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How to Fourier-transform e^(-2|t|)?

  1. Feb 11, 2017 #1
    1. The problem statement, all variables and given/known data
    Determine the Fourier-transfroms of the functions

    \begin{equation*}
    a) f : f(t) = H(t+3) - H(t-3) \text{ and } g : g(t) = \cos(5t) f(t)
    \end{equation*}
    and
    \begin{equation*}
    b) f : f(t) = e^{-2|t|} \text{ and } g : g(t) = \cos(3t) f(t)
    \end{equation*}


    2. Relevant equations
    The Fourier transformation definition:
    \begin{equation}
    \digamma\{ f(t) \}(\omega) = F(\omega) = \int_{-\infty}^{\infty} f(u)e^{-j\omega u}\,du
    \end{equation}
    What to do with cosines:
    \begin{equation}

    \digamma\{f(t)\cos(\alpha t)\}(\omega) = \frac{1}{2} F(\omega - \alpha) + \frac{1}{2}F(\omega + \alpha)

    \end{equation}

    3. The attempt at a solution
    I managed part a) somewhat:
    \begin{align*}
    \digamma\{ f(t) \}(\omega) &= \int_{-\infty}^{\infty} \left( H(t+3) - H(t-3) \right) e^{-j\omega u} \,du \\
    %
    &= \int_{-3}^{3} e^{-j\omega u} \,du = \left[ \frac{e^{-j\omega u}}{-j\omega}\right]_{-3}^{3}
    = \left( \frac{e^{-j3\omega }}{-j\omega} - \frac{e^{j3\omega}}{-j\omega} \right)\\
    %
    &= \left( \frac{ e^{j3\omega}}{j\omega} - \frac{e^{-j3\omega }}{j\omega} \right) = \frac{1}{2j\omega}(e^{j3\omega } - e^{-j3\omega })\\
    %
    &= \frac{\sin(3\omega)}{\omega} = 3 sinc(3\omega)
    \end{align*}
    Sinc is continuous at ##\omega = 0##, so we don't need to handle that case separately.
    Then, using equation (2) I wrote
    \begin{align*}
    \digamma\{ f(t) \cos(5t) \}(\omega) &= \frac{3}{2} sinc(3(\omega-5)) + \frac{3}{2} sinc(3(\omega+5))
    \end{align*}

    Part b) has me completely stumped. Since
    \begin{equation}
    |u|=
    \begin{cases}
    u &, u\geq 0\\
    -u &, u < 0,
    \end{cases}
    \end{equation}
    we write, according to equation (1):
    \begin{align*}
    \digamma\{ f(t) \}(\omega) &= \int_{-\infty}^{\infty} e^{-2|u|} e^{-j\omega u} \,du = \int_{-\infty}^{\infty} e^{-2|u|-j\omega u} e^{} \,du\\
    %
    &= \int_{-\infty}^{0} e^{2u-j\omega u} \,du + \int_{0}^{\infty} e^{-2u-j\omega u} \,du\\
    %
    &= \lim_{a \to \infty} \int_{-a}^{0} e^{(2-j\omega) u} \,du + \int_{0}^{a} e^{-(2+j\omega) u} \,du\\

    %

    &= \lim_{a \to \infty} \sij{-a}{0} \frac{e^{(2-j\omega) u}}{(2-j\omega)} + \sij{0}{a} \frac{e^{-(2+j\omega)u}}{-(2+j\omega)}\\

    %

    &= \lim_{a \to \infty} \left( \frac{e^0}{(2-j\omega)} - \frac{e^{-a(2-j\omega)}}{(2-j\omega)} \right)

    + \left( \frac{e^{-a(2-j\omega) }}{-(2+j\omega)} - \frac{e^0}{-(2+j\omega)} \right)\\

    %

    &= \lim_{a \to \infty} \left( \frac{1}{(2-j\omega)} - \frac{e^{-a(2-j\omega)}}{(2-j\omega)} \right)

    + \left( \frac{e^{-a(2-j\omega) }}{-(2+j\omega)} + \frac{1}{(2+j\omega)} \right) \\

    %

    &= \lim_{a \to \infty} \frac{1}{(2-j\omega)} + \frac{1}{(2+j\omega)} - \frac{e^{-a(2-j\omega) }}{(2+j\omega)} - \frac{e^{-a(2-j\omega)}}{(2-j\omega)}\\

    %

    &= \lim_{a \to \infty} \frac{(2+j\omega)}{(2+j\omega)(2-j\omega)} + \frac{(2-j\omega)}{(2+j\omega)(2-j\omega)}\\

    &- \frac{(2-j\omega)e^{-a(2-j\omega) }}{(2+j\omega)(2-j\omega)} - \frac{(2+j\omega)e^{-a(2-j\omega)}}{(2+j\omega)(2-j\omega)}\\
    \end{align*}

    This isn't looking good. Is there a trick that I'm missing? Help would be appreciated.
     
    Last edited: Feb 11, 2017
  2. jcsd
  3. Feb 11, 2017 #2

    Ray Vickson

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    If you write ##\cos(3t)## in terms of ##e^{\pm 3 j t}## you have
    $$\int_{-\infty}^0 e^{-j w t} e^{\pm 3 j t} e^{2t} \, dt = \frac{1}{2 - j w \pm 3 j} \left. e^{-j w t} e^{\pm 3 j t} e^{2t}\right|_{-\infty}^0,$$
    which is convergent because of the exponential damping factor ##e^{2t}## when ##t \to -\infty##. Similarly for the other part.
     
  4. Feb 11, 2017 #3

    haruspex

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    In the ##\left[\frac{etc.}{etc.}\right]_0^a## term you omitted the u from the exponent, but that's just a typo. What I do not get is what happened to that term in the next line. How did (2+jω) turn into (2-jω)? Maybe just another typo, rectified later, but this is making it hard to follow.
     
  5. Feb 11, 2017 #4
    Yhea, I just noticed. Fixing.
     
  6. Feb 11, 2017 #5
    Ok, so right now I have
    \begin{align*}
    \digamma\{f(t)\}(\omega) = \lim_{a \to \infty} \frac{(2+j\omega)e^{a(2-j\omega) }}{(2+j\omega)(2-j\omega)}
    + \frac{(2-j\omega)e^{-a(2-j\omega)}}{(2+j\omega)(2-j\omega)}
    \end{align*}
    Not compeltely sure what to do here. I'm trying to edit my offline LaTeX document at the same time, so mistakes are bound to be made.
     
  7. Feb 11, 2017 #6

    haruspex

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    You still have 2-jω in both exponents.
     
  8. Feb 11, 2017 #7
    \begin{align*}
    \digamma\{f(t)\}(\omega) &= \lim_{a \to \infty} \frac{(2+j\omega)e^{a(2-j\omega) }}{(2+j\omega)(2-j\omega)}
    + \frac{(2-j\omega)e^{-a(2+j\omega)}}{(2+j\omega)(2-j\omega)}\\
    &= \lim_{a \to \infty} \frac{(2+j\omega)e^{a(2-j\omega) } + (2-j\omega)e^{-a(2+j\omega)}}{(2+j\omega)(2-j\omega)}
    \end{align*}
    There. I think it might have been a bad idea to try and learn LaTeX proper by doing all of my coursework with it. Copy-paste can be treacherous.
     
  9. Feb 11, 2017 #8
    It looks like I have a sum of complex conjugates in the numerator.
     
  10. Feb 11, 2017 #9
    It's not the cosine, that scares me. What I'm ahving trouble with is ##f(t)##. I also made a few mistakes in the opening post because I'm copyin stuff from my offline LaTeX-document. They should be fixed now.
     
  11. Feb 11, 2017 #10

    haruspex

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    No, there's still a problem.
    Look at the last two lines of your original post. In the left hand integral, you had -a in the exponent, but it becomes +a.
    As Ray says, it should converge.
    I would not have introduced the lim operators, though I know that is technically correct. I would have just integrated still with infinities in the bounds, and used the fact that ex+iy tends to zero as x tends to minus infinity regardless of what y does.
     
  12. Feb 11, 2017 #11
    Alright, let me go over it once more. This might take a moment.
     
  13. Feb 11, 2017 #12
    I got it. Just a sec, and I'll clean up my derivation.
     
  14. Feb 11, 2017 #13

    Ray Vickson

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    Your ##f(t) = e^{-2|t|}## is no problem; that is why you used ##e^{2t}## for ##t < 0## and ##e^{-2t}## for ##t >0##. You get two convergent integrals. It is even faster and easier to use standard expressions for ##\int e^{at} \cos(bt) \, dt##, then use ##b = 3## and
    $$ a = \begin{cases} -j\omega + 2, & t < 0\\
    -j \omega - 2,& t > 0
    \end{cases}
    $$
     
    Last edited: Feb 11, 2017
  15. Feb 11, 2017 #14
    Alright, here's the complete derivation this time (hopefully) without errors:

    \begin{align*}
    \digamma\{ f(t) \}(\omega) &= \int_{-\infty}^{\infty} e^{-2|u|} e^{-j\omega u} \,du = \int_{-\infty}^{\infty} e^{-2|u|-j\omega u} \,du\\
    &= \int_{-\infty}^{0} e^{2u-j\omega u} \,du + \int_{0}^{\infty} e^{-2u-j\omega u} \,du\\
    &= \lim_{a \to \infty} \int_{-a}^{0} e^{(2-j\omega) u} \,du + \int_{0}^{a} e^{-(2+j\omega) u} \,du\\
    &= \lim_{a \to \infty} \left[ \frac{e^{(2-j\omega) u}}{(2-j\omega)}\right]_{-a}^{0} + \left[ \frac{e^{-(2+j\omega)u}}{-(2+j\omega)}\right]_{0}^{a}\\
    &= \lim_{a \to \infty} \left( \frac{e^0}{(2-j\omega)} - \frac{e^{-a(2-j\omega)}}{(2-j\omega)} \right)
    + \left( \frac{e^{-a(2-j\omega) }}{-(2+j\omega)} - \frac{e^0}{-(2+j\omega)} \right)\\
    &= \lim_{a \to \infty} \left( \frac{1}{(2-j\omega)} - \frac{e^{-a(2-j\omega)}}{(2-j\omega)} \right)
    + \left( \frac{e^{-a(2-j\omega) }}{-(2+j\omega)} + \frac{1}{(2+j\omega)} \right)\\
    &= \lim_{a \to \infty} \frac{1}{(2-j\omega)} + \frac{1}{(2+j\omega)} - \frac{e^{-a(2-j\omega) }}{(2+j\omega)} - \frac{e^{-a(2-j\omega)}{(2-j\omega)}\\
    &= \lim_{a \to \infty} \frac{(2+j\omega)}{(2+j\omega)(2-j\omega)} + \frac{(2-j\omega)}{(2+j\omega)(2-j\omega)}\\
    &- \frac{(2-j\omega) \overbrace{e^{-a(2-j\omega) }}^{\text{Real coefficient goes to zero...}}}{(2+j\omega)(2-j\omega)}
    - \frac{(2+j\omega)\overbrace{e^{-a(2-j\omega) }^{\text{... specifically exp(-2a)}}}{(2+j\omega)(2-j\omega)}\\
    &= \frac{(2-j\omega) + (2+j\omega)}{(2+j\omega)(2-j\omega)} = \frac{4}{4 + \omega^2}
    \end{align*}

    Then, if we apply equation ##(2)## to the case where we multiply ##f(t)## by ##\cos(3t)##:
    \begin{align*}
    \digamma\{f(t)\cos(3t)\}(\omega) &= \frac{1}{2} \frac{4}{4 + \left( \omega -3 \right)^2} + \frac{1}{2} \frac{4}{4 + \left( \omega + 3 \right)^2}\\
    &= \frac{2}{4 + \left( \omega -3 \right)^2} + \frac{2}{4 + \left( \omega + 3 \right)^2}\\
    \end{align*}
     
    Last edited: Feb 11, 2017
  16. Feb 11, 2017 #15

    Ray Vickson

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    Answer is correct; derivation did not render properly.
     
  17. Feb 11, 2017 #16
    There seems to be a math processing error preventing the rendering of the biggest part of this exercise. Looking at the source code, I can't spot the error on my part.

    Regradless, if it doesn't decide to resolve itself, the correct answer was
    \begin{equation}
    \digamma\{ f(t) \} = \frac{4}{4 + \omega^2}
    \end{equation}

    Thanks for the help, everyone.
     
    Last edited: Feb 11, 2017
  18. Feb 11, 2017 #17
    Yeah, I'm wondering why that is. I tried to switch any dollar signs to double pound signs and such, but its still not rendering.
     
  19. Feb 11, 2017 #18

    haruspex

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    Doesn't seem to like the text imbeds
     
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