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Closed-loop TF with digital PID, plant 1/(s(s+1))

Short answer

Compute G_D(z) = (1.4z^2 − 1.4z + 0.2) / (z(z−1)) and G(z) = (0.3679z + 0.2642) / ((z−1)(z−0.3679)). Then C(z)/R(z) = G_DG / (1 + G_DG) = (0.5151z^3 − 0.1451z^2 − 0.2964z + 0.0528) / (z^4 − 1.8528z^3 + 1.5907z^2 − 0.6642z + 0.0528).

Obtain the closed loop transfer function of the system shown in figure below. Assume proportional gain Kₚ=1, integral gain Kᵢ=0.2, derivative gain Kd=0.2. [Take T = 1 sec] [Figure: unity-feedback loop: R(z) → summing junction → E(z) → sampler (T = 1 s) → digital PID controller → ZOH → 1/s(s+1) → C(z)]

The controller and the ZOH + plant are in cascade with only one sampler in front, so C(z)/R(z)=GD(z)G(z)/[1+GD(z)G(z)]C(z)/R(z) = G_D(z)G(z)/[1+G_D(z)G(z)].

Step 1: Pulse transfer function of controller

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

GD(z)=KP+KI1−z−1+KD(1−z−1)=(KP+KI+KD)−(KP+2KD)z−1+KDz−21−z−1=1.4−1.4z−1+0.2z−21−z−1=1.4z2−1.4z+0.2z(z−1)\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}

Step 2: Pulse transfer function of ZOH + plant

G(z)=Z[1−e−Tss⋅1s(s+1)]=(1−z−1) Z[1s2(s+1)]1s2(s+1)=1s2−1s+1s+1G(z)=(1−z−1)[Tz(z−1)2−zz−1+zz−e−T]\begin{aligned} G(z) &= \mathcal{Z}\left[\frac{1-e^{-Ts}}{s}\cdot\frac{1}{s(s+1)}\right] = (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] \\ \frac{1}{s^2(s+1)} &= \frac{1}{s^2} - \frac{1}{s} + \frac{1}{s+1} \\ G(z) &= (1-z^{-1})\left[\frac{Tz}{(z-1)^2} - \frac{z}{z-1} + \frac{z}{z-e^{-T}}\right] \end{aligned}

With T=1T = 1 s and e−1=0.3679e^{-1} = 0.3679:

G(z)=0.3679z+0.2642(z−1)(z−0.3679)G(z) = \frac{0.3679z + 0.2642}{(z-1)(z-0.3679)}

Step 3: Open-loop pulse transfer function

GD(z)G(z)=(1.4z2−1.4z+0.2)(0.3679z+0.2642)z(z−1)2(z−0.3679)=0.5151z3−0.1451z2−0.2964z+0.0528z4−2.3679z3+1.7358z2−0.3679z\begin{aligned} G_D(z)G(z) &= \frac{(1.4z^2-1.4z+0.2)(0.3679z+0.2642)}{z(z-1)^2(z-0.3679)} \\ &= \frac{0.5151z^3 - 0.1451z^2 - 0.2964z + 0.0528}{z^4 - 2.3679z^3 + 1.7358z^2 - 0.3679z} \end{aligned}

Step 4: Closed-loop pulse transfer function

Adding numerator and denominator gives the characteristic polynomial:

C(z)R(z)=GDG1+GDG=0.5151z3−0.1451z2−0.2964z+0.0528z4−1.8528z3+1.5907z2−0.6642z+0.0528\frac{C(z)}{R(z)} = \frac{G_DG}{1+G_DG} = \frac{0.5151z^3 - 0.1451z^2 - 0.2964z + 0.0528}{z^4 - 1.8528z^3 + 1.5907z^2 - 0.6642z + 0.0528}

or, in powers of z−1z^{-1}:

C(z)R(z)=0.5151z−1−0.1451z−2−0.2964z−3+0.0528z−41−1.8528z−1+1.5907z−2−0.6642z−3+0.0528z−4\frac{C(z)}{R(z)} = \frac{0.5151z^{-1} - 0.1451z^{-2} - 0.2964z^{-3} + 0.0528z^{-4}}{1 - 1.8528z^{-1} + 1.5907z^{-2} - 0.6642z^{-3} + 0.0528z^{-4}}

Answer: C(z)R(z)=0.5151z3−0.1451z2−0.2964z+0.0528z4−1.8528z3+1.5907z2−0.6642z+0.0528\dfrac{C(z)}{R(z)} = \dfrac{0.5151z^3 - 0.1451z^2 - 0.2964z + 0.0528}{z^4 - 1.8528z^3 + 1.5907z^2 - 0.6642z + 0.0528}

Check: At z=1z = 1 the numerator is 0.12640.1264 and the denominator is 0.12650.1265, so the ratio is about 1. Because the controller has integral action, the steady-state error for a step input is zero, as expected.

✓ Answer checked by Claude Opus 5.5 (high effort)

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