Pulse transfer function of digital PID controller

A digital PID controller is the discrete-time version of the analog PID law. We obtain it by replacing the integral with a numerical sum (trapezoidal rule) and the derivative with a backward difference. Its pulse transfer function is $G_D(z) = U(z)/E(z)$.

Analog PID law

$$u(t) = K\left[e(t) + \frac{1}{T_i}\int_0^t e(t)\,dt + T_d\frac{de(t)}{dt}\right]$$

Here $K$ is the proportional gain, $T_i$ the integral time and $T_d$ the derivative time.

Discretization (sampling period $T$)

• Integral, trapezoidal rule: $\displaystyle\int_0^{kT} e\,dt \approx T\sum_{h=1}^{k}\frac{e((h-1)T)+e(hT)}{2}$
• Derivative, backward difference: $\displaystyle\frac{de}{dt}\Big|_{kT} \approx \frac{e(kT)-e((k-1)T)}{T}$

$$u(kT) = K\left\{e(kT) + \frac{T}{T_i}\sum_{h=1}^{k}\frac{e((h-1)T)+e(hT)}{2} + \frac{T_d}{T}\big[e(kT)-e((k-1)T)\big]\right\}$$

Taking the z-transform

Let $f(h) = \tfrac12[e((h-1)T)+e(hT)]$, so $F(z) = \tfrac12(1+z^{-1})E(z)$. A running sum transforms as $\mathcal{Z}\left[\sum_{h=0}^{k} f(h)\right] = F(z)/(1-z^{-1})$. Following Ogata, we take $f(0) = 0$ (the trapezoidal sum starts at $h = 1$), so the sum from $h = 1$ and the sum from $h = 0$ have the same transform. Therefore:

$$\begin{aligned} \mathcal{Z}\left[\sum_{h=1}^{k}\frac{e(h-1)+e(h)}{2}\right] &= \frac{1+z^{-1}}{2(1-z^{-1})}E(z) \\ \mathcal{Z}\big[e(k)-e(k-1)\big] &= (1-z^{-1})E(z) \end{aligned}$$ $$U(z) = K\left[1 + \frac{T}{2T_i}\cdot\frac{1+z^{-1}}{1-z^{-1}} + \frac{T_d}{T}(1-z^{-1})\right]E(z)$$

Rearranging into the standard form

Use $\dfrac{1+z^{-1}}{2(1-z^{-1})} = \dfrac{1}{1-z^{-1}} - \dfrac{1}{2}$:

$$\begin{aligned} U(z) &= \left[K - \frac{KT}{2T_i} + \frac{KT}{T_i}\cdot\frac{1}{1-z^{-1}} + \frac{KT_d}{T}(1-z^{-1})\right]E(z) \\ G_D(z) &= \frac{U(z)}{E(z)} = K_P + \frac{K_I}{1-z^{-1}} + K_D(1-z^{-1}) \end{aligned}$$

where

Single-fraction form

$$G_D(z) = \frac{(K_P+K_I+K_D) - (K_P+2K_D)z^{-1} + K_D z^{-2}}{1-z^{-1}}$$

Remarks

• The pole at $z = 1$ comes from the integral action. It makes the steady-state error to a step input zero.
• $K_D(1-z^{-1})$ adds a zero and gives phase lead (anticipation).
• Setting $K_D = 0$ gives the PI controller $K_P + K_I/(1-z^{-1})$. Setting $K_I = 0$ gives the PD controller $K_P + K_D(1-z^{-1})$.

Example: For $K_P = 1$, $K_I = 0.2$, $K_D = 0.2$: $G_D(z) = (1.4 - 1.4z^{-1} + 0.2z^{-2})/(1-z^{-1})$.