Fractional order transfer function
27 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
I tried to write a matlab script for the computation of this transfer function but I keep receving an error message saying that the exponent must a scalar integer.
How can I overcome this issue?
Thank you in advance.
4 commentaires
Jakob B. Nielsen
le 13 Déc 2019
Modifié(e) : Jakob B. Nielsen
le 13 Déc 2019
That is odd. It runs just fine for me, and looking at it it shouldnt give an error...
However, you do make a mistake in the evaluation so lets take that; the transfer function is one divided by your sum. Your function is the sum of one divided by each individual component. I think you are looking for this;
v=20;
a=5;
s=2;
clear denom
denom=0;
for k=1:v
denom=denom+((s/a)^((k-1)/v));
end
C=1/denom;
Abdelkarim Jabrane
le 13 Déc 2019
Modifié(e) : Abdelkarim Jabrane
le 13 Déc 2019
Réponses (1)
Jyothis Gireesh
le 2 Jan 2020
A possible workaround to this scenario may to be define the transfer function using symbolic variables. You may use the following code to implement the same.
syms s k;
v = 20;
a = 5;
C = 1/symsum((s/a)^((k-1)/v),k,1,v);
The above symbolic function can be provided as input to “ilaplace ()” to get the time-domain representation of the same. However do note that due to the presence of fractional order terms in the denominator, the final result may be a piecewise approximation of the function.
1 commentaire
Walter Roberson
le 3 Jan 2020
The piecewise() approximation is more to avoid a singularity than because of fractional order terms.
Voir également
Catégories
En savoir plus sur Calculus dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!