Nyquist stability criterion — encirclements of −1: Question bank

3.5.43 · D5 · Physics › Guidance, Navigation & Control (GNC) › Nyquist stability criterion — encirclements of −1

Reminders jo tumhe baar baar chahiye honge:

$L (s) = G (s) H (s)$ open-loop transfer function hai; hum $L (j ω)$ plot karte hain.
$Z = N + P$ : $Z$ = right-half-plane (RHP) closed-loop poles, $N$ = $- 1$ ke clockwise encirclements, $P$ = RHP open-loop poles.
Stable $⟺ Z = 0 ⟺ N = - P$ .

True or false — justify karo

Nyquist plot hamesha origin ko encircle karta hai jab loop unstable ho

False. Hum

== - 1 ==

ke wraps count karte hain, origin ke nahi. Origin ke encirclements

1 + L

se correspond karte hain, lekin hum

L

draw karte hain, isliye danger point ek left shift hokar

- 1

pe aa jaata hai.

Agar kisi system mein

P = 0

open-loop RHP poles hain, toh

- 1

ka koi bhi encirclement instability matlab hai

True $P = 0$ ke liye. Tab

Z = N

hoga, toh koi bhi

N \neq = 0

matlab

Z \neq = 0

. Yeh claim quietly

P = 0

assume karta hai;

P > 0

ke saath yeh collapse ho jaata hai.

- 1

ka ek counter-clockwise encirclement hamesha stability mein madad karta hai

Saamanya mein False. CCW se

N < 0

milta hai, jo

Z = N + P

ko kam karta hai. Yeh tabhi madad karta hai jab positive

P

cancel ho raha ho;

P = 0

plant ke liye ek CCW wrap

Z < 0

bana deta hai, jo impossible hai aur plotting ya counting error indicate karta hai.

Nyquist criterion bata deta hai ki unstable poles kahan hain

False. Yeh sirf unhe count karta hai (

Z

). Yeh answer karta hai "RHP mein kitne closed-loop poles hain?" bina kisi ko locate kiye — yahi iska power hai (koi root-solving nahi).

Gain

K

add karna poora Nyquist plot ko origin ke baare mein radially scale kar deta hai

True.

L \to K L

har point ki distance origin se

K

se multiply karta hai aur uska angle preserve karta hai, toh curve origin ke around stretch/shrink hoti hai — jo change karta hai ki fixed point

- 1

andar aata hai ya nahi.

Ek stable open-loop plant stable closed loop guarantee karta hai

False.

P = 0

ka sirf matlab hai

Z = N

; agar plot

- 1

ko wrap kare (

N > 0

) toh closed loop unstable hai. Stable-open ka matlab stable-closed nahi hota.

Ek unstable open-loop plant ko kabhi bhi feedback se stabilise nahi kiya ja sakta

False. Example 3 (

L = K / (s - 1)

P = 1

) tab stabilise hota hai jab plot

- 1

ko ek baar CCW encircle kare (

N = - 1

), jisse

Z = 0

milta hai. Feedback RHP poles ko LHP mein shift kar sakta hai.

Nyquist contour counter-clockwise traverse hota hai

False (is vault ki convention mein). Yeh clockwise traverse hota hai taaki positive

N

clockwise image encirclements count kare; argument principle tab seedha

N = Z - P

padhta hai.

L (- j ω)

ek independent curve hai jise alag se compute karna padta hai

False.

L (- j ω) = \overline{L (j ω)}

, isliye negative-frequency branch sirf positive branch ka real axis ke baare mein mirror image hai — koi nayi computation nahi.

Strictly proper

L

ke liye, contour ka infinite semicircle ek nonzero arc mein map hota hai jise dhyan se draw karna padta hai

False. Strictly proper ka matlab hai

∣ L ∣ \to 0

jab

∣ s ∣ \to \infty

, isliye bada semicircle origin pe collapse ho jaata hai — ek arc nahi, ek single point.

Error dhundo

L (0) = 1 > 0

, isliye plot positive real axis pe start karta hai aur kabhi

- 1

tak nahi pahunch sakta; therefore stable."

Error: starting point ka positive hona kuch nahi kehta ki curve baad mein

- 1

ke around loop karega ya nahi. Stability poori closed curve ke

- 1

ke encirclement ke baare mein hai,

L (0)

ke sign ke baare mein nahi.

"Plot exactly

- 1

se guzarta hai, isliye

N = 1

aur system unstable hai."

Error:

- 1

se guzarna na andar hai na bahar — iska matlab hai ek closed-loop pole imaginary axis pe baitha hai (marginal).

N

wahan undefined hai; system stability boundary pe hai, cleanly unstable nahi.

P =

RHP mein

L

ke poles ki sankhya, aur

L = K / (s (s + 1) (s + 2))

s = 0

pe pole hai, isliye

P = 1

Error:

s = 0

open RHP mein nahi, imaginary axis pe hai. Hum contour ko right mein indent karte hain toh yeh pole bahar rehta hai; yeh count nahi hota, isliye

P = 0

"Humein

N = - 2

aur

P = 0

mila, isliye

Z = - 2

: do extra-stable poles."

Error:

Z

poles count karta hai aur negative nahi ho sakta.

Z = - 2

impossible hai, isliye ya toh encirclement direction ya count galat padha gaya — traversal sense dobara check karo.

"Gain margin curve ki origin se kitni door hai, yeh hai."

Error: margins critical point

- 1

se distances hain, origin se nahi. Gain margin wahan measure hota hai jahan plot negative real axis ko cross karta hai

- 1

ke relative se (dekho Bode Plot & Gain/Phase Margins).

"Nyquist aur Routh–Hurwitz dono ko closed-loop characteristic polynomial chahiye, isliye yeh same tool hain."

Error: Routh algebraically closed-loop polynomial ke coefficients pe kaam karta hai; Nyquist graphically open-loop frequency response

L (j ω)

se kaam karta hai aur kabhi closed-loop polynomial nahi banata. Different inputs, same question.

Why questions

Hum

L

- 1

ke relative kyun study karte hain,

1 + L

0

ke relative kyun nahi?

Dono ek hi map hain ek se shift hokar.

L

directly plot karna (jo data hamare paas pehle se hai) aur critical point ko

- 1

pe move karna humein

1 + L

har jagah compute karke draw karne se bachata hai.

Clockwise contour ke andar

1 + L

ka zero

ar g (1 + L)

mein

- 2 π

kyun add karta hai?

Simple zero

z_{0}

ke paas,

1 + L \approx a (s - z_{0})

; jab

s

clockwise ek baar

z_{0}

ke around loop karta hai, factor

(s - z_{0})

apna angle

- 2 π

se rotate karta hai, toh image origin ke around ek baar clockwise wind karta hai (dekho Argument Principle (Cauchy)).

L

ka RHP pole opposite sign, yaani

+ 2 π

kyun contribute karta hai?

Pole ka matlab hai

1 + L

ka behaviour

1/ (s - p)

jaisa hai; reciprocal rotation ka sense reverse kar deta hai, isliye

p

ke around clockwise loop image ko

+ 2 π

se unwind karta hai. Yahi opposite sign hai kyun

P

Z = N + P

mein

+

ke saath appear karta hai.

j ω

-axis pe pole ke around hum left nahi right mein kyun indent karte hain?

Right-indenting axis pole ko bahar RHP region ke rakhta hai jise contour enclose karta hai, isliye yeh unstable open-loop pole ke roop mein galat count nahi hota. Left-indenting ise enclose kar leta aur

P

corrupt kar deta.

- 1

(aur koi nahi, jaise

- 2

) reinforcement point kyun hai?

1 + L = 0

ka matlab hai

L = - 1

: returned signal sign mein flip hai aur magnitude mein equal hai, toh yeh perfectly apne aap mein add ho jaata hai. Koi aur point characteristic equation solve nahi karta.

Hum roots solve kiye bina stability determine kyun kar sakte hain?

Argument principle RHP roots count karne ko ek drawn curve kisi point ke around kitni baar wind karti hai count karne mein convert karta hai — geometric tally algebra replace karta hai. Yahi poora reason hai ki Root Locus ke saath Nyquist kyun exist karta hai.

L

ke har pole se high frequency pe asymptotically

- 9 0^{\circ}

phase kyun aata hai?

Ek first-order factor

1/ (s + p)

ka phase

\to - 9 0^{\circ}

jab

ω \to \infty

;

n

poles stack hokar

- 90 n^{\circ}

dete hain, jo set karta hai ki plot origin mein kis direction mein spiral karta hai.

Edge cases

Agar plot

- 1

ko tangentially graze kare bina enclose kiye, toh system stable hai?

Yeh marginally stable hai: ek closed-loop pole imaginary axis pe hai. Encirclement count us waqt ill-defined hai; koi bhi perturbation decide karti hai kis side pe padega.

N

kya hai agar Nyquist plot negative real axis ko bilkul cross na kare?

Tab yeh

- 1

ke around loop nahi kar sakta, isliye

N = 0

. Stability reduce ho jaati hai ki

P = 0

hai ya nahi.

Ek plant mein

P = 2

RHP poles hain; closed-loop stability ke liye kaunsa

N

chahiye?

Humein

Z = 0

chahiye, isliye

N = - P = - 2

: exactly do counter-clockwise encirclements of

- 1

. Kuch bhi aur

Z > 0

chhod deta hai.

Pure integrator loop

L = K / s

ke liye,

ω = 0

point ko specially kyun handle karna padta hai?

L

ω = 0

pe blow up hota hai (axis pe pole). Hum right mein tiny semicircle se indent karte hain; uska image ek bada clockwise arc hai jo infinity ke around sweep karta hai, jise

- 1

wraps count karne se pehle include karna padta hai.

Strictly proper

L

ke liye

ω \to \infty

pe curve kahan khatam hoti hai, aur counting ke liye yeh kyun matter karta hai?

Yeh origin pe khatam hoti hai. Kyunki dono

ω \to \pm \infty

branches cleanly

0

pe milti hain (

- 1

se door), closure koi spurious wraps add nahi karta — encirclement count sirf finite part se aata hai.

Agar tum accidentally closed-loop poles ko

P

ke roop mein use karo toh

Z = N + P

ka kya hoga?

Tum answer (

Z

-related data) ko count mein feed kar rahe hote aur double-count kar rahe hote.

P

open-loop RHP poles of

L

hone chahiye; unhe mix karna criterion ko self-referential aur galat bana deta hai.

Agar

L = K / (s - 1)

mein

K \to 0

(

P = 1

), toh kya loop stable hai?

Nahi. Plot ek tiny circle mein shrink ho jaata hai origin ke paas jo ab

- 1

encircle nahi karta, isliye

N = 0

aur

Z = N + P = 1

— ek RHP pole rehta hai. Unstable plant ko

- 1

wrap karne ke liye enough gain (

K > 1

) chahiye.

Connections

Argument Principle (Cauchy) — in sab traps ke peeche har sign ka "kyun" provide karta hai.
Bode Plot & Gain/Phase Margins — " $- 1$ se distance" ko margins ke roop mein reframe karta hai.
Routh–Hurwitz Criterion — algebraic cross-check jab koi plot ambiguous ho.
Root Locus — woh poles dikhata hai jinhe encirclement count silently track kar raha hai.
Feedback Control Basics — $T = G / (1 + L)$ ka origin.
Stability Margins in GNC Loops — real autopilots mein yeh traps kahan bite karte hain.