Fourier cosine and sine series even odd

Transcription

1 Fourier cosine and sine series Evaluation of the coefficients in the Fourier series of a function f is considerably simler is the function is even or odd. A function is even if f ( x) = f (x) Examle: x 2, x 4, cos x A function is odd if Examle: x, x 3, sin x f ( x) = f (x) A function can be neither even nor odd. For examle e x, e x.

2 Theorem: Proerties of even/odd functions (a) The roduct of two even functions is even. (b) The roduct of two odd functions is even. (c) The roduct of an even function and an odd function is odd. (d) The sum (difference) of two even functions is even. (e) The sum (difference) of two odd functions is odd. (f) If f is even, then a a f (x)dx = 2 a f (x)dx. (g) If f is odd, then a a f (x)dx =.

3 Cosine and sine series If f is an even function on (, ) then the Fourier coefficients are a = 1 a n = 1 b n = 1 f (x)dx = 2 f (x)dx f (x) cos nπ xdx= 2 f (x) sin nπ xdx= f (x) cos nπ xdx

4 Similarly is f is odd on (, ) a = 1 a n = 1 b n = 1 f (x)dx = f (x) cos nπ xdx= f (x) sin nπ xdx= 2 f (x) sin nπ xdx

5 Definition: Fourier cosine and sine series (i) The Fourier series of an even function on the interval (, ) is the cosine series where f (x) = a 2 + a = 2 a n = 2 a n cos nπ x (1) f (x)dx (2) f (x) cos nπ xdx (3)

6 (ii) The Fourier series of an odd function on the interval (, ) is the sine series f (x) = b n sin nπ x (4) where b n = 2 f (x) sin nπ xdx (5)

7 Examle 1: Exansion in a sine series Exand f (x) = x, 2 < x < 2 in a Fourier series. Solution: The given function is odd on the interval ( 2, 2), so we exand it in a sine series with = 2. Using the integration by arts we get Therefore b n = 2 f (x) = 4 π x sin nπ 2 xdx= 4( 1)n+1 nπ ( 1) n+1 sin nπ n 2 x The series converges to the function on ( 2, 2) and the eriodic extension (of eriod 4).

8 Examle 2: Exansion in a sine series 1, π < x < f (x) = 1, x < π The function is odd on the interval ( π, π). With = π we have and so b n = 2 π π f (x) = 2 π (1) sin nx dx = 2 π 1 ( 1) n n 1 ( 1) n sin nx n

9 Gibbs henomenon If a function has discontinuities, then its Fourier series exansion exhibits overshooting by the artial sums from the functional values near a oint of discontinuity. This does not smooth out and remains fairly constant even for large N. This behavior of a Fourier series near a oint of discontinuity at which f is discontinuous is known as the Gibbs henomenon.

10 Half-range exansions Note that the cosine and sine series utilize the definition of a function only on < x <. So the cosine and sine exansions of a function defined on (, ) are half-range exansions. The choice of cosine or sine function determines whether the eriodic extension of the function behaves as even or odd function on the full interval (, ).

11 Examle 3: Exansion in three series Exand f (x) = x 2, < x < L (a) in a cosine series, (b) in a sine series, (c) in a Fourier series. (a) a = 2 L x 2 dx = 2 L 3 L2 a n = 2 L x 2 cos nπ L L xdx= 4L2 ( 1) n n 2 π 2 where the integration by arts was used twice to get a n. Thus f (x) = L L2 π 2 ( 1) n cos nπ L x The series converges to the 2L-eriodic even extension of f. n 2

12 (b) b n = 2 L L f (x) = 2L2 π x 2 sin nπ L xdx= 2L2 ( 1) n+1 nπ + 4L2 n 3 π 3[( 1)n 1] ( 1) n n n 3 π 2[( 1)n 1] sin nπ L x The series converges to the 2L-eriodic odd extension of f.

13 (c) with = L/2, 1/ = 2/L, and nπ/ = 2nπ/L, we have Therefore a = 2 L a n = 2 L b n = 2 L f (x) = L2 3 + L2 π L L L 1 x 2 dx = 2 3 L2 x 2 cos nπ L xdx= x 2 sin nπ L n 2 π L2 n 2 π 2 xdx= L2 nπ 2nπ cos L x 1 2nπ sin n L x The series converges to the L-eriodic extension of f.

14 Periodic driving force Fourier series can be useful in determining a articular solution of a DE describing a hysical system in which the inut or driving force is eriodic. Examle 4: We will find a articular solution of the DE describing an undamed sring/mass system m d2 x + kx = f (t) dt2 by first reresenting f by a half-range sine exansion and then assuming a articular solution of the form x (t) = B n sin nπ t

15 Consider the sring/mass system characterized by m = 1/16 kg, and k = 4 N/m and driven by the eriodic force defined as πt, < t < 1 f (t) = π(t 2) 1 t < 2 Note that we only need the half-range sine exansion of f (t) = πt, < t < 1. With = 1 it follows using integration by arts that b n = 2 1 πt sin nπt dt= 2( 1)n+1 n The differential equation of motion is then 1 d 2 x 16 dt 2 + 4x = 2( 1) n+1 sin nπt n

16 To find a articular solution x (t) we substitute x (t) = B n sin nπ t into the equation and equate the coefficients of sin nπt. This yields x (t) = 32( 1) n+1 n(64 n 2 π 2 sin nπt ) Observe that in the solution there is no integer n 1 for which the denominator 64 n 2 π 2 of B n is zero. That means that in this system we will not observe the henomenon of ure resonance. In general, we observe ure resonance if there is a value of n = N for which Nπ/ = ω where ω is the natural angular frequency of the system ω = k/m. In other words we have ure resonance if the Fourier exansion of the driving force f (t) contains a term sin(nπ/l)t or cos(nπ/l)t that has the same frequency as the free vibrations.

17 Comlex Fourier series We will now recast the definition of the Fourier series into a comlex form or exonential form. A comlex number can be written as z = a + ib where a and b are real numbers and i 2 = 1. Also a comlex number z = a ib is the comlex conjugate of the number z. Recall Euler s formula e ix = cos x + i sin x e ix = cos x i sin x

18 Comlex Fourier series Using Euler s formulas above we can exress the cosine and sine functions as cos x = eix + e ix, and sin x = eix e ix 2 2i Using these we can write the Fourier series of a function f as a e + a inπx/ + e inπx/ e inπx/ e inπx/ n + b n 2 2 2i = a (a n ib n )e inπx/ (a n + ib n )e inπx/ = c + c n e inπx/ + c n e inπx/ where c = a /2, c n = (a n ib n )/2, c n = (a n + ib n )/2.

19 When the function f is real, c n and c n are comlex conjugates and can also be written in terms of the comlex exonential functions: c = 1 1 f (x) dx 2 c n = 1 2 (a n ib n ) = 1 1 f (x) cos nπ 2 xdx i1 f (x) sin nπ xdx = 1 f (x) cos nπ 2 x i sin nπ x dx = 1 f (x) e inπx/ dx 2 c n = 1 2 (a n + ib n ) = 1 1 f (x) cos nπ 2 xdx+ i1 f (x) sin nπ xdx = 1 f (x) cos nπ 2 x + i sin nπ x dx = 1 f (x) e inπx/ dx 2

20 Definition: Comlex Fourier series The comlex Fourier series of functions f defined on an inteval (, ) is given by f (x) = n= c n e inπx/, (6) where c n = 1 2 f (x) e inπx/ dx, n =, ±1, ±2,... (7)

21 Examle 1: Comlex Fourier series Exand f (x) = e x, π < x < π in a comlex Fourier series. Solution: With = π, we get π π c n = 1 e x e inx dx = 1 e (in+1)x dx 2π π 2π π 1 = e (in+1)π e (in+1)π 2π(in + 1) We can simlify the coefficients using Euler s formula Since cos nπ = ( 1) n and sin nπ =. e (in+1)π = e π (cos nπ i sin nπ) = ( 1) n e π e (in+1)π = e π (cos nπ + i sin nπ) = ( 1) n e π

22 Hence c n = ( 1) n(eπ e π ) 2(in + 1)π = ( 1)n sinh π π 1 in n The comlex Fourier series is then f (x) = sinh π π n= ( 1) n 1 in n einx The series converges to the 2π-eriodic extension of f.

23 Fundamental frequency The Fourier series in both the real and comlex definitions define a eriodic function and the fundamental eriod of that function (i.e. the eriodic extension of f ) is T = 2. Since = T/2 the definitions become resectively a 2 + (a n cos nωx + b n sin nωx) and n= c n e inωx where the number ω = 2π/T is called the fundamental angular frequency. In Examle 1, the eriodic extension of the function has eriod T = 2π and thus the fundamental angular frequency is ω = 2π/2π = 1.

24 Frequency sectrum If f is eriodic and has fundamental eriod T, the lot of the oints (nω, c n ), where ω is the fundamental angular frequency, and c n are the coefficients of the comlex Fourier series, is called the frequency sectrum. Examle 2: Frequency sectrum In Examle 1, ω = 1, so that nω =, ±1, ±2,... Using a + ib = a 2 + b 2, we get c n = sinh π π 1 n 2 + 1

25 Examle 3: Frequency sectrum f (x) =, 1 2 < x < 1 4 1, 1 4 < x < 1 4, 1 4 < x < 1 2 Here T = 1 = 2, so = 1 2. The coefficients c n are c n = 1/2 1/2 f (x) e 2inπx dx = = 1 e inπ/2 e inπ/2 nπ 2i 1/4 1/4 = 1 nπ sin nπ 2 1. e 2inπx dx = e2inπx 2inπ 1/4 1/4 Since this result is not valid for n =, we comute c searately: c = 1/4 1/4 dx = 1 2

26 The following table shows some of the values of c n : n c n 1 5π 1 3π 1 π π 1 3π 1 5π