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I am working on drive to control the speed of motor using software PID. the output of PID will generate the reference for drive to run at particular speed. Here Kp and Kd will come from drive it self. The problem I am facing is that drive is giving integral time(Ti) instead of integral gain(Ki. How do I do the integral part if I have integral time(Ti) but not the integral gain(Ki)?

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Various forms of the PID equation are possible, but from your description it appears you have: \$(K_p + 1/T_i s + K_ds)\$. Generally, \$K_i = 1/T_i\$.

Other common forms are: \$(K_p + K_i/s + K_ds)\$ ; \$K_p(1 + 1/T_i s + T_d s)\$

Arguably, the most useful form is: \$ \frac{K_p (1 + T_i s + T_i T_d s^2)} {T_i s} \$ as this is the easiest to build into a transfer function, and also the proportional gain operates on all three terms and is identifiable as a 'loop gain'. This form is also very useful in root locus analysis as it clearly identifies the PID's open loop pole, at \$s=0\$, and two open loop zeros, with the integral and derivative times, \$T_i\$ and \$T_d\$, allowing the zeros to be placed at appropriate positions in the s-plane.

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  • \$\begingroup\$ Can I use Kp(Ts/Ti) . Where Ts is my sampling time and Ti is my integral time. \$\endgroup\$
    – spp
    Jul 5, 2015 at 5:25
  • \$\begingroup\$ I'm not sure what you mean. If you want to implement this digitally, convert the s-TF into a z-TF and then form a difference equation. The bilinear transform is a good method for \$s\rightarrow z\$ transformation. Alternatively, term-by-term discretization also works well for PID. \$\endgroup\$
    – Chu
    Jul 5, 2015 at 6:55

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