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Laplace Transforms and its applications

  1. Oct 10, 2005 #1
    Hi, I am taking a circuits analysis course currently. We are studying Laplace Transforms.
    I was wondering if anybody out there knows a way of bypassing all of the cumbersome complex algebra to find the partial fraction decomposition of a given function with respect to s.
    If not maybe some pointers.:smile:

  2. jcsd
  3. Oct 10, 2005 #2
    Quoting my TA "TI-89"
  4. Oct 10, 2005 #3


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    I'm not sure if you would regard it as "bypassing all the cumbersome complex algebra" but a shortcut I use often is to recognize that the coefficients are simply the residues from the zeros of the reciprocal polynomials. E.g.

    [tex]\frac {1}{(z-z_1)(z-z_2)} = \frac {1}{(z-z_1)(z_1-z_2)} + \frac {1}{(z_2-z_1)(z-z_2)}[/tex]

    In the case of a quadratic factor in the denominator the same principle applies, e.g.

    [tex]\frac {f(z)}{z^2 + bz + c}[/itex]

    can be written as

    [tex]\frac {f(z)}{(z-z_1)(z-z_2)}[/tex]

    and, with the results of the previous sample,

    [tex]\frac {f(z)}{z^2 + bz + c} \rightarrow \frac {(z-z_2)f(z_1) - (z-z_1)f(z_2)}{z_1-z_2} \times \frac {1}{z^2 + bz + c}[/tex]

    for the [itex]\frac {1}{z^2 + bz + c}[/itex]term. Note that I am abbreviating all the other polynomial factors in the denominator with f(z) and you would have to find their respective residues to completely specify the full partial fraction decomposition of

    [tex]\frac {f(z)}{z^2 + bz + c}[/itex]

    It's really eaiser than it looks!
    Last edited: Oct 10, 2005
  5. Oct 12, 2005 #4

    Tom Mattson

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    Yeah, get a better table of transforms. The more you have in your table, the fewer you'll have to break down into terms with 1st and 2nd degree denominators.

    Another thing you can do is compile the results that you've already derived, and use them in the future. No need to reinvent the wheel.
  6. Oct 20, 2005 #5
    partial fractions are alot easier than contour integration...
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