Inverse Laplace Transform

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Inverse Laplace Transform

aheak123
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

*0.24*exp(-3.8*t)+exp(-0.34*t)(0.26*cos(0.53*t)-0.46*sin(0.53*t)*

In the book, he got answer of

*0.48*exp(-3.8*t)+exp(-0.34*t)(0.52*cos(0.53*t)-0.92*sin(0.53*t)
*

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) =
r1/(s+3.817)+(r2*s+r3)/(s^2+0.683*s+0.393)

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

0.48/(s+3.817)+(0.52*s-0.31)/(s^2+0.683*s+0.393)

He got the answer as (a scale factor of 2 from my solution).
0.48*exp(-3.82*t)-0.93*exp(-0.34*t)*sin(0.53*t)+0.53*exp(-0.34*t)*cos(0.53*t)

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

Thanks for the help in advance.
Anthony
































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Re: Inverse Laplace Transform

Colin Macdonald-2
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
>
> *0.24*exp(-3.8*t)+exp(-0.34*t)(0.26*cos(0.53*t)-0.46*sin(0.53*t)*
>
> In the book, he got answer of
>
> *0.48*exp(-3.8*t)+exp(-0.34*t)(0.52*cos(0.53*t)-0.92*sin(0.53*t)
> *
>
> 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) =
> r1/(s+3.817)+(r2*s+r3)/(s^2+0.683*s+0.393)
>
> found r1 = 0.48, r2 = 0.52, r3 = -0.31, thus
>
> 0.48/(s+3.817)+(0.52*s-0.31)/(s^2+0.683*s+0.393)
>
> He got the answer as (a scale factor of 2 from my solution).
> 0.48*exp(-3.82*t)-0.93*exp(-0.34*t)*sin(0.53*t)+0.53*exp(-0.34*t)*cos(0.53*t)
>
> 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.

best,
Colin


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Re: Inverse Laplace Transform

aheak123
Hi Colin, thanks for the reply.  I am using Octave, but basically asking a
math question.  I did a forward laplace transform on my time domain solution
and it came pretty close to the original I(s).  The answer from the book is
off by a factor of 2.

Regards,
aheak123



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Re: Inverse Laplace Transform

Thomas D. Dean-2
On 07/17/18 21:17, aheak123 wrote:
> Hi Colin, thanks for the reply.  I am using Octave, but basically asking a
> math question.  I did a forward laplace transform on my time domain solution
> and it came pretty close to the original I(s).  The answer from the book is
> off by a factor of 2.
>

I think I sent a reply to Colin, not the list.

Maple 2018 agrees with p and r.


 > lprint(evalf(soln));
.2401970034*exp(-3.816996842*t)+(.1299014983-.2324973824*I)*exp((-.3415015788-.5256955164*I)*t)+(.1299014983+.2324973824*I)*exp((-.3415015788+.5256955164*I)*t)