1. the function must be periodic; 3. it must have only a finite number of maxima and minima within one periodic;

Transcription

1 Fourier Series 1

2 Dirichlet conditions The particular conditions that a function f(x must fulfil in order that it may be expanded as a Fourier series are known as the Dirichlet conditions, and may be summarized by the following points: 1. the function must be periodic;. it must be single-valued and continuous, except possibly at a finite number of finite discontinuities; 3. it must have only a finite number of maxima and minima within one periodic; 4. the integral over one period of f(x must converge. If the above conditions are satisfied, then the Fourier series converge to f(x at all points where f(x is continuous.

3 FIG. 1: An example of a function that may, without modification, be represented as a Fourier series. 3

4 Fourier coefficients The Fourier series expansion of the function f(x is written as f(x = a + [ ( ( ] πrx πrx a r cos + b r sin r=1 (1 where a 0, a r and b r are constants called the Fourier coefficients. For a periodic function f(x of period, the coefficients are given by a r = x0 + ( πrx f(x cos dx ( b r = x0 + ( πrx f(x sin dx (3 where is arbitrary but is often taken as 0 or /. The apparent arbitrary factor 1/ which appears in the a 0 term in Eq. (1 is included so that Eq. ( may apply for r = 0 as well as r > 0. 4

5 The relations Eqs. ( and (3 may be derived as follows. Suppose the Fourier series expansion of f(x can be written as in Eq. (1, f(x = a + [ ( ( ] πrx πrx a r cos + b r sin r=1 Then multiplying by cos(πpx/, integrating over one full period in x and changing the order of summation and integration, we get x0 + ( πpx f(x cos dx = a x0 + ( 0 πpx cos x0 + ( ( πrx πpx + a r cos cos r=1 x0 + ( ( πrx πpx + b r sin cos r=1 (4 dx dx dx 5

6 Using the following orthogonality conditions, x0 + ( ( πrx πpx sin cos dx = 0 (5 x0 + ( ( πrx πpx, r = p = 0 cos cos dx = 1, r = p > 0 0, r p (6 x0 + ( ( πrx πpx sin sin dx = we find that when p = 0, Eq. (4 becomes x0 + f(x dx = a 0 0, r = p = 0 1, r = p > 0 0, r p (7 6

7 When p 0 the only non-vanishing term on the RHS of Eq. (4 occurs when r = p, and so x0 + ( πrx f(x cos dx = a r. The other coefficients b r may be found by repeating the above process but multiplying by sin(πpx/ instead of cos(πpx/. 7

8 Example Express the square-wavefunction illustrated in the figure below as a Fourier series. FIG. : A square-wavefunction. The square wave may be represented by 1 for 1 f(t = T t < 0, +1 for 0 t < 1 T. 8

9 Note that the function is an odd function and so the series will contain only sine terms. To evaluate the coefficients in the sine series, we use Eq. (3. Hence b r = T/ ( πrt f(t sin dt T T/ T = 4 T/ ( πrt sin dt T T 0 = πr [1 ( 1r ]. Thus the sine coefficients are zero if r is even and equal to 4/πr if r is odd. hence the Fourier series for the square-wavefunction may be written as f(t = 4 ( sin 3ωt sin 5ωt sin ωt + + +, π 3 5 where ω = π/t is called the angular frequency. 9

10 Discontinuous functions At a point of finite discontinuity, x d, the Fourier series converges to 1 lim ɛ 0 [f(x d + ɛ + f(x d ɛ]. At a discontinuity, the Fourier series representation of the function will overshoot its value. Although as more terms are included the overshoot moves in position arbitrarily close to the discontinuity, it never disappears even in the limit of an infinite number of terms. This behavior is known as Gibbs phenomenon. 10

11 Example Find the value to which the Fourier series of the square-wavefunction converges at t = 0. Answer The function is discontinuous at t = 0, and we expect the series to converge to a value half-way between the upper and lower values; zero in this case. Considering the Fourier series of this function, we see that all the terms are zero and hence the Fourier series converges to zero as expected. 11

12 The Gibbs phenomenon is shown below. FIG. 3: The convergence of a Fourier series expansion of a square-wavefunction, including (a one term, (b two terms, (c three terms and (d 0 terms. The overshoot δ is shown in (d. 1

13 Non-periodic functions Figure 4(b shows the simplest extension to the function shown in Figure 4(a. However, this extension has no particular symmetry. Figures 4(c, (d show extensions as odd and even functions respectively with the benefit that only sine or cosine terms appear in the resulting Fourier series. FIG. 4: Possible periodic extensions of a function. 13

14 Example Find the Fourier series of f(x = x for 0 < x. Answer We must first make the function periodic. We do this by extending the range of interest to < x in such a way that f(x = f( x and then letting f(x + 4k = f(x where k is any integer. FIG. 5: f(x = x, 0 < x, with extended range and periodicity. 14

15 Now we have an even function of period 4. Thus, all the coefficients b r will be zero. Now we apply Eqs. ( and (3 with = 4 to determine the remaining coefficients: a r = ( πrx x cos dx = 4 ( πrx x cos Thus, a r = = = = [ πr x sin 8 [ π r x cos 16 π cos πr r 16 π r ( 1r. ( πrx ] 0 ( πrx ] 4 πr π r dx, ( πrx x sin dx ( πrx cos dx 0 15

16 Since the expression for a r has r in its denominator, to evaluate a 0 we must return to the original definition, a r = 4 From this we obtain a 0 = 4 ( πrx f(x cos x dx = 4 4 The final expression for f(x is then x = r=1 ( 1 r π r 0 dx. x dx = 8 3. ( πrx cos, for 0 < x. (8 16

17 Integration and differentiation Example Find the Fourier series of f(x = x 3 for 0 < x. Answer If we integrate Eq. (8 term by term, we obtain x 3 3 = 4 3 x + 3 r=1 ( 1 r π 3 r 3 ( πrx sin + c, where c is an arbitrary constant. We have not yet found the Fourier series for x 3 because the term 4 3 x appears in the expansion. However, now differentiating our expression for x, we obtain x = 8 r=1 ( 1 r πr ( πrx sin. 17

18 We can now write the full Fourier expansion of x 3 as x 3 = r=1 r=1 ( 1 r πr ( 1 r π 3 r 3 ( πrx sin ( πrx sin + c We can find the constant c by considering f(0. At x = 0, our Fourier expansion gives x 3 = c since all sine terms are zero, and hence c = 0. 18

19 Complex Fourier series Using exp(irx = cos rx + i sin rx, the complex Fourier series expansion is written as ( πirx f(x = c r exp, (9 r= where the Fourier coefficients are given by c r = 1 x0 + ( f(x exp πirx dx (10 This relation can be derived by multiplying Eq. (9 by exp( πipx/ before integrating and using the orthogonality relation x0 + exp ( πipx exp ( πirx dx = The complex Fourier coefficients have the following relations with the real Fourier coefficients c r = 1 (a r ib r, (11 c r = 1 (a r + ib r., r = p 0, r p 19

20 Example Find a complex Fourier series for f(x = x in the range < x <. Answer Using Eq. (10, c r = 1 x exp 4 [ = x πir exp + 1 πir exp ( πirx ( πirx dx ] ( πirx dx = 1 [exp( πir + exp(πir] πir [ ( 1 + r π exp πirx ] = i πr cos πr + i = i πr ( 1r. r sin πr π 0

21 Hence x = r= i( 1 r rπ ( πirx exp. 1

22 Parseval s theorem Parseval s theorem gives a useful way of relating the Fourier coefficients to the function that they describe. Essentially a conservation law, it states that 1 x0 + f(x dx = = r= c r ( 1 a (a r + b r. r=1 This says that the sum of the moduli squared of the complex Fourier coefficients is equal to the average value of f(x over one period. (1

23 Proof of Parseval s theorem et us consider two functions f(x and g(x, which are (or can be made periodic with period, and which have Fourier series (expressed in complex form ( πirx f(x = c r exp, r= ( πirx g(x = γ r exp r= where c r and γ r are the complex Fourier coefficients of f(x and g(x respectively. Thus, ( πirx f(x g(x = c r g (x exp r= 3

24 Integrating this equation with respect to x over the interval (, +, and dividing by, we find = r= 1 c r [ 1 c r r= 1 x0 + x0 + x0 + f(xg (xdx = ( πirx g (x exp dx ( πirx g(x exp = r= dx ] c r γ r. Finally, if we let g(x = f(x, we obtain Parseval s theorem (Eq. (1. 4

25 Example Using Parseval s theorem and the Fourier series for f(x = x, calculate the sum r=1 r 4. Answer Firstly, we find the average value of f(x over the interval < x, 1 4 x 4 dx = Now we evaluate the RHS of Eq. (1: ( 1 a a r+ 1 1 b n = Equating the two expression, we find r=1 1 1 r 4 = π4 90. ( r=1 16 π 4 r 4. 5