Fourier Series Odd and Even

10 Key excerpts on "Fourier Series Odd and Even"

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  Bird's Higher Engineering Mathematics

  • John Bird(Author)
  • 2021(Publication Date)
  • Routledge

    (Publisher)

  The Fourier series for the periodic function defined by f (x) = { -3, when - π ⟨ x ⟨ 0 +3, when 0 ⟨ x ⟨ π and which has a period of 2π is: (a) an odd function and contains no cosine terms (b) an even function and contains no sine terms (c) an odd function and contains no sine terms (d) an even function and contains no cosine terms 6. The Fourier series for the periodic function defined by f (x) =    0, -2 ⟨ x ⟨ -1 2, -1 ⟨ x ⟨ 1 0, 1 ⟨ x ⟨ 2 and which has a period of 2π is: (a) an odd function and contains no cosine terms (b) an even function and contains no cosine terms (c) an odd function and contains no sine terms (d) an even function and contains no sine terms For fully worked solutions to each of the problems in Practice Exercises 267 and 268 in this chapter, go to the website: www.routledge.com/cw/bird Chapter 63 Fourier series over any range Why it is important to understand: Fourier series over any range As has been mentioned in preceding chapters, the Fourier series has many applications; in fact, any field of physical science that uses sinusoidal signals, such as engineering, applied mathematics and chemistry, will make use of the Fourier series. In communications, the Fourier series is essential to understanding how a signal behaves when it passes through filters, amplifiers and communications channels. In astron- omy, radar and digital signal processing Fourier analysis is used to map the planet. In geology, seismic research uses Fourier analysis, and in optics, Fourier analysis is used in light diffraction. This chapter explains how to determine the Fourier series of a periodic function over any range.

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  Practical Signals Theory with MATLAB Applications

  • Richard J. Tervo(Author)
  • 2013(Publication Date)
  • Wiley

    (Publisher)

  On the other hand, practical signals formed by a linear combination including both even and odd components will be neither even nor odd. It follows that any periodic signal that is not strictly even or strictly odd must have both sine and cosine components in the Fourier series. The general cosine described by s ðt Þ ¼ cosð2πf 0 t þ ΦÞ incorporates odd and even signals by varying Φ. To see the effect of Φ, the trigonometric identity cosðx þ yÞ ¼ cosx cosy  sinx siny can be applied with x ¼ 2πf 0 t and y ¼ Φ to obtain: s ðt Þ ¼ cosð2πf 0 t þ ΦÞ ¼ cosðΦÞ cosð2πf 0 t Þ  sinðΦÞ sinð2πf 0 t Þ ð4:30Þ Note that for Φ ¼ 0, s ðt Þ ¼ cosð2πf 0 t Þ, while for Φ ¼ π=2, s ðt Þ ¼ sinð2πf 0 t Þ. For most values of Φ, s ðt Þ will be neither even nor odd. This expression describes a phase-shifted sinusoid shifted by Φ rad as a linear combination of both sines and cosines of frequency f 0 . This result can be identified as a Fourier series expansion of the form: s ðt Þ ¼ cosð2πf 0 t þ ΦÞ % X þN n¼N A n cosð2πnf 0 t Þ þ B n sinð2πnf 0 t Þ ð4:31Þ with the non-zero coefficients A 1 ¼ cosðΦÞ, and B 1 ¼ sinðΦÞ. This effect is shown in Figure 4.30 for A 1 ¼ B 1 ¼ 1. Given only the Fourier series components A 1 and B 1 , the phase shift Φ may be found directly by observing: Φ ¼ tan 1 B 1 A 1   4.13 | Part Three—The Complex Fourier Series 137 and the amplitude is constant independent of Φ, given by: amplitude ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A 2 1 þ B 2 1 q The amplitude of a shifted sinusoid is unaffected by any phase change. 4.14 The Complex Fourier Series Consider the observation that real-world signals are unlikely to be exactly even or odd signals. Add to this the fact that representation of signals that are neither odd nor even requires computation of both the sine and cosine terms in the Fourier series. Think of the problem of combining cosine and sine representations in a single graphical form, or in a single expression.

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  Engineering Mathematics

  A Programmed Approach, 3th Edition

  • C W. Evans, C. Evans(Authors)
  • 2019(Publication Date)
  • Routledge

    (Publisher)

  x −

  sin  2 x

  2

  +

  sin  3 x

  3

  − … }

  to represent f(x) = x in the assumption that equality holds on the interval (−π, π) and that we can multiply the series term by term and then integrate without disturbing or distorting the convergence.

  ■

  Of course f(x) = x is a rather strange example to use to obtain a trigonometrical series because we have no difficulty whatever in dealing with polynomial functions. It is the more obscure functions that occur in practical applications which concern us. However there are two principal reasons why we have worked through this example. First it provides a relatively simple exercise for us to use to illustrate how to calculate the Fourier coefficients, and secondly it leads us into a discussion of a special feature which some functions possess and which enables us to reduce the work in finding their Fourier coefficients.

  21.2    ODD AND EVEN FUNCTIONS

  Suppose f(x) is defined on the interval −π ⩽ x ⩽ π, then f is said to be an odd function if

  f

  (

  − x

  )

  = − f

  ( x )

          whenever   − π ≤ x ≤ π

  You already know of many functions which are odd functions. Here are a few examples: x,x3 , sinx, sinhx. Odd functions are easily recognized by their graphs; they are symmetrical with respect to the origin (Fig. 21.1 ).

  Fig. 21.1: An odd function

  Suppose f(x) is defined on the interval −π ⩽ x ⩽ π, then f is said to be an even function if

  f

  (

  − x

  )

  = f

  ( x )

          whenever   − π ≤ x ≤ π

  Here are a few examples of even functions: l,x2 , cosx, coshx. Even functions are easily recognized by their graphs; they are symmetrical about the y-axis (Fig. 21.2 ).

  Fig. 21.2: An even function In fact the identity

  f

  ( x )

  ≡

  [

  f

  ( x )

  + f

  ( −

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  Partial Differential Equations

  An Introduction

  • Walter A. Strauss(Author)
  • 2012(Publication Date)
  • Wiley

    (Publisher)

  The graph of an even function φ ( x ) must cross the y axis horizontally, φ ( x ) = 0, since the derivative is odd (provided the derivative exists). Example 1. tan x is the product of an odd function (sin x ) and an even function (1 / cos x ), both of period 2 π . Therefore tan x is an odd and periodic function. But notice that its smallest period is π , not 2 π . Its derivative sec 2 x is necessarily even and periodic; it has period π . The dilated function tan ax is also odd and periodic and has period π/ a for any a > 0. 5.2 EVEN, ODD, PERIODIC, AND COMPLEX FUNCTIONS 115 Definite integrals around symmetric intervals have the useful properties: l − l (odd) dx = 0 and l − l (even) dx = 2 l 0 (even) dx . (5) Given any function defined on the interval (0, l ), it can be extended in only one way to be even or odd. The even extension of φ ( x ) is defined as φ even ( x ) = φ ( x ) for 0 < x < l φ ( − x ) for − l < x < 0 . (6) This is just the mirror image. The even extension is not necessarily defined at the origin. Its odd extension is φ odd ( x ) = ⎧ ⎪ ⎨ ⎪ ⎩ φ ( x ) for 0 < x < l − φ ( − x ) for − l < x < 0 0 for x = 0 . (7) This is its image through the origin. FOURIER SERIES AND BOUNDARY CONDITIONS Now let’s return to the Fourier sine series. Each of its terms, sin( n π x / l ), is an odd function. Therefore, its sum (if it converges) also has to be odd. Furthermore, each of its terms has period 2 l , so that the same has to be true of its sum. Therefore, the Fourier sine series can be regarded as an expansion of an arbitrary function that is odd and has period 2l defined on the whole line −∞ < x < +∞ . Similarly, since all the cosine functions are even, the Fourier cosine series can be regarded as an expansion of an arbitrary function which is even and has period 2l defined on the whole line −∞ < x < ∞ .

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  Introduction to Partial Differential Equations for Scientists and Engineers Using Mathematica

  • Kuzman Adzievski, Abul Hasan Siddiqi(Authors)
  • 2016(Publication Date)
  • Chapman and Hall/CRC

    (Publisher)

  A useful observation for computational purposes is the following result. 1.1 FOURIER SERIES OF PERIODIC FUNCTIONS 7 Lemma 1.1.4. If f is an even function, then a n = 2 L L ∫ 0 f ( x ) cos ( nπ L x ) dx, n = 0 , 1 , 2 , . . . and b n = 0 , n = 1 , 2 , . . . If f is an odd function, then b n = 2 L L ∫ 0 f ( x ) sin ( nπ L x ) dx, n = 1 , 2 , . . . and a n = 0 , n = 0 , 1 , . . .. Proof. It easily follows from Lemma 1 . 1 . 1 and Lemma 1 . 1 . 3. ■ Remark. For normalization reasons, in the first Fourier coefficient a 0 2 we have the factor 1 2 . Also notice that this first coefficient a 0 2 is nothing but the mean (average) of the function f on the interval [ -L, L ]. In the next section we will show that every periodic function of x satisfying certain very general conditions can be represented in the form (1 . 1 . 1), that is, as a Fourier series. Now let us take several examples. Example 1.1.1. Let f : R → R be the 2 π -periodic function which on the interval [ -π, π ] is defined by f ( x ) = { 0 , -π ≤ x < 0 1 , 0 ≤ x < π. Find the Fourier series of the function f . Solution. Using the formulas (1.1.2) for the Fourier coefficients in Definition 1.1.4 we have a 0 = 1 π π ∫ -π f ( x ) dx = 1 π 0 ∫ -π 0 dx + 1 π π ∫ 0 1 dx = 1 . For n = 1 , 2 , . . . we have a n = 1 π π ∫ -π f ( x ) cos nx dx = 1 π 0 ∫ -π 0 cos nx ) dx + 1 π π ∫ 0 1 cos nx dx = 0 + 1 nπ [ sin( nπ ) -sin(0) ] = 0; 8 1. FOURIER SERIES and b n = 1 π π ∫ -π f ( x ) sin nx dx = 1 π 0 ∫ -π 0 sin nx dx + 1 π π ∫ 0 1 sin nx dx = 0 -1 nπ [ cos nπ -cos 0 ] = 1 -( -1) n nπ . Therefore the Fourier series of the function is given by S f ( x ) = 1 2 + 2 π ∞ ∑ n =1 1 2 n -1 sin (2 n -1) x 2 . Figure 1 . 1 . 1 shows the graphs of the function f , together with the partial sums S N f ( x ) of the function f taking N = 1 and N = 3 terms. Minus 2 Π MinusΠ Π 2 Π Π Figure 1.1.1 Figure 1 . 1 . 2 shows the graphs of the function f , together with the partial sums of the function f taking N = 6 and N = 14 terms.

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  Differential Equations

  Theory,Technique and Practice with Boundary Value Problems

  • Steven G. Krantz(Author)
  • 2015(Publication Date)
  • Chapman and Hall/CRC

    (Publisher)

  Then f (x) · sin nx is odd, and therefore b n = 1 π ∫ - π π f (x) sin nx dx = 0. For the cosine coefficients, we have a n = 1 π ∫ - π π f (x) cos nx dx = 2 π ∫ 0 π f (x) cos nx dx. Thus the Fourier series for an even function contains only cosine terms. By the same token, suppose now that f is an odd function on the interval [− π, π ]. Then f (x) · cos nx is an odd function, and therefore a n = 1 π ∫ - π π f (x) cos nx dx = 0. For the sine coefficients, we have b n = 1 π ∫ - π π f (x) sin nx dx = 1 π ∫ 0 π f (x) sin nx dx. Thus the Fourier series for an odd function contains only sine terms. EXAMPLE 5.3.2 Examine the Fourier series of the function f (x) = x from the point of view of even/odd. Solution: The function is odd, so the Fourier series must be a sine series. We calculated in Example 5.1.1 that the Fourier series is in fact x = f (x) = 2 (sin x - sin 2 x 2 + sin 3 x 3 - +...). (5.3.2.1) The expansion is valid on (− π, π), but not at the endpoints (since the series of course sums to 0 at − π and π). EXAMPLE 5.3.3 Examine the Fourier series of the function f (x) = | x | from the point of view of even/odd. Solution: The function is even, so the Fourier series must be a cosine series. In fact we see that a 0 = 1 π ∫ - π π | x | d x = 2 π ∫ 0 π x d x = π. Also, for n ≥ 1, a n = 2 π ∫ 0 π | x | cos nx dx = 2 π ∫ 0 π x cos nx dx. An integration by parts gives. that a n = 2 π n 2 (cos n π - 1) = 2 π n 2 [ (- 1) n - 1 ]. As a result, a 2 j = 0 and a 2 j - 1 = - 4 π (2 (- 1) 2. In. conclusion, | x | = π 2 - 4 π (cos x + cos 3 x 3 2 + cos 5 x 5 2 +...). (5.3.3.1) The periodic extension of the original function f (x) = | x | on [− π, π ] is depicted in Figure 5.13. By Theorem 5.2.7 (see also Theorem 5.2.2), the series converges to f at every point of [− π, π ]. Figure 5.13: Periodic extension of f (x) = | x |. It is worth noting that x = | x | on [0, π ]. Thus the expansions (5.3.2.1) and (5.3.3.1) represent the same function on that interval

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  Differential Equations

  A Modern Approach with Wavelets

  • Steven Krantz(Author)
  • 2020(Publication Date)
  • Chapman and Hall/CRC

    (Publisher)

  6 Fourier Series: Basic Concepts •  The idea of Fourier series •  Calculating a Fourier series •  Convergence of Fourier series •  Odd and even functions •  Fourier series on arbitrary intervals •  Orthogonality 6.1    Fourier Coefflcients Trigonometric and Fourier series constitute one of the oldest parts of analysis. They arose, for instance, in classical studies of the heat and wave equations. Today they play a central role in the study of sound, heat conduction, electromagnetic waves, mechanical vibrations, signal processing, and image analysis and compression. Whereas power series (see Chapter 3) can only be used to represent very special functions (most functions, even smooth ones, do not have convergent power series), Fourier series can be used to represent very broad classes of functions. For us, a trigonometric series or Fourier series is one of the form f (x) = 1 2 a 0 + ∑ n = 1 ∞ (a n cos n x + b n sin n x). (6.1.1) We shall be concerned with three main questions: 1. Given a function f, how do we calculate the coefficients a n, b n ? 2. Once the series for f has been calculated, can we determine that it converges, and that it converges to f ? 3. How can we use Fourier series to solve a differential equation? We begin our study with some classical calculations that were first performed by Euler. (1707–1783). It is convenient to assume that our function f is defined on the interval [− π, π ] = { x ∈ R : − π ≤ x ≤ π }. We shall temporarily make the important assumption that the trigonometric series (6.1.1) for f converges uniformly. While this turns out to be true for a large class of functions (continuously differentiable functions, for example), for now this is merely a convenience so that our calculations are justified. We apply the integral to both sides of (6.1.1). The result. is ∫ − π π f (x) d x = ∫ − π π (1 2 a 0 + ∑ n = 1 ∞ (a n cos n x + b n sin n x)) d x = ∫ − π π 1 2 a 0 d x + ∑ n = 1 ∞ ∫ − π π a n cos n x d x + ∑ n = 1 ∞ ∫ − π π b n[--=PL

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  Music: A Mathematical Offering

  • Dave Benson(Author)
  • 2006(Publication Date)
  • Cambridge University Press

    (Publisher)

  Given any function f (θ ), we can obtain an even function by taking the average of f (θ ) and f (−θ ), i.e., 1 2 ( f (θ ) + f (−θ )). Similarly, 1 2 ( f (θ ) − f (−θ )) is an odd function. These add up to give the original function f (θ ), so we have written f (θ ) as a sum of its even part and its odd part, f (θ ) = f (θ ) + f (−θ ) 2 + f (θ ) − f (−θ ) 2 . To see that this is the unique way to write the function as a sum of an even function and an odd function, let us suppose that we are given two expressions f (θ ) = g 1 (θ ) + h 1 (θ ) and f (θ ) = g 2 (θ ) + h 2 (θ ) with g 1 and g 2 even, and h 1 and h 2 odd. Rearranging g 1 + h 1 = g 2 + h 2 , we get g 1 − g 2 = h 2 − h 1 . The left side is even and the right side is odd, so their common value is both even and odd, and hence zero. This means that g 1 = g 2 and h 1 = h 2 . Multiplication of even and odd functions works like addition (and not multipli- cation) of even and odd numbers: × even odd even even odd odd odd even Now for any odd function f (θ ), and for any a > 0, we have 0 −a f (θ ) dθ = − a 0 f (θ ) dθ 2.3 Even and odd functions 45 so that a −a f (θ ) dθ = 0. So, for example, if f (θ ) is even and periodic with period 2π , then sin(mθ ) f (θ ) is odd, and so the Fourier coefficients b m are zero, since b m = 1 π 2π 0 sin(mθ ) f (θ ) dθ = 1 π π −π sin(mθ ) f (θ ) dθ = 0. Similarly, if f (θ ) is odd and periodic with period 2π , then cos(mθ ) f (θ ) is odd, and so the Fourier coefficients a m are zero, since a m = 1 π 2π 0 cos(mθ ) f (θ ) dθ = 1 π π −π cos(mθ ) f (θ ) dθ = 0. This explains, for example, why a m = 0 in the example on page 41. The square wave is not quite an even function, because f (π ) = f (−π ), but changing the value of a function at a finite set of points in the interval of integration never affects the value of an integral, so we just replace f (π ) and f (−π ) by zero to obtain an even function with the same Fourier coefficients.

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  Signals and Systems Laboratory with MATLAB

  • Alex Palamides, Anastasia Veloni(Authors)
  • 2010(Publication Date)
  • CRC Press

    (Publisher)

  Regarding the coef fi cients a k of the complex exponential Fourier series, when the signal x ( t ) is even, the relationship a k ¼ a k stands, namely, the coef fi cients also have even symmetry. In order to demonstrate this property, the fi rst 11 coef fi cients a k of the signal that in one period is de fi ned as x ( t ) ¼ t 2 , 2 t 2 are computed and plotted. It is well known that x ( t ) ¼ t 2 is a signal with even symmetry. Commands Results t0 ¼ 2; T ¼ 4; w ¼ 2*pi = T; syms t x ¼ t ^ 2; k ¼ 5:5; a ¼ (1 = T)*int(x*exp( j*k*w*t),t,t0,t0 þ T); stem(k,eval(a)) legend( ' a_k ' ) −5 −4 −3 −2 −1 0 1 2 3 4 5 −1 −0.5 0 0.5 1 1.5 a k From the above graph, it is clear that the complex exponential Fourier series coef fi cients of an even signal have also even symmetry, namely, a k ¼ a k . 5.10.2 Odd Symmetry Recall that when a signal x ( t ) is an odd function of t , the relationship x ( t ) ¼ x ( t ) stands. In this case, the trigonometric Fourier series coef fi cients b k are zero. In order to demonstrate this property, the same procedure used in the previous section is followed. The coef fi cients b k of an odd signal are computed for one period time, i.e., T = 2 t T = 2 in two parts. 278 Signals and Systems Laboratory with MATLAB 1 First, the coef fi cients b k are computed for T = 2 t 0; that is, b k are computed for the signal x ( t ). However, because of the even symmetry x ( t ) ¼ x ( t ). Thus, Commands Results syms x t k T b1 ¼ sin(k*pi)*x = k = pi w ¼ 2*pi = T; b1 ¼ (2 = T)*int( x*cos(k*w*t),t, T = 2,0) Next, the coef fi cients b k are computed for 0 t T = 2, and in order to calculate the coef fi cients b k for the signal x ( t ), T = 2 t T = 2 the two parts are added. Commands Results b2 ¼ (2 = T)*int(x*cos(k*w*t),t,0,T = 2) b2 ¼ sin(k*pi)*x = k = pi b ¼ b1 þ b2 b ¼ 0 Indeed, the trigonometric Fourier series coef fi cients b k of a signal with odd symmetry are zero.

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  A Course of Mathematics for Engineerings and Scientists

  Volume 5

  • Brian H. Chirgwin, Charles Plumpton(Authors)
  • 2013(Publication Date)
  • Pergamon

    (Publisher)

  §1:2 F O U R I E R S E R I E S 13 we define a second function Φ(χ) which is identical with φ (x) for x > 0 and, for x < 0, takes values such that Φ(χ) is even, i.e., Φ(χ) = ψ(χ) for x > 0, Φ(χ) = φ ( — #) for . < 0. Then the Fourier series expansion of the even function Φ(χ), valid for — π < x < π , contains cosines only and therefore ψ (χ) is represented by this cosine series for 0 < x < π . [The values to which the series F I G . 4(i). F I G . 4(ii). F I G . 4(iii). converges at x — 0 and x = π can be shown by the use of formulae (1.6) and (1.7) to be /(0) and f(n) respectively, and so the cosine series is valid for 0 < x < π.] Similarly we can find the (half-range) Fourier sine series for ψ(χ), defined for x > 0, by finding the full-range Fourier expansion, valid for — π < x < , of the odd function Ψ(χ) which is defined by Ψ(χ) = φ(χ) for x > 0, Ψ(χ) = -φ(-χ) for x < 0. 14 A COURSE OF MATHEMATICS Note that these half-range Fourier series will not, in general, re-present the function φ(χ) outside the range 0 < x < π. For example, if the graph of φ(χ) for 0 < x < π is as shown in Fig. 4(i), then the graph of the Fourier cosine series of φ(χ) is as shown in Fig. 4(ii). Similarly the graph of the Fourier sine series of φ (χ) is shown in Fig. 4 (in). The above results show that symmetries in the graph of f(x) cor-respond to the absence of certain terms in the Fourier series. We now show that there are other symmetry properties which correspond to the absence of certain other terms. From (1.3a), π-ftx 2π+α 1 r 1 / ' a n — / f(x) voanx dx -I -/ f(x) coanxdx π J π J a a -f a = — / f( x ) + (— l ) n / ( ^ + x )) cosnxdx. 71 J a Similarly -f a b n = — f(x) + (~) n f(n + x)}sinnxdx. a It follows that: (1) if f(n + x) = f(z), then a 2n+1 = 0 -b 2n+1 for ra = 0,1,2,..., i. c , the Fourier series contains only even harmonics; (2) if f(n + x)= -f(x), then a 2n =--0 = b 2n for w = 0,1,2,..., i.e., the Fourier series contains only odd harmonics.

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