Exercises — Nyquist stability criterion — encirclements of −1

WHAT karte hain: poles dhoondho, jo woh values of $s$ hain jo denominator ko zero banate hain. WHY: $P$ sirf unhi poles ko count karta hai jinка real part positive hai.

$(s+3)(s+4)=0$ set karo $\Rightarrow s=-3$ ya $s=-4$.

Dono ka real part negative hai, toh dono left-half plane (LHP) mein hain. Koi bhi RHP mein nahi hai. $\boxed{P=0.}$

WHAT: seedha $Z=N+P$ mein plug in karo. WHY: woh formula hi Nyquist criterion hai — ek baar $N$ aur $P$ pata chalne ke baad koi drawing nahi chahiye.

Do clockwise wraps matlab $N=+2$. $P=0$ ke saath: $Z=N+P=2+0=2.$ $Z\ne 0$, toh loop mein RHP mein 2 closed-loop poles hain → unstable.

Step 1 — WHAT: $\omega =0$ pe evaluate karo. WHY: point $L(0)$ wahan hai jahan Nyquist curve real axis pe start karti hai. $L(0)=\frac{4}{(1)(2)}=2.$ Ek positive real number: curve $+2$ pe positive real axis pe start karti hai.

Step 2 — WHAT: high frequency pe phase. WHY: limit mein har pole $-90^{\circ }$ phase contribute karta hai, jo batata hai curve kahan khatam hoti hai. Do poles $\Rightarrow$ phase $\to 2\times (-90^{\circ })=-180^{\circ }$, aur magnitude $\to 0$ (denominator ${\omega }^2$ ki tarah grow karta hai).

Step 3 — quadrants. $+2$ pe start hoke (phase $0^{\circ }$) aur phase $-180^{\circ }$ tak rotate hote hue jab woh origin ki taraf shrink hoti hai, curve fourth quadrant se clockwise sweep karti hai (positive real, negative imaginary) aur negative-real side se origin ke paas pohonchti hai.

Figure kya dikhata hai (image ke bina bhi padho): solid blue curve positive real axis pe $+2$ pe start hoti hai (pale yellow mein marked), phir fourth quadrant mein neeche aur left ki taraf bow karti hai — uska imaginary part negative ho jaata hai, real part shrink hota hai — aur woh origin mein spiral ho jaati hai. Pink $\times$ negative real axis pe $-1$ mark karta hai. Sabse important baat, blue curve picture ke right half mein bilkul rehti hai (real part apne leftmost point pe roughly $>-\frac{1}{2}$) aur kabhi pink $\times$ ko surround nahi karti. Dashed lighter-blue curve $\omega <0$ ke liye mirror image hai; saath mein woh ek closed loop banate hain jo $-1$ ko bahar chhod deti hai.

Yeh kabhi $-1$ tak reach nahi karta, jo $N=0$ foreshadow karta hai.

WHAT: Ex. 2.1 ki curve ke $-1$ around wraps count karo. WHY: $N$ woh akela missing piece hai $Z=N+P$ se pehle.

Curve $+2$ se start hoti hai, fourth quadrant mein dip karti hai, aur origin mein slide ho jaati hai — point $-1$ negative real axis pe baitha hai, us chhote loop ke bahar jo curve trace karti hai. Curve kabhi usse surround nahi karti. $N=0.$ Ex. 1.1-style factoring se, $-1,-2$ pe poles $P=0$ dete hain. Isliye $Z=N+P=0+0=0\Rightarrow \text{stable.}$

Step 1 — WHAT: $s=j\omega$ substitute karo aur woh jagah dhoondo jahan curve purely real ho. WHY: real-axis crossing wahan hai jahan phase $-180^{\circ }$ hoti hai; yahi woh exact point hai jo $-1$ hit kar sakta hai.

$L(j\omega )=\frac{K}{j\omega (j\omega +1)(j\omega +2)}.$ Denominator multiply out karo: $j\omega (j\omega +1)(j\omega +2)=j\omega [(j\omega {)}^2+3j\omega +2]=j\omega (-{\omega }^2+3j\omega +2).$ $j\omega$ distribute karo: $=j\omega (2-{\omega }^2)+j\omega \cdot 3j\omega =-3{\omega }^2+j\omega (2-{\omega }^2).$ Toh $L(j\omega )=\frac{K}{-3{\omega }^2+j\omega (2-{\omega }^2)}.$

Step 2 — WHAT: denominator ko purely real banao taaki $L$ real ho. WHY: real numerator $K$ ke saath, curve real axis ko exactly tab cross karti hai jab denominator ka zero imaginary part ho: $\omega (2-{\omega }^2)=0\Rightarrow {\omega }^2=2\Rightarrow \omega =\sqrt{2}.$ ($\omega =0$, origin pole, discard karte hain.)

Step 3 — wahan $L$ evaluate karo. ${\omega }^2=2$ pe denominator purely real hai: $-3{\omega }^2=-3(2)=-6$. Toh $L(j\sqrt{2})=\frac{K}{-6}=-\frac{K}{6}.$ Yeh crossing $-1$ pe land karti hai jab $-\frac{K}{6}=-1\Rightarrow \boxed{K=6.}$

WHAT: crossing point $-K/6$ ko $-1$ se compare karo. WHY: agar crossing $-1$ se left hai (zyada negative), toh curve ab $-1$ ke around wrap karti hai, $N$ ko $0$ se $2$ kar deti hai.

• $K=3$: crossing $-3/6=-0.5$ pe. Woh $-1$ se right mein hai → $-1$ enclosed nahi → $N=0\Rightarrow Z=0$ → stable.
• $K=6$: crossing exactly $-1$ pe → curve danger point se guzarti hai → marginally stable (ek pole imaginary axis pe baitha hai; $Z=0$ lekin boundary pe).
• $K=12$: crossing $-12/6=-2$ pe, $-1$ se left mein → $-1$ ab do baar clockwise enclosed → $N=2\Rightarrow Z=2$ → unstable (2 RHP closed-loop poles).

Figure kya dikhata hai: teen coloured dots negative real axis pe $-0.5$ (blue, $K=3$), $-1$ (yellow, $K=6$), aur $-2$ (pink, $K=12$) pe baithe hain. Chalk-white $\times$ $-1$ pe fixed danger point hai. Jaise $K$ badhta hai, crossing dot steadily leftward slide karta hai; jis moment woh $\times$ se guzarta hai ($K=6$ aur uske upar ke beech) curve $-1$ ko surround karne lagti hai aur encirclements switch on ho jaate hain.

WHAT: $Z=N+P$ ko $N$ ke liye solve karo $Z=0,P=1$ ke saath. WHY: stability $Z=0$ se define hoti hai; baaki sab bookkeeping hai. $0=N+1\Rightarrow N=-1.$ Negative $N$ matlab $-1$ ka ek counter-clockwise encirclement.

Woh kaisa dikhta hai: Nyquist curve ko danger point ke around ek baar loop karna hai, counter-clockwise spin karte hue. Yeh unstable plants ke liye Nyquist ka surprising heart hai — encirclement yahan warning sign nahi hai, yeh cure hai: yeh feedback se RHP open-loop pole ke cancel hone ka hisaab lagata hai.

Step 1 — WHAT: $L(j\omega )$ rationalize karo. WHY: geometry dekhne ke liye hum real aur imaginary parts alag karte hain. $L(j\omega )=\frac{K}{j\omega -1}=\frac{K(-1-j\omega )}{(j\omega -1)(-1-j\omega )}=\frac{K(-1-j\omega )}{1+{\omega }^2}.$ Toh $\mathrm{Re}L=\frac{-K}{1+{\omega }^2},\mathrm{Im}L=\frac{-K\omega }{1+{\omega }^2}.$

Step 2 — key points.

• $\omega =0$ pe: $L=-K$ (negative real axis pe).
• Jab $\omega \to \infty$: $L\to 0$.

Full curve (negative $\omega$ ke liye uski mirror image ke saath) ek circle hai jo $-K$ aur origin se guzarti hai, plane ke left half mein baithkar.

Figure kya dikhata hai: do circles draw hain, dono origin se guzarti hain. Blue circle ($K=0.6$) real axis pe sirf $-0.6$ tak pohonchti hai — yellow $\times$ $-1$ pe uske bahar hai. Pink circle ($K=1.6$) $-1.6$ tak pohonchti hai, toh yellow $\times$ $-1$ pe ab uske andar hai. Andar = encircled = $N=-1$.

Step 3 — $-1$ circle ke andar kab hai? Circle origin ($0$) se real axis pe $-K$ tak span karti hai. Point $-1$ exactly tab andar hota hai jab $K>1.$ Tab curve $-1$ ko ek baar counter-clockwise encircle karti hai, $N=-1$, jo deta hai $Z=N+P=-1+1=0$ → stable.

Agar $K<1$, toh point $-1$ circle ke bahar hai: $N=0$, toh $Z=0+1=1$ → unstable. $\boxed{\text{Stable for }K>1.}$

Part (a) — WHAT: RHP poles count karo. WHY: hume $Z=N+P$ ke liye $P$ chahiye. Denominator $s^2(s+10)$: poles $s=0$ pe (double, imaginary axis pe — upar diye definition ke anusaar right ki taraf indent kiya gaya, toh RHP mein count nahi) aur $s=-10$ pe (LHP). Isliye $P=0.$

Part (b) — WHAT: har real-axis crossing nikalo. WHY: curve sirf wahan $-1$ ko touch kar sakti hai jahan woh purely real ho; agar woh kisi finite negative value pe real nahi hoti siwaaye ek computable ke, toh woh value decide karti hai ki $-1$ cross hua ya nahi. $s=j\omega$ substitute karo: $L(j\omega )=\frac{K(1+j\omega )}{(j\omega {)}^2(j\omega +10)}=\frac{K(1+j\omega )}{-{\omega }^2(10+j\omega )}.$ Top aur bottom ko conjugate $(10-j\omega )$ se multiply karo: $L(j\omega )=\frac{K(1+j\omega )(10-j\omega )}{-{\omega }^2(100+{\omega }^2)}.$ Numerator expand karo (zero-factor identity use karke): $(1+j\omega )(10-j\omega )=10-j\omega +10j\omega +{\omega }^2=(10+{\omega }^2)+j9\omega .$ Toh $\mathrm{Re}L=\frac{-K(10+{\omega }^2)}{{\omega }^2(100+{\omega }^2)},\mathrm{Im}L=\frac{-9K\omega }{{\omega }^2(100+{\omega }^2)}=\frac{-9K}{\omega (100+{\omega }^2)}.$ Real-axis crossing ke liye $\mathrm{Im}L=0$ chahiye. Lekin $\mathrm{Im}L=\frac{-9K}{\omega (100+{\omega }^2)}$ kisi bhi finite $\omega >0$ ke liye kabhi zero nahi hota (numerator $-9K\ne 0$). Isliye $\text{koi bhi finite frequency nahi hai jis par plot real axis cross kare, aur isliye woh kisi bhi }K\text{ ke liye }-1\text{ kabhi cross nahi karta.}$

Part (c) — WHAT: quadrants aur indent arc track karke $N$ determine karo. WHY: koi finite real-axis crossing nahi hone ke saath, $N$ phase sweep se fix ho jaata hai, toh hum poori locus chalte hain.

• Chhota $\omega \to 0^{+}$: $\mathrm{Re}L\to \frac{-K\cdot 10}{{\omega }^2\cdot 100}=\frac{-K}{10{\omega }^2}\to -\infty$ aur $\mathrm{Im}L\to \frac{-9K}{\omega \cdot 100}\to -\infty$. Curve third quadrant se aati hai (dono parts negative), $\text{phase}=-180^{\circ }$ ke paas thoda neeche tilt hoke. DC pe total phase: do integrators $-180^{\circ }$ dete hain, zero aur $-10$ pole negligible hain → phase $\approx -180^{\circ }$.
• Bada $\omega \to \infty$: magnitude $\to 0$; phase $=\underset{\text{zero }(s+1)}{\underbrace{+90^{\circ }}}\underset{s^2}{\underbrace{-180^{\circ }}}\underset{(s+10)}{\underbrace{-90^{\circ }}}=-180^{\circ }$. Toh woh khatam bhi phase $-180^{\circ }$ pe origin approach karte hue hoti hai, yani negative real axis ke along, second/third quadrant boundary se.
• Beech mein: kyunki $\mathrm{Im}L<0$ sab $\omega >0$ ke liye (woh $\frac{-9K}{\omega (100+{\omega }^2)}$ ke barabar hai, hamesha negative), $\omega >0$ ki poori branch lower half-plane mein rehti hai (third quadrant, real part bhi poore time negative kyunki $\mathrm{Re}L<0$ sab $\omega$ ke liye). Yeh kabhi real axis ke upar nahi aati, toh $-1$ ke around loop nahi kar sakti.
• $s=0$ pe indent: double pole ke around chhoti right-half semicircle ek large clockwise arc of radius $\to \infty$ pe map hoti hai jo $+360^{\circ }$ sweep karta hai ($s^2$ angle ko double karta hai isliye) infinity pe, contour ko right side pe close karte hue — $-1$ se door.
• $\omega <0$ ke liye mirror: real axis ke baare mein reflection us branch ko upper half-plane mein daal deta hai.

Saath mein: dono finite branches lower/upper left region mein hug karti hain finite point $-1$ ko kabhi encircle kiye bina; closing arc infinity pe right mein baithti hai. $-1$ ki net winding isliye hai $N=0\text{har }K>0\text{ ke liye.}$ Isliye $Z=N+P=0+0=0\Rightarrow \boxed{\text{stable for every }K>0.}$ Sabak: $s=-1$ pe zero $+90^{\circ }$ phase lead inject karta hai (upar define kiya gaya), jo ek integrator ke $-90^{\circ }$ ko cancel kar deta hai pehle koi crossing frequency form ho sake, toh imaginary part kabhi zero nahi lauta aur curve $-1$ wrap nahi karti. Ek well-placed zero (lead compensation) hi woh cheez hai jo ek two-integrator GNC loop ko useable banata hai. Dekho Stability Margins in GNC Loops aur Bode Plot & Gain/Phase Margins.

Step 1 — WHAT: characteristic polynomial banao. WHY: closed-loop poles $1+L=0$ ke roots hain, yani $s(s+1)(s+2)+K=0$ ke. $s(s+1)(s+2)+K=s^3+3s^2+2s+K=0.$

Step 2 — WHAT: Routh array banao. WHY: Routh–Hurwitz first column mein sign changes se RHP roots count karta hai; marginal stability tab hoti hai jab first-column entry zero hit kare.

$s^1$ entry hai $\frac{6-K}{3}$.

Step 3 — marginal condition. Yeh tab vanish hoti hai jab $6-K=0$: $\boxed{K=6.}$ Yeh exactly Exercise 3.1 mein nikali Nyquist real-axis crossing se match karta hai — do independent methods, ek answer. $K<6$ ke liye first column entries $1,3,\frac{6-K}{3},K$ sab positive hain (koi sign change nahi → $Z=0$, stable); $K>6$ ke liye third entry $\frac{6-K}{3}$ negative ho jaata hai jabki $K>0$ positive rehta hai, do sign changes deta hai (positive→negative→positive) → $Z=2$, unstable. Yeh Nyquist ke $N=2$ se match karta hai aur disguise mein Root Locus ka imaginary axis crossing hai.

$\frac{K}{s(s+1)(s+2)}$ ka critical gain? ::: $K=6$. $\frac{K}{s-1}$ ke liye stable range? ::: $K>1$ (needs $N=-1$ since $P=1$). $\frac{K}{s(s+1)(s+2)}$ ki real-axis crossing kaun se $\omega$ par hoti hai? ::: $\omega =\sqrt{2}$. $\frac{K}{(s+1)(s+2)}$ ke liye, kya loop $K>0$ ke liye kabhi unstable hai? ::: Nahi — $P=0$ aur curve kabhi $-1$ nahi pohonchti, toh $Z=0$ hamesha. Double integrator ke liye $-1$ pe zero kya karta hai? ::: $+90^{\circ }$ phase lead add karta hai, $\mathrm{Im}L$ ko one-signed rakhta hai toh plot kabhi $-1$ wrap nahi karta → sab $K>0$ ke liye stable.