Worked examples — Root locus — Evans' method, rules for sketching

3.5.40 · D3 · Physics › Guidance, Navigation & Control (GNC) › Root locus — Evans' method, rules for sketching

Shuru karne se pehle, ek reminder alphabet ka taaki kuch bhi bina samjhe use na ho:

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Scenario matrix

Har root-locus problem inn cells mein se kisi ek mein hota hai. Baad mein aane wale worked examples mein se har ek ek labelled cell ko hit karta hai taaki milkar poora grid cover ho jaye.

Cell	Case class	Kya special hai	Example
A	Do real poles, koi zero nahi	Simplest breakaway, vertical asymptotes	Ex 1
B	Real pole origin par ($s=0$)	Integrator; danger axis par hi shuru hota hai	Ex 1, Ex 4
C	Teen real poles → $j\omega$ crossing	Locus right-half-plane mein jaata hai; finite max gain	Ex 2
D	Finite zero add karna	Zero ek branch ko wapas kheenchta hai; re-stabilise karta hai	Ex 3
E	Complex open-loop poles	Angle-of-departure (Rule 7) chahiye	Ex 5
F	Degenerate — zero ek pole ko cancel karta hai	Pole–zero cancellation, branch count ghatta hai	Ex 6
G	Unstable open-loop pole (right-half-plane start)	Branch ko left ki taraf kheenchna padta hai stabilise karne ke liye	Ex 7
H	Real-world word problem	Hardware ko $G(s)H(s)$ mein translate karo	Ex 4 (GNC autopilot)
I	Exam twist — required damping ke liye solve karo	Locus + Damping ratio and settling time combine karo	Ex 8

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Ex 1 — Cells A + B: do real poles, ek origin par

Forecast: padhne se pehle guess karo — do branches kahaan jaati hain, aur $K$ ki kis value par poles pehli baar oscillatory ho jaate hain?

1. Poles/zeros count karo. $n=2$ (poles $s=0,-2$ par), $m=0$. Yeh step kyun? Har rule $n$ aur $m$ par depend karta hai; pehle yeh lo.
2. Real-axis segment (Rule 3). $(-2,0)$ mein ek test point lo. Sirf pole $s=0$ par uske daahine hai → count $=1$ (odd) → segment $[-2,0]$ locus par hai. Yeh step kyun? Odd-count rule fast way hai yeh jaanne ka ki branches real stretch par kahan chalti hain split hone se pehle.
3. Asymptotes (Rule 4). $n-m=2$, toh ${\theta }_a=\frac{\pm 180°(2k+1)}{2}=90°,270°$. Centroid ${\sigma }_a=\frac{(0)+(-2)-0}{2}=-1$. Yeh step kyun? Dono branches infinity ki taraf escape karti hain; asymptotes bataate hain ki escaping direction seedha upar/neeche $\sigma =-1$ se hai.
4. Breakaway (Rule 5). $K=-\frac{D}{N}=-s(s+2)$. Phir $\frac{dK}{ds}=-(2s+2)=0\Rightarrow s=-1$. Yeh step kyun? Do branches real axis par milti hain; double root par $K$ extremum par hota hai, toh $dK/ds=0$ split ko pinpoint karta hai.
5. Breakaway par Gain. $K=-(-1)(-1+2)=1$. Yeh step kyun? Yeh split point ko label karta hai — $K>1$ ke liye poles axis se nikal jaate hain aur complex (oscillatory) ho jaate hain.

Branches $0$ aur $-2$ se nikalti hain, $-1$ par milti hain, phir seedhi upar jaati hain:

Verify: $K=1$ par characteristic equation $s^2+2s+1=(s+1{)}^2=0$ → $-1$ par double pole. ✔ $K=2$ ke liye: $s^2+2s+2=0\Rightarrow s=-1\pm j$, real part exactly $-1$ (vertical asymptote par). ✔ Sabhi $K>0$ ke liye stable.

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Ex 2 — Cell C: teen real poles, $j\omega$ crossing

Forecast: teen branches ke saath sab infinity ki taraf bhaag rahe hain, kam se kam ek right-half-plane mein curve karni chahiye. Kis $K$ par?

1. Poles/zeros. $n=3$ ($0,-1,-2$), $m=0$.
2. Asymptotes. ${\theta }_a=\frac{\pm 180°(2k+1)}{3}=60°,180°,300°$; ${\sigma }_a=\frac{0-1-2}{3}=-1$. Yeh step kyun? $60°$ aur $300°$ asymptotes daahine aur upar/neeche point karte hain — ek warning ki branches unstable region ki taraf jaati hain.
3. Characteristic equation. $s(s+1)(s+2)+K=s^3+3s^2+2s+K=0$. Yeh step kyun? Routh–Hurwitz ko polynomial coefficients chahiye.
4. Routh array. $\begin{array}{ccc}{s}^3& 1& 2\\ s^2& 3& K\\ s^1& \frac{3\cdot 2-K}{3}& 0\\ s^0& K& \end{array}$ First-column positivity ke liye chahiye $\frac{6-K}{3}>0\Rightarrow K<6$ aur $K>0$. Yeh step kyun? First column mein sign change = right-half-plane pole = instability.
5. Crossing frequency. $K=6$ par $s^1$ row zero ho jaati hai; auxiliary polynomial $s^2$ row hai: $3s^2+6=0\Rightarrow s^2=-2\Rightarrow s=\pm j\sqrt{2}$. Yeh step kyun? Zero row ka matlab poles imaginary axis par bilkul baith rahe hain; auxiliary equation unki frequency read off karta hai.

Verify: $K=6,s=j\sqrt{2}$ ko $s^3+3s^2+2s+6$ mein substitute karo: $(j\sqrt{2}{)}^3=-2\sqrt{2}j$, $3(j\sqrt{2}{)}^2=-6$, $2(j\sqrt{2})=2\sqrt{2}j$, $+6$. Sum $=(-6+6)+(-2\sqrt{2}+2\sqrt{2})j=0$. ✔ Critical gain $K=6$, $\omega =\sqrt{2}\approx 1.414\text{ rad/s}$.

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Ex 3 — Cell D: ek zero system ko re-stabilise karta hai

Forecast: kya left-half-plane zero add karne se escaping branches steeper ya shallower hoti hain? Kya yeh stability mein madad karta hai?

1. Poles/zeros. $n=3$ ($0,-1,-2$), $m=1$ (zero $-3$ par).
2. Branches to infinity. $n-m=2$, toh sirf do branches escape karti hain (Ex 2 mein teen se kam). Ek branch ab $-3$ par finite zero par terminate hoti hai. Yeh step kyun? Rule 1 — har finite zero ek branch ko "pakadta" hai, toh zero ek target hai jo ek escaping branch ko hata deta hai.
3. Naye asymptotes. ${\theta }_a=\frac{\pm 180°(2k+1)}{2}=90°,270°$ (vertical, $60°$ nahi). Centroid ${\sigma }_a=\frac{(0-1-2)-(-3)}{2}=\frac{-3+3}{2}=0$. Yeh step kyun? Ex 2 ke $60°/300°$ tilt se compare karo: zero ne asymptotes ko seedha vertical kar diya — branches ab right-half-plane mein utni aggressively lean nahi karti.
4. Stability check (Routh). Char. eqn: $s^3+3s^2+2s+K(s+3)=s^3+3s^2+(2+K)s+3K=0$. Routh $s^1$ term: $\frac{3(2+K)-3K}{3}=\frac{6}{3}=2>0$ sabhi $K>0$ ke liye. Yeh step kyun? Yahi payoff hai: $s^1$ Routh entry ek positive constant hai, toh koi $K$ uska sign change nahi karta — system har gain ke liye stable hai. Zero (PD action) ne unconditional stability kharidi.

Verify: Routh first column entries: $1,3,2,3K$. $K>0$ ke liye sab positive — koi sign change nahi → sabhi $K>0$ ke liye stable. ✔ Centroid ${\sigma }_a=0$.

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Ex 4 — Cells B + H: real-world GNC autopilot (word problem)

Forecast: double integrator ka matlab hai do poles origin par chipke hue. Intuition kehta hai "unstable" — check karo.

1. Poles/zeros. $n=3$: $s=0$ par ek repeated pole (multiplicity 2) aur $-4$ par ek pole. $m=0$. Yeh step kyun? Repeated origin pole special feature hai — do branches ek hi point se shuru hoti hain danger axis par.
2. Asymptotes. ${\sigma }_a=\frac{0+0-4}{3}=-\frac{4}{3}$, ${\theta }_a=\frac{\pm 180°(2k+1)}{3}=60°,180°,300°$. Yeh step kyun? Teeno branches escape karti hain ($n-m=3$). $60°$ aur $300°$ asymptotes upar-daahine aur neeche-daahine point karte hain — seedha unstable region ki taraf — jabki $180°$ asymptote teesri branch ko negative real axis ke saath baayi taraf bhejtaa hai. Woh akela left-going branch ek maatra well-behaved wala hai; baaki do right-half-plane mein ud jaate hain. $180°$ branch kaisi dikhti hai: $-4$ par pole $K$ badhne ke saath real axis ke saath $-\infty$ ki taraf left move karta hai, safely stable rehta hai — lekin woh do branches ko rescue nahi kar sakta jo origin se nikalti hain, jo danger axis ke paar daahini taraf almost immediately curve kar jaati hain.
3. Routh test. Char. eqn: $s^3+4s^2+0\cdot s+K=0$. $s^1$ coefficient $0$ hai. $\begin{array}{ccc}{s}^3& 1& 0\\ s^2& 4& K\\ s^1& \frac{4\cdot 0-K}{4}=-\frac{K}{4}& \\ s^0& K& \end{array}$ Yeh step kyun? $s^1$ entry $-K/4$ har $K>0$ ke liye negative hai → first-column mein sign change → sabhi $K$ ke liye kam se kam ek right-half-plane pole.
4. Conclusion. Pure gain ek double integrator ko kabhi stabilise nahi kar sakta. Tumhe ek zero add karna padega (rate feedback / PD) — yahi reason hai real autopilots derivative terms use karte hain. Yeh step kyun? Yahi engineering lesson hai: matrix cell "pole on the axis + no zero" ek guaranteed-unstable trap hai.

Verify: $K=1$ ke liye, $s^3+4s^2+1=0$ ka ek root $s\approx +0.24$ par hai (positive real part) → unstable. ✔ (neeche numerically check kiya).

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Ex 5 — Cell E: complex open-loop poles, angle of departure

Forecast: branch ek slanted pole se nikalti hai — kya woh upar-baayein (stabilising) ya upar-daayein (danger ki taraf) jaati hai?

1. Poles locate karo. $s^2+2s+2=0\Rightarrow s=-1\pm j$. Plus $s=0$ par ek pole. Toh poles: $0,-1+j,-1-j$; $m=0$, $n=3$. Yeh step kyun? Rule 7 (angle of departure) sirf complex poles par apply hota hai; pehle unhe dhundho.
2. Rule 7 ko $p=-1+j$ par set up karo. ${\theta }_d=180°+\underset{=0}{\underbrace{\sum \angle (\text{zeros}\to p)}}-\sum \angle (\text{other poles}\to p).$ Yeh step kyun? $p$ se thoda door ek test point ko abhi bhi angle condition satisfy karni chahiye; departure direction ke liye solve karna ${\theta }_d$ deta hai.
3. $0$ par pole se $p$ tak angle. Vector $p-0=-1+j$ → angle $=135°$.
4. $-1-j$ par pole se $p$ tak angle. Vector $(-1+j)-(-1-j)=2j$ → angle $=90°$. Yeh steps kyun? Har "other pole" us arrow ka angle contribute karta hai jo us pole se $p$ tak kheechi jaati hai; hum unka sum subtract karte hain.
5. Combine karo. ${\theta }_d=180°-(135°+90°)=180°-225°=-45°$ (equivalently $315°$).

Verify: ${\theta }_d=-45°$ ka matlab branch $-1+j$ se neeche-daahine real axis ki taraf jaati hai — symmetric partner ke saath consistent jo $-1-j$ se $+45°$ par jaati hai; unki conjugate symmetry (Rule 2) respect hoti hai kyunki $-45°$ aur $+45°$ mirror images hain. ✔

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Ex 6 — Cell F: degenerate pole–zero cancellation

Forecast: factor $(s+2)$ upar aur neeche dono mein appear karta hai. Kya $-2$ par pole abhi bhi ek branch start karta hai?

1. Pehle cancel karo. $\frac{s+2}{s(s+2)(s+5)}=\frac{1}{s(s+5)}$ $s\ne -2$ ke liye. Yeh step kyun? Common factor ek pole–zero cancellation hai: $-2$ par pole exactly $-2$ par zero se annihilate ho jaata hai. Yeh koi moving branch contribute nahi karta.
2. Effective count. Cancellation ke baad $n=2$ ($0,-5$), $m=0$. Sirf do branches, teen nahi. Yeh step kyun? Degenerate cell branch count ghata deta hai; cancel karna bhoolne par ek fake extra branch milti.
3. Real-axis segment. $(-5,0)$ mein test point: $0$ par pole daahine → count 1 (odd) → $[-5,0]$ locus par.
4. Breakaway. $K=-s(s+5)$, $dK/ds=-(2s+5)=0\Rightarrow s=-2.5$; gain $K=-(-2.5)(2.5)=6.25$. Yeh step kyun? Same procedure jaise Ex 1 lekin reduced system par — breakaway midpoint $-2.5$ par hai.

Verify: $\frac{1}{s(s+5)}$ ka breakaway: midpoint $-2.5$, aur $K=(2.5)(2.5)=6.25$. $K=6.25$ par: $s^2+5s+6.25=(s+2.5{)}^2=0$ ✔ $-2.5$ par double pole.

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Ex 7 — Cell G: unstable open-loop pole ko baayi taraf kheenchna padta hai

Forecast: system janam se unstable hai. Kya $-3$ par zero runaway pole ko left half-plane mein kheench sakta hai, aur kaun se gain ke upar?

1. Poles/zeros. Poles $0,+1$; zero $-3$. $n=2$, $m=1$. Yeh step kyun? Ek escaping branch ($n-m=1$) aur ek branch jo $-3$ par zero par land karni chahiye.
2. Real-axis segments. $0$ se thoda daahine $x$ test karo ($(0,1)$ mein): daahine $+1$ par pole → count 1 (odd) → segment $(0,1)$ locus par. $x<-3$ test karo: $0,+1$ par poles aur $-3$ par zero sab daahine → count 3 (odd) → $(-\infty ,-3)$ locus par. Yeh step kyun? Yeh dikhata hai ek branch $+1$ se baayi taraf jaati hai aur eventually $-3$ tak pahunchti hai: zero rescue target hai.
3. Characteristic equation. $s(s-1)+K(s+3)=s^2+(K-1)s+3K=0$. Yeh step kyun? Humein conditions chahiye ki dono roots ka real part negative ho.
4. Stability conditions (2nd-order: sab coefficients positive).
   • $s^1$ coefficient $K-1>0\Rightarrow K>1$.
   • $s^0$ coefficient $3K>0\Rightarrow K>0$ (auto-satisfied). Yeh step kyun? Ek quadratic $s^2+bs+c$ ke liye, dono roots left half-plane mein hain iff $b>0$ aur $c>0$. Binding wala $K>1$ hai.
5. Conclusion. $K>1$ ke liye stable. $K=1$ par ek pole bilkul $s=0$ par hai (marginal); us se neeche, right-half-plane pole abhi capture nahi hua.

Verify: $K=1$ par: $s^2+0\cdot s+3=0\Rightarrow s=\pm j\sqrt{3}$ — marginal ($\omega =\sqrt{3}$). $K=2$ par: $s^2+s+6=0\Rightarrow s=-0.5\pm j2.398$ — dono left-half-plane, stable. ✔

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Ex 8 — Cell I: exam twist, target damping ratio hit karo

Forecast: hum pehle se jaante hain ki $K>1$ ke liye poles $-1\pm j\beta$ hain. Kaun sa $\beta$ $\zeta =0.5$ deta hai, aur kya ise breakaway se zyada ya kam $K$ chahiye?

1. Closed-loop poles. $s^2+2s+K=0\Rightarrow s=-1\pm \sqrt{1-K}$. $K>1$ ke liye: $s=-1\pm j\sqrt{K-1}$. Yeh step kyun? Humein standard 2nd-order form $s=-\zeta {\omega }_n\pm j{\omega }_n\sqrt{1-{\zeta }^2}$ mein pole chahiye.
2. Natural frequency match karo. ${\omega }_n^2=K$ (constant term) toh ${\omega }_n=\sqrt{K}$; real part $-\zeta {\omega }_n=-1\Rightarrow \zeta {\omega }_n=1$. Yeh step kyun? Yeh do identities humein directly $\zeta$ se $K$ solve karne deti hain.
3. $K$ ke liye solve karo. $\zeta =\frac{1}{{\omega }_n}=\frac{1}{\sqrt{K}}=0.5\Rightarrow \sqrt{K}=2\Rightarrow K=4$. Yeh step kyun? $\zeta =0.5$ se ${\omega }_n=2$ fix hota hai, toh $K={\omega }_n^2=4$. (Note $K=4>1$: breakaway ke past, jaise forecast kiya.)
4. Damping-ray geometry. $\zeta =\cos \theta$ jahan $\theta$ negative real axis se angle hai. $\zeta =0.5\Rightarrow \theta =60°$. Yeh step kyun? Locus par tum origin se $60°$ ray draw karte aur dekhte ki woh vertical branch ko kahan hit karta hai — gain set karne ka purely geometric tarika (dekho Damping ratio and settling time).
5. Settling time. $t_s\approx \frac{4}{\zeta {\omega }_n}=\frac{4}{1}=4\text{ s}$ (2% criterion). Yeh step kyun? $\zeta {\omega }_n=1$ pole ka real part hai; settling time sirf usi par depend karta hai.

Verify: $K=4\Rightarrow s^2+2s+4=0\Rightarrow s=-1\pm j\sqrt{3}$. Phir ${\omega }_n=|s|=\sqrt{1+3}=2$ ✔, $\zeta =1/{\omega }_n=0.5$ ✔, $t_s=4/(0.5\cdot 2)=4$ s ✔.

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Recall

Kaun sa example cell tumhe sikhata hai ki pure gain ek double integrator ko kabhi fix nahi kar sakta? ::: Ex 4 (cells B+H) — $s^1$ Routh entry $-K/4<0$ sabhi $K>0$ ke liye hai. Ex 3 mein, $-3$ par zero add karne se asymptote centroid par kya hua? ::: Woh $-1$ (Ex 2) se $0$ par shift ho gaya aur asymptotes ko vertical kar diya, sabhi $K>0$ ke liye stability dete hue. Ex 8 mein, kaun sa gain $\zeta =0.5$ deta hai aur settling time kya hai? ::: $K=4$, poles $-1\pm j\sqrt{3}$, $t_s\approx 4$ s.

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Connections

• Root locus — Evans' method, rules for sketching — parent rules jo poore mein use hue
• Characteristic equation & closed-loop poles — woh single equation jo har example ke peeche hai
• Routh–Hurwitz stability criterion — Ex 2, 3, 4 crossings & stability ranges
• PID / PD controller design — Ex 3 & 4: re-stabilise karne ke liye zero add karna
• Damping ratio and settling time — Ex 8 target-damping design
• GNC control loops — Ex 4 autopilot context
• Nyquist criterion — same $L(s)=-1$ ka encirclement view
• Bode plot & frequency response — frequency-domain companion