Fourier Series and Transform Overview Why Fourier transform? - PDF document

Fourier Series and Transform

Overview Why Fourier transform? Trigonometric functions Who is Fourier? Fourier series Fourier transform Discrete Fourier transform Fast Fourier transform 2D Fourier transform Tips

Why Fourier transform? Fourier, not being noble, could not enter the artillery, although he was a second Newton. ⎯ Francois Jean Dominique Arago For signal processing, Fourier transform is the tool to connect the time domain and frequency domain. Why would we do the exchange between time domain and frequency domain? Because we can do all kinds of useful analytical tricks in the frequency domain that are just too hard to do computationally with the original time series in the time domain. Simplify the calculation.

) t 50 * Why Fourier transform? π 2 sin( = ) t ( f

Why Fourier transform? This example is a sound record analysis. The left picture is the sound signal changing with time. However, we have no any idea about this sound by the time record. By the Fourier transform, we know that this sound is generated at 50Hz and 120Hz mixed with other noises.

Trigonometric functions (ex.1) Trigonometric system is the periodic functions as: 1, sinx, cosx, sin2x, cos2x, …, sinnx, cosnx, …. The properties of trigonometric system θ φ = θ − φ + θ + φ cos cos cos( ) / 2 cos( ) / 2 θ φ = θ − φ − θ + φ sin sin cos( ) / 2 cos( ) / 2 θ = θ θ = − θ (sin )' cos , (cos )' sin Trigonometric system is the orthogonal system π π π = = = ∫ ∫ ∫ cos nxdx 0 , sin nxdx 0 sin mx cos nxdx 0 − π − π − π ≠ ≠ ⎧ ⎧ 0 0 π m n π m n = = ∫ ∫ ⎨ ⎨ cos cos sin sin mx nxdx mx nxdx π = π = ⎩ ⎩ m n m n − π − π

In Matlab Int(f) Int(f,a,b) if f is a symbolic expression. b ∫ f ( x ) dx a q=quad(fun,a,b) q=quad(fun,a,b,tol) [q,fcnt]=quad(fun,a,b,…) Quadrature is a numerical method used to find the area under the graph of a function, that is, to compute a definite integral. b = ∫ q f ( x ) dx a q=quad(fun,a,b) tries to approximate the integral of function fun from a to b to within an error of 1e-6 using recursive adaptive Simpson quadrature. fun is a function handle for either an M-file function or an anonymous function. The function y=fun(x) should accept a vector argument x and return a vector result y, the integrand evaluated at each element of x.

Who is Fourier? Fourier is one of the France’s greatest administrators, historians, and mathematicians. He graduated with honors from the military school in Auxerre and became a teach of math when he was 16 years old. Later he joined the faculty at Ecole Normale and then the Polytechnique in Paris when he is 27. He went to Egypt with Napoleon as the Governor of Lower Egypt after the 1798 Expedition. He was secretary of the Academy of Sciences in 1816 and Fellow in 1817. Jean Baptiste Joseph Fourier Don’t believe it? (France, 1768~1830). Neither did Lagrange, Laplace, Poisson and other big wigs. Not translated into English until 1878! But it’s true!!

Fourier’s basic idea Trigonometric functions: sin(x) and cos(x) has the period 2 π . sin(nx) and cos(nx) have period 2 π /n. The linear combination of these functions or multiply each by a constant, the adding result still has a period 2 π .

Fourier series For any function f(x) with period 2 π (f(x) = f(2 π +x)), we can describe the f(x) in terms of an infinite sum of sines and cosines a ∞ = + + ∑ 0 f ( x ) ( a cos mx b sin mx ), m m 2 = m 1 To find the coefficients a, b and a, we multiply above equation by cosmx or sinmx and integrate it over interval - π <x< π . By the orthogonality relations of sin and cos functions, we can get 1 ∫ π = a m f ( x ) cos mxdx π − π 1 ∫ π = b m f ( x ) sin mxdx π − π π 1 = ∫ a f ( x ) dx π 0 − π

Fourier series � example (ex.2) < < π ⎧ 1 0 x = ⎨ f ( x ) Period function − − π < < ⎩ 1 x 0 The parameters are π 1 1 0 = + ∫ − = ∫ a 1 dx ( 1 ) dx 0 π π 0 − π 0 π 1 1 0 = + ∫ − ∫ a cos x dx ( 1 ) cos x dx π π 1 − π 0 1 ∫ π = 1 1 b m f ( x ) sin mxdx π = − − 0 π sin x sin x π π − π − π 0 = 0 4 = = b n , n 1 , 3 , 5 ,... π n π 1 1 0 = + ∫ − ∫ a cos mx dx ( 1 ) cos mx dx π π m − π = = 0 b k 0 , k 2 , 4 , 6 ... 1 1 π = − + m 0 sin mx sin mx π π − π 0 m m m = 0 4 sin x sin 3 x sin 5 x = + + + f ( x ) ( ...) π 1 3 5

Fourier series � example (cont…) < < π ⎧ 1 0 x = ⎨ f ( x ) Period function − − π < < ⎩ 1 x 0 The Fourier series is: 4 sin x sin 3 x sin 5 x = + + + f ( x ) ( ...) π 1 3 5

Fourier series � example (cont…) < < π ⎧ 1 0 x = ⎨ f ( x ) Period function − − π < < ⎩ 1 x 0 The Fourier series is: 4 sin x sin 3 x sin 5 x = + + + f ( x ) ( ...) π 1 3 5 = +

Fourier series � example (cont…) < < π ⎧ 1 0 x = ⎨ f ( x ) Period function − − π < < ⎩ 1 x 0 The Fourier series is: 4 sin x sin 3 x sin 5 x = + + + f ( x ) ( ...) π 1 3 5 = +

Fourier series � example (cont…) < < π ⎧ 1 0 x = ⎨ f ( x ) Period function − − π < < ⎩ 1 x 0 The Fourier series is: 4 sin x sin 3 x sin 5 x = + + + f ( x ) ( ...) π 1 3 5 = +

Fourier series � example (cont…) < < π ⎧ 1 0 x = ⎨ f ( x ) Period function − − π < < ⎩ 1 x 0 The Fourier series is: 4 sin x sin 3 x sin 5 x = + + + f ( x ) ( ...) π 1 3 5

Fourier series � example = < < π f ( x ) x 0 x 2 Period function The parameters are sin x sin 2 x sin 3 x = π − + + + f ( x ) 2 ( ...) 1 2 3

Fourier series � general For any function f(x’) with arbitrary period T , a simple change of variables can be used to transform the interval of integration from [- π , π ] to [-T/2,T/2] as π π 2 2 = = x x ' , dx dx ' T T The f(x’) can be described by the Fourier series as π π a ∞ 2 2 = + + = ∑ 0 f ( x ' ) ( a cos( m x ' ) b sin( m x ' ) ), m 1 , 2 ,... m m 2 T P = m 1 where 2 T / 2 = T ∫ a f ( x ' ) dx ' 0 T − / 2 π 2 2 T / 2 = ∫ ( ' ) cos( ' ) ' a f x m x dx m T T − T / 2 π 2 2 T / 2 = ∫ b f ( x ' ) sin( m x ' ) dx ' m T T − T / 2 Replace ω � 2 π /T and x’ � t, a ∞ = + ω + ω = ∑ f ( t ) 0 ( a cos m t b sin m t ), m 1 , 2 ,... m m 2 = m 1

The complex form of Fourier series Euler formulae = + = + ⎫ − ix ix ix e cos x i sin x cos x ( e e ) / 2 ⇒ ⎬ = − = − − − ⎭ ix ix ix e cos x i sin x sin x ( e e ) / 2 i Fourier series a ∞ a b = + + + − ∑ ω − ω ω − ω im t im t im t im t f ( t ) 0 ( m ( e e ) m ( e e ) ) 2 2 2 i = m 1 For certain m=k, + − ω − ω ω − ω ik t ik t ik t ik t e e e e ω + ω = + a cos k t b sin k t a b k k k k 2 2 i − + a ib a ib = ω + − ω ik t ik t k k e k k e 2 2 Denoting that as − + a a ib a ib = = = c 0 , c k k , c k k − 0 k k 2 2 2 The complex form of Fourier series ∞ ∞ = + + = ∑ ω − ω ∑ ω ik t ik t ik t f ( t ) c ( c e c e ) c e − 0 k k k = = −∞ k 1 k 2 T / 2 = − ω ∫ ik t c f ( t ) e dt k T − T / 2

Fourier transform For any non-periodic function and assume T � ∞ , rewrite previous general Fourier series equation and get: ∞ 2 / 2 T = ∑ − ω ω ∫ ik t ik t f ( t ) ( f ( t ) e dt ) e T = −∞ − k T / 2 = π ω 1 ∞ T 2 / T / 2 → ω ξ ξ ω − ξ ∑ ∫ ik ( t ) f ( ) e d π = −∞ − k T / 2 ∞ ∞ 1 → ω ξ ξ ω − ωξ ∫ ∫ i t i ( ) e d f e d π 2 − ∞ − ∞ Define ∞ ω = − ω ∫ i t ( ) ( ) F f t e dt − ∞ ∞ 1 = ω ω ω ∫ i t f ( t ) F ( ) e d π 2 − ∞ Here, F( ω ) is called as the Fourier Transform of f(t). Equation of f(t) is called the inverse Fourier Transform .

Fourier transform � Parseval’s law The time signal squared f 2 (t) represents how the energy contained in the signal distributes over time t, while its spectrum squared F 2 ( ω ) represents how the energy distributes over frequency (therefore the term power density spectrum). Obviously, the same amount of energy is contained in either time or frequency domain, as indicated by Parseval’s formula: ∞ ∞ = ω ω 2 2 ∫ ∫ f ( t ) dt F ( ) d − ∞ − ∞