Simulink Model to transfer function via simulink and Matlab

9 vues (au cours des 30 derniers jours)
Yuval
Yuval le 20 Juil 2024
Commenté : Yuval le 23 Juil 2024
hi everyone,
in the picture i uploaded you can see my closed loop model for a dynamic system.
im trying to get the transfer function between the input signal Va to the output signal Phi , while considering all the feedbacks.
when i do it by the model linearizer app or the control system designer i get strange bode plots and wrong transfer functions.
so i would like you to help me understand how can i do it correctly plese.
the seconed question is how can i get the same closed loop transfer function on simulink script by using feedback command.
thank you!
  2 commentaires
Sam Chak
Sam Chak le 21 Juil 2024
The provided block diagram suggests that you have attempted to model two coupled second-order inertial rotational systems, characterized by the variable 'Theta_dot_dot' () and the implied 'Phi_dot_dot' ()
Ideally, in the absence of any pole-zero cancellation, the resulting transfer function for should be a fourth-order system. This is due to the inherent complexity of the coupled second-order dynamics involved in the model.
Thus, instead of using the Derivative block twice to numerically derive the angular accelaration signal , you should use the knowledge of first principles to construct the mathematical equation (ODE) for .
Yuval
Yuval le 21 Juil 2024
hi sam, thnk you for your comment. i just uploaded the physical model, this is part from my project for university, i do not have the spcify relations just what described in the photo. so i do not know how to replace it. maybe you have another idea?

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Paul
Paul le 21 Juil 2024
Hi Yuval,
What is the setting for the c parameter in the Derivative blocks? If it's the default (inf), those blocks will linearize to zero and the feedback path is effectively broken when using the linearizer app or linearize.
Also, double check if the block diagram is the correct representation of the physical system.
  8 commentaires
Sam Chak
Sam Chak le 21 Juil 2024
Hi @Paul,
The provided example was intended to demonstrate to @Yuval that both configurations (a) and (b) are equivalent, and would ultimately produce the same transfer function .
It was not the purpose to suggest that the OP should modify a biproper Compensator into a configuration with a strictly proper transfer function. The intention was purely to illustrate the conceptual equivalence between the two configurations, and not to imply any recommendations for changing the Compensator design.
Yuval
Yuval le 23 Juil 2024
hi sam and paul,
thank you so much, you guys helped me a lot!

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