Trigonometric Fourier series

Transcription

1 Trigonometric Fourier series Recall that two vectors are orthogonal if their inner product is zero Suppose we consider all functions with at most a finite number of discontinuities ined on the interval [, ] In this case, we could ine an inner product on such functions by * f, g f x g x dx where f * is the function that takes on the complex conjugate of the function f In this case, consider the functions, (x), (x), (3x), (4x), and (x), (x), (3x), (4x), This is an infinite number of functions, but notice that: whenever n is a positive integer and, nx nxdx nx n, nx nxdx nx n m n x m n x mx, nx mx nxdx m n m n whenever m and n are positive integers Also, if m n, then and m n x m n x mx, nx mx nxdx m n m n m n x m n x mx, nx mxnxdx m n m n Thus, the collection of all functions, (x), (x), (3x), (4x), and (x), (x), (3x), (4x), forms an orthogonal set of functions with respect to the inner product ined as the integral Consequently, we may consider the span of all possible linear combinations of these functions Defining exactly what this span is is exceptionally difficult, but it is sufficient to say that all functions that are continuous on the interval [, ] with at most a finite number of finite jump discontinuities are included in this span

2 Thus, we may now ect any such function onto each of these functions Recall that the ection of one vector onto another is ined as: v u vu, vv, v, and therefore in this case, we have g If we calculate this with the constant function, we have while if we try this with (mx) and (mx), we get that and mx * g x f x dx g, f f x g x g x gg, * g x g x dx f x dx f x dx, f f x,, dx m, f m, m mx f f x mx mx mx dx mx m, f m, m mx f x dx x dx f x mx mx mx dx The question is, what is mx dx and (3x), (x) and (7x), we note a trend: mx dx when m is a positive integer? If we plot (x),

3 In each case, you will notice that the area in red equals the area in white, and thus, each of these integrals must be half of the area of a rectangle with height and width Therefore, all of these integrals must have area : and mx mx f x dx f x mx mx mx f x dx f x mx

4 Exponential Fourier series Another Fourier series of serious interest to engineers is the complex Fourier series on the interval, Here, the orthogonal functions are 3 jz jz jz z jz jz 3 jz, e, e, e, e, e, e, e, These are also orthogonal, for and, assuming that m n, njz * njz njz njz e e dz e e dz mjz * njz mjz njz nm jz e e dz e e dz e e njznjz dz e dz dz dz nm jz j j j e n m n m n m z for n m j e and e nm j as m and n are integers and therefore m n is an integer Therefore, jnx jnx jn e f x dx e f x dx e, f f x e e e e, e e e dx jnx e jn jn jnx jnx jnx jnx jnx for every integer n

5 Example Let us take a very simple function: f x x x x x This is the function Once again, relying on Calculus, we see that and mx f x mx f f x dx x dx 4 mx if m is odd f x mx m if m is even mx Consequently, this suggests that this tent function equals mx f x dx f x mx f x x 3x 5x 7x The coefficients, without the trigonometric functions, are called the Fourier coefficients In this case, the constant term is and the coe coefficients are ,,,,,,,, If we plot the first five ections, shown in Figures through 5, we see that get better and better approximations of the tent function

6 Figure Approximating f(x) with Figure Approximating f(x) with 4 x Figure 3 Approximating f(x) with f x x 3 x Figure 4 Approximating f(x) with f x x 3x 5x

7 Figure 5 Approximating f(x) with f x x 3x 5x 7x Important notice: Because this specific f(x) is symmetric around, you only see coe functions with odd integer coefficients and no e functions In general, this will not be the case If you wish to try out an example your own, you can observe this with x x gx x To show that the approximations are actually getting better, Figures 6 through show the absolute errors of these first five approximations In each case, the absolute error is reduced Figure 6 The absolute error of approximating f(x) with

8 4 Figure 7 The absolute error of approximating f(x) with x Figure 8 The absolute error of approximating f(x) with f x x 3 x Figure 9 The absolute error of approximating f(x) with f x x 3x 5x

9 Figure The absolute error of approximating f(x) with f x x 3x 5x 7x As we add more and more Fourier terms, the approximation must become better and better If, with the same function, you repeat this with the complex exponentials, you will find that 3jx jx jx 3jx f x e e e e 9 9 j Note, recall Euler s formula that e e je approximation, you would get, so if you were to substitute this into this 3jx jx jx 3jx e e e e jx j 3 jx jx j jx 9 9 jx j jx 3 jx j 3 jx 9 9 Now, recall that ( x) = (x) and ( x) = (x), so e e e e jx jx jx 3jx 3 jx j 3 jx jx j jx jx jx 9 jx j jx 3 jx j 3 jx and you will recognize this from above, being our trigonometric Fourier series approximation

10 Problems Find the trigonometric and exponential Fourier series of the h x x x up to the 5 th -order approximation, meaning up to and including (5x) and (5x) in the first case, and 5 jz e in the second Use Matlab to plot the first-, second-, third-, and up to the fifth-order approximations together with the function itself on the interval [, ] Next, plot the approximations on the interval [,, 4] Without finding the Fourier series of h x x x which coefficients will be zero and which will be non-zero and why? Of those that are non-zero, will they be positive or negative, and why? 5 jz e and

11 Solution The coefficients for the trigonometric Fourier series are for the constant term and,, 3 5 for (x), (3x) and (5x), respectively, and for all other trigonometric functions For the exponential coefficients, they are j,, j,, j,, j,, j,, j for e ,, e 5jz 5jz Hint: What are the symmetries? What are the inner products?, respectively