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Math Help - Fourier Series help

  1. #1
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    Fourier Series help

    Hi,

    Need help with the following question:

    f(t) = t - 1, 1<t<2
    with an even half range expansion, and f(t)=f(t+4)

    Now, I have to sketch this between -6 and 6. Is an "even extension" the same as "even half range expansion"?
    If I am to sketch this would it be
    a t-1 line from 1<t<2 and 4<t<5
    a -t-1 line from -1<t<-2 and -4<t<-5
    and 0 elsewhere in -6<t<6
    ???
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  2. #2
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    Quote Originally Posted by hashi View Post
    Hi,

    Need help with the following question:

    f(t) = t - 1, 1<t<2
    with an even half range expansion, and f(t)=f(t+4)

    Now, I have to sketch this between -6 and 6. Is an "even extension" the same as "even half range expansion"?
    If I am to sketch this would it be
    a t-1 line from 1<t<2 and 4<t<5
    a -t-1 line from -1<t<-2 and -4<t<-5
    and 0 elsewhere in -6<t<6
    ???
    As the function has period 4, it needs to be defined on a larger interval than (1,2), is it supposed to be zero on (0,1)??

    CB
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  3. #3
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    nope.. it's supposed to be 1 to 2
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  4. #4
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    Quote Originally Posted by hashi View Post
    nope.. it's supposed to be 1 to 2
    Then there is not enough information to do anything with a periodic function of period 4, the function needs to be defined over a whole period, which allowing for even symmetry still requires it be defined on an interval of length 2.

    CB
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  5. #5
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    actually, yes i am pretty sure it's supposed to be zero elsewhere..
    now what/
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  6. #6
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    Quote Originally Posted by hashi View Post
    actually, yes i am pretty sure it's supposed to be zero elsewhere..
    now what/
    OK so now we have the function:

    f(x)= \begin{cases} 0,\ \ \ \ \ \ \ 0\le x <1 \\ t-1,\ \ 1 \le x \le 2 \end{cases}

    Which we extend symmetricaly to the range -2 \le x \le 2, to get:


    f(x)= \begin{cases} -t-1, \ \ -2 \le x \le -1 \\<br />
0, \ \ \ \ \ \ \ \ \ -1 < x < 1 \\<br />
t-1,\ \ \ \ \ \ \ 1 \le x \le 2<br />
\end{cases}<br />

    CB
    Last edited by CaptainBlack; October 19th 2008 at 11:27 AM.
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  7. #7
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    Sorry, how did you get f(x) = 1 between 1 and 2 when the function is
    f(t)=t-1?


    Also, since we are sketching over a domain -6<t<6 would it be...
    Now, I have to sketch this between -6 and 6. Is an "even extension" the same as "even half range expansion"?
    If I am to sketch this would it be
    a t-1 line from 1<t<2 and 4<t<5
    a -t-1 line from -1<t<-2 and -4<t<-5
    and 0 elsewhere in -6<t<6
    ???
    (what i said before?)

    thanks
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  8. #8
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    Hey guys. This is what I think it is. Hope I don't cause confusion:

    f(x)=\left\{\begin{array}{cc} <br />
                         -t-5 & -6\leq t \leq -5 \\<br />
                         -t-1 & -2\leq t\leq -1 \\<br />
                         t-1  & 1\leq t\leq 2 \\<br />
                         t-5  & 5\leq t \leq 6 \\<br />
                         0    & \text{otherwise}<br />
                         \end{array}\right.<br />

    I've plotted f(x) and the first 50 terms of it's Fourier series superimposed over it in the plot below. If that's the function you're working with and need help figuring it's Fourier series, I can post my calculations.
    Attached Thumbnails Attached Thumbnails Fourier Series help-fourier2.jpg  
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  9. #9
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    Quote Originally Posted by hashi View Post
    Sorry, how did you get f(x) = 1 between 1 and 2 when the function is
    f(t)=t-1?


    Also, since we are sketching over a domain -6<t<6 would it be...
    Now, I have to sketch this between -6 and 6. Is an "even extension" the same as "even half range expansion"?
    If I am to sketch this would it be
    a t-1 line from 1<t<2 and 4<t<5
    a -t-1 line from -1<t<-2 and -4<t<-5
    and 0 elsewhere in -6<t<6
    ???
    (what i said before?)

    thanks
    by misremembering what the function was (since I was not looking at it at the time) fixed now.

    CB
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  10. #10
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    Quote Originally Posted by shawsend View Post
    Hey guys. This is what I think it is. Hope I don't cause confusion:

    f(x)=\left\{\begin{array}{cc} <br />
                         -t-5 & -6\leq t \leq -5 \\<br />
                         -t-1 & -2\leq t\leq -1 \\<br />
                         t-1  & 1\leq t\leq 2 \\<br />
                         t-5  & 5\leq t \leq 6 \\<br />
                         0    & \text{otherwise}<br />
                         \end{array}\right.<br />

    I've plotted f(x) and the first 50 terms of it's Fourier series superimposed over it in the plot below. If that's the function you're working with and need help figuring it's Fourier series, I can post my calculations.
    Your diagram does not show a function with period 4.

    CB
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  11. #11
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    Quote Originally Posted by hashi View Post
    Sorry, how did you get f(x) = 1 between 1 and 2 when the function is
    f(t)=t-1?


    Also, since we are sketching over a domain -6<t<6 would it be...
    Now, I have to sketch this between -6 and 6. Is an "even extension" the same as "even half range expansion"?
    If I am to sketch this would it be
    a t-1 line from 1<t<2 and 4<t<5
    a -t-1 line from -1<t<-2 and -4<t<-5
    and 0 elsewhere in -6<t<6
    ???
    (what i said before?)

    thanks
    Remember we have an even period 4 function so the plot outside the fundmental period is obtained by a periodic extension.

    CB
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  12. #12
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    hey shawsend.. how do you just plot f(x) and it's even half range expansion?

    Thanks
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  13. #13
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    Well, I wasn't planning on saying anything further since I got it wrong above and I'm not familiar with the phrase "even half extension", but since you asked, I think it's the function:

    f(x)=\left\{\begin{array}{cc} <br />
                         -t-5 & -6\leq t \leq -5 \\<br />
                         t+3  & -3\leq t\leq -2  \\<br />
                         -t-1 & -2\leq t\leq -1 \\<br />
                          t-1  & 1\leq t\leq 2 \\<br />
                         -t+3  & 2\leq t \leq 3 \\<br />
                         t-5  & 5\leq t \leq 6 \\<br />
                         0    & \text{otherwise}<br />
                         \end{array}\right.<br />

    That's the plot below and I used the following Mathematica code with the "Which" operator to plot it:

    Code:
    f[t_] := Which[-6 <= t <= -5, -t - 5, -3 <= t <= -2, t + 3, 
        -2 <= t <= -1, -t - 1, 1 <= t <= 2, t - 1, 2 <= t <= 3, -t + 3, 
        5 <= t <= 6, t - 5]; 
    Plot[f[x], {x, -6, 6}]
    Now the Fourier series for the function is easy to calculate:

    Let T=\text{Abs}(-6-6), \omega_n=\frac{2\pi n}{T}

    and:

    \text{F.S}\; f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos(\ome  ga t)+b_n\sin(\omega t)

    with:

    a_n=\frac{2}{T}\int_{-6}^{6} f(t)\cos(\omega_n t)dt

    b_n=\frac{2}{T}\int_{-6}^{6} f(t)\sin(\omega_n t)dt

    And if Captain Black reprimands me again for getting it wrong it's your fault.
    Attached Thumbnails Attached Thumbnails Fourier Series help-plot12.jpg  
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  14. #14
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    shawsend, I have a similar problem to the one the OP has. Can you explain how to get a plot of the fourier approx in matlab?

    anyone?
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  15. #15
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    I don't use Matlab. But the principle is the same: calculate the coefficients, say 25 or so, put them all together in the Fourier sum as a function, then plot the results. The following is my Mathematica code. You can probably easily convert it to Matlab code assuming Matlab can handle the conditional function definition.

    Code:
    f[t_] := Which[-6 <= t <= -5, -t - 5, -3 <= t <= -2, t + 3, 
        -2 <= t <= -1, -t - 1, 1 <= t <= 2, t - 1, 2 <= t <= 3, 
        -t + 3, 5 <= t <= 6, t - 5, -5 <= t <= -3, 0, -1 <= t <= 1, 
        0, 3 <= t <= 5, 0]; 
    fplot = Plot[f[x], {x, -6, 6}]; 
    a0 = NIntegrate[f[t], {t, -6, 6}]; 
    a = Table[(1/6)*NIntegrate[f[t]*Cos[((Pi*n)/6)*t], {t, -6, 6}], 
        {n, 1, 25}]; 
    b = Table[(1/6)*NIntegrate[f[t]*Sin[((Pi*n)/6)*t], {t, -6, 6}], 
        {n, 1, 25}]; 
    fs[x_] := N[a0/12 + Sum[a[[n]]*Cos[Pi*(n/6)*x] + 
           b[[n]]*Sin[Pi*(n/6)*x], {n, 1, 25}]]; 
    fsplot = Plot[fs[x], {x, -6, 6}, PlotStyle -> Red]; 
    Show[{fplot, fsplot}]
    Attached Thumbnails Attached Thumbnails Fourier Series help-fourier-plot.jpg  
    Last edited by shawsend; October 27th 2008 at 12:04 PM.
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