Fourier Series Expansion Deepesh K P There are many types of series expansions for functions. The Maclaurin series, Taylor series, Laurent series are some such expansions. But these expansions become valid under certain strong assumptions on the functions (those assump- tions ensure convergence of the series). Fourier series also express a function as a series and the conditions required are fairly good and suitable when we deal with signals. Suppose f is a real valued function from R to R. In this note, we deal with the following three questions: • When does f has a Fourier series expansion? • How we find the expansion? • What are the main properties of this expansion? 1 Existance of a Fourier series expansion: There are three conditions which guarantees the existance of a valid Fourier series expansion for a given function. These conditions are collectively called the Dirichlet conditions: 1. f is a periodic function on R. This means that there exists a period T ≥ 0 such that f(x) = f(x + T ) for all x 2 R: 2. f has only a finite number of maxima and minima in a period. 3. f has atmost a finite number of discontinuous points inside a period. 4. f is integrable over the period of the function. It should be noted that the second and third conditions are satisfied by many real valued functions that we deal with, inside any finite interval. But periodicity is a condition that is satisfied by very few functions, for example, constant function, sine; cos; tan and their combinations. But we can consider any function defined on a finite interval [a; b] (or (a; b)) as a periodic function on R by thinking that the function is extended to R by repeating the values in [a; b] to the remaining part of R. Thus Most of the functions, that we commonly use, defined on finite intervals can be expanded as Fourier series Figure 1 2 Derivation of Fourier series expansion of a function defined in [−π; π]: In Fourier series expansion, we would like to write the function as a series in sine and cosine terms in the form: 1 a0 X f(x) = + a cos nx + b sin nx 2 n n n=1 For finding the above unknown co-efficients a0; an and bn in the Fourier series expansion of a function, one need to recall the value of certain integrals: Z π 1. sin mx dx = 0 for any integer m. −π Z π 2. cos mx dx = 0 for any integer m. −π Z π 3. sin mxcos nx dx = 0 for any integers m and n. −π Z π 4. sin mxsin nx dx = 0 for integers m 6= n. −π Z π 5. cos mxcos nx dx = 0 for integers m 6= n. −π Z π 6. sin mxsin nx dx = π when the integers m = n. −π Z π 7. cos mxcos nx dx = π when the integers m = n. −π [All the above integrals easily follow by evaluating using integration by parts] 1 a0 X Now suppose f(x) = 2 + ajcos jx + bj sin jx. j=1 To find a0: Observe that Z π Z π 1 Z π Z π a0 X f(x) dx = dx + a cos jx dx + b sin jx dx 2 j j −π −π j=1 −π −π 1 a0 X = 2 π + (0 + 0) 2 j=1 This implies that 1 Z π a0 = f(x) dx π −π 2 To find an: Observe that Z π Z π 1 Z π Z π a0 X f(x)cos nx dx = cos nx dx + a cos nx cos jx dx + b cos nx sin jx dx 2 j j −π −π j=1 −π −π 1 a0 X = 0 + a π + b 0 2 n j j=1 This implies that 1 Z π an = f(x) cos nx dx π −π To find bn: Observe that Z π Z π 1 Z π Z π a0 X f(x)sin nx dx = sin nx dx + a sin nx cos jx dx + b sin nx sin jx dx 2 j j −π −π j=1 −π −π 1 a0 X = 0 + a 0 + b π: 2 j n j=1 This implies that 1 Z π bn = f(x) sin nx dx π −π Thus 1 a0 X f(x) = 2 + ancos nx + bn sin nx; n=1 where Z π 1 a0 = π f(x) dx −π Z π 1 an = π f(x) cos nx dx −π Z π 1 bn = π f(x) sin nx dx −π [This expansion is valid at all those points x, where f(x) is continuous.] Note: Note that the above mentioned results hold when we take any 2 π Z c+2π length intervals [This is because sin mx dx = 0;::: are true for any c]. c Result: So whenever we take a function f defined from [c; c + 2π] (any interval of length 2π) to R, satisfying the Dirichlet conditions, we have 3 1 a0 X f(x) = 2 + ancos nx + bn sin nx; n=1 where Z c+2π 1 a0 = π f(x) dx c Z c+2π 1 an = π f(x) cos nx dx c Z c+2π 1 bn = π f(x) sin nx dx c 3 Derivation of Fourier series expansion of a function defined in an arbitrary period [a; b]: Now suppose that f(x) is defined in an arbitrary interval [a; b] and satisfy the b−a Dirichlet conditions. Let us take 2 = l, half the length of the interval. Now define the new variable π z = x: l By this simple transformation, we can convert functions on any finite interval (say, [a; b]) to functions in the new variable z, whose domain is an interval of 2π length. This is because π x = a ) z = l a and π 2 π π x = b ) z = l b = b−a (b − a + a) = 2 π + l a. Thus when the variable x in f(x) moves from a to b, the new variable z in the new function F (z) (which is the same function f in the new variable) moves π from c to c+2π, where c = l a. Hence the Fourier series expansion is applicable for F (z). Thus 1 a0 X f(x) = F (z) = + a cos nz + b sin nz; 2 n n n=1 where 1 Z c+2π a0 = F (z) dz π c 1 Z c+2π an = F (z) cos nz dz π c 1 Z c+2π bn = F (z) sin nz dz π c π and changing back to the original variable x (note that dz = l dx), we have 4 1 X nπ nπ f(x) = a0 + a cos x + b sin x; 2 n l n l n=1 where Z b 1 a0 = l f(x) dx a Z b 1 nπ an = l f(x) cos x dx a l Z b 1 nπ bn = l f(x) sin x dx; a l which is the general form of Fourier series expansion for functions on any finite interval. Also note that this is applicable to the first case of our discussion, where we need to take a = −π, b = π, l = π and then everything becomes the same as in the previous section. 3.1 Illustration We now take a simple problem to demonstrate the evaluation of Fourier series. Consider the function f defined by 8 < −10 if − 2 ≤ x ≤ −1; f(x) = x if − 1 < x < 1; : 10; if 1 ≤ x ≤ 2: We shall find the Fourier series expansion of this function. Here, note that the length of the interval is 4. So 2l = 4 and l = 2. We need to write 1 a0 X nπ nπ f(x) = + a cos x + b sin x; 2 n 2 n 2 n=1 where 1 Z 2 a0 = f(x) dx 2 −2 1 Z 2 nπ an = f(x) cos x dx 2 −2 2 1 Z 2 nπ bn = f(x) sin x dx; 2 −2 2 Now 1 Z −1 Z 1 Z 2 a0 = ( −10 dx + x dx + 10 dx) 2 −2 −1 1 1 = (−10 + 0 + 10) = 0 2 1 Z −1 nπ Z 1 nπ Z 2 nπ an = ( −10 cos x dx + x cos x dx + 10 cos x dx) 2 −2 2 −1 2 1 2 5 1 Z −1 nπ Z 1 nπ Z 2 nπ = (−10 cos x dx + xcos x dx + 10 cos x dx) 2 −2 2 −1 2 1 2 1 20 nπ 20 nπ = ( (sin − sin nπ) + 0 + (sin nπ − sin )) = 0 2 nπ 2 nπ 2 1 Z −1 nπ Z 1 nπ Z 2 nπ bn = ( −10 sin x dx + x sin x dx + 10 sin x dx) 2 −2 2 −1 2 1 2 1 20 nπ −2 nπ 4 nπ 20 nπ = f (cos( ) − cos(nπ)) + 2[ cos( ) + sin( ) − 0] − (cos(nπ) − cos( )g 2 nπ 2 nπ 2 n2π2 2 nπ 2 18 nπ 20 4 nπ = cos( ) − cos(nπ) + sin( ): nπ 2 nπ n2π2 2 4 19 So when n = 1 ) b1 = π2 ; n = 2 ) b2 = π ;::: Thus the Fourier expansion of f(x) is 0 π 4 π 2π 19 2π f(x) = + 0 cos x + sin x + 0 cos x + sin x + ::: 2 2 π2 2 2 π 2 4 π 19 2π = sin x + sin x + :::; π2 2 π 2 which is valid at all points in [−2; 2] except at −1 and 1, since the function is continuous at all points except −1 and 1.