TMS320F28069M: Hi, guys, some confusion CNTL_2P2Z about the BOOSTXL_BUCKCONV demo board

The DC gain of the compensator without the integrator is z0*z1/p1.  It is dimensionless.  I think the author has divided by this term so that KDC sets the DC gain.  After everything has been multiplied out in the last equation, KDC*p1/(z0*z1) will become a common factor in the 'B' coefficients of the compensator.

Regards,

Richard

Hi, Richard

thanks for your reply. But I can not understand what you said.  sorry. would you please give me a example formula deduction to derive the equation you give.

and I try to derive the KDC/z0/z1/p1 by the a1/a2/b0/b1/b2, but it  very hard to do it. could you give me some method to derive the KDC/z0/z1/p1 by the a1/a2/b0/b1/b2 ? and what the mean of  a common factor in the 'B' coefficients of the compensator ? 

 thanks again.

Sure, let me try to make it a bit clearer.

I did not provide an equation. With reference to the equation for G(s), I am saying if you remove the 's' term in the denominator (i.e. the integrator) and set the remaining 's' terms to zero you will have the DC gain of the compensator. It's equal to KDC*z0*z1/p1.

When tuning the power supply loop, you're usually changing the compensator pole and zero frequencies and the compensator gain to achieve some target(s), such as bandwidth or phase margin. Once the gain and pole-zero frequencies are known you find the compensator coefficients using the formula in your first post. In other words, you start with the pole-zero frequencies and compute the coefficients.

If you want to go in the other direction - to compute the gain & pole-zero frequencies from the coefficients - you would proceed as follows (with reference to the last equation in your first post):
1. Normalize the numerator polynomial by dividing through by the constant term. This term is B0 in the coefficient list in the C code, and it's what I meant by the "common factor in the 'B' coefficients".
2. Solve the numerator quadratic to find the compensator zeros.
3. Solve the denominator quadratic to find the compensator poles.
4. Now you have z0, z1, p1, and Fs, you can equate B0 to the product of the two bracketed terms on the left to find KDC.

Normally, you would only want to do this if you need the pole-zero frequencies from an existing design.

Regards,

Richard

Hi, Richard

   thank you for your answer. it let me more clear about KDC. it is simple to calculate the  compensator poles by the dome existing design,  because it just refer to one parameter a1 or a2. But other compensator z0 and z1 more difficult to calculate by  the given coefficients b0/b1/b2. So I still need your help to calculate z0 and z1 by b0/b1/b2.  

the file that I use mathcad software to calculate the 2p2z. would you please help me to find a way to calculate the z0/z1 ? 

thank you for your patience

Mathcad - 2p2z.pdf

Mark,

Unfortunately the time we are allotted to help with the thread is not enough for us to solve mathematical derivations, and that to when functions such as c2d and d2c are readily available in tools like MATLAB,

https://www.mathworks.com/help/control/ref/c2d.html

https://www.mathworks.com/help/control/ref/d2c.html#mw_56bcb4c6-731f-4ead-897c-e16a818e197e

 As you may know once we do transform from s to z, it is a many to one mapping and you may or may not get a solution , for you to be able to solve this you will have the following three equations with two unknown, but they are not linear equations and you directly cannot use linear algebra. There will be a Kdc*(p1/(2Fs*(p1+2Fs) in all the three below, but as p1 can be uniquely identified from the A1 and A0 coefficients, this can be assumed as a scalar that should not change the solution, but if you want you can comprehend that to. 

Good luck with the exercise,

Bhardwaj 

       Thank you for the detail explain。 And I all the time use the same way  you said above to do calculate the z0/z1. but as you see,  the formula of the picture is just can calculate the KDC, and I didn't derive the z0/z1, may be I ignore something to calculate, or may be it is not a solution. I need TI team confirm what I did is right or not ?  and I try three times to calculate use the way you supply, but it didn't work. so I need your help. 

      thanks again.

Hi, Jack

      I want to know the z0/z1/p1 of the existing design, and I need to check the cofficient is right. so I reverse derive the z0/z1/p1 by the b0/b1/b2. 

  Hi, Manish

       I use the Matlab d2c funtion you suggest to calculate the the G(s) by 'z' tems a1/a2/b0/b1/b2 cofficient . but it didn't happen what I want. see the below picture. can you point the wrong place or other suggestion?

      thanks.

Mark,

Going from digital to analog may have one or many solutions ,

once you do d2c you will need to use zpk format to see it is (s+z0) style format, 

What we have been trying to explain to you, is we provide s-> z domain conversion, but if you do not have the s domain for the z domain coefficients then it may not be easy for you to find the orginal pole and zeroes. This is something the tool that we provide does not support either. 

-Manish 

Manish ,

Thank you for your answer.  I didn't  find the orginal pole and zeroes even if I use the exsiting design coefficients.