Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » Software » comp.soft-sys.matlab

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Steven Lord

Posts: 17,944
Registered: 12/7/04
Re: Problem with inverse laplace
Posted: Dec 7, 2012 9:58 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply



"spv " <krimen_spv@hotmail.com> wrote in message
news:k9qttc$dki$1@newscl01ah.mathworks.com...
> that the roots of the polynomial are complex is quite obvious because this
> kind of equation represent the transient of an electric system (active and
> reactive parts), the problem is that matlab is perfectly able to compute
> ilaplace over a (polynomial of degree <=2) divided by a (polynomial of
> degree <=2) but when the function on the denominator grows one degree it
> return this kind "result"


*snip*

> inc_w2_i = ilaplace(inc_w2);
>
> the result of this is:
>
> -(25132*sum(exp(r4*t)/(11000*r4 - 168000*r4^2 + 15566), r4 in
> RootOf(s4^3 - (11*s4^2)/112 - (7783*s4)/28000 - 6283/8000, s4)))/5
>
> Completely unusable, if I apply vpa() to this result I get a sum of
> exponentials, it might be ok but it is not, for this result to be usable,
> we have to apply euler's equation: exp(x+iy)=exp(x)*(cos(y) + i*sin(y))
> which is not possible because the result is a symbolic type. (at least I
> couldn't)
>
> but if we downgrade the equation:
>
> inc_w2= inc_P_L1*((-T)/((M1*s+D1)*((M2*s+D2)+T)+(M2*s+D2)*T));
> inc_w2_i = ilaplace(inc_w2)
>
> the result is now something quite usable :
>
> (25132*2712531089^(1/2)*sinh((2712531089^(1/2)*t)/56000)*exp((9033*t)/56000))/13562655445
>
> a function that effectively represents the transient of a sinusoidal wave.
>
> So after googling a lot I have not seen a proper answer to this issue.


I think most people are familiar with the quadratic formula that is taught
in schools.

http://en.wikipedia.org/wiki/Quadratic_equation#Quadratic_formula

You may even still remember it, depending on how long it's been since you
learned it in school. The formula is short and easy to read. The cubic
formula? Not so much.

http://en.wikipedia.org/wiki/Cubic_function#Roots_of_a_cubic_function

While Symbolic Math Toolbox _probably_ could expand out the formula using
the formulae on the above Wikipedia page, I highly doubt you'd find it
"usable".

The quartic case is even worse to the point where Wikipedia shows it as an
image rather than text.

http://en.wikipedia.org/wiki/Quartic_equation#Solving_a_quartic_equation

In any case, except for specific equations we'd probably have to use the
above form for quintics and higher.

http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem

Finally, just because a PROBLEM is simple to state doesn't mean the ANSWER
is simple to state. Just ask Pierre de Fermat and Andrew Wiles.

http://en.wikipedia.org/wiki/Andrew_Wiles

--
Steve Lord
slord@mathworks.com
To contact Technical Support use the Contact Us link on
http://www.mathworks.com




Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.