Question

could you give me a short explanation of Fourier series and transform

(signals and system)

Answer 1

10.1 INTRODUCTION In many engineering problems, especially in the study of periodic phenomenae* in conduction of heat, electro-dynamics and acoustics, it is necessary to express a function in a series of sines and cosines. Most of the single-valued functions which occur in applied mathematics can be expressed in the form, 29.t + a, cos x + a, cos 2x + ... +b, sin x + b, sin 2x + ...... within a desired range of values of the variable. Such a series is known as the Fourier series. 10.2 EULER'S FORMULAE The Fourier series for the function f(x) in the interval a <x<a +2t is given by ао + Ž , cos nx + E b, sin nx 409 n=1 n=1 where ao 1 +2 ta f(x) dx 1 an ...(1) T Ja | *** f(x) cos nx dx 11*** f(x) sin nx dx These values of a, a, b, are known as Euler's formulae**. b = *Periodic functions. If at equal intervals of abscissa x, the value of each ordinate f (x) repeats itself, i.e., f(x)=f(x + a), for all x, then y=f(x) is called a periodic function having period a, e.g., sin x, cos x are periodic functions having a period 21. + To write a 2 instead of a, is a conventional device to be able to get more symmetric formulae for the coefficients. $ Named after the French mathematician and physicist Jacques Fourier (1768-1830) who was first to use Fourier series in his memorable work "Theorie Analytique de la Chaleur' in which he developed the theory of heat conduction. These series had a deep influence in the further development of mathematics and mathematical physics. **See footnote p. 205. 395 396 HIGHER ENGINEERING MATHEMATICS To establish these formulae, the following definite integrals will be required: sin nx 10+ 2x 1. pt +20 cos nx dx = San = 0 In 0) n la + 2x O + 2x COS nx 2. sin nx dx =- =0 (n0) n ax + 20 3. cos mx cos nx dr (n+0) 1 pa+ 2x (cos (m + n)x + cos ( mn)x) dx Ja O + 2x sin (m + n)x sin (m - n)x m+n m-n + =0 (min) 10x + 2x 4. 2 2 + sin 2nx 4n = Tt (n + 0) sa *** cos? nx dx = 1 +27 5. sin mx cos nx dx = cos (m-nx m-n cos (m + n)x m+n = 0 ( mn) 2 CE + 2x a + 2x 6. sin nix cos nx dx sin nx 2n = 0 (n +0) Ja lar 1a + 2 pa + 2x 7. sin mx sin nx dx = sin (m-nx sin (m + n)x m- m+n = 0 ( min) pt +27 sin? nx dx = 8. sin 2nx 12+ 2x Sam = . (n+0) 2 472

do - + 2 dx + Proof. Let f (x) be represented in the interval (a, a + 2n) by the Fourier series : f (x) = Ža, cos nx + b, sin nx To find the coefficients ag, 4,, bxwe assume that the series (i) can be integrated term by term from x = 0 to x = (1 + 21. To find an, integrate both sides of (i) from x = a to x = a + 21t. Then 1 *** f(x) dx = až 40 [**** *** a, cos nx dx + $**** 3 b, sin nx dx = 240(a + 21 – a) + 0 + 0 = agt [By integrals (1) and (2) abovel Hence 71*** f(x) dx . To find an, multiply each side of (i) by cos nx and integrate from x = a to x = a + 21. Then 1*** f(x) cos nx dx = I am cos nx cos nx dx =1 1 + 2x +2 cos nx dx + [ n1 +21 + [ Eb, sin nx cos nx dx in ng cos In-1 = 0 + Tan + 0 [By integrals (1), (3), (4), (5) and (6) + 2x Hence 1 πα f(x) cos nx dx . FOURIER SERIES AND HARMONIC ANALYSIS 397 To find b, multiply each side of (i) by sin nx and integrate from x = a to x = a + 21t. Then S**** f(x) sin nx dx = žao sa*** *** Σ + 2x sin nx dx + a, cos nx sin nx dx con nesin n=1 + 2 + Seca Σ 6, sin na sin na dx -1 1 pa+2 = 0 + 0 + Tb [By integrals (2), (5), (6), (7) and (8)] Hence f(x) sin nx dx. it Ja Cor. 1. Making a = 0, the interval becomes 0 <x<21, and the formulae (I) reduce to 1 2 f(x) dx 40 TC 1 2 * f(x) cos nx de ...(II) 1 2 on = f(x) sin nx dx Cor. 2. Putting a = -1, the interval becomes - 1<x<and the formulae (I) take the form: 1 do = J- f(x)dx 4,11* f(x) cos nx dx Br = S, f(x) sin nx dx TJ-* ...(III)