Lecture Notes Fourier Analysis Prof. Xu Chen Department of Mechanical Engineering University of Washington chx AT uw.edu X. Chen Fourier Analysis September 29, 2019 Contents 1 Background 1 1.1 Periodic functions.....................................1 1.2 Orthogonal decomposition.................................1 2 Fourier series 3 2.1 Arbitrary period......................................4 2.2 Even and odd functions..................................5 2.3 Application example: ODE with special inputs......................6 3 Complex Fourier series7 4 Fourier integral8 5 Fourier transform 10 6 *Discrete Fourier analysis 11 6.1 Discrete Fourier series................................... 11 6.2 Discrete-time Fourier transform (DTFT)......................... 12 6.3 Discrete Fourier transform (DFT)............................. 13 X. Chen Fourier Analysis September 29, 2019 Notations: If x is complex, < (x) and = (x) denote, respectively, the real and imaginary parts of x. hx; yi denotes the inner product of x and y. 1 Background Fourier analysis is important in modeling and solving partial differential equations related to boundary and initial value problems of mechanics, heat flow, electrostatics, and other fields. 1.1 Periodic functions A function f(x) is called periodic if f(x) is defined for all real x, except possibly at some points, and if there is some positive number p, called a period of f(x), such that f (x + p) = f (x) Remark 1. If p is a period of f(x); then clearly 2p; 3p; : : : are also periods of f (x). The smallest positive period is called the fundamental period. 1.2 Orthogonal decomposition In vector spaces equipped with inner products, we have the following essential theorem. Theorem 2. Suppose fe1; e2; : : : ; eng is an orthogonal basis of a vector space V. Then for every v 2 V, we have hv; e1i hv; eni v = e1 + ··· + en he1; e1i hen; eni or, by using the norm notation, hv; e1i hv; eni v = 2 e1 + ··· + 2 en jje1jj jjenjj Proof. Let v 2 V. Because fe1; e2; : : : ; eng is a basis, there exists scalars α1; : : : ; αn such that v = α1e1 + ··· + αnen Taking the inner product with ej (j = 1; : : : ; n) yields hv; eji = hα1e1 + ··· + αnen; eji = α1 he1; eji + ··· + αn hen; eji But the basis is orthogonal, hence hei; eji = 0 if i 6= j. This yields hα1e1 + ··· + αnen; eji = αj hej; eji 1 X. Chen Fourier Analysis September 29, 2019 and thus hv; eji hv; eji = αj hej; eji () αj = hej; eji Recall that we have the following fact. Fact (Inner product for functions, function spaces). The set of all real-valued continuous functions f (x), g (x), : : : x 2 [α; β] is a real vector space under the usual addition of functions and multiplication by scalars. An inner product on this function space is Z β hf; gi = f (x) g (x) dx α and the norm of f is s Z β jjf (x) jj = f (x)2 dx α As a very useful special case, the trigonometric system is a function space with orthogonal basis 1 fsin nx; cos nx; 1gn=1. To check, we first notice that the functions are all periodic with period 2π. The inner product in this case is Z π hf; gi = f (x) g (x) dx −π Noting that, by checking the area under the graphs and noting the shape of sine and cosine functions, we have Z π cos nxdx = 0; 8n = 1; 2;::: (1) −π Z π sin nxdx = 0; 8n = 1; 2;::: −π This confirms the orthogonality between 1 and sin nx or cos nx. Furthermore, using (1) we immediately know Z π Z π cos (n + m) x + cos (n − m) x hcos nx; cos mxi = cos nx cos mxdx = dx −π −π 2 1 Z π 1 Z π = cos (n + m) xdx + cos (n − m) xdx 2 −π 2 −π = 0 + 0 as long as n 6= m namely, cos nx and cos mx are orthogonal if m 6= n. Similarly, as long as n 6= m (by assumption, both m and n are positive integers), sin nx and sin mx are orthogonal. This is because Z π Z π cos (n − m) x − cos (n + m) x hsin nx; sin mxi = sin nx sin mxdx = dx = 0 −π −π 2 2 X. Chen Fourier Analysis September 29, 2019 Furthermore, regardless of the values of m and n, we have Z π Z π sin (n + m) x + sin (n − m) x hsin nx; cos mxi = sin nx cos mxdx = dx = 0 −π −π 2 So sin nx and cos mx are always orthogonal. Sketch the plots of sine and cosine functions. You should find the above results intuitive. To perform the orthogonal decomposition under the trigonometric system, we need just one more thing–the norms of the basis functions. They are Z π Z π 1 − cos 2nx jj sin nxjj2 = hsin nx; sin nxi = (sin nx)2 dx = dx = π −π −π 2 Z π Z π cos 2nx + 1 jj cos nxjj2 = (cos nx)2 dx = dx = π −π −π 2 Z π jj1jj2 = 1dx = 2π −π 2 Fourier series Fourier series is an extension of the orthogonal decomposition reviewed in the last section. Under the trigonometric system, the orthogonal decomposition of a function f (x)—provided that the decompo- sition exists (we will talk about the existence very soon)—is 1 X f (x) = a0 + (an cos nx + bn sin nx) (2) n=1 where hf (x) ; 1i 1 Z π a0 = = f (x) dx (3) h1; 1i 2π −π and hf (x) ; cos nxi 1 Z π hf (x) ; sin nxi 1 Z π an = = f (x) cos nxdx; bn = = f (x) sin nxdx (4) hcos nx; cos nxi π −π hsin nx; sin nxi π −π (2) is called the Fourier series of f (x).(3) and (4) are called the corresponding Fourier coefficients. Example 3. Find the Fourier coefficients of ( −k if − π < x < 0 f (x) = ; and f (x + 2π) = f (x) k if 0 < x < π Such functions occur as external forces acting on mechanical systems, electromotive forces in electric circuits, etc. 3 X. Chen Fourier Analysis September 29, 2019 (Solution: 4k 1 1 f (x) = sin x + sin 3x + sin 5x + ::: π 3 5 ) Theorem 4 (Existance of Fourier Series). Let f (x) be periodic with period 2π and piecewise continuous in the interval −π ≤ x ≤ π. Furthermore, let f (x) have a left-hand derivative and a right-hand derivative at each point of that interval. Then the Fourier series of f (x) converges. Its sum is f (x), except at points xo where f (x) is discontinuous. There the sum of the series is the average of the left- and right-hand limits of f (x) at xo. Hence a periodic function f (x) = f (x + 2π) with f (x) = 1=x; x 2 [−π; π] does not have a Fourier series extension, as it does not have a left- or right-hand derivative at x = 0. A discontinuous periodic function, with period 2π and ( 1; −π < x < 0 f (x) = −1; π > x ≥ 0 is piecewise continuous. At x = 0, it has a left-hand derivative of 0 and a right-hand derivative of 0. Hence the function has a Fourier series expansion. 2.1 Arbitrary period The transition from period 2π to period p = 2L can be done by a suitable change of scale. If f (x) has a period 2L, we let x v = π L You can see, that x = ±L corresponds to v = ±π. So L f (x) = f v π is periodic in v with period 2π. L Now forget about f (x). Focus just on the new f π v with period 2π. The Fourier series is 1 L X f v = a + (a cos nv + b sin nv) π 0 n n n=1 where 1 Z π L 1 Z π L 1 Z π L a0 = f v dv; an = f v cos nvdv; bn = f v sin nvdv (5) 2π −π π π −π π π −π π 4 X. Chen Fourier Analysis September 29, 2019 Changing back to the x notation, by using L x π f (x) = f v ; v = π; dv = dx π L L we have 1 X nπ nπ f (x) = a + a cos x + b sin x (6) 0 n L n L n=1 with 1 Z L 1 Z L nπ 1 Z L nπ a0 = f (x) dx; an = f (x) cos xdx; bn = f (x) sin xdx (7) 2L −L L −L L L −L L Example 5. Find the Fourier series of 8 0 if − 2 < x < −1 <> f (x) = k if − 1 < x < 1 ; p = 2L = 4;L = 2 :>0 if 1 < x < 2 Answer: k 2k π 1 3π 1 5π f (x) = + cos x − cos x + cos x − + ::: 2 π 2 3 2 5 2 2.2 Even and odd functions It turns out that, if f(x) is even or odd, the Fourier series can be significantly simplified. This is because, if f (x) is an even function, then Z L f (x) sin nxdx = 0 −L as the integral of any odd function is zero. Hence in (7) bn = 0 so that (6) simplifies to 1 X nπ f (x) = a + a cos x 0 n L n=1 Furthermore, since f (x) is even, we have 1 Z L 1 Z L 1 Z L nπ 2 Z L nπ a0 = f (x) dx = f (x) dx; an = f (x) cos xdx = f (x) cos xdx 2L −L L 0 L −L L L 0 L Similarly, if f (x) is an odd function, then Z L f (x) cos nxdx = 0 −L 5 X.