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Re: Problem with inverse laplace
Posted:
Feb 13, 2011 3:17 AM


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((0.8082522380e2 + 628.3212164*I)*t)  (0.403761551e2*I) * exp((0.8082522380e2 + 628.3212164*I)*t) + .1012589532 * exp((15.00808252 + 16.96147299*I)*t) + (0.403761551e2*I) * exp((0.8082522380e2  628.3212164*I)*t) + (0.5992394267e1*I) * exp((15.00808252 + 16.96147299*I)*t)  .1012589532 * exp((0.008082522380  628.3212164*I)*t)  (0.5992394266e1*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.



