Fourier Series, Integrals, and Transforms

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1 Chap. Sec.. Fourier Series, Integrals, and Transforms Fourier Series Content: Fourier series (5) and their coefficients (6) Calculation of Fourier coefficients by integration (Example ) Reason hy (6) gives the Fourier coefficients (Theorem ) Great generality of Fourier series (Theorem ) Problem Set.. Page Linear combinations of periodic functions. Addition of periodic functions ith the same period p gives a periodic function ith that period p, and so does the multiplication of such a function by a constant. Thus all functions of period p form an important example of a vector space. This is hat you are supposed to prove. No, by assumption, given any periodic functions f and g that have period p, that is, f x + p = f x, g x + p = g x, (A) form a linear combination of them, say, h = af + bg a, b constant and sho that h is periodic ith period p; that is, you must sho that h x + p = h x. This follos by calculating and using the definition of h and then (A): h x + p = af x + p + bg x + p = af x + bg x = h x. 3. Fourier series. In Probs. 3- you first have to find a formula for the given function. Then you obtain the Fourier coefficients from (6) by integration. In Probs. 3 and the integrands of (6) are constants times cosines or sines. In Probs. 5- they are t times cosine or sine. Such integrals are evaluated by integration by parts, just as in calculus. Of course you have noticed that they depend on n, hich gives factors /n, /n etc. in the coefficients; these are essential ith respect to convergence. In Prob. 3 you have to integrate from π/ to π/ only. (Why?) From (6a) you obtain the constant term of the series (hich is the mean value of the given function over the interval from π to π. Indeed, you obtain

2 Fourier Analysis. Partial Differential Equations (PDEs) Part C Next you use (6b) to get / cos nx dx a / dx /. / a n / n sin nx / n sin n sin n n n,, 6, /n n, 5, 9, /n n 3, 7,, Hence the sequence of the cosine coefficients is,, 3,,, 5,. Finally, sho that the sine coefficients are zero, b n / sin nx dx / / n cos nx /. Conclude that the Fourier series is a Fourier cosine series n cos n cos n. fx cos x 3 cos 3x 5 cos 5x. Sec.. Functions of Any Period p L No ne ideas in the transition to an arbitrary period, just more complicated formulas. The notation p L is practical for later applications, as is mentioned in the text. Note that Example is closely related to Example in the last section. Problem Set.. Page 9 3. Fourier series ith period pl. Here L. Obtain the mean value from (6a): a x dx x No calculate the cosine coefficients from (6b) by to integrations by parts, using that the integral of an even function froml to L equals tice the integral from to L : a n x cos nx dx x cos nx dx n x sin nx n x sin nx dx. You see that the integral-free part is. To the remaining integral, taken ith the indicated sign and factor in front, apply another integration by parts, obtaining n x cos nx cos nx dx. n The remaining integral is zero (verify!), and so is the other term at the loer limit. At the upper limit

3 Chap. Fourier Series, Integrals, and Transforms calculate its value n cos n n n. With these coefficients obtain the series 3 cosx cos x cos 3x. 9 Note that the given function is continuous. Its Fourier coefficients are proportional to /n, so that the series converges faster than that in Prob. 3 of Problem Set. (just considered in this Guide), hich corresponded to a discontinuous function and had Fourier coefficients proportional to /n, so that convergence as sloer..5 - x Sec... Prob. 3. Given periodic function fx x of period p 3. Trigonometric formulas. To obtain the formula for cos 3 x from the present viepoint, calculate a, the average of cos 3 x (see the figure), then a cos x dx 3 3, a since the integrand is odd, a 3 cos3 x cos 3x dx, a since the integrand is odd, and the further Fourier coefficients are zero. For sin 3 x and cos x proceed similarly. The anser for cos x is given on p. A9 of Appendix x Sec... Prob. 3. cos 3 x Sec..3 Even and Odd Functions. Half-Range Expansions In this section e discuss systematically hat you have noticed before, namely, that an even function has a Fourier cosine series (no sine terms), and an odd function has a Fourier sine series (no cosine terms, no

4 Fourier Analysis. Partial Differential Equations (PDEs) Part C constant term). Then e take up the situation in many applications, namely, that a function is given over an interval of length L and you can develop it into a cosine series of period L (not L) or a sine series of period L; so here you have freedom to decide one ay or another. Problem Set.3. Page Odd function. Even and odd functions ere considered before, but in this section the formulas () and () for the Fourier coefficients have been adjusted, so that you integrate only over half of the interval of periodicity. In the problem the function is odd (see the figure) and of period. For x from to / it is fx x. For x from/ toit is fx x. Accordingly you have to split the integral in () from to into to integrals. Using integration by parts, you obtain b n / x sin nx dx x sin nx dx / / n x cos nx / cos nx dx n n x cos nx cos nx dx. / n / The minus sign in front of the last integral results from x. The first integral-free expression is zero at x and /n cosn/ at the upper limit of integration / (a factor drops out). The other integral-free expression is zero at x (since x there) and equal to /n cosn/ at the loer limit. You see that these to integral-free expressions are equal; hence the first expression minus the second (minus because it refers to the loer limit) equals zero. Consider the to integrals. The integral of cos nx is/n sin nx; together ith the factor /n in front this gives/n sin nx. This is at (the loer limit of the first integral) and at(the upper limit of the second integral). At the remaining to limits you obtain sin n/ n n sin n/ sin n/. n For n,, 3, the sine has the value,,,, respectively, and then repeats these values because of the periodicity. You thus obtain the Fourier sine series fx sin x 9 sin 3x 5 sin 5x 9 sin 7x. y x 3 - Sec..3. Prob. 3. Given odd periodic function

5 Chap. Fourier Series, Integrals, and Transforms 3 Sec.. Complex Fourier Series. Optional Problem Set l.. Page Complex Fourier series for functions of period have the form (6), hich also includes the Euler formulas for the complex Fourier coefficients c n. For the given function fx x you obtain c n xe inx dx. For n this is simply the integral of x from to and gives. Hence the series has no constant term. No determine the further coefficients c n ith n. Integrate by parts. Since /i i, the integral of the exponential function e inx is in einx n i einx. You thus obtain c n ix n e inx The integral-free part (times the factor /) has the value From (3a) you see that this equals i n e in i n i n e inx dx. e i n i n ei n e i n. i n cos n i n n. No sho that the remaining integral has the value for any integer n. Indeed, by (3b), e inx dx in ein e i n in i sin n (n,,. Sec..5 Forced Oscillations Figure 7 illustrates the main point of this section, the unexpected reaction of a mass-spring system to a driving force that is periodic but not just a cosine or sine term, and this reaction is explained by the use of the Fourier series of the driving force. Problem Set.5. Page 5 5. Sinusoidal driving force. y y cos t (misprintt instead of t in the first printing) is best solved by undetermined coefficients. Substitute y A cos t into the differential equation. Since y A cos t, this gives A A cos t cos t, hence A A. Thus A /, hich becomes the larger in absolute value, the more closely you approach the point of resonance. This motivates the values ofsuggested in the problem. Note also that A is negative as long as is less than and positive for values. This illustrates Figs. 53 and 57 (ith c ) in Sec Forced undamped oscillations. You need the Fourier series of the driving force rt, as explained in the text. rt is even (sketch it). Hence it is represented by a Fourier cosine series. The period is. Accordingly, the general term of that series is of the form a n cos nt. As in Example, solve the ODE ith a single term on the right, that is,

6 Fourier Analysis. Partial Differential Equations (PDEs) Part C y y a n cos nt. Substitute y A cos nt and y n A cos nt into the ODE. This gives n A A cos nt a n cos nt. Solving for A and riting A n for A, you obtain A n a n / n. (I) This shos hat you have to do next. You must find the Fourier series of rt. Since rt is even, you can use (*) in Sec..3, ith t instead of x, and rt instead of fx. You obtain a / (the mean value of rt over the interval of periodicity) and, furthermore, for n,, 3,, by integration by parts, a n t cos nt dt n t sin nt n sin nt dt. The integral-free part is zero at both limits of integration. Integrating the last integral, you obtain for the last term n cos nt cos n n n n. This is for even n, and for odd n you have a n n, 3, 5,. n Do you see something if you compare these a n ith (3) in Sec..5? These a n are the same as the coefficients in (3). Why? Well, the present rt is obtained from that in (3) by adding the constant a /. Hence you could have avoided all those calculations, hich therefore serve the purpose of filling in the details not given in Example on p. 5. Hoever, your further calculations differ from those on p. 5 because the present differential equation is different. From a n and (I) you obtain A a / / and A n n, 3, 5,. n n Hence the particular solution obtained has the form cos t cos 3t. 9 9 Adding to it the familiar general solution of the homogeneous equation, you obtain the anser given on p. A3 in Appendix of the book. Sec..6 Approximation by Trigonometric Polynomials This approximation is suitable in connection ith Fourier series, hereas polynomial approximation by polynomials (Secs. 9.3, 9.) is natural in connection ith Taylor series. Problem Set.6. Page Minimum square error. The minimum square error E* is given by (6). To compute it, you need the integral of the square of the given function fx over the interval of periodicity. In the present problem this is the integral of from to, hose value is. You further need the Fourier coefficients of fx. This function is odd. Hence the a n are all zero. The b n can be obtained from (*) in Sec..3 (or less conveniently from the Euler formulas in Sec..). This gives

7 Chap. Fourier Series, Integrals, and Transforms 5 b n sin nx dx n cos n cos n n /n for n, 3, 5, and b n for even n. (See also Example in Sec...) From this and (6) you obtain E b b b 3 6/ /9 /5. This gives the numeric values listed on p. A3 in Appendix of the book.. Monotonicity of the minimum square error. The minimum square error (6) is monotone decreasing, that is, it cannot increase if you add further terms (by choosing a larger N for the approximating polynomial). To prove this, note that the terms in the sum in (6) are squares, hence they are nonnegative. Since the sum is subtracted from the first term in (6) (the integral), the hole expression cannot increase. This is hat is meant by monotone decreasing, hich, by definition, includes the case that an expression remains constant, in our case, E N E M if N M here M and N are upper summation limits in (6). 3. Parseval s identity can be used to find the sum of certain series. In this problem you have to find the integral of f, that is, of from/ to /. This integral, divided by, equals. From the solution of the indicated Prob. 3 in Sec.. you thus obtain, by Parseval s identity, 9 5. Accordingly the sum in the parenthesis () equals 8. Sec..7 Fourier Integral Overvie. The Fourier integral is given by (5) ith A and B given in (). The integral is made plausible by Example and the discussion on pp Theorem states sufficient conditions for the existence of a Fourier integral representation of a given function fx. Example shos an application leading to the so-called sine integral Six given by (8) (e rite Siu since x is needed otherise), hich cannot be evaluated by the usual methods of calculus. For an even or odd function the Fourier integral (5) becomes a Fourier cosine integral () or a Fourier sine integral (3), respectively. These are applied in Example 3. Problem Set.7. Page 5. Fourier integral. If only the integral ere given, the problem ould be difficult. The function on the right gives the idea of ho to proceed. The function is zero for negative x. For x it is/ (the mean value of the limits from the left and right as x approaches ). Essential to you is that fx e x for x. Use ().cancels, and you have to integrate from to because fx is zero for negative x. Thus A e v cos v dv. This integral can be solved by integration by parts, as is shon in calculus. Its value is A /. Similarly, also from (),

8 6 Fourier Analysis. Partial Differential Equations (PDEs) Part C B e v sin v dv. This gives B /. Substituting A and B into (5) gives the integral shon in the problem. Integration by parts can be avoided by orking in complex. From (), using cos visin v e iv, you obtain (ith a factor resultimg from the evaluation at the loer limit) A ib e viv dv i eiv i i, here the last expression is obtained by multiplying numerator and denominator by i. Separation of the real and imaginary parts on both sides gives the integrals for A and B on the left and their values on the right, in agreement ith the previous result. 3. Fourier cosine integral. The integrand has no sine term. From () ith fx ex x you obtain (/ cancels) A e v cos v dv (see also Prob. ). From this and () it follos that cos x fx d. 5. Fourier sine integral. Calculate B by (), evaluating the integral by integration by parts and recursion. You obtain B sin v sin v dv sin. With this you obtain the anser given on p. A3 in Appendix of the book. Sec..8 Fourier Cosine and Sine Transforms Problem Set.8. Page 57. Fourier cosine transform. From () you obtain (sketch the given function if necessary) f c cos x dx cos x dx sin x sin sin sin sin sin x sin.

9 Chap. Fourier Series, Integrals, and Transforms 7. Fourier sine transform. Use (5) and integration by parts and recursion. You obtain F se x e x cos xsin x x. This approaches as x and for x you obtain e. Sec..9 Fourier Transform. Discrete and Fast Fourier Transforms This section concerns four related topics:. Derivation of the Fourier transform (6), hich is complex, from the complex Fourier integral (), the latter being obtained by Euler s formula (3) and the trick of adding the integral () that is zero (pp ).. The physical aspect of the Fourier transform and spectral representations (pp. 5-5). 3. Operational properties of the Fourier transform (pp. 5-53).. Representation of sampled values by the discrete Fourier transform (pp. 5-57). Problem Set.9. Page Calculation of Fourier transforms amounts to evaluating the defining integral (6). For the function in Prob. this simply means the integration of a complex exponential function, hich is formally the same as in calculus. You integrate from to b, the interval in hich fx k const, hereas it is zero outside this interval. According to (6) you obtain f b ke ix dx k i eix b x ik eix Here you used that /i i. Inserting the limits of integration, you get the anser given on p. A3 in Appendix of the book. 3. Table III in Sec.. contains formulas of Fourier transforms, some of hich are related. For deriving formula from formula start from formula, hich lists the transform e ib e ic i of the function if b x c and otherise. Since you need forb x b, it is clear that you should setb instead ofb in the first term, and c b in the second term. In the result, apply Euler s formula (3) on p. 58 of the book (ith x b); then the cosine terms ill cancel. Then you have in the denomiator i and in the numerator e ib e ic cos b i sin b cos b i sin b i sin b. The quotient of these to expressions gives the right side of formula in Table III, as anted. b x.

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