# Fourier Analysis and its applications

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1 Fourier Analysis and its applications Fourier analysis originated from the study of heat conduction: Jean Baptiste Joseph Fourier ( ) Fourier analysis enables a function (signal) to be decomposed into its frequency components. Wide range of applications: Spectrum analysis Digital filtering (e.g. electronics, image processing), Deconvolution Audio compression (MP3 etc.) Solving differential equations

2 Definition of Fourier series Consider a periodic function f (t) with period T = such that f (t + ) = f (t), (1) Any periodic function can be expressed as an infinite Fourier series of sin and cos functions with frequencies that are integer multiples of ω = /T = 1, the frequency of the function: f (t) = a (a n cos(nt) + b n sin(nt)), (2) n=1 where a 0, a 1,... b 1,..., b n are the Fourier coefficients.

3 Convergence Fourier Analysis Theorem Dirichlet s theorem If f (t) is periodic of period, if for π < t < π the function f (t) has a finite number of maximum and minimum values and a finite number of discontinuities, and if π π f (t)dt is finite, then the Fourier series converges to f (t) at all points where f (t) is continuous, and at discontinuities it converges to the arithmetic mean of the right-hand and left-hand limits of the function. For all practical purposes, these conditions hold and the series converges.

4 Fourier series as an expansion of orthogonal functions can be viewed as an expansion of orthogonal functions that form a complete set over any interval: π { πδm,n if m 0 sin(mt) sin(nt)dt = (3) π 0 if m = 0 π { πδm,n if m 0 cos(mt) cos(nt)dt = (4) if m = n = 0 π π π sin(mt) cos(nt)dt = 0 all integral m and n (5) From these orthogonality relations we can derive the Fourier coefficients a n and b n

5 Fourier coefficients Fourier Analysis a n = 1 π b n = 1 π π π π π f (t) cos(nt)dt (6) f (t) sin(nt)dt (7) You can prove this by substituting the Fourier series for f (t) (eqn. (2)) in to the above relations and use the orthogonality equations (eqns. (3,4,5).

6 Example - the square wave Consider the unit step function (i.e square wave): f (t) = { 1 for π < t < 0 +1 for 0 < t < π (8) Since the function is odd, all a n s are zero. The b n s are given by b n = 1 π 0 π( 1) sin(nt)dt + 1 π π = 2 π 0 = 2 [1 cos(nπ)] nπ = π sin(nt)dt = 2 nπ [ cos(nt)]π 0 { 0, n = 2, 4, 6,... 4/nπ, n = 1, 3, 5,... 0 (+1) sin(nt)dt

7 Fourier series of square wave

8 Fourier series of square wave: N=1

9 Fourier series of square wave: N=5

10 Fourier series of square wave: N=50

11 Remarks Fourier Analysis The amplitude of the high frequency sine waves falls of 1/n. Therefore convergence is slow. In general: For discontinuous functions, like the square wave, the coefficients decrease as 1/n. For continuous functions with discontinuous slopes (e.g. full wave rectifier), coefficients typically decrease as 1/n 2. Overshoot at discontinuities is called Gibbs phenomenon. In electronics, square wave pulses are common. If apparatus does not pass the high frequencies, square wave will be rounded off.

12 Fourier series of complex functions Definition of Fourier series can easily be extended to complex functions. Using the identity e ix = cos(x) + i sin(x) we can rewrite Fourier series (eqn.(2)) as f (t) = n= c n e int, (9) where (a n ib n )/2, n > 0, c n = a 0 /2 n = 0, (a n + ib n )/2 n < 0 which can be obtained by integration (10) c n = 1 π π f (t)e int dt (11)

13 Fourier series of functions with arbitrary period So far, considered functions periodic on a interval. Can easily extend to functions with any period T : Consider a periodic function f (t) with period T, and corresponding angular frequency ω = /T. Let t ωt = t/t. Then the complex Fourier series (eqn. (9)) becomes f (t) = c n e inωt, (12) where, c n = 1 T n= T /2 T /2 f (t)e inωt dt (13)

14 Fourier series of functions with arbitrary period Similarly for the Fourier series of real functions (eqn. (2)) becomes: where, f (t) = a (a n cos(nωt) + b n sin(nωt)), (14) n=1 a n = 2 T b n = 2 T T /2 T /2 T /2 T /2 f (t) cos(nωt)dt (15) f (t) sin(nωt)dt (16) Function is expressed as a series of sin and cos functions with frequencies that are integer multiples of ω, the frequency of the function.

15 Fourier series - summary Fourier series decomposes periodic signal into its frequency components. Even discontinuous functions or functions with discontinuous slopes (where Taylor expansion fails) can be expressed in a converging Fourier series. Solving differential equations. e.g. harmonic oscillator with some periodic driving force.

16 From Fourier series to Fourier transform Concept of Fourier series can be expanded to non-periodic functions. Consider the Fourier series for functions with period T in complex representation: f (t) = n= c n e in ωt, (17) with c n = ω T /2 T /2 f (t)e in ωt dt, (18) where we have written ω = ω, since the discrete frequencies nω are separated by ω = /T.

17 From Fourier series to Fourier transform Now define c n = ω g(n ω), (19) so that g(n ω) = f (t) = 1 T /2 n= T /2 f (t)e in ωt dt, (20) ωg(n ω)e in ωt. (21) Take the limit as T. Then, n ω becomes the continuous variable ω and the summation becomes an integral as ω = /T dω.

18 Fourier Analysis time domain f (t) = 1 frequency domain g(ω) = 1 We define g(ω) to be the Fourier transform of f (t): g(ω)e iωt dω, (22) f (t)e iωt dt (23) Fourier transform F[f (t)] = g(ω) (24) inverse transform F 1 [g(ω)] = f (t) (25)

19 Remarks: Fourier Analysis As with the Fourier series, the Fourier transform is used to obtain information on the frequency spectrum. However, F[f (t)] = g(ω) is complex in general, even if f (t) is real. Fourier transform is often represented in terms of its magnitude and phase: g(ω) = g(ω) e iθ(ω), where θ = tan 1 (Im(g(ω)/Re(g(ω))). Prefactor (1/ ) of the transforms depends on convention. Here we chose them to be distributed symmetrically. Often, the Fourier transform is used to go from the time domain to the frequency domain. However, mathematically, the original transform could be considered the inverse and vice versa. Therefore, the sign in the exponential also depends on the convention.

20 The Dirac delta function If Fourier transform are consistent, then inserting f (t) into g(ω) should lead to an identity: 1 ( 1 ) g(ω) = g(ω )e iω t dω e iωt dt (26) } {{ } = f (t) [ 1 ] e i(ω ω)t dt g(ω )dω (27) The term in the square brackets [...] has to be the Dirac delta function: δ(ω ω) = 1 e i(ω ω)t dt (28)

21 The Dirac delta function Consider δ(ω ω) τ = 1 τ τ e i(ω ω)t dt = sin((ω ω)τ) π(ω ω)

22 The Dirac delta function Eqns.(27) and (28) have to hold for any function g(ω). Set g(ω) = 1: 1 = However, according to eqn. (28), δ(0) =. δ(ω ω)dω (29) Delta function is a distribution, not a function, and has only a meaning inside an integrand!

23 Properties of the Fourier transform Linearity: F[f 1 (t) + f 2 (t)] = F[f 1 (t)] + F[f 2 (t)]. Shifting: F[f (t t 0 )] = = 1 1 = e iωt 0 1 = e iωt 0 g(ω), f (t t 0 )e iωt dt f (τ)e iω(τ+t 0) dτ f (τ)e iωτ dτ where τ = t t 0. Similarly, F 1 [g(ω ω 0 )] = e iω 0t f (t)

24 Properties of the Fourier transform Scaling: Let F[f (t)] = g(ω) and α > 0. Then, F[f (αt)] = = α = 1 α g(ω/α), f (αt)e iωt dt f (t )e iωt /α dt where substitution t = αt was made. For α < 0, we obtain F[f (αt)] = 1 α g(ω/α)

25 Properties of the Fourier transform Therefore, F[f (αt)] = 1 α g(ω/α) Similarly, F 1 [g(βt)] = 1 β f (t/β). This is a crucial relation!. As function f (t) becomes more localized, its Fourier transform becomes broader in frequency domain.

26 Parseval s theorem Fourier Analysis Consider the integral I = f 1(t) f 2 (t)dt. Substitute Fourier transforms of f 1 (t) and f 2 (t): [ 1 ] [ 1 ] I = g 1 (ω)e iωt dω g 2 (ω )e iω t dω dt = = = g 1 (ω)g 2 (ω ) [ 1 ] e i(ω ω)t dt dωdω g 1 (ω)g 2 (ω )δ(ω ω)dωdω g 1 (ω)g 2 (ω)dω For f 1 (t) = f 2 (t), we obtain Parseval s theorem: f (t) 2 dt = } {{ } Power g(ω) 2 } {{ } Power spectrum dω (30)

27 Higher dimensions Fourier Analysis So far considered 1D Fourier transform. Generalization to higher dimensions as follows: 2D: 3D: F[f (x, y)] = g(k x, k y ) = 1 F[f (x, y, z)] = g(k x, k y, k z ) = ( ) = g( k) = ( 1 ) 3 2 f (x, y)e i(kx x+ky y) dxdy f (x, y, z)e i(kx x+ky y+kz z) dxdydz f ( r)e i k r d r

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