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Evaluation of a sinc integral

  1. Feb 21, 2006 #1

    chroot

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    Can anyone show me how to evaluate an integral like this by hand? I believe such integrals have an analytic solution, but I can't figure out how to find them. Mathematica seems unable to help (the Integrate command runs forever) but I believe this can be done by hand. It's a sort of integral commonly found in communications theory. I actually don't think it's supposed to be very hard...

    [itex]
    \int_{ - \infty }^\infty {\left[ {\frac{1}
    {{\sqrt T }}\operatorname{sinc} \left( {\frac{t}
    {T}} \right) - \frac{1}
    {{2\sqrt T }}\operatorname{sinc} \left( {\frac{{t - T}}
    {T}} \right)} \right]^2 dt}
    [/itex]

    where

    [itex]
    \operatorname{sinc} \left( t \right) \triangleq \frac{{\sin \left( {\pi t} \right)}}
    {{\pi t}}
    [/itex]

    Obviously I can expand the binomial out, but I'm left with products of sinc's with different arguments, and I don't know how to continue.

    - Warren
     
  2. jcsd
  3. Feb 21, 2006 #2

    chroot

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    Actually, the identity

    [itex]
    \operatorname{sinc} \left( {\frac{{t + mT}}
    {T}} \right) \cdot \operatorname{sinc} \left( {\frac{{t + nT}}
    {T}} \right) = T\operatorname{sinc} \left( {\frac{{t + (m + n)T}}
    {T}} \right)
    [/itex]

    makes life much sweeter. I got 1/4. Anyone else care to check my work?

    - Warren
     
  4. Feb 22, 2006 #3

    Tom Mattson

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    Using the identities you posted I got [itex]\pi T / 4[/itex]. I squared out the integrand and split up the integral into 3 integrals. The first two exactly canceled each other out, leaving only the third one:

    [tex]\frac{1}{4}\int_{ - \infty }^\infty \operatorname{sinc}\left(\frac{t-2T}{T}\right)dt [/tex]

    U-substitution yields:

    [tex]\frac{T}{4}\int_{ - \infty }^\infty \operatorname{sinc}(u)du [/tex]

    The integral (sans outside coefficient) evaluates to [itex]\pi[/itex].

    Regarding Mathematica, MathWorld has this to say in its article on the sinc function:

    So I guess you'll have to wait a bit longer to get the computer to do it for you.
     
    Last edited: Feb 22, 2006
  5. Feb 22, 2006 #4

    chroot

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    You can define your own piecewise Sinc function in Mathematica quite easily.

    As it turns out, though, Tom, we're both wrong, but I can't exactly say why. I followed your method similarly, finding two terms which cancelled, leaving the third.

    However, there's a much simpler way, subject to fewer mistakes. Note that all terms of the form

    [itex]
    \frac{a}
    {{\sqrt T }}\operatorname{sinc} \left( {\frac{{t - nT}}
    {T}} \right)
    [/itex]

    are normalized, and integrate to a.

    You can literally read the value of this integral directly off the coefficients a, if you consider the integrand to be a vector multiplication like this:

    [itex]
    \frac{{a_1 }}
    {{\sqrt T }}\operatorname{sinc} \left( {\frac{t}
    {T}} \right) + \frac{{a_2 }}
    {{\sqrt T }}\operatorname{sinc} \left( {\frac{{t - T}}
    {T}} \right) \Leftrightarrow \left[ {\begin{array}{*{20}c}
    {a_1 } & {a_2 } \\

    \end{array} } \right] \cdot \left[ {\begin{array}{*{20}c}
    1 \\
    1 \\

    \end{array} } \right]
    [/itex]

    The value of the integral, then, is the squared norm of this vector. In this case, the result is 5/4. I'm still not exactly sure why integrating this long-hand produces a different value, but I'm pretty sure I just flubbed some algebra.

    I've learned something in this communications class: if anything looks hard, you're doing it wrong. Thanks for the help though!

    - Warren
     
  6. Feb 22, 2006 #5

    Tom Mattson

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    I am certain that I integrated correctly. That being the case I decided to take a closer look at the identity you posted, since I started from that.

    [itex]
    \operatorname{sinc} \left( {\frac{{t + mT}}
    {T}} \right) \cdot \operatorname{sinc} \left( {\frac{{t + nT}}
    {T}} \right) = T\operatorname{sinc} \left( {\frac{{t + (m + n)T}}
    {T}} \right)
    [/itex]

    Let m=0 and n=-1. Then you get:

    sinc(t/T)sinc((t-T)/T)=Tsinc((t-T)/T)

    Now let t=0:

    sinc(0)sinc(-1)=Tsinc(-1)

    Since T is arbitrary and sinc(-1) does not equal zero, the equation above is not an identity.
     
  7. Feb 22, 2006 #6

    chroot

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    Whoops! I posted my identity incorrectly.

    It should be

    [itex]
    \operatorname{sinc} \left( {\frac{{t + mT}}
    {T}} \right) \,\star\, \operatorname{sinc} \left( {\frac{{t + nT}}
    {T}} \right) = T\operatorname{sinc} \left( {\frac{{t + (m + n)T}}
    {T}} \right)
    [/itex]

    Convolution, not multiplication. :blushing:

    - Warren
     
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