Root locus — Evans' method, rules for sketching: Visual walkthrough

3.5.40 · D2 · Physics › Guidance, Navigation & Control (GNC) › Root locus — Evans' method, rules for sketching

Pehli line se pehle, char seedhe alfaaz:

Step 1 — " $L (s) = - 1$ " asal mein kya pooch raha hai

KYA. Parent note ne dikhaya ki closed-loop poles satisfy karte hain $1 + L (s) = 0$ , yaani

loop formula us point s par L (s) = ek specific complex number - 1 .

Toh hum size solve nahi kar rahe — hum pooch rahe hain: kin points $s$ par loop formula exactly $- 1$ ke barabar hota hai?

$- 1$ kyun, aur say $0$ kyun nahi? Kyunki feedback subtract karta hai. Loop ko aisa signal wapas laana chahiye jo input ko cancel kare ( $1 + L = 0$ ). Wo number jo "same size, flipped direction" represent karta hai, wo hai $- 1$ .

PICTURE. Har complex number origin se ek arrow hai: ek length aur ek direction (angle $∠$ ). $- 1$ length $1$ wala arrow hai jo seedha left point karta hai — direction $180°$ .

Figure — Root locus — Evans' method, rules for sketching

Toh " $L (s) = - 1$ " chhupe taur par do demands rakhta hai: arrow ki length $1$ honi chahiye, aur uska angle $180°$ hona chahiye. Inhe alag karo aur Evans' ke do conditions milte hain.

Step 2 — Ek demand ko do mein todna (length aur angle)

KYA. Koi bhi complex number $z$ ko aise likha ja sakta hai (length) $\times$ (unit arrow at its angle):

z = length = magnitude ∣ z ∣ \cdot unit arrow = e^{j a r g (z)} ∠ z .

$L (s) = - 1$ se match karte hue:

magnitude condition ∣ L (s) ∣ = 1 AND angle condition ∠ L (s) = 180° .

Todna kyun? Kyunki $L (s) = K G (s) H (s)$ aur $K$ ek positive real number hai — ek seedha stretch, koi spin nahi. Ek positive number ka angle $0°$ hota hai. Toh $K$ arrow ki direction kabhi nahi badal sakta; wo sirf uski length badalta hai.

$L (s)$ ka angle depend karta hai sirf $G (s) H (s)$ par, kabhi $K$ par nahi.
Length mein $K$ rehta hai, jise dial kiya jaata hai taaki total length $1$ ho jaaye.

PICTURE. Knob $K$ ghumaana arrowhead ko uski apni direction mein slide karta hai (stretch), lekin rotate kabhi nahi karta.

Yahaan $k$ sabhi integers $Z = {\dots, - 2, - 1, 0, 1, 2, \dots}$ par range karta hai. Isse dono turning directions ek saath cover hote hain: $k = 0$ deta hai $+ 180°$ , $k = - 1$ deta hai $- 180°$ , $k = 1$ deta hai $+ 540°$ , wagera. Factor $(2 k + 1)$ (ek odd number) isliye hai kyunki "seedha left" point karna har odd multiple of $180°$ par hota hai; $k$ ko sabhi integers par run karne dena hi reason hai ki hume alag $\pm$ sign ki zaroorat nahi.

Step 3 — Ek pole/zero kya contribute karta hai: ek akela angle

KYA. $G (s) H (s)$ ko uske factors ke zariye likho. Agar $z_{1}$ par zero aur $p_{1}, p_{2}$ par poles hain:

G (s) H (s) = \frac{s - z _{1}}{( s - p _{1} ) ( s - p _{2} )} .

Term $s - p_{1}$ pole $p_{1}$ se humare test point $s$ ki taraf ka arrow hai. Uska angle $∠ (s - p_{1})$ batata hai ki $p_{1}$ se $s$ tak kitna steep dekh rahe ho.

Arrows kyun? Kyunki " $s - p_{1}$ " literally matlab hai " $p_{1}$ se shuru karo, $s$ tak jao" — wo ek directed arrow hai, aur directed arrows ka angle hota hai.

Angle bookkeeping — product/quotient ke liye, angles add aur subtract hote hain:

∠ G (s) H (s) = zero apna angle add karta hai ∠ (s - z_{1}) - har pole apna angle subtract karta hai ∠ (s - p_{1}) - ∠ (s - p_{2}) .

PICTURE. Kahin bhi ek test point $s$ rakho. Har ✕ aur ◯ se $s$ ki taraf arrow kheecho. Har arrow ka angle horizontal se measure karo. Zeros plus count hote hain, poles minus count hote hain.

Step 4 — Real-axis rule khud nikal aata hai free mein

KYA. Ek aisa point test karo jo horizontal (real) axis par baitha ho. Kaunse real poles/zeros use locus par rakhte hain?

Sirf RIGHT waale kyun count karte hain? Arrows dekho:

$s$ ke left ka ek real dot rightward point karta hai → angle $0°$ → kuch contribute nahi karta (chahe wo ✕ ho ya ◯).
$s$ ke right ka ek real dot leftward point karta hai → angle $180°$ → $\pm 180°$ contribute karta hai.
Ek complex-conjugate pair (ek upar, ek neeche) do arrows bhejta hai jinke angles $+ α$ aur $- α$ hote hain → wo cancel ho jaate hain $0°$ par.

Poles ( $- 180°$ ) aur zeros ( $+ 180°$ ) dono ek hi odd test kyun pass karte hain? Step 3 se, right side ka zero $+ 180°$ add karta hai aur right side ka pole $180°$ subtract karta hai, yaani $- 180°$ contribute karta hai. Lekin $+ 180°$ aur $- 180°$ same direction hain — inme $360°$ ka fark hai, ek poora chakkar. Angle condition ke liye hume bas itna chahiye ki total odd multiple of $180°$ ho, aur dono $+ 180°$ aur $- 180°$ odd multiples hain. Toh har right-side dot — pole ya zero — parity ko ek se flip karta hai, aur sirf count (sign nahi) matter karta hai.

Toh total angle tab odd multiple of $180°$ hoga jab right side real dots (poles + zeros) ki taadaad odd ho. ✔

PICTURE. Neeche: do poles ke beech ka ek point apni right par exactly ek dot dekhta hai → total $180°$ → locus par hai. Dono ke right wala point do dekhta hai → total $360°$ → locus par nahi.

right mein odd count

point locus par hai (total angle

180°

ka odd multiple hai)

right mein even count

point locus se bahar hai (total angle

0°

360°

hai)

Step 5 — Beads axis chhodti kahaan hain: breakaway

KYA. Do poles $s = 0$ aur $s = - 2$ ke beech (parent ka Example 1), dono beads ek doosre ki taraf slide karti hain, takraati hain, phir vertically peel off karti hain. Takraav ka point breakaway hai.

$d K / d s = 0$ ise kyun mark karta hai? Definitions box se yaad karo ki $N (s)$ numerator polynomial hai (roots = zeros) aur $D (s)$ denominator polynomial hai (roots = poles); yahaan $N (s) = 1$ aur $D (s) = s (s + 2)$ . Characteristic equation $1 + K \frac{N}{D} = 0$ rearrange hoke $D + K N = 0$ deta hai, toh gain ke liye solve kar sakte hain:

K = - \frac{D ( s )}{N ( s )} = - \frac{s ( s + 2 )}{1} = - s (s + 2) .

Real segment par, jaise $s$ $0$ se andar move karta hai, $K$ badhta hai; meeting point ke baad use real poles rakhne ke liye ghatna hoga — real gain ke liye impossible, toh poles complex ho jaate hain. Rising se falling ka switch ek peak hai: $\frac{d K}{d s} = 0$ .

\frac{d K}{d s} = - (2 s + 2) = 0 \Rightarrow s = - 1, K = - (- 1) (1) = 1.

PICTURE. $K$ ko segment $[- 2, 0]$ par $s$ ke against plot karo: ek smooth hill jiska peak $s = - 1$ , $K = 1$ par hai. Wo peak hi breakaway hai; uske baad beads seedha upar aur neeche climb karti hain.

Step 6 — Escaping branches kahaan jaati hain: angle condition se asymptotes

KYA. Dono beads jo $s = - 1$ par peel off hoti hain, unhe $K \to \infty$ par kahaan jaana hai. Definitions box ke counting letters use karte hue ( $n = 2$ poles, $m = 0$ zeros), branches ki taadaad jo infinity tak escape hoti hain wo $n - m = 2$ hai. Wo straight rays approach karti hain. Is step mein unke angles aur unka launch point seedha angle condition se derive kiya gaya hai — kuch yaad nahi karna.

$n - m$ rays kyun, aur ye angles kyun? Ek test point $s$ jo bahut door ho use dekho. Itni door se saare poles aur zeros ek same spot par sikudte dikh'te hain, toh $s$ ki taraf kheecha har arrow lagbhag same angle $θ$ rakhta hai. $m$ zeros (har ek $+$ ) aur $n$ poles (har ek $-$ ) ke saath, Step 3 ki angle bookkeeping banti hai

∠ G (s) H (s) \approx zeros, + m θ - poles, - n θ = - (n - m) θ .

Ise required $180° (2 k + 1)$ ke barabar rakh ke ray angle solve karo:

θ_{a} = \frac{180° ( 2 k + 1 )}{n - m} .

$n - m = 2$ ke liye, $k$ step karte hue milta hai $k = 0 \Rightarrow 90°$ , $k = 1 \Rightarrow 270°$ , aur $k = 2$ wapas $90° + 360°$ par aa jaata hai — toh repeat hone se pehle exactly $n - m$ distinct directions milti hain. Isliye $n - m$ asymptotes hote hain.

Launch point centroid kyun hai? Rays origin se nahi, dots ke average se nikalti hain. Yeh "next-term matching" spelled out hai. Large $∣ s ∣$ ke liye escaping branches $D (s) / N (s)$ ke ratio se govern hoti hain:

\frac{D ( s )}{N ( s )} = \frac{s ^{n} + d _{1} s ^{n - 1} + \dots}{s ^{m} + e _{1} s ^{m - 1} + \dots} = s^{n - m} + (d_{1} - e_{1}) s^{n - m - 1} + \dots

jahaan $d_{1} = - \sum (poles)$ aur $e_{1} = - \sum (zeros)$ root-sums ke negatives hain (Vieta). Hum chahte hain ek hi pole of multiplicity $n - m$ kisi point $σ_{a}$ par jo is far-field se match kare, yaani $D / N \approx (s - σ_{a})^{n - m}$ . Wo binomial expand karte hue,

(s - σ_{a})^{n - m} = s^{n - m} - (n - m) σ_{a} s^{n - m - 1} + \dots

Dono expressions ka second coefficient match karte hue:

- (n - m) σ_{a} = d_{1} - e_{1} = - \sum (poles) + \sum (zeros) \Rightarrow σ_{a} = \frac{\sum ( poles ) - \sum ( zeros )}{n - m} .

Har symbol: $\sum (poles)$ har ✕-location jodhta hai, $\sum (zeros)$ har ◯-location ghataata hai, aur $n - m$ se divide karke real axis par unka balance point milta hai. Hamare liye $σ_{a} = \frac{0 + ( - 2 ) - 0}{2} = - 1$ .

PICTURE. $- 1$ par centroid aur $90°$ aur $270°$ par climb karti do rays — beads ke highway signs.

n - m

for

1/ [s (s + 2)]

2

→ do asymptotes

90°

aur

270°

par

centroid

σ_{a}

(0 + (- 2)) /2 = - 1

Step 7 — Degenerate & edge cases (kabhi koi gap mat chhodna)

KYA. Chaar scenarios jinpar beginner trip karta hai:

Ek pole bilkul test point par baitha ho ( $s = p_{1}$ ). Toh $s - p_{1} = 0$ : arrow ki zero length hai, uska angle undefined hai. Theek hai — yahi ek branch ka start hai ( $K = 0$ ), interior test point nahi.
Ek zero bilkul test point par baitha ho ( $s = z_{1}$ ). Toh numerator arrow $s - z_{1} = 0$ vanish ho jaata hai, toh $∠ (s - z_{1})$ bhi undefined hai aur $∣ G (s) H (s) ∣ = 0$ . Magnitude condition se $K = 1/∣ G (s) H (s) ∣ \to \infty$ : yahi ek branch ka end hai ( $K \to \infty$ ), wo target jis taraf bead aim kar rahi hai — ye bhi interior test point nahi.
Test point bilkul ek real singularity par. Angle discontinuously jump karta hai (arrow flip hota hai jab cross karo). Rule un points ke beech ke liye hai, unke upar nahi.
Saare poles upar aur neeche cancel ho jaate hain (pure conjugate stack, right mein koi real dot nahi). Count $= 0$ , jo even hai → us axis ki stretch locus par nahi hai. Zero even hota hai.

Ye track kyun karein? Kyunki sketch tabhi trustworthy hai jab map ka har square account ho — khaali wale bhi.

PICTURE. Ek number line jo green hai jahaan locus hai, gray jahaan nahi, har stretch ke neeche odd/even count printed, ✕ start-points explicitly plot kiye.

zero real dots to the right

even count → locus par NAHI

s = z_{1}

(zero test point par)

branch END,

K \to \infty

— interior test point nahi

Step 8 — Danger line kyun kabhi cross nahi hoti (angle + magnitude, assertion nahi)

KYA. Imaginary axis (origin se upar-neeche) border hai: iske par closed-loop pole ka matlab steady oscillation, aur iske right ka matlab blow-up. $1/ [s (s + 2)]$ ke liye hum ab prove karte hain ki locus strictly iske left par rehti hai.

Crossing kyun nahi? Do facts jo hum pehle se earn kar chuke hain, milkar kaam karte hain:

Real-axis piece: axis par sirf on-locus real stretch $[- 2, 0]$ hai (Step 4) — jo poori tarah real part $\leq 0$ par hai.
Escaping pieces: dono branches breakaway $s = - 1$ se vertical asymptote $σ_{a} = - 1$ ke saath nikalti hain (Step 6). Real part $- 1$ par ek vertical line kabhi real part $0$ tak nahi pahunchti.

Axis par ek would-be crossing point rakho, $s = j ω$ , aur magnitude condition test karo: $K = ∣ s ∣ ∣ s + 2∣ = ω^{2} \cdot ω^{2} + 4$ . Ye har $ω$ ke liye ek positive real number hai, lekin branch ka real part asymptote ke zariye $- 1$ par pin hai — pole $- 1 + j ω$ par hai, axis par nahi. Koi bhi $(s, K \geq 0)$ pair aisa nahi hai jo locus par bhi ho aur axis par bhi.

Conclusion (stability): har closed-loop pole ka real part sabhi $K > 0$ ke liye $\leq 0$ hai — system unconditionally stable hai. (Routh–Hurwitz stability criterion se contrast karo, jo ek finite critical gain dhundhta hai jab real crossing exist karti hai — jaise parent ke Example 2 mein.)

PICTURE. Red danger line draw ki, locus visibly real part $- 1$ ko hug kar rahi hai aur kabhi touch nahi karti.

Ek-picture summary

KYA. Upar ki sab cheez, ek map par $G (s) H (s) = \frac{1}{s ( s + 2 )}$ ke liye: do ✕ poles, green on-locus segment $[- 2, 0]$ (Step 4), breakaway $s = - 1, K = 1$ par (Step 5), do branches vertically peel off karti hain centroid $σ_{a} = - 1$ ke saath asymptote angles $90°/270°$ par (Step 6), aur red danger line (imaginary axis) jise locus kabhi cross nahi karti (Step 8) — toh sabhi $K > 0$ ke liye stable.

Recall Feynman: poori walk seedhe alfaaz mein

Ek feedback loop ko chahiye ki uska formula exactly number negative one ke barabar ho — length one ka arrow jo seedha left point kare. Left-pointing ek direction hai (ek angle), aur knob $K$ sirf ek positive stretch hai jo kisi arrow ko kabhi rotate nahi kar sakta. Toh direction wala part ( $180°$ ) sirf geometry se fix hota hai, aur $K$ sirf length ko $1$ tak fine-tune karta hai. Test karne ke liye ki tum kisi path par ho, har ✕ (pole) aur ◯ (zero) ki taraf peeche dekho, tumhari taraf point karne wale arrow ka angle padhlo, zeros add karo aur poles subtract karo — agar total $180°$ hai, tum locus par ho. Flat real axis par ye khoobsurati se simplify hota hai: tumhare left ke dots door point karte hain ( $0°$ , ignore karo), tumhare right ke dots tumhari taraf point karte hain ( $180°$ ), aur mirror-image complex pairs cancel ho jaate hain — toh bas right side dots count karo, aur odd number chahiye (poles aur zeros same count karte hain kyunki $+ 180°$ aur $- 180°$ same direction hai). Do beads do poles se shuru hoke saath slide karti hain, milti hain jahaan gain $K$ apni chhoti hill ki top par hota hai ( $d K / d s = 0$ ), phir straight ramps par ud jaati hain. Ye ramps kahaan aim karti hain jaanne ke liye, bahut door se khado: saare dots ek spot mein blur ho jaate hain, har arrow ek angle $θ$ share karta hai, aur angle condition force karti hai $θ = 180° (2 k + 1) / (n - m)$ — yahaan $90°$ aur $270°$ , dots ke balance-point $- 1$ se launch hoke. Kyunki wo ramp $- 1$ par ek vertical wall hai, aur only real path bhi zero se left hai, koi bead kabhi red line (imaginary axis) ko touch nahi kar sakti — machine har knob setting par calm rehti hai.

Connections

Root locus — Evans' method, rules for sketching — wo parent jise ye page derive karta hai
Characteristic equation & closed-loop poles — jahaan $1 + L (s) = 0$ aata hai
Routh–Hurwitz stability criterion — $j ω$ crossing dhundhta hai jab exist karti hai
PID / PD controller design — zeros (◯) add karta hai locus ko leftward bend karne ke liye
Damping ratio and settling time — vertical branch ka real part damping set karta hai
Nyquist criterion — wahi $L (s) = - 1$ encirclement ke roop mein dekha
Bode plot & frequency response — magnitude/angle split, frequency ke against dekha
GNC control loops — physical loops jinhe ye tune karta hai