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Inverse fourier troubles: e^(-j*infty)

  1. Dec 1, 2008 #1
    Hello,

    I am working through Signals and Systems Demystified, but I am at an impasse.

    I would like to take the inverse Fourier transform of

    [tex]
    H(f)=\begin{cases}
    -j&\text{if } f > 0\\
    j&\text{if } f<0\end{cases}
    [/tex]

    So

    [tex]
    h(t) = \int_{-\infty}^{0} je^{j2\pi f t}df + \int_{0}^{\infty} -je^{j2\pi ft}df
    = \int_{-\infty}^{0} je^{j2\pi f t}df + \int_{-\infty}^{0} je^{j2\pi f t}df
    = 2\int_{-\infty}^{0} je^{j2\pi f t}df
    = \frac{j2e^{j2\pi ft}}{j2\pi t}|_{-\infty}^{0}
    [/tex]

    So

    [tex]
    h(t) = \frac{1}{\pi t} - \lim_{f \rightarrow -\infty} \frac{e^{j2\pi ft}}{\pi t}
    [/tex]

    The book says that [tex]h(t) = \frac{1}{\pi t}[/tex] is the answer, but I don't understand how. Isn't [tex]\lim_{f \rightarrow -\infty} \frac{e^{j2\pi ft}}{\pi t}[/tex] periodic and nonconverging?
     
  2. jcsd
  3. Dec 2, 2008 #2
    You have a constant times the sign (not sine) function: sgn(f) = 1 for f >0, and -1 for f<0.

    Your problem, as you point out is convergence. Why not try multiplying your function by decaying exponentials, say exp(af) or exp(-af)?
     
  4. Dec 8, 2008 #3
    Hi Dyad,

    Thanks for your response. Could you please give me more details? Even Mathematica hates the integral.

    By the way, I am only confident that the first part is correct.

    [tex]
    h(t) = \int_{-\infty}^{0} je^{j2\pi f t}df + \int_{0}^{\infty} -je^{j2\pi ft}df
    = j(\int_{-\infty}^{0} e^{j2\pi f t}df - \int_{0}^{\infty} e^{j2\pi ft}df)
    [/tex]

    -Sam
     
  5. Dec 8, 2008 #4
    What if you had

    [tex]

    x(t) = \int_{-\infty}^{\infty} je^{j2\pi f t}e^{-b|f|}df
    [/tex]

    with b real and greater than zero? This is a doable calculation.

    Then, after you're done, what happens when you take a certain limit involving a?
     
  6. Dec 9, 2008 #5
    Thanks for the help, Dyad!

    [tex]
    \int_{-\infty}^{\infty}-j Sgn(f)e^{j2\pi ft}e^{-b|f|}df = \\
    \int_{-\infty}^{0}je^{j2\pi ft}e^{bf}df - \int_{0}^{\infty}je^{j2\pi ft}e^{bf}df
    [/tex]

    The [tex]e^{-b|f|}[/tex] added convergence.

    I ended up with

    [tex]
    j[\frac{1}{j2\pi t+b}+\frac{1}{j2\pi t-b}]
    [/tex]

    But as b approaches 0, we get

    [tex]
    j\lim_{b \rightarrow 0}[\frac{1}{j2\pi t+b}+\frac{1}{j2\pi t-b}]=\frac{1}{\pi t}
    [/tex]
     
  7. Dec 9, 2008 #6
    Ok, but just be careful; this doesn't always work.

    Check the Fourier transform of the unit step function; this method doesn't give you the total answer. Problems arise at f = 0.

    (To do this problem, might try writing the step function in terms of the sgn function.)
     
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