I saw the following formula:
the transfer function is: $$Gr(s) = K_p \bigg(1 + \frac{1}{T_n s}+ \frac{T_v s}{1 + T_d s}\bigg) $$ From my understanding:
- $K_p$ is the proportional gain
- $T_n$ is the integral gain
- $T_v$ is the derivative gain
What is $T_d$?
EDIT:
I found an explanation to the extra variable: the differential term includes a limiting low pass also. So now is $T_v$ the derivative gain and $T_d$ the low pass filter? And, what is the use of a filter on the derivative action?
The denominator $1 + T_d s$ acts as a filter to prevent the D action from actually being a derivative. Pure derivatives are not physically implementable, because the gain would go to infinity at high frequencies. (Trace the Bode plot to verify!) The extra pole at $T_d$ limits the gain and allows the use of the D term in practice.