Variation in the transfer function calculations
4 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Transfer function using matlab commands was calculated in two different methods and the answers are different. Why is it happening?
Reference: http://ctms.engin.umich.edu/CTMS/index.php?example=InvertedPendulum§ion=SystemModeling
1. Calculated by using s = tf('s'); P_cart = (((I+m*l^2)/q)*s^2 - (m*g*l/q))/(s^4 + (b*(I + m*l^2))*s^3/q - ((M + m)*m*g*l)*s^2/q - b*m*g*l*s/q); P_pend = (m*l*s/q)/(s^3 + (b*(I + m*l^2))*s^2/q - ((M + m)*m*g*l)*s/q - b*m*g*l/q); sys_tf = [P_cart ; P_pend]; inputs = {'u'}; outputs = {'x'; 'phi'}; set(sys_tf,'InputName',inputs) set(sys_tf,'OutputName',outputs) sys_tf
sys_tf =
From input "u" to output...
4.182e-06 s^2 - 0.0001025
x: ---------------------------------------------------------
2.3e-06 s^4 + 4.182e-07 s^3 - 7.172e-05 s^2 - 1.025e-05 s
1.045e-05 s
phi: -----------------------------------------------------
2.3e-06 s^3 + 4.182e-07 s^2 - 7.172e-05 s - 1.025e-05
Continuous-time transfer function.
2. Calculated by tf(sys_ss)
sys_tf =
From input "u" to output...
1.818 s^2 + 1.615e-15 s - 44.55
x: --------------------------------------
s^4 + 0.1818 s^3 - 31.18 s^2 - 4.455 s
4.545 s - 1.277e-16
phi: ----------------------------------
s^3 + 0.1818 s^2 - 31.18 s - 4.455
Both are different even if we neglect the terms with lease magnitude.
0 commentaires
Réponses (0)
Voir également
Catégories
En savoir plus sur Assembly dans Help Center et File Exchange
Produits
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!