Help with Inverse Laplace Transform Function
    7 visualizzazioni (ultimi 30 giorni)
  
       Mostra commenti meno recenti
    
If I run the following code, I get an answer.
 n = 5; 
ku = 1; 
syms s; 
HDs = (ku^n)/(s*(ku+s)^n);
sim_data = ilaplace(HDs,s,0.1);
sim_data = 1 - (265241*exp(-1/10))/240000;
However, if n is a decimal, such as 5.1, the inverse laplace function doesn't work.
   n = 5.1; 
     ku = 1; 
     syms s 
     HDs = (ku^n)/(s*(ku+s)^n)
     sim_data = ilaplace(HDs,s,0.1)
sim_data = ilaplace(1/(s*(s + 1)^(51/10)), s, 1/10)
I need to be able to calculate the inverse laplace transform for decimal values of n. Please help.
0 Commenti
Risposte (2)
  Aveek Podder
    
 il 26 Ott 2017
        Hi,
The ilaplace function computes the analytic closed inverse Laplace form of a transfer function. It seems that mathematically a closed inverse Laplace form for this function cannot be found out, so ilaplace function is returning the input transfer function.
There is a community submission at MathWorks File Exchange which numerically approximates an inverse Laplace transform for any function of "s". I will encourage you to have a look at the submission. However, in case you have any query regarding the submission you have to get in touch with the owner of the submission.
0 Commenti
  Walter Roberson
      
      
 il 26 Ott 2017
        For positive ku and n, and non-negative t, the inverse laplace apparently works out as
((-1-n)*igamma(1+n, ku*t)+ku^n*t^n*exp(-ku*t)*(1+n)+gamma(2+n))/gamma(2+n)
0 Commenti
Vedere anche
Categorie
				Scopri di più su Calculus in Help Center e File Exchange
			
	Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!


