Fourier Series Expansion

Transcription

1 Fourier Series Expansion Deepesh K P There are many types of series expansions for functions. The Maclaurin series, Taylor series, Laurent series are some such expansions. But these expansions become valid under certain strong assumptions on the functions those assumptions ensure convergence of the series). Fourier series also express a function as a series and the conditions required are fairly good and suitable when we deal with signals. Suppose f is a real valued function from R to R. In this note, we deal with the following three questions: When does f has a Fourier series expansion? How we find the expansion? What are the main properties of this expansion? Existance of a Fourier series expansion: There are three conditions which guarantees the existance of a valid Fourier series expansion for a given function. These conditions are collectively called the Dirichlet conditions:. f is a periodic function on R. This means that there exists a period T such that fx) = fx + T ) for all x R.. f has only a finite number of maxima and minima in a period. 3. f has atmost a finite number of discontinuous points inside a period.. f is integrable over the period of the function. It should be noted that the second and third conditions are satisfied by many real valued functions that we deal with, inside any finite interval. But periodicity is a condition that is satisfied by very few functions, for example, constant function, sine, cos, tan and their combinations. But we can consider any function defined on a finite interval [a, b] or a, b)) as a periodic function on R by thinking that the function is extended to R by repeating the values in [a, b] to the remaining part of R. Thus Figure Most of the functions, that we commonly use, defined on finite intervals can be expanded as Fourier series

2 Derivation of Fourier series expansion of a function defined in [, π]: In Fourier series expansion, we would like to write the function as a series in sine and cosine terms in the form: fx) = a + a n cos nx + b n sin nx For finding the above unknown co-efficients a, a n and b n in the Fourier series expansion of a function, one need to recall the value of certain integrals: sin mx dx = for any integer m. cos mx dx = for any integer m. sin mxcos nx dx = for any integers m and n. sin mxsin nx dx = for integers m n. cos mxcos nx dx = for integers m n. sin mxsin nx dx = π when the integers m = n. cos mxcos nx dx = π when the integers m = n. [All the above integrals easily follow by evaluating using integration by parts] Now suppose fx) = a + a j cos jx + b j sin jx. To find a : Observe that = a j= dx + ) a j cos jx dx + b j sin jx dx j= = a π + + ) j= This implies that a = π

3 To find a n : Observe that fx)cos nx dx = a cos nx dx + = a + a n π + ) a j cos nx cos jx dx + b j cos nx sin jx dx j= b j j= This implies that a n = π fx) cos nx dx To find b n : Observe that fx)sin nx dx = a This implies that Thus sin nx dx + ) a j sin nx cos jx dx + b j sin nx sin jx dx j= = a + a j + b n π. j= b n = π fx) sin nx dx fx) = a + a n cos nx + b n sin nx, a = π a n = π b n = π π π fx) cos nx dx fx) sin nx dx [This expansion is valid at all those points x, fx) is continuous.] Note: Note that the above mentioned results hold when we take any π length intervals [This is because +π c sin mx dx =,... are true for any c]. Result: So whenever we take a function f defined from [c, c + π] any interval of length π) to R, satisfying the Dirichlet conditions, we have 3

4 fx) = a + a n cos nx + b n sin nx, a = π a n = π b n = π +π c c+π c+π c fx) cos nx dx fx) sin nx dx 3 Derivation of Fourier series expansion of a function defined in an arbitrary period [a, b]: Now suppose that fx) is defined in an arbitrary interval [a, b] and satisfy the Dirichlet conditions. Let us take b a = l, half the length of the interval. Now define the new variable z = π l x. By this simple transformation, we can convert functions on any finite interval say, [a, b]) to functions in the new variable z, whose domain is an interval of π length. This is because x = a z = π l a and x = b z = π l b = π b a b a + a) = π + π l a. Thus when the variable x in fx) moves from a to b, the new variable z in the new function F z) which is the same function f in the new variable) moves from c to c+π, c = π l a. Hence the Fourier series expansion is applicable for F z). Thus fx) = F z) = a + a n cos nz + b n sin nz, a = π a n = π b n = π +π c +π c +π c F z) dz F z) cos nz dz F z) sin nz dz and changing back to the original variable x note that dz = π l dx), we have

5 fx) = a + a n cos nπ l x + b n sin nπ l x, a = l a n = l b n = l b a b a b a fx) cos nπ l x dx fx) sin nπ l x dx, which is the general form of Fourier series expansion for functions on any finite interval. Also note that this is applicable to the first case of our discussion, we need to take a =, b = π, l = π and then everything becomes the same as in the previous section. 3. Illustration We now take a simple problem to demonstrate the evaluation of Fourier series. Consider the function f defined by if x, fx) = x if < x <,, if x. We shall find the Fourier series expansion of this function. Here, note that the length of the interval is. So l = and l =. We need to write fx) = a + a n cos nπ x + b n sin nπ x, Now a n = b n = a = a = fx) cos nπ x dx fx) sin nπ x dx, dx + x dx + = + + ) = dx) a n = cos nπ x dx + x cos nπ x dx + cos nπ x dx) 5

6 = cos nπ x dx + xcos nπ x dx + cos nπ x dx) = nπ sin nπ sin nπ) + + nπ sin nπ sin nπ )) = b n = sin nπ x dx + x sin nπ x dx + sin nπ x dx) = { nπ cosnπ ) cosnπ)) + [ nπ cosnπ ) + n π sinnπ = 8 nπ cosnπ ) nπ cosnπ) + n π sinnπ ). So when n = b = π, n = b = 9 π,... Thus the Fourier expansion of fx) is fx) = + cos π x + π sin π x + cos π x + 9 π sin π x +... = π sin π x + 9 π sin π x +..., which is valid at all points in [, ] except at and, since the function is continuous at all points except and. When x =, the sum of the series will be equal to the value + = 5.5 and at x =, it is + = Some special cases: Suppose the function is an odd/even function in a symmetric interval [ c, c]. That is f x) = fx) for all x R [ Even] or f x) = fx) for all x R [ Odd]. Then from the results, c c = = when fx) is odd we have some specialities in the expansions. 3.. Even functions: when fx) is even, Suppose fx) is even in a domain [ c, c]. Then it can be observed that b n = and so the Fourier series becomes fx) = a + a n cos nπ c x a n = c a = c fx) cos nπ c x dx ) ] nπ cosnπ) cosnπ )} 6

7 3.. Odd functions: Similarly when fx) is odd in a domain [ c, c]. Then a = a n = and the Fourier series becomes fx) = b n sin nπ c x b n = fx) sin nπ c c x dx Note: If you observe carefully, in the illustration problem, the function is actually odd and the domain is [, ]. Note that the Fourier series contains only sine terms. But the converse is not true in general. That is, even if the Fourier series contains only sine terms, the function may not be odd! [For proving the above two cases, one should recall that the product of an odd and an even function is always odd and when both functions are even or both odd then the product is always even.] 3.3 Evaluation of series Fourier series can be used for evaluating the sum of certain numerical series related with it. For each value of fx ), x is a continuous point of the function, we get a series by putting the value x on both sides of the function. Care should be given if we substitute a discontinuous point as then instead of fx ), we should use the average value of the left and right limit at x 3.3. Illustration Suppose fx) = x, < x < π. We shall find the Fourier series of this function and use it to evaluate the sum of the series Observe that the function is an even function in the symmetric interval, π). So we immediatly get b n = for all n. So we need to find only a o, &a n. a = x dx = 3 π a n = x cos nx dx = π π Thus the Fourier series becomes x cos nx dx = ) n n x = π 3 +. cos x +. cos x +. cos 3x Now to find the sum of the given series, take x =, which is a continuous point. Substituting on both sides = π ) 3..., which gives = π. 7

8 Fourier Sine Series and Cosine Series: Sometimes we may need to expand a given function as a series in only cosine terms or only sine terms. Such series are called cosine series and sine series respectively. This can be done easily when the given function is defined on an interval like [, c] that is a positive side interval) or [ c, ] negative side interval). Hence such series are also reffered as Half range series expansion. Suppose fx) is defined in the interval [, c]. Then one can think of the extended function as either even or odd depending on the need; i.e., cosine series even and sine series odd) into the interval [ c, c]. Extension as even function - Cosine series: Suppose we think of fx) as even in the interval [ c, c]. Now expand this function. Obviously the series you obtain will have only constant and cosine terms. This is the cosine series of the function fx). That is fx) = a + a n cos nπ c x a = c a n = fx) cos nπ c c x dx This type of a series is valid only in the domain [, c] as the function on the other side was just an extension not a part of the original function). Extension as odd function - Sine series: Similarly when fx) is thought as an extended odd function in a domain [ c, c]. Then the Fourier series becomes fx) = b n sin nπ c x b n = c fx) sin nπ c x dx Note: For doing problems one need not do the extension to [ c, c]. Just applying the formula is fine. Note: Thus a function defined in [, c] will have three types of expansions: a Fourier series expansion a Cosine series expansion a Sine series expansion. 8

9 . Illustration Consider the function fx) = x, < x < π. We shall find the Fourier cosine series of this function. First, considering as an even function in, π), we get the Fourier cosine series as fx) = a + a n cos nx Note that l = π here. Let us find a and a n. a = π = π π x dx = π. a n = x cos nx dx π = [ ) π n ] n Thus the Fourier cosine series is given by fx) = π [ ] cos 3x cos 5x cos x + π In a similar way, one can apply the formula to find the Fourier sine series of the function. Note that the same function also has got a Fourier series expansion, you need to apply the formula by taking l = π. 5 Practical Harmonic Analysis In many practical problems, we get the data at discrete points of time. That is, the function will not be given explicitly but the function values will be given at, say, regular intervals of the domain. For example, π π x 3 π π fx) Here the function is actually sin x, but given at discrete points,, π, π, 3 π, π. With these point values, how to calculate the Fourier series? 5. The Trapezoidal Rule of Integration This is a rule used for finding the definite integral of a function, which is given at equally spaced, discrete points. Suppose the values of a function are given as in x x x x... x n x n fx) y y y... y n y n were the points x, x,..., x n are equally spaced; i.e., x i+ x i = h, a constant, for all i. Then the Trapezoidal rule says that xn x = h [y + y + y y n ) + y n ] 9

10 5.. Example Calculate fx)dx fx) is given at x: fx): Solution: Note that h =.5 here. Now from trapezoidal rule, fx)dx =.5 + [ ] + 68) = Calculating the Fourier series from a discrete data Suppose that the values of a function are given as in x x x x... x n x n fx) y y y... y n y n Here the length of the interval is x n x and so l = xn x. The spacing between the points is h = x i+ x i. The Fourier series of the function is fx) = a + a n cos nπ l x + b n sin nπ l x, a = l a n = l b n = l xn x xn x xn x fx) cos nπ l x dx fx) sin nπ l x dx, Since fx) is not explicitly given here, we need to use the Trapezoidal rule to find a, a n & b n : a n = l = h l xn a = l x = h l [y + y + y y n ) + y n ] xn fx) cos nπ l x dx b n = l = h l x [ y cos nπ xn x l x + fx) sin nπ l x dx [ y sin nπ l x + In problems, you can not find all the a n s. substitute in the Fourier series formula. y cos nπ l x + y cos nπ l x y n cos nπ ) l x n + y n cos nπ ] l x n y sin nπ l x + y sin nπ l x y n sin nπ ) l x n + y n sin nπ ] l x n So find a, a, a, b, b and

11 5.. Illustration Find the Fourier series of the function from the data given below: π π x 3 π π fx) 3 5 Here the interval under consideration is, π), which is of length π. Hence l = π and h = π. The Fourier series of the function is fx) = a + a n cos nx + b n sin nx, a = π a n = π b n = π π π fx) cos nx dx fx) sin nx dx, Now we calculate a, a, a, b, b using the Trapezoidal rule. a = π = { π } π 8 [ ) + 5] = 6 a = π = π { π 8 fx)cos x dx [ cos. + cos. π ) ]} + 3 cos.π + cos.3π + 5 cos π = [ ) + 5.] = a = π = π { π 8 fx)cos x dx [ cos. + cos. π ) ]} + 3 cos.π + cos.3π + 5 cos π = [ ) + 5.] = b = π = π { π 8 fx)sin x dx [ sin. + sin. π ) ]} + 3 sin.π + sin.3π + 5 sin π = [ ) + 5.] = b = π = π { π 8 fx)sin x dx [ sin. + sin. π ) ]} + 3 sin.π + sin.3π + 5 sin π = [ ) + 5.] =

12 Thus the Fourier series expansion of the function is fx) = +.cos x +.cos x sin x +.sin x Problems. Obtain the Fourier series representation of fx) = π x), < x < π. [Hint: l = π, a = π 6, a n = n, b n = ]. Expand fx) = x sin x, x π as a Fourier series. [Hint: Even, l = π, b n =, a =, a n = n, if n, a = ]