Fourier Series coefficient question

Discussion in 'Math' started by spark360z, Sep 16, 2012.

  1. spark360z

    Thread Starter New Member

    Jul 27, 2012
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    I'm learning about the Fourier Series and have a few question.

    First, we have 2 methods to find the Fourier coefficient of the Fourier Series.

    1. Use the formula Cn below.

    [​IMG]

    note that: x=2∏ft

    2. Expand the function.

    For Example I have y(t)=3cos(2∏9t)+4cos(2∏8t), let use the Expansion method

    y(t)=3cos(2∏9t)+4cos(2∏8t) <---- Already expand

    we have

    y(j2∏9t)=3 at f=9 Hz Fourier coefficient=3/2
    y(j2∏8t)=4 at f=8 Hz Fourier coefficient=4/2

    Note:I devided by 2 for two side analysis.

    This method makes sense to me.

    but when it comes to use of formula Cn, it becomes nonsense.

    [​IMG]

    note that: x=2∏ft

    My question is how come multiply the whole function with sine & cosine at some

    particular frequency and integrate it. it turns into the Fourier coefficient.

    I mean why is that actually work? It's an area, how can that related to the coefficient.

    I want some intuition.

    Thank you!
     
  2. steveb

    Senior Member

    Jul 3, 2008
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    You can draw an analogy with components of a vector. The basis vectors are all orthogonal to each other. So if you do a dot product of a vector with a basis vector, you get the component in that direction.

    For example, if v=4i+6j+7k, then each component is obtained by the dot product with each unit vector, that is, v dot i is 4, v dot j is 6 and v dot k is 7.

    So, the Fourier coefficient integral is like the dot product since if you integrate basis functions (sine and cos at different multiples of the fundamental frequency) you get one when the frequency is matched and zero orherwise. So what better way to pick off the coefficient of each component.
     
    Last edited: Sep 16, 2012
    spark360z likes this.
  3. spark360z

    Thread Starter New Member

    Jul 27, 2012
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    Thank you so much!

    I never know a signal behave like a vector!
     
    Last edited: Sep 17, 2012
  4. spark360z

    Thread Starter New Member

    Jul 27, 2012
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    After doing some practice I still confuse about the equation.

    Why we divided the result of the integration with the period of the signal.

    I know if we don't do that the coefficient won't come, but why?

    And If I take limit T-->infinity and integral from -infinity--->+infinity.

    why the equation become zero not a Fourier transform?

    This is the most difficult topic I've encountered.

    Please help again! Thank you!
     
    Last edited: Sep 18, 2012
  5. steveb

    Senior Member

    Jul 3, 2008
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    Normalization by dividing by T makes sense since the coefficents are determined by the function shape, not the relative time scale. The time scale is encoded in the sine/cosine functions by the frequency/period. That is, cos(2 PI f t) is the time dependence, and the coefficient determines the magnitude of that component which should not depend on the period, but only on the shape of the function.

    For this reason, you should be able to integrate over any multiple of T, for example 2T, 3T ... Etc. Just make sure you divide by the appropriate time period of 2T, 3T ... Etc.
     
  6. spark360z

    Thread Starter New Member

    Jul 27, 2012
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    How come something divided the period is independence to the time scale?

    why am I so stupid?!? :(
     
  7. steveb

    Senior Member

    Jul 3, 2008
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    I think the simplest answer is that the numerator is an integral over the time span of T, and the denominator is the time span T itself. Hence, if you did not divide by T, the numerator would get bigger (in magnitude) in proportion with T. (assuming, of course, that coefficient is not zero)

    Also, you need to consider units. If you don't divide by something that has units of time, then the coefficient would not have the correct units.


    You are not stupid. It is normal to go through this questioning period when you learn a new subject. I think probing questions like this are a sign of intelligence.

    Also, my answers have not been the clearest possible answers. I've been traveling and just providing very quick answers when I get into a wireless hot spot (I'm in Starbucks right now). So, some of the fault lies with me. But, this is why questioning is so important (and smart!) when learning. If you keep questioning, eventually you will find the answer.
     
  8. spark360z

    Thread Starter New Member

    Jul 27, 2012
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    0
    I think I finally got it!

    Thank you so much!! Thank you!!
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    In addition to steveb's answer, this might also help.

    Let's say that I have a signal that is a voltage applied across a resistor. If I want to know how much energy is dumped into that resistor, then I can integrate the square of the voltage over time and get the total amount of energy. But what if that signal is something like a sinewave that is defined from -infinity to +infinity? Well, I have an infinite amount of energy dumped into that resistor, which doesn't do me a lot of good. So how about, instead, we change our focus and work with the average power instead of the total energy. To find that, we integrate over a period of time and divide the result by that amount of time.

    Because this is a common occurrence with the basic, idealized signals we use as building blocks, we tend to work with 'power signals' instead of 'energy signals', hence the division by T.

    The consequence is that the Fourier transform will tell you how much of the power contained by a signal is present within a given frequency band.
     
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