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Find the fourier transform

  1. Aug 26, 2011 #1
    1. The problem statement, all variables and given/known data
    find the fourier transform of the function

    [tex]
    x(t)=\left\{\begin{matrix}
    &25 - \frac{25}{8}|t-10| &for &|t-10|<8 \\
    &0 &for& |t-10|>8
    \end{matrix}\right.
    [/tex]


    2. Relevant equations



    3. The attempt at a solution

    we know that

    [tex]
    g(t)=\left\{\begin{matrix}
    &1-|t| &for &|t|<1 \\
    &0 &for& |t|>1
    \end{matrix}\right.\leftrightarrow X(j\omega) =\left\{\begin{matrix}
    &\begin{bmatrix}
    {\frac{\frac{sin(\omega)}{2}}{\frac{\omega}{2}}}
    \end{bmatrix}^{2} &for &|\omega|<1 \\
    &0 &for& |\omega|>1
    \end{matrix}\right.
    [/tex]

    we can see that [tex]x(t) = 25g(\frac{1}{8} (t-10)) [/tex]

    now

    [tex]25g(t-10) \leftrightarrow 25 X(j \omega) e^{-j 10 \omega} [/tex]

    and

    [tex] 25g(1/8(t-10)) \leftrightarrow \frac{25}{8} X(j\frac{\omega}{8}) e^{-j 10 \frac{\omega}{8}}[/tex]

    is this correct?

    thanks in advance
     
    Last edited: Aug 26, 2011
  2. jcsd
  3. Aug 26, 2011 #2

    vela

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    Your approach looks fine, but I think you got the factors of 8 in the wrong place. Since it divides t, it should be multiplying ω. Also, I'm not sure where you got the overall factor of 1/8 in front.

    There's an error in what you wrote for X(ω), but I assume it's just a typo. It should be sin(ω/2), not (sin ω)/2.
     
    Last edited: Aug 26, 2011
  4. Aug 26, 2011 #3
    Yes you're right

    [tex] 25g(1/8(t-10)) \leftrightarrow 200 X(j 8 \omega) e^{-j 80 \omega}[/tex]

    "Also, I'm not sure where you got the overall factor of 1/8 in front."

    isn't 25g(1/8(t-10)) = x(t)?

    for g(1/8 t - 10/8) and the first interval I will have

    [tex] 1 - |\frac{t}{8} - \frac{10}{8}|[/tex] for [tex]|\frac{t}{8} - \frac{10}{8}|<1[/tex]

    for the second interval I will get zero as a result

    now if I multiply this by 25 I will get the following result for the first interval

    [tex] 25-\frac{25}{8}|t-10|[/tex] for [tex]|t-10|<8[/tex]

    and the second remains 0 but with interval [tex]|t-10|>8[/tex]

    which is essentially the x(t)
     
    Last edited: Aug 26, 2011
  5. Aug 26, 2011 #4

    vela

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    I meant the 8 that you multiplied into the 25 to get 200. I didn't think it should be there, but I just checked the tables and found you were right.
     
  6. Aug 26, 2011 #5
    thanks I have one more question

    I think I'm wrong with my FT calculations


    [tex]25g(t-10) \leftrightarrow 25 X(j \omega) e^{-j 10 \omega} [/tex]

    but I have [tex] 25g(\frac{1}{8}(t-10)) [/tex]

    hence
    [tex] 25g(1/8(t-10)) \leftrightarrow 200 X(j 8 \omega) e^{-j 10 \omega}[/tex]

    I mean I should first find 25 g(1/8 t)

    and then

    find the final result due to the shifting by 10...but I'm not sure about
     
    Last edited: Aug 26, 2011
  7. Aug 26, 2011 #6

    vela

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    Good catch. As you suggested, you can look at it as a shift by 10 of g(t/8). I think that's the simplest way.

    You could, however, perform the shift first and then the scaling, but it's a bit tricky. The shift would have to be by 10/8, giving you g(t - 10/8). Then when you scale time, you replace t by t/8 to get g(t/8 - 10/8) = g[(t-10)/8]. In the frequency domain, the shift factor would originally be e-j(10/8)ω, and then the scaling would replace ω by 8ω, giving you e-j10ω again.
     
  8. Aug 27, 2011 #7
    that was a nice alternative

    thanks a lot for your help :)
     
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