3.5.42 · D3Gain margin, phase margin — stability margins
Yeh page stability margins ke liye drill hall hai. Parent note ne do rulers banaye the — gain margin (GM) aur phase margin (PM) — critical point − 1 har tarah ke cases un rulers pe throw karte hain taaki koi bhi exam question, koi bhi sign, koi bhi degenerate loop tumhe surprise na kar sake.
Shuru karne se pehle, teen reminders plain words mein taaki koi bhi line one se lost na ho:
Definition Woh vocabulary jo hum use karte rahenge
s L ( s ) s ω s = j ω j = − 1 L ( j ω ) L ( s ) L ( j ω ) open-loop response : ek complex number jisme ek size ∣ L ∣ angle ∠ L ω g c gain crossover : woh ω 1 ω p c phase crossover : woh ω − 18 0 ∘ GM = 1/∣ L ( j ω p c ) ∣ PM = 18 0 ∘ + ∠ L ( j ω g c )
Is topic ka har question in cells mein se kisi ek mein aata hai. Neeche har worked example us cell ke saath tagged hai jo woh cover karta hai.
Cell
Kya cheez isse alag banati hai
Example
A. Clean GM Loop ka ek real phase crossover hai; GM compute karo
Ex 1
B. Clean PM Loop $
L
C. Dono margins saath Usi loop par, GM aur PM dono nikalো
Ex 3
D. Infinite GM (degenerate) Phase kabhi − 18 0 ∘ = ∞
Ex 4
E. Negative margin (unstable) Gain bahut zyada push ho gayi ⟹ GM < 0 < 0
Ex 5
F. Time-delay twist Delay phase margin kha jaati hai; τ ma x
Ex 6
G. Design / word problem Target PM hit karne ke liye K
Ex 7
H. Exam trap Conditionally stable / sign confusion
Ex 8
Worked example Ex 1 — Cell A: clean gain marginStatement. L ( s ) = s ( s + 1 ) ( s + 2 ) 4
Forecast: Andaaza lagao — kya GM yahan parent ke Worked Example 1 se bada hai ya chota (jahan K = 1 = 6
ω p c ∠ L = − 18 0 ∘ ∠ L = − 9 0 ∘ − arctan ω − arctan ( ω /2 ) = − 18 0 ∘ arctan ω + arctan ( ω /2 ) = 9 0 ∘ Yeh step kyun? GM exactly wahan padhna padta hai jahan phase already danger angle par baith chuki ho — tab bas extra size chahiye − 1 9 0 ∘ tan ( a + b ) = 1 − t a n a t a n b t a n a + t a n b 1 − ω ⋅ 2 ω = 0 ω p c = 2 Wahan size. ∣ L ∣ = ω ω 2 + 1 ω 2 + 4 4 ω = 2 = 2 ⋅ 3 ⋅ 6 = 6 ∣ L ∣ = 6 4 = 3 2 Yeh step kyun? ω p c ∣ L ∣ GM = ∣ L ∣ 1 = 2 3 = 1.5 20 log 10 1.5 ≈ 3.52 Yeh step kyun? Hum ∣ L ∣ invert karte hain kyunki GM woh factor hai jisse tum gain multiply kar sakte ho before ∣ L ∣ 1 ∣ L ∣ = 3 2 2 3 1 − 1 ∣ L ∣
Verify: Parent ke K = 1 6 4 × K ∣ L ∣ 4 6/4 = 1.5
Worked example Ex 2 — Cell B: clean phase marginStatement. L ( s ) = s ( s + 1 ) 1
Forecast: Parent ne PM ≈ 51. 8 ∘
Gain crossover. ∣ L ∣ = ω ω 2 + 1 1 = 1 ⇒ ω 4 + ω 2 − 1 = 0 u = ω 2 u = 2 − 1 + 5 ≈ 0.618 ω g c = 0.618 ≈ 0.786 Yeh step kyun? PM wahan padhte hain jahan size exactly 1 phase lag chahiye − 18 0 ∘ Wahan phase. ∠ L = − 9 0 ∘ − arctan ( 0.786 ) = − 9 0 ∘ − 38. 2 ∘ = − 128. 2 ∘ PM = 18 0 ∘ + ( − 128. 2 ∘ ) = 51. 8 ∘
Figure dekho: loop curve unit circle (size = 1 − 18 0 ∘ hi phase margin hai.
Verify: tan ( 38. 2 ∘ ) = 0.787 ≈ ω g c 3 0 ∘ 6 0 ∘
Worked example Ex 3 — Cell C: ek hi loop par dono marginsStatement. L ( s ) = s ( s + 1 ) ( s + 2 ) 2 dono GM (dB) aur PM nikalо.
Forecast: K = 2 K = 1
GM. Wahi ω p c = 2 K ∣ L ∣ = 6 2 = 3 1 = 3 = 20 log 10 3 ≈ 9.54 Yeh step kyun? K PM — ω g c Solve karo ∣ L ∣ = ω ω 2 + 1 ω 2 + 4 2 = 1 ω g c ≈ 0.749 Yeh step kyun? Size= 1 Wahan phase. ∠ L = − 9 0 ∘ − arctan ( 0.749 ) − arctan ( 0.749/2 ) = − 9 0 ∘ − 36.8 3 ∘ − 20.5 3 ∘ = − 147.3 6 ∘ PM = 18 0 ∘ − 147.3 6 ∘ = 32. 6 ∘
Verify: Dono positive ⟹ stable, GM = 3 > 1 ω = 0.749 ∣ L ∣ ≈ 1.00 Bode plots dekho.)
Worked example Ex 4 — Cell D: infinite gain margin (degenerate)Statement. First-order loop L ( s ) = s + 1 5
Forecast: Instability se pehle kitni extra gain? "Koi finite factor" vs "unlimited" guess karo.
Phase range. ∠ L = − arctan ω ω : 0 → ∞ ∠ L 0 ∘ − 9 0 ∘ kabhi − 18 0 ∘ .
Yeh step kyun? GM − 18 0 ∘ ω p c GM = ∞ ∞ − 9 0 ∘ − 1 PM. Size ∣ L ∣ = ω 2 + 1 5 = 1 ⇒ ω 2 + 1 = 25 ⇒ ω g c = 24 ≈ 4.899 = − arctan ( 4.899 ) = − 78.4 6 ∘ = 18 0 ∘ − 78.4 6 ∘ = 101. 5 ∘
Verify: Ek single pole kabhi bhi ek simple loop ko gain akele unstable nahi kar sakta — classic result, GM = ∞ > 9 0 ∘ Root locus ke saath match karta hai jo sab K
Worked example Ex 5 — Cell E: negative margin (pehle se unstable)Statement. Ex 1 ka plant lo lekin gain crank karo: L ( s ) = s ( s + 1 ) ( s + 2 ) 12
Forecast: Hum jaante hain K = 6 K = 1 = 6 K = 12
GM. ω p c = 2 ∣ L ∣ = 6 12 = 2 = 2 1 = 0.5 = 20 log 10 0.5 ≈ − 6.02 Yeh step kyun? GM < 1 − 1 PM. ∣ L ∣ = 1 ω ω 2 + 1 ω 2 + 4 12 = 1 ⇒ ω g c ≈ 1.96 = − 9 0 ∘ − arctan ( 1.96 ) − arctan ( 0.98 ) = − 9 0 ∘ − 62.9 7 ∘ − 44.4 2 ∘ = − 197. 4 ∘ = 18 0 ∘ − 197. 4 ∘ ≈ − 17. 4 ∘ Yeh step kyun? Negative PM confirm karta hai ke phase gain ke 1 Verdict. GM < 0 aur PM < 0 unstable .
Verify: Critical gain K = 6 = 1 12 = 2 × 6 1/2 Nyquist stability criterion − 1
Worked example Ex 6 — Cell F: time-delay twistStatement. Stable loop L ( s ) = s ( s + 1 ) 1 = 51. 8 ∘ ω g c = 0.786 transport delay τ τ
Forecast: Delay sirf phase lag add karta hai, size mein koi change nahi. Guess karo kya τ ma x
Delay kya karta hai. Ek delay e − s τ 1 − ω τ Yeh step kyun? Kyunki size untouched hai, ω g c 0.786 Saari margin kha jao. Instability us moment aati hai jab added lag poori PM ke barabar ho jaata hai:
τ ma x = ω g c PM in radians . = 51. 8 ∘ = 51.8 × 180 π = 0.9040 Compute karo. τ ma x = 0.786 0.9040 = 1.150
Figure mein amber PM wedge dikhti hai jo delay ke growing lag se nibble hoti rehti hai jab tak woh khatam nahi ho jaati.
Verify: Units: rad/s rad = s 0.786 × 1.150 = 0.904 = 51. 8 ∘ e − s τ Time delay and Padé approximation dekho.
Worked example Ex 7 — Cell G: design / word problem (gain choose karo)Statement. Ek satellite reaction-wheel loop model hai L ( s ) = s ( s + 2 ) K = 4 5 ∘ K
Forecast: Ek handy rule hai (neeche justify ki gayi) ke PM degrees mein 100 ζ = 4 5 ∘ ζ ≈ 0.45 K 4
PM = 4 5 ∘ PM = 18 0 ∘ + ∠ L ( j ω g c ) = 4 5 ∘ ⇒ ∠ L ( j ω g c ) = − 13 5 ∘ ∠ L = − 9 0 ∘ − arctan ( ω /2 ) = − 13 5 ∘ ⇒ arctan ( ω /2 ) = 4 5 ∘ ⇒ ω g c = 2 Yeh step kyun? Phase K target angle pehle ω g c
Us ω g c = 1 ∣ L ∣ = ω ω 2 + 4 K = 1 ω = 2 2 ⋅ 8 K = 1 ⇒ K = 2 8 = 4 2 ≈ 5.657 Yeh step kyun? Ab K K ω g c = 2
ζ ≈ PM ∘ /100 Damping ratio ζ Damping ratio and overshoot ) measure karta hai ke second-order response kitna oscillatory hai: ζ = 0 ζ = 1 ζ ≲ 6 0 ∘ 1/100 ζ ≈ PM ∘ /100 approximation hai, law nahi — gain choose karne ke liye kaafi hai, phir simulation se confirm karo.
Verify: K = 5.657 ω g c = 2 = − 9 0 ∘ − arctan ( 1 ) = − 13 5 ∘ = 4 5 ∘ ζ ≈ 0.45 20% Damping ratio and overshoot ); overshoot trim karne ke liye bina speed khoye, PID controller tuning ya Loop shaping se reshape karo. K ≈ 5.66 > 4
Worked example Ex 8 — Cell H: exam trap (sign / "which frequency" confusion)Statement. Ek student likhta hai: "L ( s ) = s ( s + 1 ) ( s + 2 ) 10 ∠ L ( j ω g c ) = − 15 0 ∘ = − 15 0 ∘
Forecast: Guess karo kya − 15 0 ∘ positive ya negative PM matlab hai.
Trap pakdo. PM phase khud nahi hai ; yeh − 18 0 ∘ gap hai: PM = 18 0 ∘ + ∠ L ( j ω g c ) Yeh step kyun? Sabse common mistake (parent ka Mistake #4) ∠ L Sahi value. PM = 18 0 ∘ + ( − 15 0 ∘ ) = + 3 0 ∘ stable (barely; ringing likely).Doosra trap — which frequency? Agar student ne PM padhne ke liye ω p c ∠ = − 18 0 ∘ 0 ∘ "GaP" yaad karo: margin wahan measure hota hai jahan doosri quantity apni critical value hit kare.
Verify: 180 + ( − 150 ) = 30 > 0 − 15 0 ∘ − 18 0 ∘ kam lag hai, toh abhi bhi positive margin honi chahiye — geometry algebra confirm karta hai.
Recall Rapid self-testPhase crossover par tum kaun sa margin padhte ho? ::: Gain margin (GM = 1/∣ L ( j ω p c ) ∣ s + 1 K − 18 0 ∘ − ω τ 1 ≈ − 17. 4 ∘ = − 6 = 4 5 ∘ K = ∠ L ( j ω g c ) = 18 0 ∘ + ∠ L ( j ω g c )
"GaP" — G M at P hase crossover, P M at G ain crossover. Angle mein 180 add karo PM paane ke liye. Delay phase khaata hai, size kabhi nahi.