Other way to a not supported inverse laplace transform with maxima?

On 04/05/2010 05:56 PM, Benjamin Vanheuverzwijn wrote:
> Hi,
>
> I'm trying to calculate the inverse laplace transform of the following 
> equation and it doesn't seem to work:
> ilt((15*%e^(-3*s)+15*%e^(-2*s)-15)/(2*s^2+4*s+34),s,t)
>
> It is said it the documentation that "expr must be a ratio of 
> polynomials whose denominator has only linear and quadratic factors.". 
> As far as i know, (2*s^2+4*s+34) doesn't have "real quadratic factor".
>
> Is there any other way to deal with that kind of operation with maxima 
> (dev. version or an extension)?
>
> Thanks,
>
> -- 
> Benjamin Vanheuverzwijn
>
> Google Talk/Jabber - bvanheu at gmail.com <mailto:bvanheu at gmail.com>
> http://vanheu.ca
>
The inverse transform for this function involves step functions 
(Heaviside functions).
When I last taught this stuff, I found that maxima did not handle those 
well, so I wrote my own little program "My_laplace.mac" to do those.

Here is an example.

Note that I have %e aliased to E for my convenience, and I use the 
notation 'lap' for the Laplace transform and 'ilap' for the inverse 
transform, no 's' or 't' appended. Also I have 'ustep' as an alias for 
'unit_step'

(%i17) display2d: false;

(%o17) false
(%i18) load(My_laplace);
(%i19) ilap((15*%e^(-3*s)+15*%e^(-2*s)-15)/(2*s^2+4*s+34));

(%o19) E^-t*(15*E^2*ustep(t-2)*sin(4*t-8)+15*E^3*ustep(t-3)*sin(4*t-12)
                                          -15*sin(4*t))
         /8
(%i20) lap(%);

(%o20) -E^-(3*s)*(15*E^(3*s)-15*E^s-15)/(2*s^2+4*s+34)

If you are interested in the code, then let me know by private email and 
I'll send it to you.
Of course, I don't guarantee that it is bug free or anything else, but 
it did all the problems in the book I used for the course just fine.

-sen