Fourier Series

Discussion in 'Math' started by Lightfire, Feb 8, 2014.

  1. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    I'm fairly new to this topic. I have not yet studied the first topics in the Differential Equation but I decided to study this first. I don't know why. But I do hope you'll help me understand this.

    Problem: Suppose that f(x) has period 2 and that



    f(x) = 0 ; -1 ≤ x ≤ 0
    = 1 ; 0 < x < 1



    What is the value of L in that? In the video it says it's 1. But how come 1 is the value of L? What is L?

    L is from the equation

    [​IMG]

    Thank you very much and I do hope you can help me.
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    Actually, it's the half period. The period is 2L since it goes from -L to +L.

    Since the signal has a period of 2 (2 what? Months? Meters? Shetland ponies?), the value of L is 1.
     
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  3. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    You are right. L is half period.
    OP, sorry about the error.
     
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  4. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    Isn't it π?
     
  5. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Likely 2 seconds, but could me something smaller.
     
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  6. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    I have another question:

    [​IMG]

    I do understand that the indefinite integral of cos is sin so okay.

    But I am confused how nπx has a indefinite integral of nπx/nπ. I know it used u-substitution. But I am somewhat confused.
     
  7. WBahn

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    No, it's not π (though, because there are no units anywhere, you could create a unit that would result in the numerical value being π in that unit compared to it being 2 in whatever units it is actually in.

    Actually, Units Nazi that I am, L just has to have the same units as x and both can carry their units without having to explicitly indicate them. So the problem, as given, is internally consistent.
     
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  8. WBahn

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    It's not that nπx has an integral of nπx/nπ. It's that what you have is

    <br />
\int \cos(u)du = \sin(u) + C<br />

    But you don't have that, you have

    <br />
\int \cos(ax)dx<br />

    So you need to get it into the form of the former in order to use that rule.

    If you have u=ax, the du = a dx. So you can rewrite your integral as

    <br />
\int \cos(ax)\frac{a dx}{a} \ = \ \frac{1}{a}\int \cos(ax) (a dx) \ = \ \frac{1}{a} \sin(ax) + C<br />
     
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  9. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    I knew differentiation but suppose I have a function ax, is its derivative a dx?

    Does ax represents the nπ?

    Can you tell me how you did go from and to \ \frac{1}{a}\int \cos(ax) (a dx) \ = \ \frac{1}{a} \sin(ax) + C<br />
?

    Thank you.
     
  10. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    He did the u thing.

    Remember, he defined u=ax. Then he took derivative, du=adx. Now replace ax with u, and adx with du. Now you have a constant, 1/a, times integral of cos(u)du, the result is constant times sin(u). Now take the result and replace u with ax, you have constant times sin(ax).
     
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  11. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    I really cannot follow. Can you give me a link that has this topic so I can write.

    What is ax? Is it the nπx, where nπ acts as the a and x as the x?
     
  12. WBahn

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    Yes. In integration tables (and most math in general) letters near the beginning of the alphabet represent constants. So 'a' is a constant that, in this case, happens to be equal to 'nπ', which is just another constant (as far as the integral is concerned, meaning that it does not depend on 'x').
     
  13. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    So do you mean that 'nπ' represents (or should be treated) as a single constant?
     
  14. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    <br />
\int \cos(ax)\frac{a dx}{a} \ = \ \frac{1}{a}\int \cos(ax) (a dx) \ = \ \frac{1}{a} \sin(ax) + C<br />

    Why did you divide adx by a? Is it because for the reason that adx should be just 'dx' so it can act as if it means with 'respect to x'?

    So am I correct with this?

    <br />
\int \cos(ax)\frac{dx}{a} \ = \ \int \frac{1}{a} \cos(ax) (dx) \ = \ \frac{1}{a}\int \cos(ax) (dx) \ = \ \frac{1}{a} \sin(ax) + C<br />

    So what happened in 1/a is it just moved outside the integral sign since it is a constant. Am I correct with this?

    Thanks.

    P.S.: Okay I misunderstood that. But I better leave my original post.

    New Question. Why 'adx' and not just 'dx', is it because of the 'a' in the ax?
     
  15. WBahn

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    You can't change the original equation but you can multiply it by 1 all day long without changing it. Well, (a/a) is 1. So I multiplied by (a/a) and then grouped the numerator with dx and brought the denominator outside the integral.

    What is the derivative of ax? It's adx, right?

    d(ax)/dx = a

    therefore

    d(ax) = a dx
     
  16. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    Okay. I have found out that for every constant a, the indefinite integral of cos(ax) is (sin ax)/a + C. The same holds true for other simple trigonometric function. Like sin(ax) is -[(cos ax)/a)].

    What is d(ax)/dx = a?

    How do you read that and what is that?

    Sorry.
     
  17. WBahn

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    The derivative of 'a' times 'x' taken with respect to 'x' is 'a'.

    I think you need to spend some time learning what derivatives and integrals are. You goal should be to be able to derive the result that the indefinite integral of cos(ax) is sin(ax)/a plus an arbitrary constant.
     
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  18. Lightfire

    Thread Starter Well-Known Member

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    Thank you.
     
  19. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    Please spare me some of your valuable time to check my solution in a fourier series problem given.

    Thank you!:D

    The problem is attached as pdf..

    EDIT: THERE ARE ERRORS

    the second to the last expression should be [​IMG]

    [​IMG]

    [​IMG]

    EDIT I Have attached the corrected format of my work http://forum.allaboutcircuits.com/attachment.php?attachmentid=65153&d=1393062237

    I graphed it and there seems to be an error. I don't know where. I'll come back as soon as possible.....
     
    Last edited: Feb 22, 2014
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