Root locus — Evans' method, rules for sketching: Question bank

3.5.40 · D5 · Physics › Guidance, Navigation & Control (GNC) › Root locus — Evans' method, rules for sketching

Do rules ki yaad dilata hoon jis par sab kuch tika hai:

Angle condition $∠ G (s) H (s) = \pm (2 k + 1) \times 180°$ jahan $k = 0, 1, 2, \dots$ — yeh locus ki shape decide karta hai (kaunse points locus par hain). Factor $(2 k + 1)$ sirf odd multiples $180°, 540°, \dots$ enumerate karta hai (sab physically same angle mod $360°$ hain); alag $k$ alag branches ko isse satisfy karne deta hai.
Magnitude condition $K = 1/∣ G (s) H (s) ∣$ — yeh har point par gain label decide karta hai.

Intuition Denominator

$n - m$ kyun hai, aur centroid average kyun hai Neeche ke figures dekho. Origin se bahut door, har pole aur zero lagbhag ek hi jagah par dikh ta hai — utni door se unki individual positions resolve nahi hoti. Toh $G (s) H (s) = \frac{N ( s )}{D ( s )}$ collapse hokar $\frac{1}{( s - σ _{a} ) ^{n - m}}$ ban jaata hai: bacha hua $n - m$ poles (poles minus woh $m$ jo zeros infinity par "cancel" kar dete hain) ek single pole of multiplicity $n - m$ ki tarah behave karta hai jo apne balance point $σ_{a}$ par rakha ho. Us akele effective pole par angle condition, $(n - m) \times ∠ (s - σ_{a}) = \pm (2 k + 1) \times 180°$ , seedha asymptote-angle formula deta hai. Balance point $σ_{a}$ average isliye hai kyunki $D / N$ ke polynomial long division ke do leading terms match karne par effective pole exactly (sum of poles − sum of zeros)/(n−m) par aata hai.

True ya false — justify karo

Root locus dikhata hai ki

K

badhne par OPEN-loop poles kaise move karte hain.

False — open-loop poles fixed hote hain (

D (s)

ke roots,

K

se independent). Locus closed-loop poles track karta hai, jo

K = 0

par open-loop poles par shuru hote hain aur

K

badhne par door jaate hain.

Locus ki har branch kisi finite open-loop zero par khatam hoti hai.

False — sirf

m

branches (finite zeros ki sankhya,

N (s)

ke roots) zeros par khatam hoti hain; baaki

n - m

branches asymptotes ke along infinity ki taraf nikal jaati hain. Agar koi finite zero na ho toh saari

n

branches infinity ki taraf chali jaati hain.

Locus hamesha real axis ke baare mein symmetric hota hai.

True — characteristic polynomial ke real coefficients hote hain, isliye koi bhi complex closed-loop pole apne conjugate ke saath aata hai. Map upar-neeche mirror image hota hai.

Agar saare open-loop poles left half plane mein hain, toh closed-loop system har

K > 0

ke liye stable hai.

False —

K

badhne par branches right ki taraf migrate karke imaginary axis cross kar sakti hain (jaise

1/ [s (s + 1) (s + 2)]

K = 6

par unstable ho jaata hai). Left-half open-loop poles large-gain behaviour ke baare mein kuch guarantee nahi karte.

Centroid

σ_{a}

woh point hai jahan locus imaginary axis cross karta hai.

False —

σ_{a}

sirf woh jagah hai jahan asymptotes real axis par milte hain. Actual

j ω

-crossing Routh–Hurwitz se milta hai aur usually alag value hoti hai.

Ek zero add karna (jaise PD controller karta hai) locus ko left half plane ki taraf wapas mod sakta hai.

True — ek finite zero (

N (s)

ka root) ek branch ko attract karta hai, asymptote count

n - m

badal deta hai aur aksar escaping branches ko stable side ki taraf kheench leta hai. Exactly aise hi PD control damping add karta hai.

Angle condition aur magnitude condition dono independently poora locus determine karte hain.

False — sirf angle condition shape determine karta hai (kaunse points locus par hain). Magnitude condition sirf un points ko unki

K

value se label karta hai; woh kuch naya nahi kheenchta.

K

badhane se response hamesha faster aur better damped hoti hai.

False — zyada

K

typically poles ko kam damping ratio ki taraf le jaata hai (branches

j ω

ki taraf curl karti hain). Yeh response speed kar sakta hai lekin damping kam kar sakta hai, jisse overshoot ya instability aa sakti hai.

Galti pakdo

"Ek real point locus par hai agar uske daayein poles hain." — kya galat hai?

Do galtiyan hain: daayein poles aur zeros dono count hone chahiye, aur count odd hona chahiye, sirf nonzero nahi. Even count (jaise daayein do poles) point ko locus se bahar rakhta hai.

m

branches isliye hain kyunki

m

zeros hain." — theek karo.

Branches ki sankhya

n

ke barabar hoti hai, open-loop poles ki sankhya (polynomial

D (s) + K N (s)

s

mein degree

n

hoti hai jab

n \geq m

). Inme se

m

zeros par khatam hoti hain aur

n - m

infinity ki taraf jaati hain.

"Breakaway point hamesha do poles ke beech ka midpoint hota hai." — yeh sirf kab sahi hai?

Symmetry breakaway (real point jahan branches axis chhod ti hain) ko do real poles ke beech hone par majboor karti hai, lekin exact location

d K / d s = 0

solve karne se aati hai. Sirf symmetric pair jaise

{0, - 2}

ke liye hi woh arithmetic midpoint par hota hai; paas ka zero ise shift kar deta hai.

K = 1/∣ G (s) H (s) ∣

negative ho sakta hai, isliye locus mein negative gains bhi shamil hain." — theek karo.

Standard root locus ke liye

K \geq 0

hai, aur

∣ G (s) H (s) ∣

ek magnitude hai (hamesha

\geq 0

), isliye

K \geq 0

automatically aata hai. Negative

K

ek complementary locus trace karta hai jo angle condition

∠ = 0°

use karta hai — yeh ek alag diagram hai.

"Breakaway point par gain

K

zero hoti hai." — asli statement kya hai?

Breakaway par do branches milti hain, matlab polynomial ka repeated root hai, isliye

K (s) = - D (s) / N (s)

ek local extremum (

d K / d s = 0

) par hota hai — zaroori nahi ki zero ho.

1/ [s (s + 2)]

ke liye breakaway gain

K = 1

hai.

"Asymptote angles is baat par depend karte hain ki poles kahan hain." — kyun galat hai?

Angles

θ_{a} = \pm (2 k + 1) \times 180°/ (n - m)

sirf count

n - m

par depend karte hain, pole positions par nahi. Pole positions centroid

σ_{a}

(jahan asymptotes hain) ko affect karti hain, unke slopes ko nahi.

Why questions

Locus par koi point angle condition se kyun govern hota hai, magnitude se kyun nahi?

Kyunki

L (s) = - 1

do parts mein split hota hai: "magnitude

= 1

" aur "angle

= 180°

". Kisi bhi point ke liye, hum hamesha magnitude fix karne ke liye ek

K

choose kar sakte hain, isliye magnitude kabhi kisi point ko rule out nahi karta — sirf fixed angle requirement karta hai.

Complex-conjugate poles kisi real test point ko net zero angle contribution kyun dete hain?

Conjugate pair real axis ke upar aur neeche symmetrically baitha hota hai. Kisi bhi real point par unke angle contributions equal aur opposite hote hain, isliye cancel ho jaate hain — yahi reason hai ki real-axis rule ke liye sirf real singularities matter karti hain.

Far-field locus centroid par multiplicity

n - m

ke single pole jaisa kyun dikhta hai?

∣ s ∣ \to \infty

ke liye individual poles aur zeros unresolvable ho jaate hain, isliye division

D (s) / N (s)

ke do leading terms match karne ke baad

G (s) H (s) \approx 1/ (s - σ_{a})^{n - m}

ho jaata hai (figure s02 dekho). Pura cluster unke weighted average

σ_{a}

par ek effective pole mein blur ho jaata hai, aur uss par angle condition apply karne se asymptote angles milte hain.

Departure angle formula

180°

se kyun shuru hota hai?

Yeh complex pole ke infinitesimally paas ek test point par angle condition

∠ G (s) H (s) = 180°

apply karne se aata hai (figure s03 dekho).

180°

target hai; doosre poles/zeros se angles ke sums woh corrections hain jo isse hit karne ke liye chahiye.

j ω

crossing ke liye angle condition ki jagah Routh–Hurwitz kyun use karte hain?

Angle condition shape batata hai lekin yeh nahi batata ki kaunsa

K

ek pole ko exactly

j ω

par push karta hai. Routh–Hurwitz directly woh critical

K

deta hai jo ek row ko vanish karta hai, aur uska auxiliary equation crossing frequency

ω

deta hai — stability boundary.

"Pehle shape, phir gain" — correct mental order kyun hai?

Geometry (real-axis segments, asymptotes, breakaways) poori tarah pole/zero positions se angle condition ke zariye decide hoti hai —

K

se independent. Sirf path kheenchne ke baad magnitude condition use karke uspar

K

-values stamp karte hain.

Edge cases

Ek single open-loop pole aur koi zero nahi,

G (s) H (s) = 1/ (s + a)

— locus kaisa dikhega?

Ek branch

s = - a

se shuru hoti hai, aur

n - m = 1

hone se akela asymptote

180°

par hai, isliye pole real axis par seedha left ki taraf

- \infty

ki taraf slide karta hai. Har

K > 0

ke liye stable.

Ek hi jagah par do real poles (repeated pole), jaise

1/ (s + a)^{2}

—

K = 0^{+}

par kya hota hai?

Dono branches double pole

s = - a

par shuru hoti hain, phir turant real axis ke perpendicular break karti hain (breakaway pole hi hai). Yeh vertically upar/neeche jaati hain kyunki

n - m = 2

asymptotes

90°, 270°

par deta hai.

Origin par imaginary axis par baitha ek pole,

G (s) H (s) = 1/ s

— kya origin locus par hai?

Origin

K = 0

par branch ka start hai (yeh open-loop pole hai).

K

badhne par closed-loop pole left ki taraf axis se door ho jaata hai, isliye system

K > 0

ke liye stable hai — marginal open-loop pole ke bawajood.

Agar ek open-loop pole aur zero exactly cancel ho jaaye, jaise

G (s) H (s) = \frac{( s + 2 )}{( s + 2 ) ( s + 5 )}

Common factor

(s + 2)

cancel ho jaata hai, isliye yeh na locus start hai na end — inke beech koi branch nahi chalti; effectively

n

aur

m

dono ek ek kam ho jaate hain aur locus

1/ (s + 5)

wala ban jaata hai. Savdhan: exact cancellation real hardware mein kabhi nahi hoti, isliye near-cancelled pole ek slow, poorly-controllable mode chhupa leta hai (hidden pole) jo reduced system ke root locus mein nahi dikhega.

Agar

n = m

ho (poles aur zeros barabar)?

Tab

n - m = 0

: koi branch infinity ki taraf nahi jaati, saari

n

branches finite zeros par khatam hoti hain. Draw karne ke liye koi asymptote nahi. Yeh proper-but-not-strictly-proper loops mein hota hai.

Ek zero jo real axis par saare poles ke daayein rakha gaya ho — kya us zero ke aage ka segment locus par hai?

Sabse daayein singularity ke daayein, daayein taraf poles+zeros ki count zero hai (even), isliye woh segment locus se bahar hai. Rule sabkuch daayein count karta hai, zeros bhi.

Agar do branches real axis par milti hain aur dono use chhod deti hain, toh wahan

K

ke baare mein kya sach hai?

Woh breakaway point par milti hain jahan

K

real axis ke along local maximum par hota hai (

d K / d s = 0

);

K

aur badhane par poles complex ho jaate hain, isliye woh conjugate pair ke roop mein axis chhod deti hain.

Kya root locus kabhi poori tarah real axis se bahar ho sakta hai, bina kisi real-axis segment ke?

Sirf tab jab real-axis odd-count rule kabhi satisfy na ho — jaise pure complex-conjugate pole pair bina kisi real pole ya zero ke. Tab branches plane mein curve karti hain real axis ko touch kiye bina, sirf symmetry considerations se.

Connections

Root locus — Evans' method, rules for sketching — woh parent jise yeh bank test karta hai
Characteristic equation & closed-loop poles
Routh–Hurwitz stability criterion — $j ω$ -crossing tool
PID / PD controller design — zeros as reshaping tools
Bode plot & frequency response
Nyquist criterion
Damping ratio and settling time
GNC control loops