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Re: Inverse Laplace Transform

From: Colin Macdonald
Subject: Re: Inverse Laplace Transform
Date: Tue, 17 Jul 2018 10:16:46 -0700
User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:52.0) Gecko/20100101 Thunderbird/52.9.1

On 2018-07-15 11:03 AM, aheak123 wrote:
Hi, I am trying to take the inverse laplace transform from (Signal and System
with Matlab Application by Steven T. Karris)

I(s) = (s^2+2*s-1)/(2*s^3+9*s^2+6*s+3)

I use the residue command
n = [1 2 -1];
d = [2 9 6 3];
[r,p,k] = residue(n,d)
r =

    0.24020 + 0.00000i
    0.12990 + 0.23250i
    0.12990 - 0.23250i

p =

   -3.81700 + 0.00000i
   -0.34150 + 0.52570i
   -0.34150 - 0.52570i

k = [](0x0)

from this I got the inverse laplace transform of


In the book, he got answer of


which is just a factor of 2 from my solution.  In the book he did not use
the residue command.

He did it partial fraction expansion:

(s^2+2*s-1)/(s+3.817)*(s^2+0.683*s+0.393) =

found r1 = 0.48, r2 = 0.52, r3 = -0.31, thus


He got the answer as (a scale factor of 2 from my solution).

But which one is correct?  I checked both ways of doing it and they seem to
be correct.  What am I doing wrong here?

I'm not sure. I think you're asking a math question rather than an Octave question.

One way to check the two solutions would be to compute the (forward) Laplace transform and check which one gives I(s).

>> pkg load symbolic
>> t
>> B = 0.48*exp(-3.8*t) + exp(-0.34*t)*(0.52*cos(0.53*t)-0.92*sin(0.53*t))

(Note this will give lots of warnings because of those decimal places, but should be close enough for a rough test).

>> laplace(B)

ans = (sym)

          ⎛       2                  ⎞
        4⋅⎝12500⋅s  + 24895⋅s - 12384⎠
    ⎛       3          2                  ⎞
  5⋅⎝10000⋅s  + 44800⋅s  + 29805⋅s + 15067⎠

Note that is very close to I. But if you do this with your 0.24*exp()..., the result is also off by a factor of two.


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