Derive the discrete transfer function for the parallel form of a PID controller. Gc(s)= Kc(1+ 1/(τ1s)+ τDs) using the BD substitution for s. Compare the result with the velocity form of the PID algorithm given in Eq. 8-27.

Question

Derive the discrete transfer function for the parallel form of a PID controller.
[latex] G_{c}(s)= K_{c}(1+ \frac {1}{τ_{1}s}+ τ_{D}s) [/latex] (1)
using the BD substitution for s. Compare the result with the velocity form of the PID algorithm given in Equation
[latex] \Delta p_{k} = p_{k}-p_{k-1}=K_{c}[(e_{k}-e_{k-1})+\frac{\Delta t}{ au _{I}}e_{k} +\frac{ au _{D}}{\Delta t}(e_{k}-2e_{k-1}+e_{k-2}) ] [/latex] (*)

Accepted Answer

Substituting [latex] s≅(1−z^{−1})/Δt[/latex] into Equation (1) gives
[latex] G_{c}(z) = \frac{K_{c}(a_{0}+a_{1}z^{-1}+a_{2}z^{-2})}{1-z^{-1}} [/latex] (2)
where
[latex] a_{0} = 1+\frac{\Delta t}{ au _{I}}+\frac{ au _{D}}{\Delta t}, \quad a_{1} = -(1+\frac{2 au _{D}}{\Delta t}), \quad a_{2} = \frac{ au _{D}}{\Delta t} [/latex]
Because E(z) is the error signal (input) and P(z) is the controller output, [latex] P(z) = G_{c}(z)E(z). [/latex]
Multiplying both sides of Equation (2) by [latex] (1−z^{−1}) [/latex] yields
[latex] (1−z^{−1})P(z)=K_{c}(a_{0} +a_{1}z^{−1} +a_{2}z^{−2})E(z) [/latex]
Converting the controller transfer function into difference equation form gives
[latex] p(k)−p(k−1)=K_{c}a_{0}e(k)+K_{c}a_{1}e(k−1) +K_{c}a_{2}e(k−2) [/latex]
Substituting for [latex] a_{0},\space a_{1} [/latex] , and [latex]a_{2} [/latex] and collecting terms with respect to the controller settings [latex] K_{c},\space τ_{I}, \space and \space τ_{D} [/latex] gives

[latex] p(k)−p(k−1)=K_{c}[(e(k)−e(k−1))+ \frac{\Delta t}{ au _{I}}e(k)+\frac{ au _{D}}{\Delta t}(e(k)−2e(k−1)+e(k−2)) ] [/latex]
Note that this equation is identical to Equation (*) , which was derived using a finite-difference approximation in the time domain.

Question 17.4: Derive the discrete transfer function for the parallel form ...

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