13-1(c) is neither odd nor even.In the Fourier series corresponding to an odd function, only sine terms can be present. In theFourier series corresponding to an even function, only cosine terms (and possibly a constant which weshall consider a cosine term) can be present.HALF RANGE FOURIER SINE OR COSINE SERIESA half range Fourier sine or cosine series is a series in which only sine terms or only cosine terms arepresent, respectively. When a half range series corresponding to a given function is desired, the functionis generally defined in the interval ð0; LÞ [which is half of the interval ð#L; LÞ, thus accounting for thename half range] and then the function is specified as odd or even, so that it is clearly defined in the otherhalf of the interval, namely, ð#L; 0Þ. In such case, we have8ð2 Ln"x adx for half range sine series¼0;b¼f ðxÞ sinn nL 0Lð4Þð 2 Ln"x dx for half range cosine seriesf ðxÞ cos: bn ¼ 0; an ¼L 0LPARSEVAL’S IDENTITYIf an and bn are the Fourier coefficients corresponding to f ðxÞ and if f ðxÞ satisfies the Dirichletconditions.Then(See Problem 13.13.)1LðL#Lf f ðxÞg2 dx ¼1a20 Xþða2 þ b2n Þ2 n¼1 n(5)
CHAP. 13]FOURIER SERIES339DIFFERENTIATION AND INTEGRATION OF FOURIER SERIESDifferentiation and integration of Fourier series can be justified by using the theorems on Pages 271and 272, which hold for series in general. It must be emphasized, however, that those theorems providesufficient conditions and are not necessary. The following theorem for integration is especially useful.Theorem.The Fourier series corresponding toð f ðxÞ may be integrated term by term from a to x, and thexresulting series will converge uniformly toaf ðxÞ dx provided that f ðxÞ is piecewise continuous in#L @ x @ L and both a and x are in this interval.COMPLEX NOTATION FOR FOURIER SERIESUsing Euler’s identities,ei! ¼ cos ! þ i sin !;e#i! ¼ cos ! # i sin !ð6Þpffiffiffiffiffiffiwhere i ¼ #1 (see Problem 11.48, Chapter 11, Page 295), the Fourier series for f ðxÞ can be written asf ðxÞ ¼1X12Lð7Þf ðxÞe#in"x L dxð8Þn¼#1wherecn ¼cn ein"x LðL#LIn writing the equality (7), we are supposing that the Dirichlet conditions are satisfied and furtherthat f ðxÞ is continuous at x. If f ðxÞ is discontinuous at x, the left side of (7) should be replaced byðf ðx þ 0Þ þ f ðx # 0Þ:2BOUNDARY-VALUE PROBLEMSBoundary-value problems seek to determine solutions of partial differential equations satisfyingcertain prescribed conditions called boundary conditions. Some of these problems can be solved byuse of Fourier series (see Problem 13.24).EXAMPLE.The classical problem of a vibrating string may be idealized in the following way. See Fig. 13-2.Suppose a string is tautly stretched between points ð0; 0Þ and ðL; 0Þ. Suppose the tension, F, is thesame at every point of the string. The string is made tovibrate in the xy plane by pulling it to the parabolicposition gðxÞ ¼ mðLx # x2 Þ and releasing it. (m is anumerically small positive constant.) Its equation willbe of the form y ¼ f ðx; tÞ. The problem of establishingthis equation is idealized by (a) assuming that the constant tension, F, is so large as compared to the weight wLof the string that the gravitational force can be neglected,(b) the displacement at any point of the string is so smallthat the length of the string may be taken as L for any ofits positions, and (c) the vibrations are purely transverse.w@2 yThe force acting on a segment PQ is!x 2 ;g@tx x1 x þ !x; g & 32 ft per sec:2 . If # and are theangles that F makes with the horizontal, then the verticalFig. 13-2
340FOURIER SERIES[CHAP. 13difference in tensions is Fðsin ! " sin "Þ. This is the force producing the acceleration that accounts for()"the vibratory motion.tan !tan "@yðx þ !x; tÞ"NowFfsin ! " sin "g ¼ F fiffiffiffi " fiffiffiffi & Fftan ! " tan "g ¼ F2#2@x1 þ tan "1 þ tan !@yðx; tÞ , where the squared terms in the denominator are neglected because the vibrations are small.@xNext, equate the two forms of the force, i.e.,"#@y@yw@2 yðx þ !x; tÞ " ðx; tÞ ¼ !x 2F@x@xg@trffiffiffiffiffiFg, the resulting equation isdivide by !x, and then let !x ! 0. After letting ! ¼w2@2 y2@ y¼!@t2@x2This homogeneous second partial derivative equation is the classical equation for the vibratingstring. Associated boundary conditions areyð0; tÞ ¼ 0; yðL; tÞ ¼ 0; t 0The initial conditions areyðx; 0Þ ¼ mðLx " x2 Þ;@yðx; 0Þ ¼ 0; 0 x L@tThe method of solution is to separate variables, i.e., assumeyðx; tÞ ¼ GðxÞHðtÞThen upon substitutingGðxÞ H 00 ðtÞ ¼ !2 G 00 ðxÞ HðtÞSeparating variables yieldsG 00H 00¼ k;¼ !2 k; where k is an arbitrary constantGHSince the solution must be periodic, trial solutions arepffiffiffiffiffiffipffiffiffiffiffiffiGðxÞ ¼ c1 sin "k x þ c2 cos "k x; 0pffiffiffiffiffiffipffiffiffiffiffiffiHðtÞ ¼ c3 sin ! "k t þ c4 cos ! "k tThereforey ¼ GH ¼ ½c1 iffiffiffipffiffiffiffiffiffi"k x þ c2 cos "k x(½c3 sin ! "k t þ c4 cos ! "k t(The initial condition y ¼ 0 at x ¼ 0 for all t leads to the evaluation c2 ¼ iffiffiffiffiy ¼ ½c1 sin "k x(½c3 sin ! "k t þ c4 cos ! "k t(pffiffiffiffiffiffipffiffiffiffiffiffiNowthe boundary condition y ¼ 0 at x ¼ L, thus 0 ¼ ½c1 sin "k L(½c3 sin ! "k t þpffiffiffiimposeffiffiffic4 cos ! "k t(:c1 6¼ 0 as that would imply y ¼ 0 and a trivial solution. The next simplest solution results from thehpffiffiffiffiffiffi n#n# ihn#n# it þ c4 cos !t and the first factor is zero whenchoice "k ¼ , since y ¼ c1 sin x c3 sin !LLlLx ¼ L.
CHAP. 13]341FOURIER SERIES@yðx; 0Þ ¼ 0, 0 x L can be considered.@t@y hn! ihn!n!n!n! i¼ c1 sinx c3 "cos "t c4 "sin "t@tLLLLLWith this equation in place the boundary conditionAt t ¼ 0Since c1 6¼ 0 and sinhn! in!0 ¼ c1 sinx c3 "LLn!x is not identically zero, it follows that c3 ¼ 0 and thatLhn! ihn!n! ix c4 "cos "ty ¼ c1 sinLLLThe remaining initial condition isyðx; 0Þ ¼ mðLx x2 Þ; 0 x LWhen it is imposedmðLx x2 Þ ¼ c1 c4 "n!n!sinxLLHowever, this relation cannot be satisfied for all x on the interval ð0; LÞ. Thus, the precedingextensive analysis of the problem of the vibrating string has led us to an inadequate formn!n!n!y ¼ c1 c4 "sinx cos "tLLLand an initial condition that is not satisfied. At this point the power of Fourier series is employed. Inparticular, a theorem of differential equations states that any finite sum of a particular solution also is asolution. Generalize this to infinite sum and considery¼1Xbn sinn¼1n!n!x cos "tLLwith the initial condition expressed through a half range sine series, i.e.,1Xbn sinn¼1n!x ¼ mðLx x2 Þ;Lt¼0According to the formula of Page 338 for coefficient of a half range sine seriesðLLn!xbn ¼ ðLx x2 Þ sindx2mL0That isLb ¼2m nðLLx sin0n!xdx LðLx2 sin0n!xdxLApplication of integration by parts to the second integral yieldsðLðLLn!xL3Ln!xbn ¼ L x sindx þcos2x dxcos n! þ2mLLn!00 n!When integration by parts is applied to the two integrals of this expression and a little algebra isemployed the result isbn ¼4L2ð1 cos n!Þðn!Þ3
342FOURIER SERIES[CHAP. 13Therefore,y¼1Xbn sinn¼1n!n!x cos "tLLwith the coefficients bn defined above.ORTHOGONAL FUNCTIONSTwo vectors A and B are called orthogonal (perpendicular) if A " B ¼ 0 or A1 B1 þ A2 B2 þ A3 B3 ¼ 0,where A ¼ A1 i þ A2 j þ A3 k and B ¼ B1 i þ B2 j þ B3 k. Although not geometrically or physically evident, these ideas can be generalized to include vectors with more than three components. In particular,we can think of a function, say, AðxÞ, as being a vector with an infinity of components (i.e., an infinitedimensional vector), the value of each component being specified by substituting a particular value of x insome interval ða; bÞ. It is natural in such case to define two functions, AðxÞ and BðxÞ, as orthogonal inða; bÞ ifðbaAðxÞ BðxÞ dx ¼ 0ð9ÞA vector A is called a unit vector or normalized vector if its magnitude is unity, i.e., if A " A ¼ A2 ¼ 1.Extending the concept, we say that the function AðxÞ is normal or normalized in ða; bÞ ifðbafAðxÞg2 dx ¼ 1ð10ÞFrom the above it is clear that we can consider a set of functions f#k ðxÞg; k ¼ 1; 2; 3; . . . ; having thepropertiesðbaðba#m ðxÞ#n ðxÞ dx ¼ 0f#m ðxÞg2 dx ¼ 1m 6¼ nm ¼ 1; 2; 3; . . .ð11Þð12ÞIn such case, each member of the set is orthogonal to every other member of the set and is alsonormalized. We call such a set of functions an orthonormal set.The equations (11) and (12) can be summarized by writingðba#m ðxÞ#n ðxÞ dx ¼ mnð13Þwhere mn , called Kronecker’s symbol, is defined as 0 if m 6¼ n and 1 if m ¼ n.Just as any vector r in three dimensions can be expanded in a set of mutually orthogonal unit vectorsi; j; k in the form r ¼ c1 i þ c2 j þ c3 k, so we consider the possibility of expanding a function f ðxÞ in a setof orthonormal functions, i.e.,f ðxÞ ¼1Xn¼1cn #n ðxÞa@x@bð14ÞAs we have seen, Fourier series are constructed from orthogonal functions. Generalizations ofFourier series are of great interest and utility both from theoretical and applied viewpoints.
CHAP. 13]343FOURIER SERIESSolved ProblemsFOURIER SERIES13.1. Graph each of the following functions.!30 x 5Period ¼ 10ðaÞ f ðxÞ ¼ 3 5 x 0f (x)Period325201510x505310152025Fig. 13-3Since the period is 10, that portion of the graph in 5 x 5 (indicated heavy in Fig. 13-3 above) isextended periodically outside this range (indicated dashed).Note that f ðxÞ is not defined atx ¼ 0; 5; 5; 10; 10; 15; 15, and so on. These values are the discontinuities of f ðxÞ.ðbÞ f ðxÞ ¼!sin x0@x@!0! x 2!Period ¼ 2!f (x)3p2ppPeriod02pp3p4pxFig. 13-4Refer to Fig. 13-4 above.8 0ðcÞ f ðxÞ ¼ 1 :0Note that f ðxÞ is defined for all x and is continuous everywhere.0@x 22@x 44@x 6Period ¼ 6f (x)Period11210864202468101214xFig. 13-5Refer to Fig. 13-5 above. Note that f ðxÞ is defined for all x and is discontinuous at x ¼ %2; %4; %8;%10; %14; . . . .
344FOURIER SERIES13.2. ProveðLsin!LðLcos!Lk!xdx ¼ 0Lif k ¼ 1; 2; 3; . . . ."k!xLk!x ""LLLdx ¼ !cos" ¼ ! k! cos k! þ k! cosð!k!Þ ¼ 0Lk!L!L!L"LðL"k!xLk!x "LLdx ¼sinsin k! !sinð!k!Þ ¼ 0cos¼Lk!L "!L k!k!!LðL13.3. Provek!xdx ¼L[CHAP. 13sinðL#ðLm!xn!xm!xn!x0cosdx ¼sindx ¼(a)cossinLLLLL!L!LðLm!xn!x(b)cosdx ¼ 0sinLL!Lm 6¼ nm¼nwhere m and n can assume any of the values 1; 2; 3; . . . .(a) From trigonometry: cos A cos B ¼ 12 fcosðA ! BÞ þ cosðA þ BÞg; sin A sin B ¼ 12 fcosðA ! BÞ ! cosðA þ BÞg:Then, if m 6¼ n, by Problem 13.2, ðLð #m!xn!x1 Lðm ! nÞ!xðm þ nÞ!xcosdx ¼þ coscoscosdx ¼ 0LL2 !LLL!LSimilarly, if m 6¼ n, ðLð #m!xn!x1 Lðm ! nÞ!xðm þ nÞ!xsindx ¼! cosdx ¼ 0sincosLL2 !LLL!LIf m ¼ n, we haveðL&ðL %2n!xdx ¼ L1 þ cosL!L!L&ðLð %m!xn!x1 L2n!xsindx ¼dx ¼ Lsin1 ! cosLL2 !LL!Lcosm!xn!x1cosdx ¼LL2Note that if m ¼ n these integrals are equal to 2L and 0 respectively.(b) We have sin A cos B ¼ 12 fsinðA ! BÞ þ sinðA þ BÞg. Then by Problem 13.2, if m 6¼ n, ðLð #m!xn!x1 Lðm ! nÞ!xðm þ nÞ!xsinsincosdx ¼þ sindx ¼ 0LL2 !LLL!LIf m ¼ n,ðLsin!Lm!xn!x1cosdx ¼LL2ðLsin!L2n!xdx ¼ 0LThe results of parts (a) and (b) remain valid even when the limits of integration !L; L are replacedby c; c þ 2L, respectively.13.4. If the series A þ1 'Xn¼1an cosn!xn!x(þ bn sinconverges uniformly to f ðxÞ in ð!L; LÞ, show thatLLfor n ¼ 1; 2; 3; . . . ;ð1 Ln!xdx;f ðxÞ cosðaÞ an ¼L !LLðbÞ bn ¼1LðL!Lf ðxÞ sinn!xdx;LðcÞ A ¼a0:2
CHAP. 13]345FOURIER SERIES(a) Multiplyingf ðxÞ ¼ A þ1 !Xan cosn¼1n!xn!x"þ bn sinLLð1Þm!xand integrating from %L to L, using Problem 13.3, we haveLðLðLm!xm!xdx ¼ Adxf ðxÞ coscosLL%L%L%ðLðL1 Xm!xn!xm!xn!xcosdx þ bnsindxþancoscosLLLL%L%Ln¼1by cos¼ am Lam ¼Thus1Lif m 6¼ 0ðL%Lf ðxÞ cosm!xdxLif m ¼ 1; 2; 3; . . .m!xand integrating from %L to L, using Problem 13.3, we haveLðLðLm!xm!xdx ¼ Adxf ðxÞ sinsinLL%L%L %ððL1LXm!xn!xm!xn!xcosdx þ bnsindxþansinsinLLLL%L%Ln¼1(b) Multiplying (1) by sin¼ bm LThus(c)1bm ¼LðL%Lf ðxÞ sinm!xdxLif m ¼ 1; 2; 3; . . .Integrating of (1) from %L to L, using Problem 13.2, givesðL%Lf ðxÞ dx ¼ 2ALorA¼12LðL%Lf ðxÞ dxð1 Laf ðxÞ dx and so A ¼ 0 .L %L2The above results also hold when the integration limits %L; L are replaced by c; c þ 2L:Note that in all parts above, interchange of summation and integration is valid because the series isassumed to converge uniformly to f ðxÞ in ð%L; LÞ. Even when this assumption is not warranted, thecoefficients am and bm as obtained above are called Fourier coefficients corresponding to f ðxÞ, and thecorresponding series with these values of am and bm is called the Fourier series corresponding to f ðxÞ.An important problem in this case is to investigate conditions under which this series actually convergesto f ðxÞ. Sufficient conditions for this convergence are the Dirichlet conditions established in Problems13.18 through 13.23.Putting m ¼ 0 in the result of part (a), we find a0 ¼13.5. (a) Find the Fourier coefficients corresponding to the function 0 %5 x 0f ðxÞ ¼Period ¼ 1030 x 5(b) Write the corresponding Fourier series.(c) How should f ðxÞ be defined at x ¼ %5; x ¼ 0; and x ¼ 5 in order that the Fourier series willconverge to f ðxÞ for %5 @ x @ 5?The graph of f ðxÞ is shown in Fig. 13-6.
346FOURIER SERIES[CHAP. 13f (x)Period31510x551015Fig. 13-6(a) Period ¼ 2L ¼ 10 and L ¼ 5. Choose the interval c to c þ 2L as #5 to 5, so that c ¼ #5.ðð1 cþ2Ln!x1 5n!xf ðxÞ cosf ðxÞ cosan ¼dx ¼dxL cL5 #55"ð 0#ð5ð1n!xn!x3 5n!xð0Þ coscosdx þ ð3Þ cosdx ¼dx¼5 #5555500 %&3 5n!x &&5sin¼¼0if n 6¼ 05 n!5 &0If n ¼ 0; an ¼ a0 ¼35ð50cos0!x3dx ¼55ð50Thendx ¼ 3:ðð1 cþ2Ln!x1 5n!xdx ¼dxf ðxÞ sinf ðxÞ sinL cL5 #55"ð#ðð51 0n!xn!x3 5n!x¼dx þ ð3Þ sindx ¼dxð0Þ sinsin5 #5555 050 %&35n!x &&5 3ð1 # cos n!Þ#cos¼¼5n!5 &n!bn ¼0(b) The corresponding Fourier series is1 '1a0 Xn!xn!x( 3 X3ð1 # cos n!Þn!xþ bn sinsin¼ þþan cosLL2 n¼1n!52 n¼1 %3 6!x 13!x 15!x¼ þsinþ sinþ sinþ &&&2 !53555(c)Since f ðxÞ satisfies the Dirichlet conditions, we can say that the series converges to f ðxÞ at all points off ðx þ 0Þ þ f ðx # 0Þcontinuity and toat points of discontinuity. At x ¼ #5, 0, and 5, which are points2of discontinuity, the series converges to ð3 þ 0Þ 2 ¼ 3 2 as seen from the graph. If we redefine f ðxÞ asfollows,83 2x ¼ #5 #5 x 0 0f ðxÞ ¼ 3 2x¼0Period ¼ 10 30 x 5 :3 2x¼5then the series will converge to f ðxÞ for #5 @ x @ 5.13.6. Expand f ðxÞ ¼ x2 ; 0 x 2! in a Fourier series if (a) the period is 2!, (b) the period is notspecified.(a) The graph of f ðxÞ with period 2! is shown in Fig. 13-7 below.
CHAP. 13]347FOURIER SERIESf (x)4p 26p4p2pxO2p4p6pFig. 13-7Period ¼ 2L ¼ 2! and L ¼ !. Choosing c ¼ 0, we haveðð1 cþ2Ln!x1 2! 2f ðxÞ cosx cos nx dxan ¼dx ¼L cL! 0 "#"#%&"#1sin nx% cos nx% sin nx &&2!42ðx Þþ2¼% ð2xÞ& ¼ n2 ;!nn2n30If n ¼ 0; a0 ¼1!ð 2!0x2 dx ¼8!2:3ðn!x1 2! 2dx ¼x sin nx dxL! 0c "#"#%&2!' cos nx(1sin nxcos nx &&%4!¼¼þ ð2Þ% ð2xÞ % 2ðx2 Þ %&3!nnnn0bn ¼1Lð cþ2Ln 6¼ 0f ðxÞ sin#1 "4!2 X44!sin nx :þcos nx %Then f ðxÞ ¼ x ¼n3n2n¼12This is valid for 0 x 2!. At x ¼ 0 and x ¼ 2! the series converges to 2!2 .(b) If the period is not specified, the Fourier series cannot be determined uniquely in general.13.7. Using the results of Problem 13.6, prove that111!2.þþþ&&&¼612 22 32At x ¼ 0 the Fourier series of Problem 13.6 reduces to14!2 X4þ.23nn¼1By the Dirichlet conditions, the series converges at x ¼ 0 to 12 ð0 þ 4!2 Þ ¼ 2!2 .Then11X4!2 X41!2þ¼ 2!2 , and so¼ .2236nnn¼1n¼1ODD AND EVEN FUNCTIONS, HALF RANGE FOURIER SERIES13.8. Classify each of the following functions according as they are even, odd, or neither even nor odd. 20 x 3Period ¼ 6ðaÞ f ðxÞ ¼%2 %3 x 0From Fig. 13-8 below it is seen that f ð%xÞ ¼ %f ðxÞ, so that the function is odd.ðbÞ f ðxÞ ¼ cos x00 x !! x 2!Period ¼ 2!
348FOURIER SERIES[CHAP. 13f (x)26x3362Fig. 13-8From Fig. 13-9 below it is seen that the function is neither even nor odd.f (x)12ppO2ppx3pFig. 13-9ðcÞ f ðxÞ ¼ xð10 xÞ; 0 x 10; Period ¼ 10:From Fig. 13-10 below the function is seen to be even.f (x)2510Ox510Fig. 13-1013.9. Show that an even function can have no sine terms in its Fourier expansion.Method 1:No sine terms appear if bn ¼ 0; n ¼ 1; 2; 3; . . . .1bn ¼LðLn!x1f ðxÞ sindx ¼LL Lð0To show this, let us writen!x1f ðxÞ sindx þLL LðL0f ðxÞ sinn!xdxLð1ÞIf we make the transformation x ¼ u in the first integral on the right of (1), we obtain1Lð0ð" n!u#1 Ln!uf ð uÞ sin f ð uÞ sindudu ¼ LLL00ðLðL1n!u1n!x¼ f ðuÞ sinf ðxÞ sindu ¼ dxL 0LL 0Ln!x1f ðxÞ sindx ¼LL LðLð2Þwhere we have used the fact that for an even function f ð uÞ ¼ f ðuÞ and in the last step that the dummyvariable of integration u can be replaced by any other symbol, in particular x. Thus, from (1), using (2), wehave
CHAP. 13]349FOURIER SERIES1bn ¼ "LðL0n!x1dx þf ðxÞ sinLL1 "a0 Xþa cos2 n¼1 n1 "Xaa cosf ð"xÞ ¼ 0 þ2 n¼1 nf ðxÞ ¼Method 2: AssumeThenIf f ðxÞ is even, f ð"xÞ ¼ f ðxÞ.ðL0f ðxÞ sinn!xdx ¼ 0Ln!xn!x#þ bn sin:LLn!xn!x#" bN sin:LLHence,1 "1 "a0 Xn!xn!x# a0 Xn!xn!x#þ bn sin" bn sin¼ þþan cosan cosLLLL2 n¼12 n¼11Xand sobn sinn¼1n!x¼ 0;Li.e., f ðxÞ ¼1a0 Xn!xþan cos2Ln¼1and no sine terms appear.In a similar manner we can show that an odd function has no cosine terms (or constant term) in itsFourier expansion.13.10. If f ðxÞ is even, show thatðaÞan ¼1LðL"L(a) an ¼f ðxÞ cos2LðL0f ðxÞ cosn!x1dx ¼LLð0"Ln!xdx;Lf ðxÞ cosðbÞ bn ¼ 0.n!x1dx þLLðL0f ðxÞ cosn!xdxLLetting x ¼ "u,ððð""n!u#1 0n!x1 L1 Ln!uf ðxÞ cosdx ¼f ð"uÞ cosf ðuÞ cosdudu ¼L "LLL 0LL 0Lsince by definition of an even function f ð"uÞ ¼ f ðuÞ. Thenððð1 Ln!u1 Ln!x2 Ln!xdu þdx ¼dxf ðuÞ cosf ðxÞ cosf ðxÞ cosan ¼L 0LL 0LL 0L(b) This follows by Method 1 of Problem 13.9.13.11. Expand f ðxÞ ¼ sin x; 0 x !, in a Fourier cosine series.A Fourier series consisting of cosine terms alone is obtained only for an even function. Hence, weextend the definition of f ðxÞ so that it becomes even (dashed part of Fig. 13-11 below). With this extension,f ðxÞ is then defined in an interval of length 2!. Taking the period as 2!, we have 2L ¼ 2! so that L ¼ !.f (x)2pxOppFig. 13-11By Problem 13.10, bn ¼ 0 andan ¼2LðL0f ðxÞ cosn!x2dx ¼L!ð!0sin x cos nx dx2p
350FOURIER SERIES[CHAP. 13"# ð1 !1cosðn þ 1Þx cosðn % 1Þx !fsinðx þ nxÞ þ sinðx % nxÞg ¼%þ ! 0!nþ1n%10"#"#1 1 % cosðn þ 1Þ! cosðn % 1Þ! % 11 1 þ cos n! 1 þ cos n!þ%¼¼!nþ1n%1!nþ1n%1%2ð1 þ cos n!Þ¼if n 6¼ 1:!ðn2 % 1Þ ð2 !2 sin2 x !sin x cos x dx ¼¼ 0:a1 ¼! 0! 2 0¼For n ¼ 1;For n ¼ 0;2a0 ¼!ð!0 ! 24sin x dx ¼ ð% cos xÞ ¼ :!!012 2Xð1 þ cos n!Þ%cos nx! ! n¼2 n2 % 1%&2 4 cos 2x cos 4x cos 6xþþþ &&&¼ %! ! 22 % 1 42 % 1 62 % 1f ðxÞ ¼Then13.12. Expand f ðxÞ ¼ x; 0 x 2, in a half range(a) sine series,(b) cosine series.(a) Extend the definition of the given function to that of the odd function of period 4 shown in Fig. 13-12below. This is sometimes called the odd extension of f ðxÞ. Then 2L ¼ 4; L ¼ 2.f (x)O64x2246Fig. 13-12Thus an ¼ 0 andðð2 Ln!x2 2n!xf ðxÞ sindx ¼x sindxL 0L2 02" %&%&# 2%2n!x%4n!x %4cos% ð1Þ 2 2 sin¼ ðxÞ ¼ n! cos n!n!22n !0bn ¼Thenf ðxÞ ¼1X%4cos n! sinn!x2n!%&4!x 12!x 13!x¼sin% sinþ sin% &&&!22232n¼1(b) Extend the definition of f ðxÞ to that of the even function of period 4 shown in Fig. 13-13 below. This isthe even extension of f ðxÞ. Then 2L ¼ 4; L ¼ 2.
CHAP. 13]351FOURIER SERIESf (x)64xO2246Fig. 13-13Thus bn ¼ 0,ðð2 Ln!x2 2n!xdx ¼dxf ðxÞ cosx cosL 0L2 02 "#"#%&2n!x 4n!x &&2¼ ðxÞsin ð1Þ 2 2 cos&n!22n !04¼ 2 2 ðcos n! 1ÞIf n 6¼ 0n !an ¼If n ¼ 0; a0 ¼Thenð20x dx ¼ 2:1X4n!xðcos n! 1Þ cos222n !n¼1"#8!x 13!x 15!x¼ 1 2 cosþ 2 cosþ 2 cosþ &&&222!35f ðxÞ ¼ 1 þIt should be noted that the given function f ðxÞ ¼ x, 0 x 2, is represented equally well by thetwo different series in (a) and (b).PARSEVAL’S IDENTITY13.13. Assuming that the Fourier series corresponding to f ðxÞ converges uniformly to f ðxÞ in ð L; LÞ,prove Parseval’s identityð1 La2f f ðxÞg2 dx ¼ 0 þ !ða2n þ b2n ÞL L2where the integral is assumed to exist.1 'a0 Xn!xn!x(, then multiplying by f ðxÞ and integrating term by termþ bn sinþan cosLL2 n¼1from L to L (which is justified since the series is uniformly convergent) we obtain%ðLððLðL1 Xa Ln!xn!xdx þ bndxf f ðxÞg2 dx ¼ 0f ðxÞ dx þanf ðxÞ cosf ðxÞ sinLL2 L L L Ln¼1If f ðxÞ ¼¼1Xa20LþLða2n þ b2n Þ2n¼1where we have used the resultsðLn!xf ðxÞ cosdx ¼ Lan ;L Lobtained from the Fourier coefficients.ðL Lð1Þf ðxÞ sinn!xdx ¼ Lbn ;LðL Lf ðxÞ dx ¼ La0ð2Þ
352FOURIER SERIES[CHAP. 13The required result follows on dividing both sides of (1) by L. Parseval’s identity is valid under lessrestrictive conditions than that imposed here.13.14. (a) Write Parseval’s identity corresponding to the Fourier series of Problem 13.12(b).1111(b) Determine from (a) the sum S of the series 4 þ 4 þ 4 þ " " " þ 4 þ " " " .123n4ðcos n! % 1Þ; n 6¼ 0; bn ¼ 0.n2 !2Then Parseval’s identity becomesðð11 21 2 2ð2Þ2 X16f f ðxÞg2 dx ¼x dx ¼ðcos n! % 1Þ2þ4 !42 %22 %22nn¼1(a) Here L ¼ 2; a0 ¼ 2; an ¼or"#864 111111!4¼ 2 þ 4 4 þ 4 þ 4 þ " " " ; i.e., 4 þ 4 þ 4 þ " " " ¼396:! 135135"# "#111111111ðbÞ S ¼ 4 þ 4 þ 4 þ " " " ¼ 4 þ 4 þ 4 þ " " " þ 4 þ 4 þ 4 þ " " "123135246"#"#1111 111¼ 4 þ 4 þ 4 þ """ þ 4 4 þ 4 þ 4 þ """1352 123¼!4 Sþ ;96 16from which S ¼!49013.15. Prove that for all positive integers M,Ma20 X1þða2n þ b2n Þ @L2 n¼1ðL%Lf f ðxÞg2 dxwhere an and bn are the Fourier coefficients corresponding to f ðxÞ, and f ðxÞ is assumed piecewisecontinuous in ð%L; LÞ.SM ðxÞ ¼LetM a0 Xn!xn!x%þan cosþ bn sin2LLn¼1For M ¼ 1; 2; 3; . . . this is the sequence of partial sums of the Fourier series corresponding to f ðxÞ.We haveðLf f ðxÞ % SM ðxÞg2 dx A 0%Lsince the integrand is non-negative. Expanding the integrand, we obtainðLðLðL2f ðxÞ SM ðxÞ dx %SMðxÞ dx @f f ðxÞg2 dx2%L%L%L(1)ð2Þð3ÞMultiplying both sides of (1) by 2 f ðxÞ and integrating from %L to L, using equations (2) of Problem13.13, gives()ðLMa20 X222þf ðxÞ SM ðxÞ dx ¼ 2Lða þ bn Þð4Þ2 n¼1 n%LAlso, squaring (1) and integrating from %L to L, using Problem 13.3, we find()ðLMa20 X222þSM ðxÞ dx ¼ Lða þ bn Þ2 n¼1 n%LSubstitution of (4) and (5) into (3) and dividing by L yields the required result.ð5Þ
CHAP. 13]353FOURIER SERIESTaking the limit as M ! 1, we obtain Bessel’s inequality1a20 X1þða2n þ b2n Þ @2Ln¼1ðL%Lf f ðxÞg2 dxð6ÞIf the equality holds, we have Parseval’s identity (Problem 13.13).We can think of SM ðxÞ as representing an approximation to f ðxÞ, while the left-hand side of (2), dividedby 2L, represents the mean square error of the approximation. Parseval’s identity indicates that as M ! 1the mean square error approaches zero, while Bessels’ inequality indicates the possibility that this meansquare error does not approach zero.The results are connected with the idea of completeness of an orthonormal set. If, for example, we wereto leave out one or more terms in a Fourier series (say cos 4!x L, for example), we could never get the meansquare error to approach zero no matter how many terms we took. For an analogy with three-dimensionalvectors, see Problem 13.60.DIFFERENTIATION AND INTEGRATION OF FOURIER SERIES13.16. (a) Find a Fourier series for f ðxÞ ¼ x2 ; 0 x 2, by integrating the series of Problem 13.12(a).1Xð%1Þn%1(b) Use (a) to evaluate the series.n2n¼1(a) From Problem 13.12(a),x¼"#4!x 12!x 13!xsin% sinþ sin% &&&!22232ð1ÞIntegrating both sides from 0 to x (applying the theorem of Page 339) and multiplying by 2, we find"#16!x 12!x 13!xx ¼ C % 2 cos% 2 cosþ 2 cos% &&&222!232where C ¼ð2Þ"#161111%þ%þ&&&:!222 32 42(b) To determine C in another way, note that (2) represents the Fourier cosine series for x2 in 0 x 2.Then since L ¼ 2 in this case,C¼a0 1¼2LðL0f ðxÞ ¼12ð20x2 dx ¼43Then from the value of C in (a), we have1Xð%1Þn%1n¼1n2¼1%111!2 4 !2& ¼þ¼þ&&&¼16 3 1222 32 4213.17. Show that term by term differentiation of the series in Problem 13.12(a) is not valid."#!x2!x3!xTerm by term differentiation yields 2 cos% cosþ cos% &&& :222Since the nth term of this series does not approach 0, the series does not converge for any value of x.
354FOURIER SERIES[CHAP. 13CONVERGENCE OF FOURIER SERIES13.18. Prove that(a)12(b)1!þ cos t þ cos 2t þ " " " þ cos Mt ¼ð!0sinðM þ 12Þt1dt ¼ ;122 sin 2 t1!sinðM þ 12Þt2 sin 12 tsinðM þ 12Þt1dt ¼ :122 sin 2 t&!ð0(a) We have cos nt sin 12 t ¼ 12 fsinðn þ 12Þt & sinðn & 12Þtg.Then summing from n ¼ 1 to M,sin 12 tfcos t þ cos 2t þ " " " þ cos Mtg ¼ ðsin 32 t & sin 12 tÞ þ ðsin 52 t & sin 32 tÞ"#þ " " " þ sinðM þ 12Þt & sinðM & 12Þt¼ 12 fsinðM þ 12Þt & sin 12 tgOn dividing by sin 12 t and adding 12, the required result follows.(b) Integrating the result in (a) from &! to 0 and 0 to !, respectively. This gives the required results, sincethe integrals of all the cosine terms are zero.ð!13.19. Prove that limn!1 &!f ðxÞ sin nx dx ¼ limð!n!1 &!f ðxÞ cos nx dx ¼ 0 if f ðxÞ is piecewise continuous.1a2 XThis follows at once from Problem 13.15, since if the series 0 þða2 þ b2n Þ is convergent, lim an ¼n!12 n¼1 nlim bn ¼ 0.n!1The result is sometimes called Riemann’s theorem.13.20. Prove that limð!M!1 &!We haveð!&!f ðxÞ sinðM þ 12Þx dx ¼ 0 if f ðxÞ is piecewise continuous.f ðxÞ sinðM þ 12Þx dx ¼ð!&!f f ðxÞ sin 12 xg cos Mx dx þð!&!f f ðxÞ cos 12 xg sin Mx dxThen the required result follows at once by using the result of Problem 13.19, with f ðxÞ replaced byf ðxÞ sin 12 x and f ðxÞ cos 12 x respectively, which are piecewise continuous if f ðxÞ is.The result can also be proved when the integration limits are a and b instead of &! and !.13.21. Assuming that L ¼ !, i.e., that the Fourier series corresponding to f ðxÞ has period 2L ¼ 2!, showth
Fourier series corresponding to an even function, only cosine terms (and possibly a constant which we shall consider a cosine term) can be present. HALF RANGE FOURIER SINE OR COSINE SERIES A half range Fourier sine or cosine series i
Gambar 5. Koefisien Deret Fourier untuk isyarat kotak diskret dengan (2N1 1) 5, dan (a) N 10, (b) N 20, dan (c) N 40. 1.2 Transformasi Fourier 1.2.1 Transformasi Fourier untuk isyarat kontinyu Sebagaimana pada uraian tentang Deret Fourier, fungsi periodis yang memenuhi persamaan (1) dapat dinyatakan dengan superposisi fungsi sinus dan kosinus.File Size: 568KB
straightforward. The Fourier transform and inverse Fourier transform formulas for functions f: Rn!C are given by f ( ) Z Rn f(x)e ix dx; 2Rn; f(x) (2ˇ) n Z Rn f ( )eix d ; x2Rn: Like in the case of Fourier series, also the Fourier transform can be de ned on a large class of generalized functions (the space of tempered
(d) Fourier transform in the complex domain (for those who took "Complex Variables") is discussed in Appendix 5.2.5. (e) Fourier Series interpreted as Discrete Fourier transform are discussed in Appendix 5.2.5. 5.1.3 cos- and sin-Fourier transform and integral Applying the same arguments as in Section 4.5 we can rewrite formulae (5.1.8 .
Expression (1.2.2) is called the Fourier integral or Fourier transform of f. Expression (1.2.1) is called the inverse Fourier integral for f. The Plancherel identity suggests that the Fourier transform is a one-to-one norm preserving map of the Hilbert space L2[1 ;1] onto itself (or to anoth
Deret Fourier Arjuni Budi P Jurusan Pendidikan Teknik Elektro FPTK-Universitas Pendidikan Indonesia Gambar 5. Deret Fourier dari Gelombang Gigi Gergaji 3. Deret Fourier Eksponensial Kompleks Deret Fourier eksponensial kompleks menggambarkan respon frekuensi dan mengandung seluruh komponen frekuensi (harmonisa dari frekuensi dasar) dari sinyal.File Size: 416KB
Deret dan Transformasi Fourier Deret Fourier Koefisien Fourier. Suatu fungsi periodik dapat diuraikan menjadi komponen-komponen sinus. Penguraian ini tidak lain adalah pernyataan fungsi periodik kedalam deret Fourier. Jika f(t) adalah fungsi periodik yang memenuhi persyaratan Dirichlet
Alternatively, the 3D Fourier slice theorem makes it pos-sible to compute the 3D Fourier transform of the unknown radioactive distribution from the set of 2D Fourier transforms of the projection data[20 25]. These direct Fourier meth-ods (DFM) have the potential to substantially speed up the reconstruction process (when a simple 3D .
Because black holes have no hair, as the scientist John Wheeler put it, one can't tell from the outside what is inside a black hole, apart from its mass, electric charge, and rotation. This means that a black hole contains a lot of information that is hidden from the outside world. If the amount of hidden information inside a black hole depends on the size of the hole, one would expect from .
