22-Mec-A3 System Analysis and Control · December 2016

Question 4 of 6: Root-Locus Design for a Damping Ratio of 0.5

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 4: Root-Locus Design for a Damping Ratio of 0.5 (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two unity-negative-feedback loops whose forward paths are $K^{\prime}/[(s+1)(s+5)]$ and $K^{\prime}/[s(s+2)(s+6)]$, with $K^{\prime}$ the adjustable gain. Design target: closed-loop damping ratio $\zeta=0.5$.

Find. For each system: the root-locus sketch, the gain $K^{\prime}$ giving $\zeta=0.5$, and — at that gain — the largest closed-loop time constant and the damped natural frequency.

Approach. Form the characteristic equation $1+G(s)=0$, sketch the locus from the standard construction rules, then impose $\zeta=0.5$. For the second-order case that fixes $\omega_{n}$ directly from the (gain-independent) $s$ coefficient; for the third-order case the complex pair must be written as $-a\pm ja\sqrt{3}$ and Vieta's relations solved for $a$, the third real pole, and $K^{\prime}$.

Part (a): $K^{\prime}/[(s+1)(s+5)]$

1. Write the characteristic equation. $$1+\frac{K^{\prime}}{(s+1)(s+5)}=0\;\Longrightarrow\;s^{2}+6s+(5+K^{\prime})=0$$
2. Sketch the locus. There are two open-loop poles at $s=-1$ and $s=-5$ and no finite zeros, so both branches run to infinity along asymptotes at $\pm90^{\circ}$ centred on the centroid $\sigma_{a}=(-1-5)/2=-3$. The real-axis segment between $-5$ and $-1$ belongs to the locus (one pole lies to its right at $-1$ — an odd count). Setting $dK^{\prime}/ds=0$ on $K^{\prime}=-(s^{2}+6s+5)$ gives $2s+6=0$, so the branches meet and break away at $s=-3$, where $K^{\prime}=(3-1)(5-3)=4$. Beyond that gain the roots are complex and travel vertically, their real part locked at $-3$.
3. Impose the damping ratio. Comparing with $s^{2}+2\zeta\omega_{n}s+\omega_{n}^{2}$, the $s$ coefficient is fixed at 6 by the plant and does not depend on gain, so$$2\zeta\omega_{n}=6\;\Longrightarrow\;\omega_{n}=\frac{6}{2(0.5)}=6\ \text{rad/s}$$
4. Solve for the gain. The constant term must equal $\omega_{n}^{2}$:$$5+K^{\prime}=\omega_{n}^{2}=36\;\Longrightarrow\;\boxed{K^{\prime}=31}$$
5. Locate the closed-loop poles and extract the answers. With $K^{\prime}=31$ the roots of $s^{2}+6s+36=0$ are$$s=-3\pm j3\sqrt{3}=-3\pm j5.196$$Both poles have the same real part $\sigma=3$, so there is a single time constant:$$\boxed{\tau_{\max}=\frac{1}{\zeta\omega_{n}}=\frac{1}{3}=0.333\ \text{s}}\qquad\boxed{\omega_{d}=\omega_{n}\sqrt{1-\zeta^{2}}=6\sqrt{0.75}=5.196\ \text{rad/s}}$$As a check, $\zeta=\sigma/\omega_{n}=3/6=0.5$ as required.

Part (b): $K^{\prime}/[s(s+2)(s+6)]$

1. Write the characteristic equation. $$1+\frac{K^{\prime}}{s(s+2)(s+6)}=0\;\Longrightarrow\;s^{3}+8s^{2}+12s+K^{\prime}=0$$
2. Sketch the locus. Three poles ($0,-2,-6$), no finite zeros, so $n-m=3$ asymptotes at $60^{\circ},180^{\circ},300^{\circ}$ leaving the centroid$$\sigma_{a}=\frac{0+(-2)+(-6)}{3}=-2.667$$Real-axis segments are $[-2,0]$ and $(-\infty,-6]$. Breakaway from $dK^{\prime}/ds=0$ on $K^{\prime}=-(s^{3}+8s^{2}+12s)$ requires $3s^{2}+16s+12=0$, whose roots are $-0.903$ and $-4.431$; only $-0.903$ lies on a $K^{\prime}>0$ segment, so that is the breakaway, at $K^{\prime}=5.05$. The locus crosses the imaginary axis where the Routh $s^{1}$ row vanishes, at $K^{\prime}=(8)(12)=96$ and $\omega=\sqrt{12}=3.464\ \text{rad/s}$; the loop is therefore stable for $0<K^{\prime}<96$.
3. Parametrise the design point. A damping ratio of 0.5 places the complex pair on rays at $60^{\circ}$ from the negative real axis, because $\cos^{-1}(0.5)=60^{\circ}$. Writing the pair as $s=-a\pm ja\sqrt{3}$ and the unknown third (real) pole as $s=-c$, the cubic factors as$$(s+c)\left[(s+a)^{2}+3a^{2}\right]=s^{3}+8s^{2}+12s+K^{\prime}$$Note that $\zeta=a/\sqrt{a^{2}+3a^{2}}=0.5$ automatically, for any $a$.
4. Match coefficients (Vieta). Equating the $s^{2}$, $s^{1}$ and constant terms:$$2a+c=8,\qquad 4a^{2}+2ac=12,\qquad K^{\prime}=4a^{2}c$$Substituting $c=8-2a$ into the middle relation collapses the quadratic terms:$$4a^{2}+2a(8-2a)=4a^{2}+16a-4a^{2}=16a=12\;\Longrightarrow\;a=0.75$$
5. Back-substitute for the third pole and the gain. $$c=8-2(0.75)=6.5,\qquad K^{\prime}=4(0.75)^{2}(6.5)=\boxed{14.625}$$The closed-loop poles are therefore $s=-0.75\pm j1.299$ and $s=-6.5$. It is worth stressing that $K^{\prime}$ is the product of the three pole magnitudes measured from the open-loop poles — forgetting the far-off real pole is the standard way to get this wrong.
6. Extract the time constant and damped frequency. The two candidate time constants are $1/0.75=1.333\ \text{s}$ for the complex pair and $1/6.5=0.154\ \text{s}$ for the real pole. The maximum — and hence the one that governs settling — is the dominant pair:$$\boxed{\tau_{\max}=\frac{1}{0.75}=1.333\ \text{s}}\qquad\boxed{\omega_{d}=0.75\sqrt{3}=1.299\ \text{rad/s}}$$The real pole is more than eight times faster, so the complex pair genuinely dominates and the second-order approximation is sound. Note also $\omega_{n}=\sqrt{a^{2}+3a^{2}}=2a=1.5\ \text{rad/s}$.