Pulse TF with PID controller, plant 1/(s+1)

There is one sampler in front of the controller, so the closed-loop pulse transfer function is $C(z)/R(z) = G_D(z)G(z)/[1+G_D(z)G(z)]$, where $G(z)$ is the z-transform of ZOH and plant together.

Step 1: Digital PID controller

Digital PID controller. Using the positional PID form (Ogata), with $K_P = 1$, $K_I = 0.2$, $K_D = 0.2$:

$$\begin{aligned} G_D(z) &= K_P + \frac{K_I}{1-z^{-1}} + K_D(1-z^{-1}) \\ &= \frac{(K_P+K_I+K_D) - (K_P+2K_D)z^{-1} + K_D z^{-2}}{1-z^{-1}} \\ &= \frac{1.4 - 1.4z^{-1} + 0.2z^{-2}}{1-z^{-1}} = \frac{1.4z^2 - 1.4z + 0.2}{z(z-1)} \end{aligned}$$

Here $K_P = K - K_I/2$, $K_I = KT/T_i$ and $K_D = KT_d/T$ are the usual digital PID gains. Trapezoidal integration and a backward difference for the derivative give this form.

Step 2: ZOH + plant $1/(s+1)$

$$\begin{aligned} G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s(s+1)}\right] = (1-z^{-1})\left[\frac{z}{z-1} - \frac{z}{z-e^{-T}}\right] \\ &= \frac{1-e^{-T}}{z-e^{-T}} = \frac{0.6321}{z-0.3679} \quad (T = 1) \end{aligned}$$

Step 3: Open-loop pulse transfer function

$$\begin{aligned} G_D(z)G(z) &= \frac{0.6321(1.4z^2-1.4z+0.2)}{z(z-1)(z-0.3679)} \\ &= \frac{0.8850z^2 - 0.8850z + 0.1264}{z^3 - 1.3679z^2 + 0.3679z} \end{aligned}$$

Step 4: Closed-loop pulse transfer function

$$\begin{aligned} \frac{C(z)}{R(z)} &= \frac{0.8850z^2 - 0.8850z + 0.1264}{(z^3 - 1.3679z^2 + 0.3679z) + (0.8850z^2 - 0.8850z + 0.1264)} \\ &= \frac{0.8850z^2 - 0.8850z + 0.1264}{z^3 - 0.4829z^2 - 0.5171z + 0.1264} \end{aligned}$$

In powers of $z^{-1}$:

$$\frac{C(z)}{R(z)} = \frac{0.8850z^{-1} - 0.8850z^{-2} + 0.1264z^{-3}}{1 - 0.4829z^{-1} - 0.5171z^{-2} + 0.1264z^{-3}}$$

Answer: $\dfrac{C(z)}{R(z)} = \dfrac{0.8850z^2 - 0.8850z + 0.1264}{z^3 - 0.4829z^2 - 0.5171z + 0.1264}$

Check: At $z = 1$ both numerator and denominator equal $0.1264$. The DC gain is 1, so there is zero steady-state error to a step input because of the integral term.