slow ilapalce calculation for systems with higher polynomial orders
7 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Is it possible to calculate the output o(t) of a system h(s) for the input x(s) faster than following codes? Why the following codes run extremely slow? What is wrong?
syms s t
time=linspace(0,4/1e09,401);
h=((5034923382327397*s^3)/627710173538668076383578942320766641610235544446403451289600 + (6766076405664713*s^2)/71362384635297994052914298472474756819137331200 + (210759975314673*s)/64903710731685345356631204115251200 + 376686340223973/281474976710656000000000000)/((5034923382327397*s^3)/12554203470773361527671578846415332832204710888928069025792 + (2284590225627523*s^2)/356811923176489970264571492362373784095686656 + (7512718920571563*s)/649037107316853453566312041152512 + 765850114395651/37778931862957161709568);
x=(100e-03/s)*(1-exp(-s/(2*1e09)))/(1+exp(-s/(2*1e09)));
y=ilaplace(x*h,s,t);
o=subs(y,t,time);
plot(time,o)
2 commentaires
Walter Roberson
le 6 Fév 2019
your expression for h is either incorrect or meaningless. Numbers with that many digits will be approximated with floating point which will introduce false behaviour with roots and poles in different places than you expect .
Chances are high that you should be using symbolic constants not floating point .
sym('26271892563873553613')
Réponse acceptée
Walter Roberson
le 6 Fév 2019
subs(vpa(y,16),t,time)
No point in calculating with large numbers to indefinite precision only to throw away that precision since plot() only has single precision resolution according to some tests I have done.
0 commentaires
Plus de réponses (0)
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!