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Pulse TF with digital PID, plant 1/(s(s+2))

Short answer

Digital PID with Kp=1, Ki=0.2, Kd=0.2 gives G_D(z)=(1.4z²-1.4z+0.2)/(z(z-1)). ZOH plus plant 1/s(s+2) with T=1 yields G(z)=(0.2838z+0.1485)/((z-1)(z-0.1353)). Closed-loop: C/R = (0.3974z³-0.1895z²-0.1511z+0.0297)/(z⁴-1.7380z³+1.0812z²-0.2865z+0.0297).

For the control system with digital PID controller as shown in the figure, obtain the pulse transfer function. Assume that the sampling period T is 1 second. Consider Kₚ=1, KI=0.2 and Kd=0.2 of PID controller. [Figure: unity-feedback loop: r(t) → summing junction → e(t) → sampler (T = 1) → e(kT) → digital controller → ZOH → 1/s(s+2) → C(t)]

The digital controller is a PID controller. It is followed by a ZOH and the plant, with one sampler on the error signal, so:

C(z)R(z)=GD(z)G(z)1+GD(z)G(z)\frac{C(z)}{R(z)} = \frac{G_D(z)G(z)}{1+G_D(z)G(z)}

Step 1: Digital PID 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: ZOH + plant 1/[s(s+2)]1/[s(s+2)]

G(z)=(1−z−1) Z[1s2(s+2)]1s2(s+2)=1/2s2−1/4s+1/4s+2G(z)=(1−z−1)[0.5Tz(z−1)2−0.25zz−1+0.25zz−e−2T]=(2T−1+e−2T)z+(1−e−2T−2Te−2T)4(z−1)(z−e−2T)\begin{aligned} G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+2)}\right] \\ \frac{1}{s^2(s+2)} &= \frac{1/2}{s^2} - \frac{1/4}{s} + \frac{1/4}{s+2} \\ G(z) &= (1-z^{-1})\left[\frac{0.5Tz}{(z-1)^2} - \frac{0.25z}{z-1} + \frac{0.25z}{z-e^{-2T}}\right] \\ &= \frac{(2T-1+e^{-2T})z + (1-e^{-2T}-2Te^{-2T})}{4(z-1)(z-e^{-2T})} \end{aligned}

With T=1T = 1 and e−2=0.1353e^{-2} = 0.1353:

G(z)=(1+0.1353)z+(1−0.4060)4(z−1)(z−0.1353)=0.2838z+0.1485(z−1)(z−0.1353)G(z) = \frac{(1+0.1353)z + (1-0.4060)}{4(z-1)(z-0.1353)} = \frac{0.2838z + 0.1485}{(z-1)(z-0.1353)}

Step 3: Open-loop pulse transfer function

GD(z)G(z)=(1.4z2−1.4z+0.2)(0.2838z+0.1485)z(z−1)2(z−0.1353)=0.3974z3−0.1895z2−0.1511z+0.0297z4−2.1353z3+1.2707z2−0.1353z\begin{aligned} G_D(z)G(z) &= \frac{(1.4z^2 - 1.4z + 0.2)(0.2838z + 0.1485)}{z(z-1)^2(z-0.1353)} \\ &= \frac{0.3974z^3 - 0.1895z^2 - 0.1511z + 0.0297}{z^4 - 2.1353z^3 + 1.2707z^2 - 0.1353z} \end{aligned}

Step 4: Closed-loop pulse transfer function

C(z)R(z)=0.3974z3−0.1895z2−0.1511z+0.0297z4−1.7380z3+1.0812z2−0.2865z+0.0297\frac{C(z)}{R(z)} = \frac{0.3974z^3 - 0.1895z^2 - 0.1511z + 0.0297}{z^4 - 1.7380z^3 + 1.0812z^2 - 0.2865z + 0.0297}

or

C(z)R(z)=0.3974z−1−0.1895z−2−0.1511z−3+0.0297z−41−1.7380z−1+1.0812z−2−0.2865z−3+0.0297z−4\frac{C(z)}{R(z)} = \frac{0.3974z^{-1} - 0.1895z^{-2} - 0.1511z^{-3} + 0.0297z^{-4}}{1 - 1.7380z^{-1} + 1.0812z^{-2} - 0.2865z^{-3} + 0.0297z^{-4}}

Answer: The closed-loop pulse transfer function is as above, with characteristic equation z4−1.7380z3+1.0812z2−0.2865z+0.0297=0z^4 - 1.7380z^3 + 1.0812z^2 - 0.2865z + 0.0297 = 0.

Check: At z=1z = 1 the numerator is 0.08650.0865 and the denominator is 0.08640.0864, so the DC gain is about 1, as expected with integral action.

✓ Answer checked by Claude Opus 5.5 (high effort)

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