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CCS/LAUNCHXL-F28069M: Applicaton question about the instaspin-motion lab12a

ming chen3
Intellectual 775 points

Part Number: LAUNCHXL-F28069M
Other Parts Discussed in Thread: MOTORWARE

Tool/software: Code Composer Studio

Hello everyone.

I use the lab12a to identify the inertia of my motor with 1000 line Inc encoder.

Because I want to use the inertia parameter to calculate the speed pid according to lab05b.

but I find that the inertia value  unit  identify by lab12a  is A / Krpm/s.

the inertia SI unit is Kg * m^2.

1.So how to change the inertia value from A / Krpm/s to the SI unit  Kg * m^2.

2.I found that the speed filter pole is 100 rad/s in the lab05b guide,but I have not found it in lab05b program.

So I want to know all the speed filter pole default value is 100rad/s? and where I can find it in lab05b program?


thanks.

best regards.

  • 0
    TI__Guru** 117160 points

    1. You should easy to find a tool to do the conversion from the website, even you don't need to calculate it using a formula.

    2. That depends on your system and the control bandwidth you want. All instaSPIN-FOC labs in motorWare don't use inertia to calculate the gains of speed regulators, only intsaSPIN-motion uses the inertia.

  • 0 in reply to Yanming Luo
    Intellectual 775 points

    Thanks yanming.

    I have solved the  the question 1.

    About the question 2 :

    I think the speed filter  is just a LPF. and the LPF pole is equal to 1/τ.

    The lab05b has used the speed filter pole and the Inertia J to calculate the speed PID .just as fellow in <lab user guide>

    Although the calculation speed pid method  has not implement in lab05b program.

    and I just want to use the  speed filter pole parameter to  simulation the system in MATLAB.

    So  the  speed filter pole parameter is necessary.

    Please give me a help.

    Thanks

    Best Regards

  • 0 in reply to ming chen3
    TI__Guru** 117160 points

    As replied to you, you might take a look at any digital control papers or books that should have a detailed description of these content. Unfortunately, we are difficult to explain the foundational knowledge to the designer limit to the resource.