Fourier Series and the Discrete Fourier Expansion

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1 Fourier Series and the Discrete Fourier Expansion Matthew Lincoln Adrienne Carter December 5, 2 Abstract This article is intended to introduce the Fourier series and the Discrete Fourier Expansion and demonstrate their power through examples. Page of 35. Like the function in Figure, any function can be approximated by an infinite series of sine and cosine functions. As we can see from the Figure 2, the green, red, cyan, and violet are the basis for the blue function. A different view is provided in Figure 3.

2 Page 2 of 35 Figure : This complex wave is a combination of five independent sinusoidal functions varying in amplitude and frequency.

3 Figure 2: The blue wave is the sum of the four simple harmonic functions. Page 3 of 35

4 Page 4 of 35 Figure 3: A three dimensional view of the functions summed and the resulting complex wave.

5 2 2.. The Inner Product Before we begin our discussion of a basis, we need to broaden our definition of the inner product. The inner product, or dot product, of two vectors results in a scalar. Similarly, the dot product of two functions, given the limits, will also yield a number. The inner product of two functions is defined as the integral of the product of those functions. In symbols, f(x), g(x) = b a f(x)g(x) dx. For this to be an inner product, it must satisfy these five restraints that define the dot product:. f, f ; 2. f, f =, iff f = ; 3. f, g = g, f ; 4. αf, g = f, αg = α f, g ; 5. f, g + h = f, g + f, h. Each of these properties can be easily proven.. 2. f, f = f, f = b a b a [f(x)] 2 dx [f(x)] 2 dx =, iff f(x) =.5.5 Page 5 of 35

6 f, g + h = = = f, g = = αf, g = = b a b a b b a b a = g, f b a b a = f, αg = b a = α f, g f(x)g(x) dx g(x)f(x) dx αf(x)g(x) dx f, αgdx α f, g dx f(x) (g(x) + h(x)) dx f(x)g(x) + f(x)h(x) dx f(x)g(x)dx + a a = f, g + f, h b f(x)h(x) dx.5.5 Page 6 of 35

7 Orthogonality of the Basis We will use.5.5 {, cos(x), sin(x), cos(2x), sin(2x),..., cos(nx), sin(nx),...}, n =,, 2,... as a basis. Our first task is to show that distinct basis elements are orthogonal. That is, we must show that, cos nx =,, sin nx =, cos mx, cos nx =, m n; sin mx, sin nx =, m n; cos mx, sin nx =. Note n and m are integers. It is important that when we dot a cosine function with another cosine function, or sine with sine, we have different frequencies. These different frequencies must be integer multiples of the original frequency. The inner product of a cosine function and a sine function has no restriction on the frequency. We perform all the integrations over the period, represented here from [, 2π]. Page 7 of 35

8 2 2. If nɛ{, 2, 3,..., }, then, cos nx = = cos nx dx cos nx dx = n sin nx 2π.5.5 = [sin(2nπ) sin ] n = ( ) n = Therefore, and cos nx are orthogonal. 2. If nɛ{, 2, 3,..., }, then, sin nx = Therefore, is orthogonal to sin nx. sin nx dx = n cos nx 2 π = [cos(2nπ) cos ] n = ( ) n = n () = Page 8 of 35

9 If n m, then cos mx, cos nx = cos mx cos nx dx sin((n m)x) = + 2(n m) [ sin(2π(n m)) = + 2(n m) [ sin((n m)) + 2(n m) = ( + ) ( + ) = sin((n + m)x) 2(n + m) 2π ] sin 2π(n + m) 2(n + m) sin((n + m)) 2(n + m) ].5.5 Therefore, cos mx and cos nx are orthogonal, provided m n. 4. If m n, then sin mx, sin nx = a sin mx sin nx dx sin((n m)x) = 2(n m) = + [ sin(2π(n m)) 2(n m) + [ sin((n m)) + 2(n m) = ( + ) ( + ) = sin(n + m)x 2(n + m) 2π sin(2π(n + m)) 2(n + m) ] sin((n + m)) 2(n + m) ] Page 9 of 35

10 2 2 Therefore, sin mx and sin nx are orthogonal if m n. 5. If n, m are any positive integers, then.5.5 cos mx, sin nx = cos mx sin nx dx 2π cos((n m)x) cos((n + m)x) = 2(n m) 2(n + m) [ ] cos(2π(n m)) cos(2π(n + m)) = + 2(n m) 2(n + m) [ ] cos((n m)) cos((n + m)) + 2(n m) 2(n + m) ( ) ( ) = 2(n m) + 2(n + m) 2(n m) + 2(n + m) = Therefore, cos mx and sin nx are orthogonal at any frequency. Thus, distinct members of the basis are orthogonal. Now, what happens when you take the inner product of a basis element with itself? Page of 35. First, compute,., = dx = [x 2π = 2π = 2π

11 Secondly, compute cos nx, cos nx. cos nx, cos nx = = = [ 2 cos 2 nx dx ( ) + cos 2nx dx 2 ( ) cos 2nx + dx 2 2 2π sin 2nx x + 4n = = [ sin 2n(2π) 2 2π + 4n ] sin 2n() 4n.5.5 = 2π 2 + ( ) = π Page of 35

12 Finally, compute sin nx, sin nx. sin nx, sin nx = 2. = = [ 2 sin 2 nx dx cos 2nx dx 2 ( ) cos 2nx 2 2 2π sin 2n = x 4n = [ sin 2n(2π) 2 (2π) 4n = 2π ( ) 2 = π ] sin 2n() 4n Any function can be written as an infinite sum of sine and cosine functions. Jean Baptist Fourier was the first mathematician to write this as an expression. It is now called the Fourier series and is defined as f(x) = a + (a n cos(nx) + b n sin(nx)), n=.5.5 Page 2 of 35

13 2 2 where a = 2π a n = π b n = π f(x) dx, f(x) cos(nx) dx, and, f(x) sin(nx) dx. The integration is performed over the period, stated in the individual terms as [, 2π]. It can just as easily be represented as [ π, π]. We will use both expressions of the period in the later examples Derivation of the Fourier Coefficients We now have f(x) = a + a n cos nx + b n sin nx. Take the inner product of both sides with respect to the basis element. f, = (a + n= a n cos nx + b n sin nx), n= Now, using the properties of the inner product, f, = a, + (a n cos nx, + b n sin nx, ) n= But, due to the orthogonality between the basis elements, all but one of the terms go to zero. Thus, we are left with f, = a,. Page 3 of 35

14 2 2 Recall that we established, = π. Therefore, f, = a (2π), a = f,, 2π a = f(x) dx. 2π Now, take the inner product of both sides with respect to the basis element cos nx. f, cos nx = (a + a n cos nx + b n sin nx), cos nx n= f, cos nx = a, cos nx + (a n cos nx, cos nx + b n sin nx, cos nx ) n= Again, the orthogonality of the elements simplifies this equation to f, cos nx = a n cos nx, cos nx. But, we previously found the value of cos nx, cos nx. f, cos nx = a n (π) a n = f, cos nx π = π f(x) cos nx dx Finally, repeat the procedure taking the inner product of both sides with respect to.5.5 Page 4 of 35

15 2 2 sin nx, the remaining basis element. f, sin nx = (a + a n cos nx + b n sin nx), sin nx n= f, sin nx = a, sin nx + (a n cos nx, sin nx + b n sin nx, sin nx ) n= Remember, sin nx is orthogonal to every basis element but itself..5.5 f, sin nx = b n sin nx, sin nx f, sin nx = b n π b n = f, sin nx π = π f(x) sin nx dx 2.2. Using the Fourier Series To demonstrate the Fourier series, we will expand the function f(x) = x, π x π. See Figure 4. We being by finding the coefficients. Start with a. If we think graphically, a represents the initial weight or displacement. We should look at the area under this curve. The same amount of area exists to the left and right of the y-axis. But, the area on the left is negative, so our initial area under the curve, a is zero. To prove it, the Page 5 of 35

16 Figure 4: The graph of f(x) = x from [ π, π]. Page 6 of 35

17 2 2 expression for a is evaluated from [ π, π]. a = π x dx 2π π = [ x 2 π 2π π = 2π [π2 ( π) 2 ] = 2π (π2 π 2 ) = 2π () =.5.5 Now, we will calculate a n. π a n = x cos nx dx π π = [ π x sin nx π n nπ π = nπ [x sin nx π π + nπ π [ n π sin nx dx cos nx π π = (π sin nπ π sin( nπ)) + nπ n 2 π [cos nx π π = ( ) + nπ n 2 (cos nπ cos( nπ)) π = + n 2 ( ) π = Page 7 of 35

18 2 2 If we think of this problem graphically, we notice that f(x) = x is an odd function, so cosine would not have any terms because it is an even function. Finally, we will find b n..5.5 π b n = x sin nx dx π π = [ n π π x cos nx π nπ = nπ [x cos nx π π + nπ π π [ nπ cos nx dx sin nx π π = (π cos nπ + π cos( nπ)) + nπ n 2 π [sin nx π π = (2π) cos nπ + nπ n 2 (sin nπ sin( nπ)) π = 2π cos nπ + nπ n 2 π () = 2 n ( )n+ Combining these three values into the Fourier series, we get f(x) = x = a + = + = 2 (a n cos nx + b n sin nx), π x π n= n= ( cos(nx) + ( ( ) n+ n= n ( 2n ) ) ( )n+ sin nx, π x π ) sin nx, π x π. Page 8 of 35

19 2 2 Expand this summation a few terms, ( ) sin 2x sin 3x f(x) = x = 2 sin x +, π x π. 2 3 Notice the signs alternate and the frequency of each term is equal to the denominator of each term. Now, we will graph the summation with an increasing number of terms. We start with four terms. The equation for Figure 5 follows. f(x) = x = 2 ( sin 2x sin 3x sin x ) sin 4x, π x π. 4 As we increase the number of terms in our series to ten, we see the improvement in the graph (see Figure 6), and the separation between the original function and modeled function decrease. But the new graph still doesn t look like f(x) = x. The middle of the graph of fifty terms shown in Figure 7 looks very close. But, even at n = 5, there appears to be some error at the very edges of the graph. This ringing is known as Gibb s phenomenon The Discrete Fourier Expansion Our first derivation led to the Fourier coefficients a = 2π a n = π b n = π f(x) dx, f(x) cos(nx) dx, and, f(x) sin(nx) dx. Page 9 of 35

20 Figure 5: Here, n = 4. Page 2 of 35

21 Figure 6: n =. Page 2 of 35

22 Figure 7: n = 5. Page 22 of 35

23 2 2 In the Fourier approximation of f(x) = x on [ π, π], we saw that equivalent representations for the Fourier were a = π f(x)d x, 2π a n = π b n = π π π π π π f(x) cos(nx) dx, and, f(x) sin(nx) dx. In general, the Fourier coefficients for the expansion of a function within period T are given by and the series is defined by a = T a n = 2 T b n = 2 T f(x) = a + T T T n= f(x) dx, f(x) cos 2πnx T f(x) sin 2πnx T a n cos 2πnx T dx, and, dx + b n sin 2πnx T. If and when f(x) is unknown or unable to be integrated, the Fourier series expansion doesn t work, but f(x) can still be examined point by point with the Discrete Fourier Expansion. For example, consider the periodic wave in Figure 8. This is an analog sample that could represent a musical tone of some type or a periodic signal. First, we will segregate a single period in Figure Page 23 of 35

24 Page 24 of 35 Figure 8: Periodic wave.

25 Figure 9: A single period of the wave. Page 25 of 35

26 Figure : The left-endpoints used for the discrete values are circled in red. Again, a will be calculated first. a = T T f(x) dx The function of this single period is unknown, so instead of taking the integral over the period, we will use the left-endpoint method by selecting sample points to find the area under the curve. See Figure..5 Page 26 of 35

27 2 2 a T = n= f(n x) x 9 f(n.). n= = (f() + f(.) + f(.2) + f(.3) + f(.4) + f(.5) + f(.6) + f(.7) + f(.8) + f(.9))(.) = ( = )(.) A more complete understanding of a will send ambiguity forth to funerals. Cosine and sine waves osculate about the x-axis with a total sum of zero underneath their curves. In the Fourier series, a behaves as a constant much like b in the formula y = mx + b. Every scaled sine and cosine wave (that makes up the character of the wave) will contain sums of zero underneath their curves. The addition of a prevents a zero integral, provided a nonzero total area exists under the original function. Thusly, the value of a is the area underneath the original and model wave. Now, a n = 2 T f(x) cos 2πnx dx T T a = 2 = 2 f(x) cos 2π()x f(x) cos(2πx) dx Again, we must use the left-endpoint method to find the area under the product curve dx.5.5 Page 27 of 35

28 Page 28 of Figure : The original function (top left), the cosine (top right), and their product (bottom).

29 2 2.5 of f(x) cos 2πx. a = 2(f(x) cos(2π()) + f(.) cos(2π(.)) + f(.2) cos(2π(.2))+ + f(.9) cos(2π(.9))) x = 2( )(.) = We can store this information in a table. a a 2 a 3 a 4 a 5 x f(x) f(x) cos ωx f(x) cos 2ωx f(x) cos 3ωx f(x) cos 4ωx f(x) cos 5ωx Total area 5.2 a n Then we proceed to find a 2, a 3, a 4, and a 5 by summing the products of f(x) cos 2πnx (of their respective frequencies) and multiplying by x and 2/T. Then, store this information into a table. Page 29 of 35

30 2 2.5 a a 2 a 3 a 4 a 5 x f(x) f(x) cos ωx f(x) cos 2ωx f(x) cos 3ωx f(x) cos 4ωx f(x) cos 5ωx Total area a n We can find b in a similar manner. b n = 2 T b = 2 b = 2 T f(x) sin 2πnx T f(x) sin 2π()x f(x) sin 2πx dx dx dx 2(f() sin(2π()) + f(.) sin(2π(.)) + f(.2) sin(2π(.2)) + = f(.9) sin(2π(.9)))(.) The product of f(x) and sin 2πx is demonstrated graphically in Figure 2. Discrete values from the product function are summed and then multiplied by 2/T and x to yield b. Page 3 of 35

31 Page 3 of Figure 2: The original function (top left), the sine function (top right), and their product (bottom).

32 2 2 Then this information is placed in a table where the values of b 2, b 3, b 4, and b 5 respectively placed as well. are.5.5 b b 2 b 3 b 4 b 5 x f(x) f(x) sin ωx f(x) sin 2ωx f(x) sin 3ωx f(x) sin 4ωx f(x) sin 5ωx Total area b n Now, we can list the coefficients that were found discretely. a = 8. a = 3.22 b = 2.38 a 2 = 2.2 b 2 = 4.23 a 3 =.988 b 3 = a 4 =.22 b 4 =.4 a 5 =.2 b 5 =. Page 32 of 35

33 Figure 3: The original function (left) and our model with ten terms (right). And, using these numbers, we can write f(x) a + a cos( 2πx) + b sin( 2πx + a 2 cos(2 2πx) + b 2 sin(2 2πx) a 5 cos(5 2πx) + b 5 sin(5 2πx), = cos(2πx) sin(2πx) cos(4πx) sin(4πx) cos(6πx) sin(6πx).22 cos(8πx) +.4 sin(8πx).2 cos(πx) + sin(πx). Now, we can compare the original and modeled functions in Figure Interpreting Sound: An Application Hearing is a complicated process that combines the structure of the ear with our newly understood Fourier coefficients. Sound waves are differences in air pressure through a period of time. The waves enter the ear through the ear canal and are channeled through Page 33 of 35

34 Figure 4: Three simple harmonic functions combine to make a noise that enters the ear. As the vibrations cause waves in the fluid of the cochlea, each cilia picks up the amplitude of a specific frequency and reports to the brain. the outer ear to the ear drum. The ear drum s movements vibrate the three tiny bones that connect the ear drum to the cochlea. The cochlea is a spiral shaped sensory organ that contains fluid and cilia, the little hair-like structures attached to nerve cells. While the ear drum is causing the connecting bones to move, those bones cause the fluid to oscillate. The cilia move with the fluid and send information to the nerve cells which pass it along to neurons, and finally the brain received the information. We must think of a noise as a combination of sinusoidal functions. These functions vary in frequency and in amplitude. Each cilia has an assigned frequency that causes it to respond to the fluid s movement. The higher frequencies are located at the outer end of the cochlea while the lower frequencies are located towards the inner part. The amount the cilia vibrates depends on the amplitude from the wave of that specific frequency. The information that is reported to the brain is the amplitude. Page 34 of 35

35 2 2 If we had a graph of a sound wave, we could use the Discrete Fourier Expansion to approximate the coefficients of the Fourier series. Then, by adding enough terms from the expansion, we could approximate the sound wave..5.5 References [] Anton, Howard and Chris Rorres, Elementary Linear Algebra: Applications Version, John Wiley & Sons, New York, 2, p [2] Ramirez, Robert W., The FFT: Fundamentals and Concepts, Prentice Hall PTR, New York, 985, p.23 [3] Tolstov, Georgi P., Fourier Series, Dover Publications, New York, 962, p.-4 [4] Transnational College of LEX, Who is Fourier?: A Mathematical Adventure, Language Research Foundation, Boston, 995 Page 35 of 35

Weighted SS(E) = w 2 1( y 1 y1) y n. w n. Wy W y 2 = Wy WAx 2. WAx = Wy. (WA) T WAx = (WA) T b

Weighted SS(E) = w 2 1( y 1 y1) y n. w n. Wy W y 2 = Wy WAx 2. WAx = Wy. (WA) T WAx = (WA) T b 6.8 - Applications of Inner Product Spaces Weighted Least-Squares Sometimes you want to get a least-squares solution to a problem where some of the data points are less reliable than others. In this case,