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Thread: Unit step function/Heaviside function.

  1. #1
    Member Danshader's Avatar
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    Unit step function/Heaviside function.

    well i have 2 questions.

    1) Consider the function given below:

    f(t) = 0, 0 < t < 2
    sin(t) t > 2

    Find the Laplace transform of the function,
    Well what i found was:

    L[f(t)] = L { 0.[H(t-0) - H(t-2)] + sin(t).H(t-2)}
    = integration from 0 to 2 for 0.exp^(-st) + integration from 2 to
    infinity for sin(t).exp(-st)
    = 0 + integration from 2 to infinity for sin(t).exp(-st)

    if i were to use integration by parts will end up with:
    = const + integration from 2 to infinity for cos (t).exp(-st)/s

    which brings me back to my initial problem of integrating the trigonometry.

    So my question now is how do i solve this unending integration?




    2) Doing some exercise in a maths book i found another similar question:

    f(t) = sin(t), 0 <= t <= p
    0 t >= pi

    and the goven answer was:
    [ 1+ exp(-s.pi) ]/ (s^2 +1)

    so what i did for this question was:

    L{f(t)} = L{ sin(t) [H(t-0) - H(t-pi)] + 0.H(t-pi)}
    = L{ sin(t).H(t)} - L{sin(t).H(t-pi)}
    = 1/(s^2 +1) - [exp(-s.pi)] / [(s^2 +1)]
    = [ 1+ exp(-s.pi) ]/ (s^2 +1)

    the formula given for the heaviside function was:
    L{f(t-a)H(t-a)} =exp(-as).F(s)

    In the above solution i made i didn't used the general form whereby i was suppose to use L{sin(t-pi).H(t-pi)} instead of L{sin(t).H(t-pi)}. still i get the same answer.

    then i look at other examples like:
    L{exp(-t).H(t-6)}
    the answer was exp[-6(s+1)]/(s+1)

    if i were to do the same thing:
    a = 6
    L{exp(-t)} = 1/(s+1)
    since it is a shifting function the solution for the heaviside will be exp[6(s+1)]
    combining both i will get exp[-6(s+1)]/(s^2 +1)

    So my question is it correct in what i am doing where i straight away perform Laplace transform on the function before the heaviside without converting (taking the last example) exp(-t) to exp[-(t-6)] before doing the calculations? my lecturer taught me to convert it but i am not sure about the way i am solving it now

    well as a matter of fact i dont really understand how heaviside works yet so this questions could have already been solved or part of the theory itself o.o"

    All help is appreciated.
    Last edited by Danshader; Apr 22nd 2008 at 02:07 AM.
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  2. #2
    Flow Master
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    Quote Originally Posted by Danshader View Post
    well i have 2 questions.

    1) Consider the function given below:

    f(t) = 0, 0 < t < 2
    sin(t) t > 2

    Find the Laplace transform of the function,
    Well what i found was:

    L[f(t)] = L { 0.[H(t-0) - H(t-2)] + sin(t).H(t-2)}
    = integration from 0 to 2 for 0.exp^(-st) + integration from 2 to
    infinity for sin(t).exp(-st)
    = 0 + integration from 2 to infinity for sin(t).exp(-st)

    if i were to use integration by parts will end up with:
    = const + integration from 2 to infinity for cos (t).exp(-st)/s

    which brings me back to my initial problem of integrating the trigonometry.

    So my question now is how do i solve this unending integration?



    2) Doing some exercise in a maths book i found another similar question:

    f(t) = sin(t), 0 <= t <= p
    0 t >= pi

    and the goven answer was:
    [ 1+ exp(-s.pi) ]/ (s^2 +1)

    so what i did for this question was:

    L{f(t)} = L{ sin(t) [H(t-0) - H(t-pi)] + 0.H(t-pi)}
    = L{ sin(t).H(t)} - L{sin(t).H(t-pi)}
    = 1/(s^2 +1) - [exp(-s.pi)] / [(s^2 +1)]
    = [ 1+ exp(-s.pi) ]/ (s^2 +1)

    the formula given for the heaviside function was:
    L{f(t-a)H(t-a)} =exp(-as).F(s)

    In the above solution i made i didn't used the general form whereby i was suppose to use L{sin(t-pi).H(t-pi)} instead of L{sin(t).H(t-pi)}. still i get the same answer.

    then i look at other examples like:
    L{exp(-t).H(t-6)}
    the answer was exp[-6(s+1)]/(s+1)

    if i were to do the same thing:
    a = 6
    L{exp(-t)} = 1/(s+1)
    since it is a shifting function the solution for the heaviside will be exp[6(s+1)]
    combining both i will get exp[-6(s+1)]/(s^2 +1)

    So my question is it correct in what i am doing where i straight away perform Laplace transform on the function before the heaviside without converting (taking the last example) exp(-t) to exp[-(t-6)] before doing the calculations? my lecturer taught me to convert it but i am not sure about the way i am solving it now

    well as a matter of fact i dont really understand how heaviside works yet so this questions could have already been solved or part of the theory itself o.o"

    All help is appreciated.
    I have no time now but will look at it later if no-one else has.
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  3. #3
    Senior Member Peritus's Avatar
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    $\displaystyle
    \int\limits_2^\infty {\sin (t)e^{ - st} dt = \left. { - \frac{{\sin (t)}}
    {s}e^{ - st} } \right|} _2^\infty + \int\limits_0^\infty {\frac{{\cos (t)}}
    {s}e^{ - st} dt = }
    $

    $\displaystyle = \frac{{\sin 2}}
    {s}e^{ - 2s} - \left. {\frac{{\cos (t)}}
    {{s^2 }}e^{ - st} } \right|_2^\infty - \int\limits_0^\infty {\frac{{\sin (t)}}
    {{s^2 }}e^{ - st} dt = } $


    $\displaystyle = \frac{{\sin 2}}
    {s}e^{ - 2s} + \frac{{\cos 2}}
    {{s^2 }}e^{ - 2s} - \frac{1}
    {{s^2 }}\int\limits_0^\infty {\sin (t)e^{ - st} dt = } $

    and now the "tricky" part:


    $\displaystyle

    \Leftrightarrow \left( {1 + \frac{1}
    {{s^2 }}} \right)\int\limits_2^\infty {\sin (t)e^{ - st} dt = } \left( {\frac{{\sin 2}}
    {s} + \frac{{\cos 2}}
    {{s^2 }}} \right)e^{ - 2s}

    $
    $\displaystyle

    \Leftrightarrow \int\limits_2^\infty {\sin (t)e^{ - st} dt = } \frac{{s\left( {\sin (2)} \right) + \cos (2)}}
    {{s^2 + 1}}e^{ - 2s} $

    -------------------------------------------------------------------------------
    2. can be done the same way

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~

    1. can be solved in a much simpler way if we use the well known Laplace transforms:

    $\displaystyle
    \begin{gathered}
    L\left\{ {\sin (\omega t)u(t)} \right\} = \frac{\omega }
    {{s^2 + \omega ^2 }} \hfill \\
    L\left\{ {\cos (\omega t)u(t)} \right\} = \frac{s}
    {{s^2 + \omega ^2 }} \hfill \\
    \end{gathered}
    $

    $\displaystyle
    \int\limits_2^\infty {\sin (t)e^{ - st} dt} = \int\limits_0^\infty {\sin (t)u(t - 2)e^{ - st} dt}
    $

    which is equivalent to:

    $\displaystyle
    \int\limits_0^\infty {\sin (t + 2)u(t)e^{ - s(t + 2)} dt}
    $

    and now for a little trigonometry:

    $\displaystyle
    \begin{gathered}
    e^{ - 2s} \int\limits_0^\infty {\left[ {\sin (t)\cos (2) + \sin (2)\cos (t)} \right]} e^{ - st} dt = \hfill \\
    \hfill \\
    \cos (2)e^{ - 2s} L\left\{ {\sin (t)} \right\} + \sin (2)e^{ - 2s} L\left\{ {\cos (t)} \right\} = ... \hfill \\
    \end{gathered}

    $
    Last edited by Peritus; Apr 22nd 2008 at 10:17 AM.
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  4. #4
    Lord of certain Rings
    Isomorphism's Avatar
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    Quote Originally Posted by Danshader View Post
    well i have 2 questions.

    1) Consider the function given below:

    f(t) = 0, 0 < t < 2
    sin(t) t > 2

    Find the Laplace transform of the function,
    Well what i found was:

    L[f(t)] = L { 0.[H(t-0) - H(t-2)] + sin(t).H(t-2)}
    = integration from 0 to 2 for 0.exp^(-st) + integration from 2 to
    infinity for sin(t).exp(-st)
    = 0 + integration from 2 to infinity for sin(t).exp(-st)

    if i were to use integration by parts will end up with:
    = const + integration from 2 to infinity for cos (t).exp(-st)/s

    which brings me back to my initial problem of integrating the trigonometry.

    So my question now is how do i solve this unending integration?
    Dont guess it is unending with just one try. As Peritus showed, with a couple of try you can get it in terms of sin again. This is a useful trick.

    However you can try doing this from a different angle. Use the definition $\displaystyle \sin t = \frac{e^{jt} - e^{-jt}}{2j}$
    $\displaystyle \int_2^{\infty} \sin t \, e^{-st} \, dt = \frac1{2j} \int_2^{\infty} e^{jt-st} - e^{-jt-st}\, dt$

    $\displaystyle \frac1{2j} \left(\int_2^{\infty} e^{jt-st} \, dt - \int_2^{\infty} e^{-jt-st} \, dt\right) = \frac1{2j} \left(\frac1{j-s}\,e^{jt-st} + \frac1{j+s}\,e^{-jt-st} \right)$ evaluated between 2 and infinity.

    This is where ROC hops in. To make this transform meaningful we have to choose Re(s) > 0.

    Now because of this $\displaystyle t \rightarrow \infty \Rightarrow e^{jt-st} \rightarrow 0 $. Similar story happens with the other integral. Now evaluate and do some manipulations.
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  5. #5
    Member Danshader's Avatar
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    xD ahhh that make sense. I also found that same method in a book ^-^ thanks for your time
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  6. #6
    Lord of certain Rings
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    Quote Originally Posted by Danshader View Post
    xD ahhh that make sense. I also found that same method in a book ^-^ thanks for your time
    Just curious
    Which method are you referring to? Peritus' or mine?
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