Worked examples — Nyquist stability criterion — encirclements of −1

3.5.43 · D3 · Physics › Guidance, Navigation & Control (GNC) › Nyquist stability criterion — encirclements of −1

Yeh parent Nyquist note ka "har case grind karo" wala page hai. Agar koi symbol unfamiliar lage, to parent note usse scratch se build karta hai — hum assume karte hain ki tum jaante ho ki $Z=N+P$ hai aur kyun $-1$ danger point hai. Abhi humara kaam hai: har scenario class ko cover karo taaki koi bhi exam problem surprise na kare.

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Scenario matrix

Yahan $P$ = RHP open-loop poles, $N$ = $-1$ ke clockwise encirclements, $Z=N+P$ = RHP closed-loop poles (unstable iff $Z>0$).

Cell	$P$	Plot ka behaviour $-1$ ke paas	$N$	Result	Example
A	$0$	kabhi $-1$ wrap nahi karta	$0$	stable	Ex 1
B	$0$, saare poles LHP	high gain $-1$ ko do baar CW wrap karta hai	$+2$	unstable	Ex 2
C	$0$ (integrator)	tangent — $-1$ se guzarta hai	boundary	marginal	Ex 3
D	$1$	ek counter-clockwise wrap	$-1$	stable	Ex 4 (high gain)
E	$1$	koi wrap nahi	$0$	unstable	Ex 4 (low gain)
F	$0$, $s=0$ par pole	integrator → contour indent karo	(mechanics)	B/C set up karta hai	Ex 5
G	$0$	limiting endpoints $\omega \to 0,\infty$; koi finite crossing nahi	$0$	stable (ends padho)	Ex 6
H	$0$	degenerate pure gain, single point	$0$	trivially stable	Ex 7
I	$0$ (integrator)	real-world GNC autopilot, koi finite crossing nahi	$0$	sab $K>0$ ke liye stable	Ex 8
J	$0$ ya $2$	same plot, ek CW wrap	$+1$	$P$ par depend karta hai	Ex 9

Ab hum cells A→J walk karte hain.

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Example 2 mein Example 5 ka integrator mechanics use hota hai

Kyunki Example 2 aur Example 3 dono $L(s)=\frac{K}{s(s+1)(s+2)}$ use karte hain (jisme $s=0$ par integrator pole hai), Example 5 pehle present karna zaroori nahi — lekin Example 2 ka Step 1 neeche self-contained likhaa gaya hai: woh $P=0$ spot par derive karta hai aur Ex 5 ko sirf picture ke liye point karta hai.

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Example 1 — Cell A: $P=0$, koi wrap nahi → stable

Step 1 — $P$ nikalo. Yeh step kyun? $Z=N+P$ ko pehle $P$ chahiye. Poles solve karo $(s+1)(s+2)=0\Rightarrow s=-1,-2$, dono LHP. To $==P=0==$.

Step 2 — Curve ke do endpoints evaluate karo. Yeh step kyun? Start ($\omega =0$) aur end ($\omega \to \infty$) yeh frame karte hain ki plot kahan rehta hai. $L(0)=\frac{2}{1\cdot 2}=1(\text{on the }+\text{real axis}).$ Jab $\omega \to \infty$, do poles mein se har ek $-90^{\circ }$ phase contribute karta hai, to phase $\to -180^{\circ }$ aur magnitude $\to 0$: plot neeche se origin mein die ho jaata hai.

Step 3 — Shape trace karo (figure dekho). Yeh step kyun? Hame $-1$ se compare karne ke liye leftmost real-axis crossing chahiye. Curve $+1$ se start hoti hai, fourth quadrant mein swing karti hai, aur $0$ par return ho jaati hai.

Kya observe karna hai: blue curve $L(j\omega )$ $+1$ (green dot) se start hoti hai, lower half-plane mein clockwise loop karti hai, aur origin par die ho jaati hai. Red ✗ $-1$ par blue curve ke har point se kaafi left baitha hai — woh kabhi surrounded nahi hota.

Step 4 — $N$ count karo aur $Z$ compute karo. Red $-1$ point kabhi enclosed nahi hota, isliye $N=0$. Tab $Z=N+P=0+0=0$ → stable.

Recall Verify

Closed-loop denominator hai $1+L=1+\frac{2}{(s+1)(s+2)}=\frac{s^2+3s+4}{(s+1)(s+2)}$. $s^2+3s+4$ ke roots: $s=\frac{-3\pm \sqrt{9-16}}{2}=-1.5\pm 1.32i$ — dono LHP. $Z=0$ confirmed. ✓

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Example 2 — Cell B: $P=0$, gain bahut zyada → do CW wraps

Step 1 — $P$ nikalo (self-contained). Yeh step kyun? $Z=N+P$ ko $P$ chahiye, aur $s=0$ par exactly imaginary axis par ek pole hai, isliye decide karna hoga ki woh count hota hai ya nahi. Nyquist contour $j\omega$-axis ke saath hugs karta hai lekin sirf RHP enclose karta hai. Hum ise ek tiny semicircle se deform karte hain jo $s=0$ ke around RHP mein bulge karta hai, to woh pole enclosed region ke just bahar reh jaata hai. Isliye $s=0$ pole hamare count ke liye RHP pole nahi hai: enclosed poles $\{0,-1,-2\}$ mein se koi nahi, isliye $==P=0==$ milta hai. (Ex 5 yeh indentation draw karta hai; $P=0$ ki algebra ke liye is paragraph se zyada kuch nahi chahiye.)

Step 2 — $\mathrm{Im}\{L(j\omega )\}$ derive karo aur pata karo kahan zero hai. Yeh step kyun? Negative-real-axis crossing exactly wahan hoti hai jahan $L$ ka imaginary part zero ho aur real part negative ho — yeh phase $=-180^{\circ }$ condition hai, aur iska location decide karta hai ki $-1$ enclosed hai ya nahi. $s=j\omega$ substitute karo aur denominator expand karo: $s(s+1)(s+2){|}_{s=j\omega }=j\omega (j\omega +1)(j\omega +2)=j\omega [(j\omega {)}^2+3j\omega +2]=j\omega (2-{\omega }^2)+3j^2{\omega }^2.$ Denominator ke real aur imaginary parts collect karo: $D(j\omega )=\underset{\text{real}}{\underbrace{-3{\omega }^2}}+j\underset{\text{imag}}{\underbrace{\omega (2-{\omega }^2)}}.$ Tab $L(j\omega )=\frac{K}{D}=\frac{K\overline{D}}{|D{|}^2}=\frac{K[-3{\omega }^2-j\omega (2-{\omega }^2)]}{|D{|}^2}.$ $L$ ka imaginary part tab vanish hoga jab numerator ka imaginary part zero ho: $-K\omega (2-{\omega }^2)=0\Rightarrow {\omega }^2=2(\text{taking }\omega >0).$

Step 3 — Crossing magnitude evaluate karo. Yeh step kyun? Ab hum jaante hain kahan ($\omega =\sqrt{2}$) curve real axis hit karta hai; hame jaanna hai kitna left. $\omega =\sqrt{2}$ par imaginary part gone hai, isliye $L$ purely real part hai: $L(j\sqrt{2})=\frac{K(-3\cdot 2)}{|D{|}^2}.|D{|}^2=(-3\cdot 2{)}^2+0=36.$ $L(j\sqrt{2})=\frac{-6K}{36}=-\frac{K}{6}.$ To crossing negative real axis par $-K/6$ par baithti hai (real part negative ✓ — sach mein $-180^{\circ }$ crossing).

Step 4 — $-1$ se compare karo aur $N$ count karo (figure dekho). $K=12$ ke liye: crossing $-12/6=-2$ par, jo $-1$ se left hai, isliye $-1$ enclosed hai.

Kya observe karna hai: finite blue curve real axis ko $-2$ par cross karti hai (red ✗ $-1$ se left). Bada orange arc tiny right-indentation ka image hai $s=0$ ke around (Ex 5): jab $s$ $-90^{\circ }\to +90^{\circ }$ sweep karta hai $0$ ke paas, $L\approx K/s$ ek giant clockwise arc trace karta hai jo, finite lobe ke saath milkar, $-1$ ko do baar clockwise wrap karta hai. Orange direction arrows follow karo: $-1$ point ko big arc se ek baar aur returning lobe se ek baar circle hota hai → $N=+2$.

Step 5 — $Z$ compute karo. $Z=N+P=2+0=2$ → unstable, do RHP poles.

Recall Verify

$1+L=0\Rightarrow s(s+1)(s+2)+12=0\Rightarrow s^3+3s^2+2s+12=0$. Routh test: $s^3+3s^2+2s+K$ ke liye, row entry hai $(3\cdot 2-K)/3=(6-K)/3$; $K=12$ ke saath yeh $-2<0$ hai → do sign changes → 2 RHP roots. $Z=2$ se match karta hai. ✓

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Example 3 — Cell C: critical gain, $-1$ ko tangent

Step 1 — Crossing ko $-1$ ke equal set karo. Yeh step kyun? Marginal stability = curve danger point se guzarti hai. Ex 2 Step 3 se crossing $-K/6$ par hai: $-\frac{K}{6}=-1\Rightarrow K=6.$

Step 2 — Interpret karo. Yeh step kyun? Geometric threshold ko stable/unstable ranges mein translate karte hain. $K<6$ ke liye crossing $-1$ se right mein → $N=0$ → stable. $K>6$ ke liye → $N=2$ → unstable. $K=6$ par plot $-1$ ko graze karta hai: do closed-loop poles imaginary axis par baithte hain (sustained oscillation).

Recall Verify

$K=6$ par: $s^3+3s^2+2s+6=(s+3)(s^2+2)=0\Rightarrow s=-3,\pm j\sqrt{2}$. $\pm j\sqrt{2}$ par pure imaginary roots → marginal, oscillation frequency $\omega =\sqrt{2}$ (Ex 2 ki crossing frequency se match karta hai). ✓

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Example 4 — Cells D & E: unstable plant, ek CCW wrap zaroori

Step 1 — $P$ confirm karo. Yeh step kyun? Poora verdict $P$ par hinge karta hai. $s=+1$ par pole RHP mein hai → $==P=1==$.

Step 2 — $L(j\omega )$ explicitly likho. Yeh step kyun? Actual curve chahiye taaki wrap dekh sakein. Conjugate se upar aur neeche multiply karo: $L(j\omega )=\frac{K}{j\omega -1}=\frac{K(-1-j\omega )}{(-1{)}^2+{\omega }^2}=\frac{K(-1-j\omega )}{1+{\omega }^2}.$

Step 3 — Endpoints. Yeh step kyun? Do ends circle ko pin down karte hain. $\omega =0$ par: $L=-K$ (negative real axis). Jab $\omega \to \infty$: $L\to 0$. Poori curve ek circle hai jo $-K$ aur origin se guzarti hai (figure dekho).

Kya observe karna hai: orange circle $-K=-2$ (green dot) se guzarti hai $\omega =0$ par aur $\omega \to \infty$ ke saath origin tak shrink hoti hai. Direction arrow counter-clockwise sweep dikhata hai, aur red ✗ $-1$ par circle ke andar baitha hai — to exactly ek baar CCW wrap hota hai.

Step 4 — $-1$ kahan hai? Yeh step kyun? $-1$ ka enclosure $-K$ aur $-1$ compare karke decide hota hai. Agar $K>1$ hai, to $-K<-1$, isliye $-1$ circle ke andar hai → ek baar encircled, aur traverse direction ise counter-clockwise banata hai: $N=-1$. Tab $Z=N+P=-1+1=0$ → stable (Cell D). Agar $K<1$ hai, to $-1$ bahar hai → $N=0$, $Z=0+1=1$ → unstable (Cell E).

Recall Verify

$1+L=0\Rightarrow s-1+K=0\Rightarrow s=1-K$. Stable iff $s<0$ iff $K>1$. Boundary $K=1$ par (origin par pole). Geometric threshold se match karta hai. ✓

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Example 5 — Cell F: integrator, contour indent karo

Step 1 — Right mein indent karo. Yeh step kyun? Nyquist contour sirf RHP enclose karna chahiye. Hum ise $s=0$ ke around $\epsilon \to 0$ radius ki tiny semicircle se deform karte hain jo RHP mein bulge karti hai, pole ko enclosed region ke bahar rakhte hain. Isliye woh $P$ mein count nahi hota: $P=0$.

Step 2 — Small semicircle ko map karo. Yeh step kyun? Hame jaanna hai woh tiny detour $L$-plot par kya banta hai, kyunki woh big arc contribute karta hai jo $-1$ ko wrap karne mein madad karta hai. Uspar $s=\epsilon e^{j\varphi }$ hai jahan $\varphi$ $-90^{\circ }\to +90^{\circ }$ run karta hai. Tab $L\approx \frac{K}{s}=\frac{K}{\epsilon }{e}^{-j\varphi }$ — radius $K/\epsilon \to \infty$ ka large arc, $+90^{\circ }\to -90^{\circ }$ (clockwise) sweep karta hai. Figure dekho.

Kya observe karna hai: blue contour $j\omega$-axis ke upar run karta hai lekin red ✗ pole $s=0$ ke around ek small right-hand bump (green) banata hai, use bahar rakhta hai. $L$-plane mein yeh tiny bump ek giant clockwise arc mein balloon ho jaata hai jaisa Ex 2 ke figure mein dikhta hai.

Step 3 — Consequence. Yeh step kyun? Yeh plot ko consistently close karta hai taaki encirclement count well-defined ho. Yeh giant clockwise arc wahi hai jo plot ko infinity par close karta hai aur hame $-1$ ke wraps consistently count karne deta hai (Ex 2 & 3 mein use hota hai).

Recall Verify

Indentation ke saath, enclosed RHP mein poles $-1,-2$ hain? Nahi — woh LHP mein hain. $s=0$ hai? Nahi — bump se exclude kiya gaya. To $P=0$, jo Ex 2 ($Z=2$) aur Ex 3 ($K=6$ par boundary) mein use kiye gaye count ko confirm karta hai. ✓ (dono upar Routh se cross-checked hain.)

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Example 6 — Cell G: limiting endpoints padhna

Step 1 — $\omega =0$ endpoint. Yeh step kyun? Curve ka start ek anchor fix karta hai. $L(0)=\frac{5}{1}=5$ positive real axis par.

Step 2 — $\omega \to \infty$ endpoint. Yeh step kyun? End doosra anchor fix karta hai. Magnitude $\to 0$, phase $\to -180^{\circ }$: negative-real side se origin approach karta hai.

Step 3 — $\mathrm{Im}\{L(j\omega )\}$ derive karo aur crossing ke liye solve karo. Yeh step kyun? Finite negative-real crossing ke liye $\mathrm{Im}\{L\}=0$ finite $\omega$ par chahiye; algebraically test karte hain guess ki jagah. Expand karo: $(j\omega +1{)}^2=(1-{\omega }^2)+2j\omega ,L(j\omega )=\frac{5}{(1-{\omega }^2)+2j\omega }=\frac{5[(1-{\omega }^2)-2j\omega ]}{(1-{\omega }^2{)}^2+4{\omega }^2}.$ To $\mathrm{Im}\{L(j\omega )\}=\frac{-10\omega }{(1-{\omega }^2{)}^2+4{\omega }^2}.$ Yeh sirf $\omega =0$ par zero hai (jahan $L=+5$, positive axis par) aur limit $\omega \to \infty$ mein (jahan $|L|\to 0$). Negative-real crossing dene wala koi finite $\omega >0$ nahi hai. Spot-check ke liye, $\omega =1$ par: $L(j)=\frac{5}{2j}=-2.5j$ — purely imaginary, real axis par nahi.

Step 4 — Conclusion. Yeh step kyun? "No finite crossing" ko encirclement count mein convert karte hain. Kyunki curve sirf origin par limit mein negative real axis touch karti hai, is $L$ ke liye woh kabhi $-1$ enclose nahi kar sakti: $P=0$, $N=0$, $Z=0$ → unconditionally stable.

Recall Verify

$1+L=0\Rightarrow (s+1{)}^2+5=0\Rightarrow s=-1\pm j\sqrt{5}$ — dono LHP. $Z=0$. Aur $|L(j)|=5/|(j+1{)}^2|=5/2=2.5$. ✓

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Example 7 — Cell H: degenerate pure-gain plant

Step 1 — Poles? Yeh step kyun? $P$ poles se aata hai, aur yahan koi poles nahi hain → $P=0$. Nyquist "plot" $K$ par ek single point hai (har $\omega$ ke liye same).

Step 2 — $-1$ encircle karo? Yeh step kyun? Ek single point ka koi winding nahi hota. Woh kuch bhi wrap nahi kar sakta → $N=0$ → $Z=0$: encirclement sense mein hamesha "stable".

Step 3 — Actual closed-loop pole check karo. Yeh step kyun? Encirclement bookkeeping algebra se agree karni chahiye. $1+L=1+K=0\Rightarrow$ closed-loop pole sirf tab hai jab $K=-1$ (tab $1+K=0$ identically — degenerate). $K\ne -1$ ke liye loop gain sirf $\frac{G}{1+K}$ hai, ek static number — destabilize karne ke liye koi pole nahi.

Step 4 — Edge case $K=-1$. Yeh step kyun? Har case cover karna hoga, degenerate singular wala bhi. Denominator $1+L=0$ har jagah: transfer function undefined hai (har $s$ par division by zero). Yeh degenerate "$-1$ par exactly baitha" limit hai.

Recall Verify

$1+K=0\Rightarrow K=-1$ ek hi special value hai. Kisi bhi $K=2$ ke liye, closed-loop static gain $=G/(1+2)=G/3$, finite → stable. ✓

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Example 8 — Cell I: real-world GNC autopilot

Step 1 — Poles. Yeh step kyun? Kuch aur karne se pehle $P$ fix karo. $s=0$ (indent, count nahi) aur $s=-4$. $P=0$.

Step 2 — Finite $\omega$ par phase. Yeh step kyun? Finite negative-real crossing ke liye phase $=-180^{\circ }$ finite $\omega$ par chahiye; test karte hain. $\angle L(j\omega )=-90^{\circ }-\arctan \frac{\omega }{4}.$ Yeh $-180^{\circ }$ sirf $\omega \to \infty$ par reach karta hai, jahan $|L|\to 0$. To plot ka negative-real crossing origin par hai — origin se left kisi finite point par kabhi nahi.

Step 3 — Conclusion. Yeh step kyun? "No finite crossing" ko count mein convert karo. $-1$ kabhi enclosed nahi hota kisi bhi $K>0$ ke liye: $N=0$, $Z=0$. Loop sab positive gains ke liye stable hai — classic type-1 second-order shape.

Step 4 — Sanity via characteristic eq. Yeh step kyun? Encirclement verdict ko algebra se cross-check karo. $s^2+4s+K=0$ ke roots $s=-2\pm \sqrt{4-K}$ hain; real part $=-2<0$ sab $K>0$ ke liye (complex jab $K>4$, phir bhi stable). To "sabse bada stable $K$" unbounded hai — design limit encirclement se nahi aati, margins/noise se aati hai (dekho Stability Margins in GNC Loops aur Bode Plot & Gain/Phase Margins).

Recall Verify

$s^2+4s+K$: roots ka sum $=-4<0$, product $=K>0$ ⇒ har $K>0$ ke liye dono roots LHP (Routh: saare coefficients positive, second-order ⇒ stable). ✓

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Example 9 — Cell J: exam twist — same plot, do alag $P$

Step 1 — Student A ke liye $Z=N+P$ apply karo. Yeh step kyun? Diye gaye $N$ aur har claimed $P$ ko master formula mein plug karo. $Z=1+0=1$ → ek unstable pole → unstable.

Step 2 — Student B ke liye apply karo. Yeh step kyun? Same formula, alag $P$. $Z=1+2=3$ → teen unstable poles → bahut unstable.

Step 3 — Twist resolved. Yeh step kyun? Dikhao ki unstable poles ki sankhya, sirf yes/no nahi, $P$ par depend karti hai. Yahan koi bhi reading stable nahi hai, lekin $Z$ alag hai. $P=2$ wale plant ko stable karne ke liye tumhe $N=-2$ chahiye hoga (do CCW wraps). Ek single CW wrap ($N=+1$) kisi bhi plant ko kabhi stabilize nahi kar sakta.

Step 4 — General rule. Yeh step kyun? Reusable law extract karo. Stable rehne ke liye $N=-P$ chahiye. Sirf $N\le 0$ (net CCW) ek unstable plant ko rescue kar sakta hai; $N>0$ hamesha instability add karta hai.

Recall Verify

$Z_A=N+P=1+0=1$; $Z_B=1+2=3$. Stability ke liye $Z=0$ chahiye; dono $0$ nahi hain. ✓

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Recall

Recall Quick self-test

Kisi bhi Nyquist problem mein sabse pehle kya compute karo? ::: $P$ — RHP open-loop poles ki sankhya. Ex 2 mein $K/[s(s+1)(s+2)]$ ka critical gain? ::: $K=6$ (crossing $-K/6=-1$ par). Ex 4: $K/(s-1)$ ke liye threshold gain? ::: $K=1$; $K>1$ ke liye stable. $L=K/(s+1{)}^2$ kyun hamesha stable hai? ::: Iska $\mathrm{Im}\{L\}=0$ sirf $\omega =0$ par (positive axis) aur $\omega \to \infty$ par (magnitude zero) hota hai — curve negative real axis kabhi cross nahi karti, isliye $-1$ kabhi enclosed nahi hota. Same plot ($N=+1$), kya $P$ answer change karta hai? ::: Haan — $Z=N+P$, isliye $Z$ (aur kitna unstable) $P$ par depend karta hai.

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Connections

• Argument Principle (Cauchy) — wraps count karna poles kyun count karta hai.
• Routh–Hurwitz Criterion — yahan har "Verify" ke peeche ki algebra.
• Root Locus — dekho closed-loop poles kaise move karte hain jab $K$ badhta hai (Ex 2, 3).
• Bode Plot & Gain/Phase Margins — Ex 8 ka "kitna $K$" wala sawaal yahan rehta hai.
• Feedback Control Basics — $T=G/(1+L)$ ka origin.
• Stability Margins in GNC Loops — applied autopilot design (Ex 8).