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Root-locus design: 8 samples per cycle, T = 0.2 s

Short answer

Digital controller: GD(z)=13.93(z-0.8187)/(z-0.1596), a lead compensator. Static velocity error constant Kv=3.0 s^-1.

Using the root locus method in z-plane determine a digital controller for a system shown below so that the dominant closed loop poles of the compensated system will have damping ratio of 0.5 and number of samples per cycle of damped sinusoidal oscillation to be 8. Assume that the sampling period T as 0.2 sec. [Figure: unity-feedback loop: R(z) → summing junction → sampler → digital controller → zero order hold → 1/s(s+1) → C(z)]

Step 1: Desired dominant poles

With 8 samples per cycle, ωdT=2π/8=45∘\omega_dT = 2\pi/8 = 45^\circ, and

∣z∣=exp⁡(−2πζ1−ζ2⋅ωdωs)=exp⁡(−2π(0.5)0.866⋅18)=0.6354z1,2=0.4493±j0.4493\begin{aligned} |z| &= \exp\left(-\frac{2\pi\zeta}{\sqrt{1-\zeta^2}}\cdot\frac{\omega_d}{\omega_s}\right) = \exp\left(-\frac{2\pi(0.5)}{0.866}\cdot\frac18\right) = 0.6354 \\ z_{1,2} &= 0.4493 \pm j0.4493 \end{aligned}

With T=0.2T = 0.2 s, this means ωd=ωs/8=3.927\omega_d = \omega_s/8 = 3.927 rad/s, ωn=4.534\omega_n = 4.534 rad/s, and ts≈4/(ζωn)=1.76t_s \approx 4/(\zeta\omega_n) = 1.76 s.

Step 2: Plant with ZOH (T=0.2T = 0.2 s)

G(z)=(T−1+e−T)z+(1−e−T−Te−T)(z−1)(z−e−T)=0.01873z+0.01752(z−1)(z−0.8187)=0.01873(z+0.9355)(z−1)(z−0.8187)\begin{aligned} G(z) &= \frac{(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})}{(z-1)(z-e^{-T})} = \frac{0.01873z + 0.01752}{(z-1)(z-0.8187)} \\ &= \frac{0.01873(z + 0.9355)}{(z-1)(z-0.8187)} \end{aligned}

Step 3: Angle deficiency at z1z_1

FactorAngle
zero at −0.9355-0.9355+17.98∘+17.98^\circ
pole at 11−140.79∘-140.79^\circ
pole at 0.81870.8187−129.43∘-129.43^\circ
Total−252.24∘-252.24^\circ

The deficiency is 72.24∘72.24^\circ, so a lead compensator is needed.

Step 4: Controller GD(z)=Kz−0.8187z−βG_D(z) = K\dfrac{z-0.8187}{z-\beta}

The zero cancels the pole at 0.8187 and adds +129.43∘+129.43^\circ. The controller pole must then contribute 129.43∘−72.24∘=57.19∘129.43^\circ - 72.24^\circ = 57.19^\circ:

β=0.4493−0.4493tan⁡57.19∘=0.1596\beta = 0.4493 - \frac{0.4493}{\tan 57.19^\circ} = 0.1596

Step 5: Gain

K=∣z1−1∣ ∣z1−0.1596∣0.01873 ∣z1+0.9355∣=0.7107×0.53460.01873×1.4559=13.93K = \frac{|z_1-1|\,|z_1-0.1596|}{0.01873\,|z_1+0.9355|} = \frac{0.7107 \times 0.5346}{0.01873 \times 1.4559} = 13.93 GD(z)=13.93 z−0.8187z−0.1596G_D(z) = 13.93\,\frac{z - 0.8187}{z - 0.1596}

Step 6: Check

GDG=0.2610(z+0.9355)(z−1)(z−0.1596)G_DG = \frac{0.2610(z + 0.9355)}{(z-1)(z-0.1596)}

The characteristic equation is z2−0.8986z+0.4038=0z^2 - 0.8986z + 0.4038 = 0, with roots 0.4493±j0.44930.4493 \pm j0.4493. ✔

Kv=1T⋅0.2610×1.93551−0.1596=3.01 s−1K_v = \frac{1}{T}\cdot\frac{0.2610 \times 1.9355}{1 - 0.1596} = 3.01\ \text{s}^{-1}

Answer: GD(z)=13.93z−0.8187z−0.1596G_D(z) = 13.93\dfrac{z-0.8187}{z-0.1596}, a lead compensator, with Kv≈3.0 s−1K_v \approx 3.0\ \text{s}^{-1}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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