Nyquist stability criterion — encirclements of −1: Visual walkthrough

3.5.43 · D2 · Physics › Guidance, Navigation & Control (GNC) › Nyquist stability criterion — encirclements of −1

Hum parent note ke central result ko bilkul scratch se, visually, re-derive kar rahe hain.

Step 1 — Ek complex number ek arrow hota hai jiska ek length aur ek angle hota hai

KYA. Sabse pehle, hume ek idea chahiye: ek complex number $w = x + i y$ bas ek arrow hai jo origin se point $(x, y)$ tak ek flat sheet (yani "complex plane") pe drawn hota hai. Yahan $x$ hai kitna right, $y$ hai kitna upar, aur symbol $i$ ek bookkeeping tag hai jiska matlab hai "yeh part upar point karta hai, right mein nahi."

Har arrow ke do facts hote hain jo humein chahiye:

uski length $∣ w ∣ = x^{2} + y^{2}$ — literally ruler-distance origin se tip tak;
uska angle $ar g w$ — arrow kitna counter-clockwise ghuma hai seedha right (positive $x$ -axis) se.

YEH DO KYUN. Kyunki poori Nyquist ki kahani angles ke add hone ke baare mein hai jab hum ek loop pe chalte hain. Length muskil se matter karegi; angle ka ghoomna hi poora game hai. Isliye hum angle ko abhi isolate karte hain.

PICTURE. Neeche orange arrow $w$ hai. Uski length magenta ruler hai; uska angle violet wedge hai jo dashed rightward axis se measure hota hai.

Step 2 — Arrows ko multiply karna unke angles ko ADD karta hai

KYA. Do arrows lo $w_{1}$ aur $w_{2}$ . Jab hum inhe multiply karte hain, product $w_{1} w_{2}$ ek naya arrow hota hai jiska length lengths ka product hai aur jiska angle angles ka sum hai:

ar g (w_{1} w_{2}) = pehle ka angle ar g w_{1} + doosre ka angle ar g w_{2} .

Har symbol: $ar g w_{1}$ hai pehla arrow kitna ghuma; $ar g w_{2}$ doosra; multiply karna ghoomaavon ko stack karta hai.

HUMEIN YEH KYUN CHAHIYE. Jo bhi transfer function hum milenge woh factors jaise $(s - z)$ aur $(s - p)$ ko multiply aur divide karke bani hogi. Agar multiplication angles add karta hai, toh ek bade product ka total angle bas chhote angles ka sum hai — aur sum ek aisi cheez hai jise hum track kar sakte hain jab hum ek loop pe chalte hain. Yahi tool "encirclements count karo" ko "angle changes add karo" mein convert karta hai.

PICTURE. Do chhote arrows milke ek lamba arrow banate hain jiska angle do wedges ek ke baad ek rakh ke aaya hai.

Step 3 — Ek loop pe chalo; dekho ek factor ka angle kaise turn karta hai

KYA. Ab input point $s$ ko apne plane mein ek closed loop ke around ek baar, clockwise, walk karao. Ek single factor $(s - z_{0})$ chuno, jahan $z_{0}$ ek fixed point hai. Yeh factor khud ek arrow hai jiska tail $z_{0}$ pe aur tip $s$ pe hai (Step 1 mein general "arrow" definition yaad karo — same rules, bas moved tail): yeh $z_{0}$ se $s$ ki taraf point karta hai. Jaise jaise $s$ apne loop pe chalta hai, woh arrow pivot karta hai.

Do cases, aur sirf do:

Agar $z_{0}$ loop ke andar baitha hai, toh $z_{0}$ se $s$ tak ka arrow ek full turn sweep karta hai: uska angle $- 2 π$ badalta hai (minus kyunki hum clockwise chalte hain).
Agar $z_{0}$ loop ke bahar baitha hai, toh arrow aage peeche hiltaa hai lekin wapas wahi aata hai jahan se shuru kiya tha: net angle change $= 0$ .

YEH CRUX KYUN HAI. Yahi encirclement counting ka poora mechanism hai, bilkul saaf. "Enclosed → ek full turn; not enclosed → zero turns." Baki sab kuch kai aese factors pe bookkeeping hai.

PICTURE. Left: $z_{0}$ andar — arrow ek complete revolution karta hai. Right: $z_{0}$ bahar — arrow bas rocks karta hai aur wapas aata hai.

Step 4 — Saare factors add karo: Argument Principle aata hai

KYA. Ek characteristic-type function aese factors ka ratio hoti hai:

F (s) = \frac{( s - z _{1} ) ( s - z _{2} ) \dots}{( s - p _{1} ) ( s - p _{2} ) \dots} .

Har $z_{k}$ ek zero hai (numerator arrow → angle add karta hai).
Har $p_{k}$ ek pole hai (denominator arrow → angle subtract karta hai).

$s$ ko clockwise loop $C$ ke around walk karao. $F$ ka total angle change har factor ke angle change ka sum hai (Step 2: angles add hote hain). Step 3 se, ek factor $- 2 π$ contribute karta hai sirf tabhi jab uska point $C$ ke andar ho, warna $0$ :

Δ_{C} (ar g F) = andar ke zeros, har ek - 2 π (- 2 π) Z - andar ke poles, sign flip hoga (- 2 π) P = - 2 π (Z - P) .

Yahan $Z$ = loop ke andar $F$ ke zeros ki sankhya, $P$ = andar ke poles ki sankhya.

YEH KYUN MATTER KARTA HAI. $F$ ka origin ke around net turning kuch nahi bas ek headcount hai: (jo cheezein turn add karti hain) minus (jo cheezein turn subtract karti hain). Humne ek geometric winding ko poles aur zeros ki ek pure integer count se connect kar diya — roots ke liye koi algebra solve kiye bina.

PICTURE. Image curve $F (C)$ $F$ -plane mein draw ki gayi; origin ke around uska arrow $(Z - P)$ net clockwise loops banata hai.

Step 5 — Woh loop chuno jo poora danger zone trap kare

KYA. "Unstable" ka matlab hai right-half plane (RHP) mein ek closed-loop pole — woh region jahan $s = σ + j ω$ mein $σ > 0$ ho. Real part positive ka matlab ek aisa signal jo $e^{σ t}$ ki tarah grow karta hai instead of decay karne ke.

Toh hum chahte hain ek loop $C$ jo poore RHP ko enclose kare. Woh loop, Nyquist contour, hai:

Imaginary axis ke straight upar chalo, $s = j ω$ , $ω = - \infty$ se $ω = + \infty$ tak — ruko, hum clockwise jaate hain: $+ \infty$ se $- \infty$ tak neeche jaana CCW hai, toh hum $s = j ω$ upward lete hain phir right mein close karte hain;
RHP mein bulge karte hue ek infinite semicircle se close karo.

Clockwise traverse karne par, yeh fence har RHP point ko corrals karta hai.

IMAGINARY AXIS KYUN. Kyunki yahi woh ek hissa hai contour ka jo hum actually measure kar sakte hain: $s = j ω$ pe, transfer function $L (j ω)$ loop ka real, physical frequency response hai — wohi cheez jo Bode plot show karta hai. Infinite arc, kisi bhi real proper system ke liye, apni image ko ek single point tak shrink kar deta hai (usually origin), toh isse koi encirclements cost nahi hota.

Axis pe baitha ek pole (axis pole). Agar $L$ ka ek pole exactly $j ω$ -axis pe ho (jaise $s = 0$ pe), toh contour us blow-up point se nahi guzar sakta. Hum ek tiny semicircle se jo RIGHT bulge kare, RHP mein — wohi side jahan bada arc hai — uske around steer karte hain. Right kyun: yeh axis pole ko enclosed region ke bahar rakhta hai (contour uske around detour leta hai), toh pole $P$ mein count nahi hota. Yeh right-bulging indentation hamari single, consistent convention hai; phrase "indent right of the axis pole" bilkul yehi choice hai (ise left bulging mat samajhna). Is tiny right-hand half-loop pe ride karte hue, $s$ counter-clockwise $+ π$ angle se turn karta hai (half-turn); isliye factor $1/ s$ clockwise $- π$ se turn karta hai jab uski magnitude explode karti hai, jo $L$ -plane mein ek huge clockwise half-sweep produce karta hai. Woh sweep $N$ mein clockwise wraps add kar sakta hai, isliye uski direction matter karti hai — hum Step 7 mein detail karte hain.

PICTURE. Left: $s$ -plane shaded RHP ke saath aur clockwise D-shaped Nyquist contour. Chhota dent right-bulging indentation show karta hai axis pe baithey pole ke around (Step 7).

Recall Bada semicircle usually kyun vanish ho jaata hai?

Question ::: Strictly proper $L (s)$ ke liye (denominator degree numerator se zyada), $∣ L (s) ∣ \to 0$ jab $∣ s ∣ \to \infty$ , toh poora infinite arc single point $0$ pe map ho jaata hai — koi angle change contribute nahi karta.

Step 6 — Target shift karo: origin ban jaata hai $- 1$

KYA. Hum Step 4 apply karte hain specific choice ke saath

F (s) = 1 + L (s),

jahan $L (s)$ feedback loop ka open-loop transfer function hai: yeh woh total gain hai jo ek signal comparison point se pehle ek baar loop ke around jaane par pick up karta hai. Ek standard loop mein $L (s) = G (s) H (s)$ , jahan $G (s)$ forward-path transfer function hai (controller + plant, woh block jise error signal drive karta hai) aur $H (s)$ feedback-path transfer function hai (sensor / measurement block jo output ko wapas compare karne ke liye return karta hai). Inhe multiply karna do blocks ko series mein chain karta hai — exactly Step 2 ka "angles add" rule. Phir:

$F$ ke zeros = $1 + L = 0$ ke solutions = closed-loop poles → RHP wale exactly $Z$ hain (woh zombies jinse hum darte hain);
$F$ ke poles = $L$ ke poles → RHP wale $P$ hain.

Step 4 se, curve $1 + L (j ω)$ origin ke around $(Z - P)$ baar clockwise wind karta hai.

Lekin hum $1 + L$ kabhi plot nahi karte; hum $L$ plot karte hain. Aur " $1 + L$ around $0$ " wohi picture hai left mein ek shift ke saath: woh hai " $L$ around $- 1$ ." 1 subtract karna poore plot ko left slide karta hai, target point ko $0$ se $- 1$ par le jaata hai.

CW wraps of L around - 1 N = RHP closed-loop poles Z - RHP open-loop poles P ⟹ Z = N + P

Har symbol, ek aakhri baar: $N$ = drawn $L (j ω)$ curve ke $- 1$ ke around net clockwise loops; $P$ = RHP mein already existing $L$ ke kitne poles hain (plot karne se pehle pata hota hai); $Z$ = woh count jo hum actually chahte hain — RHP closed-loop poles. Stable iff $Z = 0$ , yani iff $N = - P$ .

$- 1$ KYUN AUR KAHIN AUR NAHI. $1 + L = 0$ ka matlab hai $L = - 1$ : open-loop response sign mein flip hokar aur size mein equal hokar wapas aata hai — perfect self-reinforcement (canyon echo exactly utni hi force se wapas dhakelta hai jitni tumne dhakelaa). $- 1$ ko encircle karna exactly count karta hai kitne poles "grows forever" region mein ghus gaye.

PICTURE. $1 + L$ ka origin-winding (left) aur $L$ ka identical $- 1$ -winding (right) — same shape, ek se left slide kiya.

Common mistake

$j ω$ -axis pe baitha ek closed-loop pole ( $F$ ka zero) Kyun miss karna aasaan hai: hum axis pe $L$ ke poles ke around indent karte hain, lekin kya ho agar $F = 1 + L$ ka ek zero — yaani ek closed-loop pole — exactly imaginary axis pe land kare? Tab Nyquist curve $L (j ω)$ us frequency pe exactly $- 1$ se guzarti hai, winding number $N$ undefined ho jaata hai, aur system marginally stable hota hai (ek sustained, non-growing oscillation) saaf stable ya unstable hone ki bajaye. Fix: $- 1$ ko touch karne wali curve ko ek boundary/degenerate case maano — gain ko thoda sa perturb karo (nudge $K$ ) us zero ko axis se push karne ke liye, phir $N$ recount karo yeh dekhne ke liye ki tum actually stability ke kis side pe ho. Yahi critical-gain crossing hai (jaise next step mein $K = 6$ ): crossing par curve $- 1$ ko kiss karti hai; dono taraf woh clear karti hai.

Step 7 — Degenerate case: fence pe baitha ek pole

KYA. Wapas aao $L$ ke paas jiska pole exactly $s = 0$ pe hai (ya $j ω$ -axis pe kahin bhi). Hamara contour warna seedha ek aisi point se guzarta jahan $L$ blow up ho jaata hai — wahan picture undefined hai.

Fix (Step 5 ki same convention): contour mein ek tiny semicircle dalo jo RIGHT mein bulge kare, RHP mein, taaki axis pole enclosed region ke bahar rahe (isliye woh $P$ mein count nahi hota). Jab $s$ is tiny right-hand half-loop of radius $ε \to 0$ pe ride karta hai, toh woh counter-clockwise $+ π$ se turn karta hai; factor $1/ s$ ki magnitude $1/ ε \to \infty$ hai aur uska angle clockwise $- π$ se turn karta hai, toh image infinite radius pe ek huge clockwise semicircular arc trace karta hai.

DIRECTION KYUN MATTER KARTI HAI. Boundary pe baitha ek pole clearly "in" ya "out" nahi hai; right-bulging indentation choice ko explicit aur consistent banata hai — pole excluded hai, isliye woh kabhi $P$ inflate nahi karta. Aur kyunki image half-sweep clockwise hai, yeh $N$ mein clockwise wraps contribute kar sakta hai; direction galat karo aur tum $P$ aur $N$ dono miscount kar loge.

Yaad rakhne wale cases:

Axis pole ke right mein indent karo (RHP mein) → tiny detour pole ko enclosed region se exclude karta hai → $P$ mein count nahi hota. (Standard convention, is page pe har jagah use hoti hai.)
Image arc huge aur clockwise hai, aur yahi woh jagah hai jahan $L = K / [s (s + 1) (s + 2)]$ jaise plots mein bade sweeps aate hain.

PICTURE. Left: $s$ -plane mein small right-bulging indentation. Right: $L$ -plane mein woh giant clockwise arc jo woh produce karta hai.

Step 8 — Teen outcomes, teen pictures (saare cases)

KYA. Boxed formula ko parent note ke teen archetypes mein plug karo.

$P = 0$ , no wraps — $L = \frac{2}{( s + 1 ) ( s + 2 )}$ . $+ 1$ se start karta hai, fourth quadrant se origin mein sink ho jaata hai, $- 1$ tak kabhi nahi pahunchta. $N = 0, P = 0 \Rightarrow Z = 0$ : stable.
$P = 0$ , gain too high — $L = \frac{K}{s ( s + 1 ) ( s + 2 )}$ . Negative real axis ka uska crossing left ki taraf chalti hai jaise $K$ badhta hai; $K = 6$ ke baad woh $- 1$ se aage khisakti hai, $N = 2$ deta hai. Tab $Z = 2$ : unstable (do poles RHP mein chale gaye).
$P = 1$ , stabilised by feedback — $L = \frac{K}{s - 1}$ . Yahan hume $N = - 1$ chahiye (ek counter-clockwise wrap of $- 1$ ) $Z = 0$ hit karne ke liye. Agar $K > 1$ , circle $- 1$ ko CCW mein swallow karta hai → stable; agar $K < 1$ nahi karta → unstable.

TEEN KYUN DIKHATE HAIN. Yeh sign logic exhaust karte hain: $N$ positive (bad when $P = 0$ ), $N = 0$ (safe when $P = 0$ ), aur $N$ negative (required when $P > 0$ ). Sign aur count dono matter karte hain — ise Routh–Hurwitz aur root locus ke saath mirror karo, jo same $Z$ doosre tareekon se dete hain.

PICTURE. Teen mini-plots side by side, har ek mein $- 1$ marked aur uska verdict.

Ek-picture summary

Sab kuch ek single flow mein collapse ho jaata hai: RHP fence pe chalo → factors angles add karte hain → net turns poles count karte hain → target $- 1$ pe shift karo.

Recall Poore walkthrough ka Feynman retelling

Loop ke har ingredient ko ek chhota sa arrow samjho, aur ek magic rule yaad rakho: jab arrows multiply hote hain, unke angles bas add ho jaate hain. Ab ek walker $s$ ko ek baar ek bade fence ke around bhejo jo poore "signals that grow" region ko pen mein rakhta hai. Har trapped point ke liye, walker ko point karne wala arrow ek full turn ghoomta hai; har free point ke liye, woh bas wapas ghar wiggle karta hai. Toh poore product ka total spin simply hai (trapped zeros) minus (trapped poles) — ek headcount, koi root-solving nahi. Hum yeh function $1 + L$ pe aim karte hain, jiske trapped zeros exactly unstable modes $Z$ hain aur jiske trapped poles plant ke apne bure poles $P$ hain. Kyunki $1 + L$ ko $0$ ke around draw karna $L$ ko $- 1$ ke around draw karne jaisa hai, hum bas count karte hain ki hamari measured frequency-response curve danger point $- 1$ ko kitni baar loop karti hai, aur woh kis direction mein ghoomti hai. Woh count $N$ hai, aur $Z = N + P$ nikal aata hai. Agar $Z = 0$ , echo marta hai — stable. (Aur agar curve $- 1$ se exactly guzarti hai, hum knife-edge pe hain — gain nudge karo yeh dekhne ke liye ki woh kis taraf girti hai.)

Connections

Argument Principle (Cauchy) — Steps 3–4 ke peeche ka angle-counting engine.
Bode Plot & Gain/Phase Margins — wohi $L (j ω)$ data, aur curve $- 1$ ke kitne paas se guzarti hai.
Routh–Hurwitz Criterion — algebraic tarika same $Z$ paane ka (jaise Case 2 mein $K = 6$ ).
Root Locus — dekh ta hai closed-loop poles $K$ vary hone par RHP mein kaise cross karte hain.
Feedback Control Basics — jahan se $T = G / (1 + L)$ aur $L = G H$ aate hain.
Stability Margins in GNC Loops — autopilot/attitude loops pe $- 1$ -distance apply karna.