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Thread: Calculus explanation of integral

  1. #1
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    Question Calculus explanation of integral

    Hi,

    can someone explain, what is this f(x) function?


    Why are we multiplicating with e^{-jnx}?

    And how do we get the result with Kronecker delta? A little less math and more of explanation part.

    Thank you.
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  2. #2
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    Re: Calculus explanation of integral

    $f(x)$ is any periodic function of $x$. It is being described here by it's Fourier Series representation.

    You don't seem to have any idea what you're reading. The whole idea is to derive the form of the Fourier series coefficients.

    as for the last question

    Suppose $k \neq n$

    $\begin{align*}
    &\displaystyle \int_{-\pi}^{\pi}~e^{j(k-n)x}~dx = \\ \\

    &\left . \dfrac{e^{j(k-n)x}}{j(k-n)}\right|_{-\pi}^{\pi} = \\ \\

    &\dfrac{2}{k-n} \dfrac{e^{j(k-n)\pi}-e^{-j(k-n)\pi}}{2j} = \\ \\

    &\dfrac{2}{k-n}\sin((k-n)\pi) = 0,~k\neq n, ~k,n\in \mathbb{Z}

    \end{align*}$


    for $k=n$

    $\displaystyle \int_{-\pi}^{\pi}~e^{j(0)x}~dx =\displaystyle \int_{-\pi}^{\pi}~1~dx = 2\pi$



    $\displaystyle \int_{-\pi}^{\pi}~e^{j(k-n)x}~dx = \begin{cases} 2\pi &k=n \\ 0 &k \neq n \end{cases} = 2\pi \delta(x)$
    Last edited by romsek; Dec 21st 2017 at 02:24 PM.
    Thanks from HallsofIvy and Nforce
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  3. #3
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    Re: Calculus explanation of integral

    Quote Originally Posted by romsek View Post

    $\displaystyle \int_{-\pi}^{\pi}~e^{j(k-n)x}~dx = \begin{cases} 2\pi &k=n \\ 0 &k \neq n \end{cases} = 2\pi \delta(x)$
    this should read

    $\displaystyle \int_{-\pi}^{\pi}~e^{j(k-n)x}~dx = \begin{cases} 2\pi &k=n \\ 0 &k \neq n \end{cases} = 2\pi \delta_{kn}$
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  4. #4
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    Re: Calculus explanation of integral

    So this is a derivation of Fourier exponential series? For calculating $c_n$.

    What about zero'th Fourier exponential coefficient? It's missing here, that's just an average.

    What if we get complex number as a $c_n$ coefficient? If we calculate the amplitude of this complex number, what does it say about the system?

    Thanks.
    Last edited by skeeter; Dec 22nd 2017 at 07:35 AM. Reason: edited Latex
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  5. #5
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    Re: Calculus explanation of integral

    First the summation is over all integers so does include k= 0
    In your first post, you have $c_k= \int_{-\pi}^\pi f(x)e^{-jkx}$ (Of course, it should be $c_k= \int_{-\pi}^\pi f(x)e^{-jkx}dx$). Taking k= 0, $c_0= \int_{-\pi}^\pi f(x) dx$.

    Further, as romsek showed, that $c_k$ can't be non-real as long as f(x) is real.

    You can think of this as writing the function f(x), in the "vector space" of square integrable functions, in terms of the orthogonal basis, $\{e^{jkx}\}$. To write a vector in terms of a given ortho-normal basis, take the dot product of the vector with each of the basis vectors. The integral given is the dot product in that vector space.
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  6. #6
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    Re: Calculus explanation of integral

    How $c_k$ can't be unreal?

    Look, at the calculation of $c_1$:

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    Re: Calculus explanation of integral

    Do you understand what is my question?
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  8. #8
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    Re: Calculus explanation of integral

    Am I missing something?
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    Re: Calculus explanation of integral

    Quote Originally Posted by Nforce View Post
    Am I missing something?
    I don't follow what you've written, I don't know what $U_{AN}$ and $U_d$ are,

    but Hall's was slightly mistaken.

    Even real functions are guaranteed to have real Fourier coefficients and Odd real functions are guaranteed to have imaginary ones.

    Other than that the coefficients can be any complex numbers even for real functions.
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    Re: Calculus explanation of integral

    $U_an$ and $U_d$ are just constants.

    Let say $U_an$ is 5 Volts.
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  11. #11
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    Re: Calculus explanation of integral

    Quote Originally Posted by Nforce View Post
    How $c_k$ can't be unreal?

    Look, at the calculation of $c_1$:

    So then, this calculation is correct? (If $U_{AN}$ is just a constant let's say a voltage).
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    Re: Calculus explanation of integral

    Quote Originally Posted by Nforce View Post
    So then, this calculation is correct? (If $U_{AN}$ is just a constant let's say a voltage).
    No.

    $\dfrac {1}{2\pi} \displaystyle \int \limits_0^{2\pi}~U_{AN} e^{-j x}~dx = 0$

    Just as you'd expect the integral of a sinusoid over a full period to be.

    The multiplicative constant has no effect on this.

    If you think about what you are doing, you are finding the Fourier coefficients of a constant value $U_{AN}$.

    Well this is pretty clearly $c_0 = U_{AN}, ~c_k = 0,~k\neq 0$
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    Re: Calculus explanation of integral

    Ok thank you for the answer.

    Maybe I have misunderstand the problem. Because it's a problem from electrical engineering. Let's say that it's harmonically constant.
    If I say that it's for example sin(x). The amplitude does not change, is it then the calculation correct?

    Because we have 60 Hz sinusoidal voltage.
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  14. #14
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    Re: Calculus explanation of integral

    Quote Originally Posted by Nforce View Post
    Ok thank you for the answer.

    Maybe I have misunderstand the problem. Because it's a problem from electrical engineering. Let's say that it's harmonically constant.
    If I say that it's for example sin(x). The amplitude does not change, is it then the calculation correct?

    Because we have 60 Hz sinusoidal voltage.
    No.

    The Fourier coefficients of $\sin(x)$ can be easily seen by rewriting

    $\sin(x) = \dfrac{e^{j x} - e^{-j x}}{2j}$

    $c_1 = \dfrac{1}{2j}$

    $c_{-1} = -\dfrac{1}{2j}$

    btw a sinusoid with a 60 Hz frequency is $\sin(120\pi x)$ not $\sin(x)$
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  15. #15
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    Re: Calculus explanation of integral

    Thanks for your time and effort.

    This is something I have missed:
    Even real functions are guaranteed to have real Fourier coefficients and Odd real functions are guaranteed to have imaginary ones.
    So this is always true; If I have an odd real function then I always get imaginary complex Fourier coefficients? And for the trigonometric series, I always get for cosine part = 0.
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