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Math Help - Laplace transform problem

  1. #1
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    [SOLVED]Laplace transform problem

    Compute the Laplace transform of cos(at-b), where a,b are real numbers.

    I've tried this several times and I am ending up with an answer that I know is wrong. I seem to be making the same mistake over and over.

    I set up the problem like this:

    \int{e^{-st}cos(at-b)}

    and then use integration by parts.

    Judging by WolframAlpha my answer should be

    \frac{s}{s^2+1}

    but I keep getting answers similar to this:

    \frac{-s}{s^2-a^2}e^{-st}[cos(at-b)+\frac{a}{s}sin(at-b)]

    Help would be greatly appreciated.
    Last edited by pkrogel; September 7th 2009 at 11:11 AM.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Two things
    1. the deniminator of your antiderivative should be s^2 + a^2

    2. You nedd to evaluate this between 0 and infinity

    See attachment
    Attached Files Attached Files
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  3. #3
    MHF Contributor Calculus26's Avatar
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    Alternatively

    cos(at-b) = cos(at)cos(b) + sin(at)sin(b)

    We can use the basic formulas for cos(at) and sin(at)

    L{cos(at-b)} = L{cos(at)cos(b)} + L {sin(at)sin(b)}

    = cos(b)s/(s^2+a^2) + asin(b)/(s^2+a^2)
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  4. #4
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    Thanks so much for the help!
    I'm confused about two things with your attached document.

    When evaluating from 0 to infinity:
    <br /> <br />
\frac {-scos(at-b)e^{-st} + asin(b-at)e^{-st}}{a^2+s^2}<br />

    1. Why does the first part of the evaluation (with infinity) simply get discarded?

    2. When evaluating 0 with
    scos(at-b), why does it become -scos(b) and not scos(-b)

    I attatched a picture to help explain where I am confused.
    Attached Thumbnails Attached Thumbnails Laplace transform problem-laplace2.jpg  
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  5. #5
    MHF Contributor Calculus26's Avatar
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    neglecting the constants lim e^(-t) cos(t) or lim e^(-t) sin(t) are 0


    This comes from squeezing theorem - e^(-t) < e^(-t) cos(t) < e^(-t)


    lim -e^(-t) = lime^(-t) = 0

    Remember your under damped harmonic oscillator

    2. you have - s cos(at-b)

    a. because you are evaluating at the lower limit you subtract

    you have scos(-b)

    b cos(-b) = cos(b) you have scos(b)
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  6. #6
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    Quote Originally Posted by pkrogel View Post
    Compute the Laplace transform of cos(at-b), where a,b are real numbers.

    I've tried this several times and I am ending up with an answer that I know is wrong. I seem to be making the same mistake over and over.

    I set up the problem like this:

    \int{e^{-st}cos(at-b)}

    and then use integration by parts.

    Judging by WolframAlpha my answer should be

    \frac{s}{s^2+1}

    but I keep getting answers similar to this:

    \frac{-s}{s^2-a^2}e^{-st}[cos(at-b)+\frac{a}{s}sin(at-b)]

    Help would be greatly appreciated.
    Well, that can't be right because the Laplace transform is a function of s only, not t. Did you not evaluate at s= 0 and \infty?

    To integrate by parts, let u= e^{-st} and dv= cos(at- b)dt. Then du= -se^{-st}dt and v= \frac{1}{a}sin(at- b). \int_0^\infty e^{-st}cos(at-b)dt= \frac{s}{a}e^{-st}\right|_0^\infty+ \frac{s}{a}\int_0^\infty e^{-st}sin(at-b)dt = -\frac{s}{a}sin(b)+ \frac{s}{a}\int_0^\infty e^{-st}sin(at-b)dt.

    To do [/b]that[/b] integral, let u= e^{-st} again and let dv= sin(at-b)dt. Then du= -se^{-st}dt and [tex]v= -\frac{1}{a}cos(at- b). \int_0^\infty e^{-st}sin(at-b)dt=frac{1}{a}e^{-st}cos(at-b)\right|_0^\infty- \frac{s}{a}\int_0^\infty e^{-st}cos(at-b)dt = \frac{1}{a}cos(b)-  \frac{s}{a}\int_0^\infty e^{-st}cos(at-b)dt.

    Putting that into the first integral, \int_0^\infty e^{-st}cos(at-b)= -\frac{s}{a}sin(b))-  \frac{s}{a}\int_0^\infty e^{-st}cos(at-b)dt.
    Adding that integral on the right to the left side gives (1+ \frac{s}{a})\int_0^\infty e^{-st}cos(at-b)dt= \frac{a+s}{a}\int_0^\infty e^{-st}cos(at-b)dt= -\frac{s}{a}sin(b)) so \int_0^\infty e^{-st}cos(at-b)dt= -\frac{a}{a+s}\frac{s}{a}sin(b)= \frac{s}{s+a}sin(b).

    That's the basic idea- better check for errors.
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  7. #7
    MHF Contributor Calculus26's Avatar
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    Alternatively

    cos(at-b) = cos(at)cos(b) + sin(at)sin(b)

    We can use the basic formulas for cos(at) and sin(at)

    L{cos(at-b)} = L{cos(at)cos(b)} + L {sin(at)sin(b)}

    = cos(b)s/(s^2+a^2) + asin(b)/(s^2+a^2)

    Gives the correct Laplace transform -- the attachment of the post
    previous to this one does so using the definition of laplace transform
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  8. #8
    MHF Contributor Calculus26's Avatar
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    For the details of the integration see the attachment
    Attached Files Attached Files
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  9. #9
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    thanks so much guys I finally got it!
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