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Topic: Problem with inverse laplace
Replies: 5   Last Post: Mar 19, 2014 11:05 AM

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Walter Roberson

Posts: 7,673
Registered: 1/20/06
Re: Problem with inverse laplace
Posted: Feb 13, 2011 3:17 AM
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On 12/02/11 11:57 PM, Chaitanya Vichare wrote:
> Hi all I am trying to get inverse laplace of following expression
>> syms s
>>> Ys3 = (80000*s^2 + 400000*s)/(s^5 + 30*s^4 + 395300*s^3 +
>>> 11850000*s^2 + 202500000*s);

>
> so I did in the following way:
> yt3 = ilaplace(Ys3);
>
> I am getting the following answer! I expect a numeric value as a answer.

>>> yt3 = sum((400000*exp(r3*t) + 80000*r3*exp(r3*t))/(4*r3^3 + 90*r3^2 +
>>> 790600*r3 + 11850000), r3 in RootOf(s3^4 + 30*s3^3 + 395300*s3^2 +
>>> 11850000*s3 + 202500000, s3))

>
> r3 and s3 are not defined by me!
>
> Please help to understand this expression.


You should not be expecting a pure numeric answer. If a pure numeric
answer were to result, then it would imply that yt3 = laplace(c,t,s) for
some numeric constant c, but the laplace transform of a numeric constant
is just c/s, a *much* simpler expression than what you got. Thus, you
should be expecting that the inverse laplace will be an expression in a
variable.

The calculation involves finding the roots of a quartic equation. The
quartic equation is the first argument of the RootOf() and it needs to
be solved with respect to the variable given by the second argument of
RootOf, s3. Once you have that set of roots, then you need to do the sum
indicated, with r3 taking on the value of each of the roots in turn.

There are exact (algebraic) solutions to quartic equations, but they are
almost always very long and messy. Automatic expansion of the roots
would make it almost impossible for a human to get a sense of what the
expanded solution "meant", so by default quartics are not expanded.

You can use vpa() or double() to request that the numeric values be
converted to decimal as far as possible. The solution comes out as
approximately

-.1012589531 * exp((0.8082522380e-2 + 628.3212164*I)*t) -
(0.403761551e-2*I) * exp((0.8082522380e-2 + 628.3212164*I)*t) +
.1012589532 * exp((-15.00808252 + 16.96147299*I)*t) + (0.403761551e-2*I)
* exp((0.8082522380e-2 - 628.3212164*I)*t) + (0.5992394267e-1*I) *
exp((-15.00808252 + 16.96147299*I)*t) - .1012589532 *
exp((0.008082522380 - 628.3212164*I)*t) - (0.5992394266e-1*I) *
exp((-15.00808252 - 16.96147299*I)*t) + .1012589532*exp((-15.00808252 -
16.96147299*I)*t)

Yes, that does have complex numbers piled upon complex numbers. All of
the roots of the denominator of your original polynomial are complex.




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