Fourier transformation with finite integration limits. Sufficient conditions for the representability of the Fourier integral function

Which already fed up. And I feel that the moment has come when it is time to extract new canned food theory from strategic reserves. Is it possible to decompose the function in a row any different way? For example, express a cut of a straight line through sinuses and cosines? It seems incredible, but such, it would seem distant from each other.
"Reunion". In addition to the dyed degrees in theory and practice, there are other approaches to the decomposition of a function in a row.

At this lesson, we will get acquainted with the trigonometric Near Fourier, taking the question of its convergence and sums and, of course, we will analyze numerous examples on the decomposition of functions in a Fourier series. I sincerely, I wanted to call the article "Fourier Rows for Doodles", but it would be a lucavism, because to solve problems, the knowledge of other sections of mathematical analysis and some practical experience will be required. Therefore, the preamble will resemble the training of astronauts \u003d)

First, the study of the materials of the page should be approached in excellent form. I survived, rested and sober. Without strong emotions about the broken hamster's paws and obsessive thoughts about the life of aquarium fish. Fourier series is not complicated from the point of view of understanding, however, practical tasks require simply increased concentration of attention - ideally should be completely deducted from external stimuli. The situation is aggravated by the fact that there is no easiest way to verify the solution and answer. Thus, if your state of health is below average, then it is better to do something easier. Truth.

Secondly, before flying into space, it is necessary to study the dashboard of the spacecraft. Let's start with the values \u200b\u200bof the functions that should be closed on the machine:

With any natural meaning:

one) . And in fact, the sinusoid "stitches" the abscissa axis through each "PI":
. In the case of negative values \u200b\u200bof the argument, the result, of course, will be the same :.

2). But it was not everyone knew. Cosine "PI En" is the equivalent of "flashers":

Negative business argument does not change: .

Perhaps enough.

And, thirdly, dear cosmonaut detachment, you need to be able to ... integrate.
In particular, confidently make a function under the sign of differential, integrate in parts and be in freaks with newton Labeau Formula. Let's start important assumed exercises. I categorically I do not recommend passing, so that it does not flatter in weightlessness:

Example 1.

Calculate certain integrals

where receives natural values.

Decision: Integration is carried out according to the "X" variable and at this stage the discrete variable "En" is considered a constant. In all integrals sweep the function under the sign of the differential:

A short version of the solution to which it would be nice to shoot, looks like this:

Get used to:

Four remaining item alone. Try to conscientiously treat the task and arrange the integrals in a short way. Sample solutions at the end of the lesson.

After high-quality exercise exercises, we put on the spacesuit
and get ready for the start!

Decomposition of a function in a row of Fourier on the interval

Consider some function that defined At least in the interval (A, perhaps, at the greater interval). If this function is integrated on the segment, it can be decomposed into trigonometric fourier row:
where - the so-called fourier coefficients.

With the number called decomposition period, and number - semi-period decomposition.

Obviously, in the general case, the Fourier series consists of sinuses and cosine:

Indeed, we are writing it in detail:

The zero member of the series is customary in the form.

Fourier coefficients are calculated according to the following formulas:

I understand perfectly well that new terms are lowered beginners to study the topic: period of decomposition, semiphery, fourier coefficients et al. without panic, it is not comparable to excitement before entering outdoor space. We will understand everything in the near example, before performing which it is logical to ask the urgent practical issues:

What needs to be done in the following tasks?

Dispatch the function in a row of Fourier. Additionally, it is often necessary to portray the graph of the function, the chart of the sum of the row, partial amount and in the case of sophisticated professorship fantasies - to do something else.

How to decompose the function in a row of Fourier?

Essentially need to find fourier coefficients , that is, make up and calculate three certain integral.

Please rewrite the general view of a Fourier series and three working formulas to yourself in a notebook. I am very glad that some visitors of the site right on my eyes the children's dream is carried out to become an astronaut \u003d)

Example 2.

Dispatch the function in the Fourier series on the interval. Build a schedule, graph sum of a number and partial amount.

Decision: The first part of the task is to decompose the function in the Fourier series.

Starting start, be sure to write down that:

In this problem, the period of decomposition, half-period.

Spread the function in Fourier series at the interval:

Using the appropriate formulas, we will find fourier coefficients. Now you need to make up and calculate three certain integral. For convenience, I will numbered items:

1) The first integral is the easiest, however, and it requires an eye yes eye:

2) We use the second formula:

This integral is well sign and he is taken in parts:

When used, used method of summing up a function under the sign of differential.

In this task, it is more handy to immediately use integration Integration Formula in a specific integral :

A pair of technical comments. First, after the application of the formula all expression needs to enter into large bracketsSince the constant is located in front of the original integral. Do not lose it! Brackets can be revealed at any further step, I did it in the very least. In the first "piece" We show extreme accuracy in the substitution, as you can see, the constant is not in cases, and the integration limits are substituted into the work. This action is highlighted by square brackets. Well, the integral of the second "piece" of the formula is well familiar with the training task ;-)

And the most important thing is the maximum concentration of attention!

3) We are looking for a third Fourier coefficient:

Received a relative of the previous integral, which too integrates in parts:

This instance is slightly complicated, comment on further actions step by step:

(1) Expression fully conclude in large brackets. I did not want to seem boring, too often lose the constant.

(2) In this case, I immediately revealed these large brackets. Special attention We pay the first "piece": the constant smokes on the sidelines and does not participate in the substitution of the integration limits (and) into the work. Due to the clusterness of the recording, this action is again advisable to highlight with square brackets. With the second "piece" Everything is easier: here the fraction appeared after the disclosure of large brackets, and the constant - as a result of the integration of the familiar integral ;-)

(3) In square brackets, we carry out conversion, and in the right integral - substitution of the integration limits.

(4) We carry out "flasher" from square brackets:, after which we reveal the internal brackets :.

(5) mutually suitable 1 and -1 in brackets and conduct final simplifications.

Finally, all three Fourier coefficients were found:

Substitute them in the formula :

At the same time, do not forget to divide in half. At the last step of the constant ("minus two"), independent of "En", is carried out beyond the amount.

Thus, we received a decomposition of a function in a row of Fourier on the interval:

Let's study the issue of the convergence of the Fourier series. I will explain the theory, in particular theorem Dirichlet., literally "on your fingers", so if you need strict wording, please refer to the textbook on mathematical analysis (for example, the 2nd Tom Buchana; or the 3rd volume of Fihtendulz, but it is harder in it).

In the second part of the task you need to portray the schedule, the number of the sum of the range and the partial sum graph.

The graph of the function is the usual straight on the planewhich was carried out by a black dotted line:

We deal with the sum of the row. As you know, the functional series converge to functions. In our case, a built Fourier series with any meaning "X" It will take place to the function that is depicted in red. This feature is tolerate rasp 1 At points, but defined in them (red dots in the drawing)

In this way: . It is easy to see what is noticeably different from the source function, which is why in the record The "Tilda" icon is set, not a sign of equality.

We study the algorithm, according to which it is convenient to build the sum of the row.

At the central interval, the Fourier series converges to the function itself (the central red cut coincides with the black dotted line of the linear function).

Now we ripen about the nature of the trigonometric decomposition under consideration. In a row Fourier only periodic functions (constant, sinuses and cosine), so the sum of the row also represents a periodic function.

What does this mean in our particular example? And this indicates that the sum of the row – certainly periodic And the red interval cut is obliged to repeatedly repeated on the left and right.

I think now I finally cleared the meaning of the phrase "period of decomposition". Simplistic speaking, through every situation again and again repeats.

In practice, it is usually enough to portray three decomposition periods, as is done in the drawing. Well, more "hardware" of the neighboring periods - to understand that the schedule continues.

Of particular interest are point of gap of the 1st kind. At such points, the Fourier row converges to isolated values \u200b\u200bthat are located Rasnohonko in the middle of the "jump" of the break (red dots in the drawing). How to learn the ordinate of these points? First, find the ordinate "top floor": for this we calculate the value of the function at the extreme right point of the central decomposition period :. To calculate the "lower floor" ordinate, the easiest way to take the leftmost value of the same period is the easiest: . The ordinate of the average value is the arithmetic amount of the "Verkh and Niza" :. Pleasant is the fact that when building a drawing, you will immediately see, correctly or incorrectly calculated the middle.

We construct the partial sum of the row and at the same time we repeat the meaning of the term "convergence". Motive is known from the lesson about sum of numerical row. Sick our wealth detail:

To compile a partial amount, it is necessary to record zero + two more members of the row. I.e,

In the drawing, the graph of the function is depicted by green color, and, as you can see, he grudges tightly "wraps" the full amount. If you consider a partial amount out of five members of a series, then the graph of this feature will be even more accurate to bring red lines, if one hundred members - then the "green sniped" actually completely populates with red segments, etc. Thus, the Fourier series converges to its amount.

It is interesting to note that any partial amount is continuous functionHowever, the total amount of the row is still breaking.

In practice, it is not so rarely required to build a partial sum graph. How to do it? In our case, it is necessary to consider the function on the segment, calculate its values \u200b\u200bat the ends of the segment and at intermediate points (the more points consider - the more accurate the schedule). Then, you should mark these points in the drawing and gently depict the schedule on the period, after which it is "rubbing" it to adjacent intervals. How else? After all, the approach is also a periodic function ... ... something of her graphic reminds me of even heart rhythm on the displays of the medical device.

It is not very convenient to perform the construction, as it is necessary to exercise supercharge, withstanding accuracy is no less than half a millimeter. However, readers who are not in freaks with drawing, I will do - in the "real" task to do a drawing, it is not always possible, somewhere in 50% of cases it is necessary to decompose the function in the Fourier series and that's it.

After completing the drawing, we complete the task:

Answer:

In many tasks, the function tolerates gap Right on the period of decomposition:

Example 3.

Scroll into a Fourier series a function specified on the segment. Draw a graph of the function and the total amount of the row.

The proposed function is set piecewise (and notice, only on the segment) And tolerate gap At point. Is it possible to calculate Fourier coefficients? No problem. And the left and right parts of the function are integrable at their intervals, so the integrals in each of the three formulas should be represented as the sum of two integrals. Let us see, for example, how this is done at the zero coefficient:

The second integral turned out to be equal to zero, which was lost operation, but it happens not always.

Similarly, two other Fourier coefficients are described.

How to portray the sum of the row? On the left interval, the segment of the straight line, and on the interval - the straight line (boldly weighing the axis section). That is, in the interval of the decomposition, the sum of the row coincides with the function everywhere, apart from three "bad" points. At the break point of the function, the Fourier row will take place to an isolated value that is located exactly in the middle "jump" of the break. It is easy to see and orally: Left-sided limit:, right-sided limit: And, it is obvious that the ordinate of the midpoint is 0.5.

Due to the frequency of the amount, the picture must be "propagated" to neighboring periods, in particular, to portray the same at intervals and. At the same time, at points, the Fourier row will come down to the median values.

In fact, there is nothing new here.

Try yourself to cope with this task. Exemplary sample design and drawing at the end of the lesson.

Decomposition of the function in a row of Fourier on an arbitrary period

For an arbitrary period of decomposition, where "El" - any positive number, formulas of a series of Fourier and Fourier coefficients are distinguished by a slightly complicated argument of sine and cosine:

If, the gap formulas are obtained from which we started.

The algorithm and principles of solving the problem are fully preserved, but the technical complexity of calculations increases:

Example 4.

Eliminate the function in a Fourier series and build a summary.

Decision: actually analogue of Example No. 3 with gap At point. In this problem, the period of decomposition, half-period. The function is defined only on the semi-interval, but this does not change the case - it is important that both pieces of functions are integrable.

Spread the function in Fourier series:

Since the function is breaking at the beginning of the coordinates, then each Fourier coefficient is obviously written in the form of the sum of two integrals:

1) the first integral will write as much as possible:

2) carefully peer into the surface of the moon:

Second integral take in parts:

What should be paid attention to, after we ask the stars to open the continuation of the solution?

First, do not lose the first integral where immediately perform summing up a differential sign. Secondly, do not forget the ill-fated constant to large brackets and do not confuse in signs When using formula . Large brackets, after all, it is more convenient to reveal immediately at the next step.

The rest of the technology, difficulties can only cause the lack of experience with solutions.

Yes, no wonder the famous colleagues of French Mathematics Fourier was indignant - how did this dare spread the functions into trigonometric rows?! \u003d) For the word, probably everyone is interesting practical meaning The task under consideration. Fourier himself worked on the mathematical model of thermal conductivity, and subsequently a number called by his name began to be used to study many periodic processes, which in the surrounding world apparently invisible. By the way, by the way, I caught myself thinking that it did not accidentally compare the graph of the second example with the periodic rhythm of the heart. Those who wish can familiarize themselves with practical application fourier transform In third-party sources. ... although it is better not necessary - I will remember how first love \u003d)

3) Given the repeatedly mentioned weak links, we deal with the third coefficient:

We integrate in parts:

Substitut the found Fourier coefficients in the formula , not forgetting to divide the zero coefficient in half:

Build a schedule of the sum of a number. Briefly repeat the procedure: on the interval, we build a straight line, and on the interval is straight. At the zero value of "X", we put the point in the middle of the "jump" of the gap and "reproach" the schedule for neighboring periods:


At the "joints" of periods, the amount will also be equal to the middle of the "jump" of the gap.

Ready. I remind you that the function itself is determined by the condition only on the half-interval and, obviously, coincides with the sum of the range at the intervals

Answer:

Sometimes a piecewise specified function is continuous on the period of decomposition. The simplest sample: . Decision (see the 2nd Tom Buchana) The same as the two previous examples: despite continuity function At the point, each Fourier coefficient is expressed by the sum of two integrals.

On the expansion interval growing points of the 1st kind and / or "joint" points of the schedule can be more (two, three and in general any finite quantity). If the function is integrated at each part, it is also decomposed in a Fourier series. But from the practical experience, such a story I do not remember something. Nevertheless, there are more difficult tasks than just considered, and at the end of the article for everyone there are references to the Fourier series of elevated complexity.

In the meantime, we will relax, leaning in the chairs and contemplating the endless starry spaces:

Example 5.

Dispatch the function in a Fourier series on the interval and build a schedule of the sum of the row.

In this task, the function continuous On the semi-interval of decomposition, which simplifies the solution. Everything is very similar to example No. 2. From the spacecraft, it is not anywhere - it will have to decide \u003d) an exemplary sample of the design at the end of the lesson, the schedule is attached.

Decomposition in a series of Fourier for read and odd functions

With even and odd functions, the process of solving the problem is simplified noticeably. And that's why. Let's return to the decomposition of the function in a series of Fourier on the period "Two PI" and arbitrary period "Two El" .

Suppose our functions are black. The general member of the series, as you see, contains koshinuses and odd sines. And if we lay an even function, then why do we need odd sines?! Let's reset the unnecessary coefficient :.

In this way, an easy function unfolds in a row of Fourier only by cosine:

Insofar as integrals from what functionsaccording to a symmetrical relative zero, the segment of the integration can be doubled, the other Fourier coefficients are simplified.

For gap:

For an arbitrary gap:

To the shittomatic examples that are practically in any textbook on matanaliz include decomposition of read functions. . In addition, they have repeatedly met in my personal practice:

Example 6.

Dana feature. Requires:

1) decompose the function in a Fourier series with a period where - an arbitrary positive number;

2) Record decomposition on the interval, build a function and graph of the total amount of the row.

Decision: In the first paragraph it is proposed to solve the problem in general, and it is very convenient! Implementation will appear - just substitute your value.

1) In this problem, the period of decomposition, half-period. In the course of further actions, in particular, when integrating, El "is considered a constant

The function is even, and therefore, it is laid out in a Fourier series only by cosine: .

Fourier coefficients are looking for formulas . Pay attention to their unconditional advantages. First, the integration is carried out on a positive segment of the decomposition, and therefore we are safely getting rid of the module , examining from two pieces only "X". And, secondly, integration is noticeably simplified.

Two:

We integrate in parts:

In this way:
At the same time, the constant, which does not depend on the "En", we take out the amount.

Answer:

2) We write decomposition on the interval, for this, in the general formula we substitute the desired value of the half-period:

I. Fourier transformations.

Definition 1. Function

Called fourier transformation Functions.

The integral here is understood in the sense of main importance

And it is believed that it exists.

If - the function is absolutely integrated, since When, for any such function, the Fourier transformation (1) makes sense, and the integral (1) converges absolutely and evenly on the whole line.

Definition 2.. If a - Fourier transformation function
, then compaled integral

Understood in the sense of main importance is called fourier integral function .

Example 1. Find Fourier Conversion Functions

The specified function is absolutely integrable on, indeed,

Definition 3. Understood in the sense of the main importance integral

Called respectively cosine- and fourier sinus transformations .

Believed , , , get partly familiar to us by the ranks of Fourier

As can be seen from relations (3), (4),

Formulas (5), (6) show that Fourier transforms are fully determined throughout the direct, if known only for non-negative values \u200b\u200bof the argument.

Example 2. Find cosine - and sinus - Fourier transform function

As shown in Example 1, the specified function is absolutely integrated on.

We will find its cosine - Fourier transformation by formula (3):

Similarly, it is not difficult to find sinus - Fourier transformation function f.(x.) By formula (4):

Using examples 1 and 2, it is not difficult to make sure that for f.(x.) Ratio (5) is performed.

If the function is really recognized, then from formulas (5), (6) in this case should

Since in this case, the real functions on R, which can be seen from their definitions (3), (4). However, equality (7) provided It turns out and directly from the definition (1) of the Fourier transformation, if we consider that the conjugation sign can be made under the sign of the integral. Recent observation allows you to conclude that equality is fair for any feature.

It is also useful to note that if it is a real and even function, i.e. T.

if it is a real and odd function, i.e. T.

And if it is purely imaginary function, i.e. . T.

Note that if it is a real-valued function, then the Fourier integral can also be written as

Where

Example 3.
(counting )

Since we know the value of the Dirichlet integral

The function considered in the example is not absolutely integrable on and its Fourier transformation has breaks. That the Fourier transformation is absolutely integrable functions does not have breaks, says the following

Lemma 1. If the function locally integrable and absolutely integrable on T.

a) her Fourier transformation defined with any meaning

b)

Recall that if - real or complex-valued function defined in the open set that function called locally integrable by if anyone point It has a neighborhood in which the function is integrated. In particular, if, the condition of local integrability of the function is obviously equivalent to the fact that For any segment.

Example 4.Find the Fourier transformation function :

Differentiating the last integral by parameter and integrating then in parts, we find that

or

It means , where is the constant, which, using the Euler-Poisson integral, we find from the ratio

So, we found that, and at the same time showed that, and .

Definition 4. It is said that the function defined in the punctured neighborhood of the point satisfies Dini at the point if

a) At the point there are both one-sided limits

b) both integral

converge absolutely.

Absolute convergence integral means the absolute convergence of the integral at least with some value.

Sufficient conditions Representability of the Fourier integral function.

Theorem 1.If absolutely integrated on and locally piecewise continuous function satisfies at point dini's conditions, then its Fourier integral converges at this point, and to the meaning

equal to the left and right limits of the function values \u200b\u200bat this point.

Corollary 1.If the function continuous, has at each point finite one-sided derivatives and absolutely integrated on then it seems to with its integral Fourier

where fourier transformation function .

The function of the Fourier integral function can be rewritten in the form:

Comment. Formulated in Theorem 1 and Conduction 1 The conditions for the function are sufficient, but are not necessary for the possibility of such a submission.

Example 5. Present the Fourier integral function if

This function is odd and continuous on ℝ, except for points ,.

Due to the oddness and materiality of the function, we have:

and from equalities (5) and (10) it follows that

At the continuity points, we have:

But the function is odd, therefore

since the integral is calculated in the sense of main importance.

Function is even, so

if a , . When equality should be performed

Believing from here we find

If in the last expression for putting, then

Believing here we will find

If the function is real-valid piecewise continuous on any segment of the actual direct absolutely integrable on and has the final one-sided derivatives at each point at the continuity points of the function appears as a Fourier integral

and at the point of break points, the left part of equality (1) should be replaced by

If continuous, absolutely integrated on the function has finite one-sided derivatives at each point, then in the case when this function is even, equality is fair

and in the case when - an odd function, equality is performed

Example 5 '. Present a Fourier integral feature if:

Since it is continuous on an even function, then using formulas (13.2), (13.2 '), we have

Denote by the symbol understood in the sense of the main importance integral

Corollary 2.For any function satisfying the conditions of Corollary 1, there are all transformations , , , and have equality

Having in mind these ratios, transformation (14) is often called reverse Fourier transformationand instead they write, and equality themselves (15) are called fourier conversion formula.

Example 6.Let I.

Note that if , then with any function

Take now the function. Then

If you take a function that is an odd continuation of the function , on the whole numerical axis, then

Using theorem 1, we get that

All integrals here are understood in the sense of main importance,

Separating in the last two integrals valid and imaginary parts, find Laplace integrals

Definition . Function

let's call the normalized Fourier transformation.

Definition . If it is a normalized Fourier transformation function, then the compaled integral

We will call the normalized Fourier integral function.

We will consider the normalized Fourier transformation (16).

We introduce the following notation for convenience:

(those. ).

In comparison with the previous designations, this is just a renormalization: it means, in particular, relations (15) make it possible to conclude that

or in a shorter record

Definition 5. Operator We will be called the normalized Fourier transformation, and the operator will be called the reverse rationed Fourier transformation.

In Lemma 1, it was noted that the Fourier transformation of any absolutely integrated on function tends to infinity to zero. In the following two statements, it is stated that, like Fourier coefficients, the Fourier transformation is stronger to zero, the smoost of the function from which it is taken (in the first approval); Mutual with this fact will be that the faster the function is striving for zero, from which the Fourier transformation is taken, the harden its Fourier transformation (second approval).

Approval 1.(On the connection of the smoothness of the function and the rate of decrease in its Fourier transformation). If a and all functions absolutely integrable by , that:

but) with any

b)

Approval 2.(On the connection of the speed of descending function and smoothness of its Fourier transformation). If locally integrable function : such is the function absolutely integrable N.but , that:

but) fourier transformation function belong class

b) there is an inequality

We give the main hardware properties of Fourier transform.

Lemma 2. Suppose for functions and there is a Fourier transformation (respectively, the reverse Fourier transformation), then, what would any numbers and, there is a Fourier transformation (respectively, the Fourier reverse transformation) and for the function , and

(respectively).

This property is called the Fourier transformation linearity, (respectively, the Fourier reverse transformation).

Corollary. .

Lemma 3.Fourier transformation, as well as the reverse transformation, is a mutually unambiguous conversion on a set of continuously integrable functions on the entire axis of functions having one-sided derivatives at each point.

This means that if both are two functions of the specified type and if (respectively, if ), then on the entire axis.

From the approval of Lemma 1, you can get the following lemma.

Lemma 4.If the sequence of absolutely integrable functions and the absolutely integrable function is such that

the sequence evenly on the entire axis converges to the function.

Now we will now study the Fourier conversion of a bunch of two functions. For convenience, we see the definition of a convolution, adding an additional multiplier

Theorem 2. Let the functions and are limited, are continuous and absolutely integrated on the real axis, then

those. Fourier conversion of two functions is equal to the product of Fourier transformations of these functions.

We will make a consolidated table No. 1 of the properties of the normalized Fourier transformation, useful when solving the tasks below.

Table №1

Function	Normated Fourier transformation

Using properties 1-4 and 6, we get

Example 7.Find the normalized Fourier transformation function

In example 4, it was shown that

as if

According to this, we have:

Similarly, you can make a table number 2 for the normalized Fourier conversion:

Table # 2.

Function	Normated Fourier Reverse Transformation

As before, using properties 1-4 and 6 we get that

Example 8.Find a normalized Fourier Function Conversion

As follows from Example 6

When we have:

Presenting a function in the form

we use property 6 when

Options for setting and graphic work

1. Find Sinus - Fourier Convert Functions

2. Find sinus - Fourier transformation function

3. Find cosine - Fourier transformation function

4. Find cosine - Fourier transformation function

5. Find sinus - Fourier transformation function

6.Night kosinus - Fourier transformation function

7.Night sinus - Fourier transformation function

8. Find cosine - Fourier transformation function

9. Find cosine - Fourier transformation function

10. Find Sinus - Fourier Convert Functions

11. Find Sinus - Fourier Transformation Functions

12. Find sinus - function conversion

13. Find Sinus - Function Conversion

14. Find cosine - function conversion

15. Find cosine - function conversion

16. Find Fourier Convert Functions if:

17. Find Fourier Convert Functions if:

18. Find Fourier Convert Functions if:

19. Find Fourier Convert Functions if:

20. Find Fourier Convert Functions if:

21. Find Fourier Convert Functions if:

22. Find the normalized Fourier Function Conversion

Using the formula

24. Find the normalized Fourier Function Conversion

Using the formula

26. Find the normalized Fourier Function Conversion

Using the formula

28. Find the normalized Fourier Function Conversion

Using the formula

30. Find the normalized reverse Fourier transformation function

Using the formula

23. Find the normalized Fourier Function Conversion

Using the formula

25. Find the normalized reverse Fourier Function Conversion

Using the formula

27. Find the normalized Fourier Fourier Conversion

Using the formula

29. Find a normalized Fourier Function Conversion

Using the formula

31. Find the normalized Fourier conversion function

Using the formula

32. Submit the Fourier integral function

33. Present the Fourier integral function

34. Present the Fourier integral function

35. Present the Fourier integral function

36. Present a Fourier integral function

37. Present the Fourier integral function

38. Present Fourier integral function

39. Present the Fourier integral function

40. Present the Fourier integral function

41. Present a Fourier integral function

42. Present a Fourier integral function

43. Present a Fourier integral function by continuing it to an odd way to the interval if:

44. Present a Fourier integral function by continuing it to an odd way to the interval if.

One of the powerful means of studying the problems of mathematical physics is the method of integral transformations. Let the function f (x) be set on the interval (A, 6), finite or infinite. The integral conversion of the function f (x) is the function where K (x, w) is fixed for this conversion function called the core of the conversion (it is assumed that the integral (*) exist own or incomprehensible). §one. Fourier integral Any function f (x), which on the segment [-f, I] satisfies the conditions of decomposability in a row of Fourier, may be presented by trigonometric Number of coefficients A *, and 6 "row (1) are determined by Euler Fourier formulas : Fourier transformation Integrated Fourier Complex form of integral transformation Fourier cosine and sine conversion amplitude and phase spectra of the application properties A number in the right part of the equality (1) can be recorded in a different form. To this end, we will make it from formulas (2) the values \u200b\u200bof the coefficients A "and the OP, we will submit under the signs of COS ^ x and SIN integrals (which is possible, since the integration variable is T) O) and use the formula for the causing difference. We will have if the function / (g) was initially determined on the interval of the numerical axis, greater than the segment [-1,1] (for example, on the entire axis), then the decomposition (3) will reproduce the values \u200b\u200bof this function only on the segment [-1, 1] and will continue on the entire numerical axis as a periodic function with a period 21 (Fig. 1). Therefore, if the function f (x) (generally speaking, non-periodic) is defined on the entire numerical axis, in formula (3) you can try to go to the limit at I + OO. It is natural to require the following conditions: 1. F (x) satisfies the conditions of decomposability in a row of Fourier on any finite segment OH \\ 2. The function f (x) is absolutely integrated on the entire numeric axis, when performing condition 2, the first term of the right part of equality (3) with i - * + oo tends to zero. In fact, we will try to establish what will go to the limit at the I + OO sum in the right, part (3). It is so that then the sum in the right side (3) will take a form due to the absolute convergence of the integral this amount at large I differs little from the expression that resembles an integrated amount for the variable function £-composed for the interval (0, + oo) changes so naturally expect, As for the sum (5), it turns into the integral of the source side, with fixed) from formula (3) it follows that we obtain equality sufficient conditions for the equity of formula (7) is expressed by the following theorem. Theorem 1. If the function f (x) is absolutely integrated on the entire numeric axis, together with its derivative, a finite number of the first-order break points on any segment [A, 6], then the equality is right at the same time in any point of XQ, which is a break point 1 "The genus function / (g), the value of the integral in the right-hand side (7) is equal to the formula (7) are called the Fourier integral formula, and the integral of the Fourier integral is called in its right part. If you take advantage of the formula of the diversity of the difference, then formula (7) can be written in the form of function A (£), b (£) are the analogues of the corresponding Fourier coefficients UP 2TG-periodic function, but the latter are defined for discrete values \u200b\u200bof P, WTO A (0\u003e But definite definitive values \u200b\u200bof £ G (-OO, + oo). The complex form of the Fourier integral is assumed / (x) absolutely integrated through the entire axis oh, consider the integral this integral is evenly converged for, since therefore is a continuous and Obviously, an odd function from but then on the other hand, the integral is an even function of the variable so that therefore the integral Fourier formula can be written as: I will multiply the equality on the imaginary unit I and add to the equality (10). We will get from where the Euler formula will have This is a comprehensive form of the Fourier integral. Here, the external integration of £ is understood in the sense of the main value of Cauchy: §2. Fourier transformation. Cosine and sinus-transform Fourier let fun The F (X) is a piecewise smooth on any finite segment of the axis oh and absolutely integrable through the entire axis. Definition. The function from where, by virtue of the Euler formula, will be called the Fourier transformation of the function / (d) (spectral function). This is an integral function transformation / (d) on the interval (-o, + oo) with the kernel using the Fourier integral formula, we obtain this so-called Fourier conversion, which gives the transition from F (£) to / (x). Sometimes a direct Fourier transformation is specified like this: Then the reverse Fourier transformation is determined by the Fourier transformation formula function / (g) is also determined as follows: Fourier transformation Integral Fourier integrated form of the Fourier transformation Cosinous and sinus conversion amplitude and phase spectra of the application then, in turn, At the same time, the position of the multiplier ^ is quite arbitrary: it can enter either in formula (1 "), or in formula (2"). Example 1. Find the Fourier transformation function -4 We have this equality to make differentiation of £ under the integral sign (the integral resulting after differentiation is uniformly converges when (belongs to any final segment): Integrating in parts, we will have an off-integrated term accepted to zero, and we We get from where (C is constant integration). Believing in (4) £ \u003d 0, we find C \u003d F (0). By virtue of (3), we know that in particular, for) we obtain that Example 2 (Democked Kokdemsetor through SPO). Consider the function 4 for the spectra of the function f (£) we get from here (Fig. 2). The condition of the absolute integrity of the function f (x) on the entire numerical axis is very tough. It eliminates, for example, such elementary functions, as) \u003d \u200b\u200bcos, f (x) \u003d e1, for which the Fourier transforms (in the classical form under consideration) does not exist. Fourier-image have only those functions that quickly strive for zero at | x | - + + oo (as in Examples 1 and 2). 2.1. Fourier cosine and sinus conversion using a cosine formula, a difference that rewrite the Fourier integral formula in the following form: Let F (x) be an even function. Then, so that agems (5) have in the case of an odd F (x), in the case of the odd F (x), it is similar to when f (x) is given only to (0, -foo), then formula (6) continues F (x) to the entire axis oh evenly, and Formula (7) is odd. (7) Definition. The function is called the Fourier transformation cosine function F (X). From (6) it follows that for an even function f (x), this means that F (x), in turn, is a cosine-conversion for Fc (£). In other words, functions / and fc are mutual cosine transformations. Definition. The function is called the Fourier sinus conversion function f (x). From (7) we obtain that for an odd function f (x) i.e. F and FS are mutual sinus transformations. Example 3 (Retropular Pulse). Let f (t) be an even function defined as follows: (Fig. 3). We use the resulting result to calculate the integral by force of formula (9) we have Fig.3 0 0 at point T \u003d 0 The function f (t) is continuous and equal to one. Therefore, from (12 ") we obtain 2.2. The amplitude and phase spectra of the Fourier integral Let Periodic with the period 2t function / (x) decomposes into the Fourier series, this equality can be written in the form where - the amplitude of oscillations with the frequency of P, - phase. On this path we We come to the concepts of the amplitude and phase spectra of the periodic function. For the non-periodic function f (x) specified on (-OO, + oo), under certain conditions it turns out to be possible to present it to the Fourier integral by the expansion of this function in all frequencies (decomposition on continuous frequency spectrum ). Definition. The spectral function, or the spectral density of the Fourier integral, is called expression (the direct transformation of the Fourier function F is called an amplitude spectrum, and the function F ") \u003d -AggSFC) is a phase spectrum of the function / ("). The amplitude spectrum. And (£) serves as a measure of the frequency contribution £ into the function / (g). Example 4. Find the amplitude and phase spectra of the function 4 we find the spectral function from here the graphs of these functions are depicted in Fig. 4. §3. Fourier transform properties 1. Lineness. If G (0 is the Fourier transforms of functions / (x) and d (x), respectively, then with any permanent A and P Fourier transformation of the function AF (X) + R D (x), there will be a function A using the property of the integral linearity, we have such The Fourier transformation has a linear operator. Denoting it through we will write. If f (£) is the Fourier transformation is absolutely integrated on the entire numeric axis of the function / (g), then f (() is limited at all. Let the function f (x) absolutely Integrable on the entire axis - Fourier transformation function f (x). Then 3 "FLTSJ. Let F (X) be a function, the admission of the KNCEA transformation of Fourier, L - fusion numbers. Foundation Fh (x) \u003d f (zh) is called the Puejdi shift f (x). Using the Fourier transformation is defined, to show that the task. Let the F (Z) function of Fourier F (0\u003e H is a valid number. Show that 3. Fourier transformation and processing of differentiation. Let an absolutely integrable function f (x) has a derivative F "(x), also absolutely integrated through the entire axis Oh, so / (I) seeks zero at | f | - "+ oo. Considering f "(x) a smooth function, write integrating in parts, we will have an off-inhibitically addressed to zero (since, and we thus get the differentiation of the function / (x) corresponds to the multiplication of its Fourier image ^ n /] to the multiplier if the function f (x) has a glad * "E absolutely intended derivatives to order M inclusive and all of them, as well as the function f (x), strive for zero at that, integrating in parts the desired number of times, we get the Fourier transformation is very useful precisely because It replaces the differentiation operation of multiplying operation and thus simplifies taskintegration of certain types of differential equations. Since the Fourier transformation is absolutely integrated function F ^ k \\ x) there is a limited function from (Property 2), then from the ratio (2) we get for the following rating : Fourier transformation Integrated Fourier complex form integral Fourier transformation cosine and sine conversion amplitude and phase spectra of the application properties from this rating with Ice: The greater the function f (x) has absolutely integrable derivatives, the faster its Fourier transform tends to zero at. Comment. The condition is quite natural, since the usual 1Etheria of Fourier integrals deals with processes that in one way or another have the beginning and horses, but do not continue indefinitely with about the same intensity. 4. Communication between the decrease rate of the function f (x) at | z | - "-f oo and smoothness of its conversion Furm. Suppose that not only / (x), but also its product XF (x) is an absolutely integrable function on the entire axis oh. Then the Fourier transformation) will be differentiated by the function. Indeed, the formal differentiation by parameter £ of the integrated function leads to an integral that is absolutely and evenly converging relative to the parameter consequently, differentiation is possible, and thus, i.e., the operation of multiplication F (X) on the argument x goes after the Fourier transformation . If, together with the function f (x), the functions are absolutely integrable through the entire axis, the differentiation process can be continued. We obtain that the function has derived to order M inclusive, and thus, the faster the function f (x) decreases at the more smooth the function of Theorem 2 (about the drilling). Easually transforming Fourier functions /, (g) and f2 (x), respectively. Then, the double integral in the right part converges absolutely. Put - x. Then we will have either, changing the order of integration, the function is called a convolution of functions and is indicated by the symbol of formula (1) can be written now: it can be seen that the Fourier conversion of the functions f \\ (x) and f2 (x) is not multiplied by y / 2 The product of the Fourier transformations of the coagulated functions, remark. It is not difficult to install the following convolution properties: 1) linearity: 2) commutativity: §4. Fourier transform applications 1. Let P (^) be a linear differential operator of order M with constant coefficients using the formula for converting Fourier derivatives of functions in (x), we find "Consider the differential equation where P is the differential operator introduced above. Suppose that the sought decision (x) has a Fourier transformation (O. and the function f (x) has a transformation / (£) using the Fourier transformation to equation (1), we obtain instead of the differential algebraic equation on the axis relative to where so that formally where the symbol indicates the Fourier reverse transformation . The basic limitation of the applicability of this method is associated with the following fact. Common decision differential equation with constant coefficients contains the functions of the form ate *, EAZ COS Fix, eah sin. Px. They are not absolutely integrable on the axis -oo< х < 4-оо, и преобразование Фурье для них не определено, так что, строго говоря, применятьданный метод нельзя. Это ограничение можно обойти, если ввести в рассмотрение так называемые обобщенные функции. Однако в ряде случаев преобразование Фурье все же применимо в своей классической форме. Пример. Найти решение а = а(х, t) уравнения (а = const), при начальных условиях Это - задача о свободных колебаниях бесконечной однородной струны, когда задано начальное отклонение <р(х) точек сгруны, а начальные скорости отсутствуют. 4 Поскольку пространственная переменная х изменяется в пределах от -оо до +оо, подвергнем уравнение и начальные условия преобразованию Фурье по переменной х. Будем предполагать, что 1) функции и(х, t) и