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Root-locus design: ζ = 0.5, ts = 2 s, plant 1/(s(s+1))

Short answer

With T=0.2s, desired z-domain poles are 0.5158±j0.4281. Plant G(z)=0.01873(z+0.9355)/[(z-1)(z-0.8187)]. A lead compensator cancels the plant pole at 0.8187, adds a pole at 0.2474, and gain 11.52: G_D(z)=11.52(z-0.8187)/(z-0.2474).

Design a digital controller such that the compensated system should have damping ratio ζ=0.5 and settling time = 2 seconds of the following system using root locus approach. [Figure: unity-feedback loop: R(s) → summing junction → E(s) → sampler (period T) → digital controller G₁*(s) → ZOH → 1/s(s+1) → C(s)]

The sampling period is not given, so we choose it first.

Step 1: Desired poles and choice of TT

From ζ=0.5\zeta = 0.5 and ts=4/(ζωn)=2t_s = 4/(\zeta\omega_n) = 2 s:

ζωn=2,ωn=4 rad/s,ωd=41−0.25=3.464 rad/s\zeta\omega_n = 2, \quad \omega_n = 4\ \text{rad/s}, \quad \omega_d = 4\sqrt{1-0.25} = 3.464\ \text{rad/s}

Choose T=0.2T = 0.2 s. Then ωs=2π/T=31.4\omega_s = 2\pi/T = 31.4 rad/s, giving about 9 samples per damped cycle. This is fast enough (the usual rule is at least 8–10), and the ZOH adds little phase lag.

∣z∣=e−ζωnT=e−0.4=0.6703∠z=ωdT=0.6928 rad=39.69∘z1,2=0.5158±j0.4281\begin{aligned} |z| &= e^{-\zeta\omega_nT} = e^{-0.4} = 0.6703 \\ \angle z &= \omega_dT = 0.6928\ \text{rad} = 39.69^\circ \\ z_{1,2} &= 0.5158 \pm j0.4281 \end{aligned}

Step 2: Plant with ZOH

G(z)=(1−z−1) Z[1s2(s+1)]=(T−1+e−T)z+(1−e−T−Te−T)(z−1)(z−e−T)=0.01873(z+0.9355)(z−1)(z−0.8187)\begin{aligned} G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] = \frac{(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})}{(z-1)(z-e^{-T})} \\ &= \frac{0.01873(z + 0.9355)}{(z-1)(z-0.8187)} \end{aligned}

Step 3: Angle condition at z1z_1

FactorAngle
zero at −0.9355-0.9355+16.44∘+16.44^\circ
pole at 11−138.52∘-138.52^\circ
pole at 0.81870.8187−125.28∘-125.28^\circ
Total−247.36∘-247.36^\circ

The angle deficiency is 247.36∘−180∘=67.36∘247.36^\circ - 180^\circ = 67.36^\circ, so a lead controller is needed.

Step 4: Lead controller GD(z)=Kz−αz−βG_D(z) = K\dfrac{z-\alpha}{z-\beta}

Cancel the slow plant pole with the zero: α=0.8187\alpha = 0.8187, which adds +125.28∘+125.28^\circ. The controller pole must give 125.28∘−67.36∘=57.92∘125.28^\circ - 67.36^\circ = 57.92^\circ:

β=0.5158−0.4281tan⁡57.92∘=0.2474\beta = 0.5158 - \frac{0.4281}{\tan 57.92^\circ} = 0.2474

Step 5: Gain

K=∣z1−1∣ ∣z1−0.2474∣0.01873 ∣z1+0.9355∣=0.6464×0.50530.01873×1.5131=11.52K = \frac{|z_1-1|\,|z_1-0.2474|}{0.01873\,|z_1+0.9355|} = \frac{0.6464 \times 0.5053}{0.01873 \times 1.5131} = 11.52 GD(z)=11.52 z−0.8187z−0.2474G_D(z) = 11.52\,\frac{z - 0.8187}{z - 0.2474}

Step 6: Check

The open loop is 0.2158(z+0.9355)(z−1)(z−0.2474)\dfrac{0.2158(z+0.9355)}{(z-1)(z-0.2474)}. The characteristic equation is z2−1.0316z+0.4493=0z^2 - 1.0316z + 0.4493 = 0, whose roots are 0.5158±j0.42810.5158 \pm j0.4281. ✔

Kv=1Tlim⁡z→1(1−z−1)GDG=0.2158×1.93550.2×0.7526=2.78 s−1K_v = \frac{1}{T}\lim_{z\to1}(1-z^{-1})G_DG = \frac{0.2158 \times 1.9355}{0.2 \times 0.7526} = 2.78\ \text{s}^{-1}

Answer (for T=0.2T = 0.2 s): GD(z)=11.52z−0.8187z−0.2474G_D(z) = 11.52\dfrac{z-0.8187}{z-0.2474}. The closed-loop poles are at 0.5158±j0.42810.5158 \pm j0.4281 (ζ=0.5\zeta = 0.5, ts=2t_s = 2 s), and Kv=2.78 s−1K_v = 2.78\ \text{s}^{-1}. A different TT changes the numbers but not the method.

✓ Answer checked by Claude Opus 5.5 (high effort)

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