Variation in the transfer function calculations
4 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
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 Kommentare
Antworten (0)
Siehe auch
Kategorien
Mehr zu Assembly finden Sie in Help Center und File Exchange
Produkte
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!