nπt a n cos y(x) satisfies d 2 y dx 2 + P EI y = 0 If the column is constrained at both ends then we can apply the boundary conditions

Transcription

1 Chapter 6: Page 6- Fourier Series 6. INTRODUCTION Fourier Anaysis breaks down functions into their frequency components. In particuar, a function can be written as a series invoving trigonometric functions. For exampe, et f(t) be a function defined on the interva a < t < a + then f(t) can be written in terms of a series expansion. that is, f(t) = a + ( a n cos ( ) ( )) nπt nπt + b n sin where a, a n and b n are caed the co-efficients which are constant and are to be determined. The frequency components are nπ for n =,,.... In many instances, the magnitude of the co-efficients (ie a n + b n ) is graphed against the frequencies. This graph is caed the spectra graph of the function and the named given to the anaysis of data is caed spectra anaysis. Engineers use spectra anaysis to determine the dominant frequency for a particuar probem. For exampe, consider sender coumn of uniform cross-section as described in the figure beow. P (Load) x= x y Here y(x) is the defection curve for the coumn. Therefore, y(x) satisfies d y dx + P EI y = x=l where P is the oad on the coumn, E is Young s moduus and I is the moment of inertia for the coumn. If the coumn is constrained at both ends then we can appy the boundary conditions y() = y(l) =. It can be shown that the defection curve of the coumn is y(x) = = y n (x) c n sin P n x

2 Page 6- where P n = n π EI, where n =,, 3,.... L This particuar probem is caed the eigenvaue probem and wi be ater discussed in the next chapter. It is easiy seen that y(x) is now in terms of trigonometric functions. the particuar series is caed Fourier Series. Aso, the frequency component is P n When n =, P = π L EI. This is caed the Euer oad or dominant oad. Aso, is caed the first bucking oad. y (x) = c sin πx L which is caed the critica oads. 6. FUNCTIONS Before the introduction to Fourier Series it is worthwhie refreshing some ideas and concepts about functions. In particuar, the properties of odd and even functions. We sha use these properties quite frequenty when discussing Fourier Series ater on. 6.. Properties of Even Functions Reca that if f(x) is an even function then f( x) = +f(x). (a) The graph of an even function is symmetric about the y-axis (or vertica). Exampes f(x) f(x) f(x) f(a) f(a) -a a -a a -a a (a) (b) (c) Figure (a) coud be expressed as the function rue f(x) = x 4 x, Figure (b) with rue f(x) = ( x) for < x < and Figure (c) with rue f(x) = for xɛ[ a, a]. It can be seen from these graphs that even functions have the property that a a a f(x) dx = f(x) dx. (b) The Macaurin series of an even function contains even powers of x ony. Exampes of even functions:, x, x 4, cos x, cosh x,....

3 Page Properties of Odd Functions Reca that if f(x) is an odd function then f( x) = f(x). (a) The graph of an odd function is skew-symmetric through the origin. Exampes f(x) f(x) f(x) f(a) -a a -a -a a f(-a) - (a) (b) (c) a The function rue for Figure (a) coud be written as f(x) = x 3, whereas Figure (b) coud have the rue f(x) = x. Figure (c) coud be written in function form as f(x) = for x > and f(x) = for x <. Odd functions have the property that a a f(x) dx =. (b) Odd functions have the property that their Macaurin series contains odd powers of x ony. Exampes of odd functions: x, x 3, sin x, sinh x, erf(x),.... (c) If f(x) is continuous at x = then f() = Products of Odd and Even Functions The property of the product of even and odd functions can be summarised as foows: (even function) (even function) = (even function) (odd function) (odd function) = (even function) (even function) (odd function) = (odd function). Exampe The function x is an odd function and cos x is an even function, and hence their product is an even function. That is, x cos x = (odd function) (even function) = (odd function). Thus, x cos x dx =.

4 Page 6-4 Simiary, Thus, x sin x = (odd function) (odd function) = (even function). x sin x dx = x sin x dx Trigonometry Formuae sin(a + B) = sin A cos B + cos A sin B sin(a B) = sin A cos B cos A sin B cos(a B) = cos A cos B + sin A sin B cos(a + B) = cos A cos B sin A sin B sin A cos B = sin(a + B) + sin(a B) cos A sin B = sin(a + B) sin(a B) cos A cos B = cos(a B) + cos(a + B) sin A sin B = cos(a B) cos(a + B) sin A = sin A cos A cos A = cos A sin A = cos A Note: When n is an integer = sin A sin nπ = and cos nπ = ( ) n Orthogona Functions Two functions f m and f n are said to be orthogona on an interva [a, b] if b a f m (x) f n (x) dx =. Exampes of such orthogona functions are the trigonometric functions Sine and Cosine and the Besse function. In particuar, the set {, cos x, cos x....} form a set of orthogona functions. Thus if m and n are integers, then the foowing integration properties hod for both the sine and cosine functions: cos mx cos nx dx = cos mx sin nx dx = sin mx sin nx dx = cos mx dx = m n sin mx dx = π, m.

5 Page Periodic Functions A function f(t) is caed periodic if it is defined for a rea t and if there is a positive number τ that f(t + τ) = f(t). This number τ is caed the period of f(t). The graph of such a function is obtained by the periodic repetition of its graph in any interva of ength τ, as shown in the figures beow. f(t) f(t) t t!! The trigonometric functions are the most common periodic functions of period π. Note: The function f(t) = c = constant is a periodic function Sinusoids Sometimes a function occurs as a inear combination of sine functions or sinusoids. For instance, consider the function f(t) = a sin ωt + b sin ωt + c sin 3ωt + d sin 4ωt. The fundamenta anguar frequency is ω as this is the owest frequency (or owest vaued co-efficient of t ) for each of the sin terms. Thus the frequency of osciation is defined as ω π which is usuay measured in cyces per second or Hertz. Hence, the argest period of the sin terms is Period = τ = π ω. The other frequencies are integer mutipes of this owest frequency. Thus the fundamenta or first harmonic is the term that invoves the fundamenta frequency. That is, a sin ωt is the fundamenta or first harmonic. The next term, b sin ωt is caed the second harmonic, and so on. The ampitude of the first harmonic is a and the ampitude of the second harmonic is b.

6 Page 6-6 Note: Reca the trigonometric identity that R sin(ωt + φ) = R sin ωt cos φ + R cos ωt sin φ = A sin ωt + B cos ωt where Therefore, A = R cos φ B = R sin φ. R = A + B φ = tan B A Thus a function that has a inear combination of sine s and cosine s can be expressed as a sum of sinusoids with different ampitudes and phase anges. That is, f(t) = = a n sin nωt + b n cos nωt R n sin (nωt + φ n ) Exampe (a) Describe the frequency and ampitude characteristics of the different components of the function f(t) = sin 3πt.7 sin 9πt +.3 sin 5πt. Method The fundamenta anguar frequency is 3π or frequecny of osciation is 3 hertz. The ampitude of the first harmonic is. The fundamenta or first harmonic is sin 3πt. The second and forth harmonics are missing, that is, the terms sin 6πt and sin πt. The third harmonic ampitude is.7 (b) If f(t) = sin t + cos t, express f(t) as a singe sinusoid and hence determine its ampitude and phase. Method Reca the trigonometric identity then R = + = 5 φ = tan = tan. rad. Hence, f(t) = 5 sin(t +.). Thus the ampitude is 5 and the phase ange is. radians.

7 Page 6-7 Note: Our function f(t) coud have been in terms of the cos function rather than the sin function. We coud use the same techniques as above except that our phase ange woud differ by π Lapace Transform of a Periodic Function As an engineer or an appied mathematician, it is important to know about periodic functions and their behaviour. This is due to the fact that these functions arise in many practica probems. Periodic functions can be more compicated then the norma periodic functions of sines and cosines. One important aspect is know how to take Lapace transforms of periodic functions. If f(t) is a piecewise continuous function of interva ength τ, then its Lapace transform exists. That is, L { f(t) } = e st f(t) dt. We can rewrite the integra on the right hand side into a series of integras over successive periods, namey, L { f(t) } = τ τ e st f(t) dt + τ 3τ e st f(t) dt + τ e st f(t) dt Substituting z = t τ in the second integra, z = t τ in the third integra and, consequentiay, z = t nτ for the nth integra etc, then integras become τ L { f(t) } = τ e sz f(z) dz + = ( + e sτ + e sτ +...) τ e s(z+τ) f(z) dz + τ e sz f(z) dz e s(z+τ) f(z) dz Hence, by the binomia expansion we find that L { f(t) } = e sτ τ e sz f(z) dz. Exampe Find the Lapace transform f where f(t) is the square wave as shown. f(t) t a b a b

8 Page 6-8 Method Here τ = b then b L { f(t) } = e sb e sz f(z) dz ( a ) b = e sb e sz f(z) dz + e sz f(z) dz a = e sb b a e sz = e sb s e sz dz ] b a = e sa e sb s( e sb ) Exercise 6A Determine if the foowing functions are even or odd. Sketch their graphs. (a) sin mx dx = (a) f(x) = x (b) f(x) = x + (c) f(x) = x 3 (b) (c) cos mx dx = {, m π, m = cos mx sin nx dx = (d) (e) (f) f(x) = x + sin x f(x) = e x f(x) = x State whether the foowing functions are even, odd or neither. (a) (b) (c) (d) (e) (e) f(x) = x sin x f(x) = x sin x f(x) = sin 3x sin x f(x) = x sin x f(x) = e x sin x f(x) = x x 3 Prove the foowing orthogonaity resuts for trigonometric functions. (d) (e) cos mx cos nxdx = sin mx sin nx dx = (f) cos mx dx = {, m n π, m = n and m π, m = n = {, m n π, m = n and m, m = n = sin mx dx = π, m. 4 Sketch the graph of the foowing periodic functions. (a) f(x) = x, x and f(x + ) = f(x). continued next page...

9 Page 6-9 { x π x (b) f(x) = x < x π and (a) f(t) f(x + π) = f(x). - t (c) f(x) = { sin x x < π π x π and f(x + π) = f(x). (b) f(x)! "3! "! "!!! 3! x 5 Express the foowing functions as a singe sinusoid and hence find their ampitudes and phases. (a) f(x) = 3 sin x cos x (c) "! 3 f(x) (b) f(x) = sin 3x x (c) f(x) = 3 sin x + cos x Find the Lapace transform of the foowing periodic functions that are shown beow. 6.3 DEFINITION An important ski for engineers is to be abe to anayse certain wave forms. In particuar the frequency components of a wave. A mathematica too that enabes an engineer to break a wave to its frequency components is caed Fourier Anaysis. This section wi ook at the essentia requirements to break down a wave into its frequency components by using Fourier Series Fourier Series Generay, a Fourier series for a function f(x) defined on an interva a < x < a + is of the form f(x) S(x) = a + a n cos nπ x + b n sin nπ x, where and a n = b n = a+ a a+ a f(x) cos nπ x dx n =,,,..., f(x) sin nπ xdx n =,, 3,.... A property of this series is that it wi converge to the mean vaue of the function at any discontinuity either in the interva or at an endpoint.

10 Page 6- If the defining function f(x) is itsef ony defined on a given interva a < x < a + (and not extended to be periodic by some part of its definition), then the Fourier series which converges to the function f(x) in the defining interva may be extended to represent a function F (x) such that f(x), a < x < a + F (x) = f(x + ), esewhere. An advantage of studying Fourier series is that the sum of some infinite rea series can be obtained. This can be achieved by substituting vaues of x into the series and the function. An exampe of this wi be seen ater. Specificay, a Fourier series for a function f(x) defined on an interva < x < π is of the form f(x) S(x) = a + a n cos nx + b n sin nx, where and Note: a n = π b n = π f(x) cos nx dx n =,,,..., f(x) sin nxdx n =,, 3,.... Here a = and a + = π, hence = π and the function is assumed to have period of π. Exampe Find the Fourier series, S(x), of the function f(x) = x on the interva < x < π. Method Here a = and a + = π. That is, = π. Aso, it is assume that the function f(x) is periodic with period equa to the ength of the interva ( π ). The Fourier series for f(x) is where the co-efficents are S(x) = a + { a n cos(nx) + b n sin(nx) } a = π a n = π b n = π f(x) dx, f(x) cos(nx) dx, n =,, 3,..., f(x) sin(nx) dx n =,, 3,...,. At this stage it is beneficia to draw a graph of the function to aid in determining the Fourier co-efficients and therefore the Fourier series. This is due to the fact that some of the co-efficents may be automaticay zero and thus the Fourier series woud be simpified.

11 Page 6- The graph of f(x) is f(x)! "!! x "! From the graph of f(x), it can be seen that f(x) is an odd function. Since, the sine function is an odd function, it can be easiy shown that a = a n = automaticay and therefore we need ony find the b n co-efficient. Here, b n = π f(x) sin nx dx = x sin nx dx π cos(nπ) = n = ( )n n = ( )n+ n (integrate by parts) where cos(nπ) = ( ) n Thus, the Fourier series for f(x) is given by S(x) = ( ) n+ sin(nx). n Note: This series is an infinite series. We can ook at the partia sum of this series to detemine how we it approximates the given function. The first few terms of this series are sin(x) sin(x) + 3 sin(3x) sin(4x) +... The foowing figure compares the function f(x) to the partia sums S, S and S, respectivey. f(x)=x f(x)=x f(x)=x S() S() S()

12 Page 6- It can be seen that the higher number of terms in the partia sum, the better the approximation to the given function Periodicity and Discontinuities Reca f(x). That is, f(x) = x π < x < π. Then, et F (x) be the periodic function of f(x) then the graph of F (x) over the interva ( 3π, 3π) is f(x)! "3! "! "!!! 3! x "! It can be seen from the above figure that F (x) has discontinuities at..., 3π,, and π, 3π,.... For instance, when x = π (say), F (x) is not defined. Aso, from the graph of periodic function F (x) we can see that, If we consider the Fourier series we find that Note: im F (x) = π and im F (x) = x π x π + im S(x) = im x π x π n= ( ) n+ sin(nx) =. n S() is the average of the imit vaues obtained above. That is, { im F (x) + im F (x) x π x π + } = = S(π). Generay, at a point of discontinuity of F (x), the Fourier series averages the imiting vaues of the function at the discontinuous point. That is, if x = a be a discontinuous point of F (x) then { im F (x) + im F (x) x a x a + } = S(a).

13 Page Finding the Sum of a Series We can use the Fourier series approximation to a given function to determine the sum of an infinite series. For instance, reca that f(x) = x, < x < π and f(x) S(x) = n= ( ) n+ sin(nx). n If we et x = π (say), then f( Upon substituting x = π π ) = π. into the Fourier series we obtain S( π ) = ( ) n+ sin( nπ n ). The infinite series can be simpified further due to the fact that when n is an even integer sin( nπ ) =. That is, sin nπ = for n =, 4, 6,.... Aso, when n is an odd integer we can simpify the term sin( nπ ) as foows. Firsty, we rewrite the odd vaues of n in the form of n where n =,, 3,..., then sin( nπ ) can be written as sin((n ) π ( ) = sin nπ π ) where n =,, 3... = sin nπ cos π cos nπ sin π =( ) n+. As a resut, S( π ) = = ( ) n n ( )n ( ) n n upon simpification Since f( π ) S(π ) then or π = ( ) n n. ( ) n n = π 4. Hence, with the aid of Fourier series, we have found the sum of an infinite series.

14 Page 6-4 Exampe Find the Fourier series S(x) for the foowing function f(x) = {, 3 x x, x 3. Here a = 3 and a + = 3, therefore, = 3. The function F (x) is assume to be the periodic extension of f(x) with period 6 (ie the ength of the interva). The graph of F (x) over the interva [ 9, 9] is 3 f(x) x From this graph it can be seen that the function f(x) is neither an even or odd function. Therefore, a Fourier co-efficent need to be evauated. The Fourier co-efficients of f(x) on the interva [ 3, 3] are a = 3 = f(x) dx x dx = 3. and a n = 3 = f(x) cos nπx 3 dx x cos nπx 3 dx = 3 n [cos(nπ) ] π = 3 n π [( )n ]. cos(nπ) = ( )n Now a n can be further simpified due to the fact that when n is an even integer a n =. However, when n is an odd integer a n = 6 n π. That is, the odd co-efficients of a n can be expressed as a n = 6 (n ) π for n =,, 3,...,.

15 Page 6-5 Now b n = 3 = f(x) sin nπx 3 dx x sin nπx 3 dx = 3 nπ cos(nπ) = 3 nπ ( )n+. cos(nπ) = ( )n Therefore the Fourier series of f(x) on the interva [ 3, 3] is S(x) = ( ( 6 (n )πx (n ) π cos 3 ) + 3 nπ ( )n+ sin ( nπx ) ). 3 Exercise 6B For each of the functions f(x) beow, et F (x) be the periodic extension of f(x) then (i) (ii) (iii) (iv) Sketch the graph of the function f(x) on (, π). Find the Fourier series, S(x), for the function on (, π). Sketch the graph of the function F (x) (which is represented by the series S(x) ) on the interva ( 5π, 5π), Investigate the identities obtained using ( ) im F (x) + im F (x) x π x π + = S(π). That is, show that at the point of discontinuity (here x = π ), the Fourier series takes the average of the imit vaues of F (x). The functions are: {, < x < (a) f(x) =, < x < π (c) f(x) = x, < x < π (d) f(x) = x, < x < π (e) f(x) = sin x, < x < π. (f) f(x) = h(x), < x <. Find the Fourier series for each of the foowing periodic functions f(x). In each case, make sketches of the functions represented by the series S(x) on the interva 6 < x < 6. {, < x < (a) f(x) =, < x < where f(x + ) = f(x) (b) f(x) = cos x, < x < 3 where f(x + 3) = f(x). 3 For each of the given functions in (), et x = π then find the sum of an infinite series. {, < x < (b) f(x) = x, < x < π

16 Page FOURIER SINE AND COSINE SERIES Sometimes it is possibe to represent a function as a Fourier Sine or Cosine series. To do this we use the properties of even and odd functions as defined in section 6... To determine a series we usuay extend the interva of definition (backwards) to create a new function that is either even or odd depending on the type of series required. If we require a Fourier sine series then the new function that is created is chosen to be an odd function. Simiary, if we require a Fourier cosine series then the new function created is chosen to be an even function. For exampe, et f(x) be defined on the interva [, ]. (a) If we require a Fourier sine series then we create a new function, g(x) (say), which is an odd function over the interva [, ]. That is, we et g(x) = { f(x) x f( x) x. We ca g an odd extension of f to [, ]. The Fourier series for g(x) over the interva [, ] is a Fourier sine series due to the fact that g(x) is an odd function. Hence, the Fourier co-efficients are and b n = g(x) sin( nπx = as g(x) = f(x) over the interva [, ]. a = a n = ) dx ( f(x) sin( nπx ) even function on [, ]) f(x) sin( nπx ) dx, n =,, 3,.... The construction of g(x) is simpy an aid to determine the required Fourier series for f(x) on [, ]. (b) If we require a Fourier cosine series then we create a new function, g(x) (say), which is an even function over the interva [, ]. That is, we et g(x) = { f(x) x f( x) x. We ca g an even extension of f to [, ]. The Fourier series for g(x) over the interva [, ] is a Fourier cosine series due to the fact that g(x) is an even function. Hence, the Fourier co-efficents are b n = and and a n = = g(x) cos( nπx as g(x) = f(x) over the interva [, ]. a = = g(x) dx f(x) dx. ) dx g(x) cos( nπx ) even funtion on [, ]) f(x) cos( nπ x) dx, n =,, 3,.... Once again, the construction of g(x) is just an aid to determine the required Fourier series for f(x) on [, ].

17 Page Fourier Cosine Series If f(x) is a function defined on the interva < x <, then it has a Fourier Cosine series f(x) S(x) = a + a n cos nπx where a n = f(x) cos nπx dx, n =,,,.... Exampe Find the Fourier cosine series for the function f(x) = x where x. Method We make an even extension of f to form the new function g so that f(x) = x x g(x) =. f( x) = x x Graphicay, we have f(-x)=-x f(x)=x - x Since g(x) is an even function then the Fourier co-efficients are b n =, and a = = = g(x) dx = x dx f(x) dx a n = = g(x) cos( nπx f(x) cos( nπx ) dx ) dx g(x) cos( nπx ) is an even function on [, ]

18 Page 6-8 That is, a n = x cos( nπx ) dx integrating by parts = 4 n [cos nπ ] π = 4 n π [( )n ] cos nπ = ( )n The co-efficient a n can be further simpified by noting that when n is an even integer a n =. Aso, when n is an odd integer we find that a n = 8 n. We can rewrite the odd co-efficients in the form π 8 a n = (n ) π n =,, 3,...,. Hence, the Fourier cosine series for f(x) on the interva [, ] is S(x) = 8 π Note: The a term is divided by. (n ) cos ( (n )πx ) Fourier Sine Series If f(x) is a function defined on the interva < x <, then it has a Fourier Sine series f(x) S(x) = b n sin nπx where b n = f(x) sin nπx dx, n =,,,...,. Exampe Find the Fourier sine series for the function f(x) = x where x. Method We make an odd extension of f to form the new function g so that f(x) = x x g(x) =. f( x) = x x

19 Page 6-9 Graphicay, we have f(x)=x - -f(-x)=x Since g(x) is an odd function then the Fourier co-efficients are a and a n = and b n = = = g(x) sin( nπx n+ 4 =( ) nπ. f(x) sin( nπx ) dx ) dx g(x) sin( nπx ) even function on [, ] x sin( nπx ) dx integrating by parts Therefore, the Fourier sine series, S(x), for f(x) on [, ] is f(x) S(x) = = 4 π b n sin( nπx ) ( ) n+ sin( nπx ). n Note The Fourier Cosine and Sine series represent, respectivey, an even and an odd function on the interva < x <, which correspond to the defined function f(x) on the haf interva < x <. These series are often caed haf interva expansions. Exercise 6C (a) Expand each of the foowing functions as indicated over the interva given. Aso, for each case, make sketches to show the function F(x) (the periodic extension of f(x) ) which is represented by your series S(x) on 3π < x < 3π. (i) f(x) = as a sine series on < x < π. (ii) f(x) = x as a cosine series on < x < π. (iii) f(x) = cos x as a sine series on < x < π. continued next page...

20 Page 6- (b) Find the identities which occur when you make the foowing substitutions in your answers to part (a) of this question. (i) x = π in (a)(i) (c) function represented by the series in (a). What series identity can be found by substituting x = in the series in (a)? (ii) (iii) (iv) x = in (a)(ii). x = π in (a)(iii) x = π in (a)(iii). 4 3 The function f is defined on the interva (, ), by f(x) = x, < x <. (a) Find its Fourier cosine series. The function f is defined on the interva (, ) by f(x) =, < x <. (b) On the interva ( 3, 3), sketch the function represented by the series in (a). (a) Show that its Fourier sine series is 4 {sin πx + 3 π sin 3πx + 5 } sin5πx +... (b) On the interva ( 3, 3), sketch the (c) Use the series in (a), or otherwise, to prove that (n ) +... = π 8.