Bode plot — magnitude and phase vs frequency: Worked examples

3.5.41 · D3 · Physics › Guidance, Navigation & Control (GNC) › Bode plot — magnitude and phase vs frequency

Yeh page Bode plots ke liye "no surprises" drill hai. Hum har tarah ke case enumerate karte hain jo ek Bode problem mein aa sakte hain, phir ek example per case work karte hain taaki jab tum naya problem dekho toh bol sako "ah, yeh cell X hai, maine yeh dekha hai."

Yahan sab kuch sirf un tools se hai jo parent note mein pehle se banaye gaye hain: Transfer function $G (s)$ , uska frequency response $G (j ω)$ , magnitude in decibels $∣ G ∣_{dB} = 20 lo g_{10} ∣ G (j ω) ∣$ , aur phase $∠ G (j ω)$ . Agar koi symbol yahan aata hai, toh woh wahan define tha ya neeche spot pe define kiya gaya hai.

Scenario matrix

Ek Bode problem ko socho jaise har column mein se ek option choose karna. Neeche har example cell (row label) ke saath tagged hai jo woh cover karta hai.

Cell	Case class	Kya alag banata hai ise	Covered by
A	Sirf constant gain	Flat line, zero phase — "do-nothing" baseline	Ex 1
B	Single real pole (lag)	Ek −20 dB/dec corner, phase → −90°	Ex 2
C	Single real zero (lead)	Ek +20 dB/dec corner, phase → +90°	Ex 3
D	Pole at the origin (integrator)	Degenerate: koi corner nahi, slope from $ω \to 0$ , phase pinned at −90°	Ex 4
E	Repeated / high-order factor	Slopes aur phases multiply hote hain (−40 dB/dec, −180°)	Ex 5
F	Pole aur zero saath mein (lead–lag)	Do corners; slope inke beech mein phir flat ho jaati hai	Ex 6
G	Gain sign negative hai ( $K < 0$ )	Magnitude unchanged, phase shifted by ±180°	Ex 7
H	Limiting behaviour ( $ω \to 0$ and $ω \to \infty$ )	DC gain aur final slope calculator ke bina padh lo	Ex 8
I	Real-world word problem	Physics ko translate karo → $G (s)$ → margins	Ex 9
J	Exam twist: crossover & margins numerically find karo	$	L
K	Underdamped 2nd-order (complex poles)	Resonant peak aur −180° ka sharp phase drop	Ex 11

Inhi rows ke andar chuppe columns hain sign cases (positive/negative gain, poles vs zeros → phase up vs down) aur degenerate cases (pole at origin, $ω = 0$ , $ω = \infty$ , complex resonance). Har ek ko apna worked cell milta hai.

Example 1 — Cell A: constant gain

Yeh baseline hai jiske against har doosra plot measure hota hai: har jagah $12$ dB add karo aur phase ko $0$ se rotate karo.

Example 2 — Cell B: single real pole (lag)

Worked example

$G (s) = \frac{1}{1 + s /20}$ , corner $ω_{c} = 20$ rad/s Forecast: exactly $ω = 20$ pe dB value aur wahan phase guess karo.

Figure mein magnitude (dB, blue) $ω$ ke against log axis pe plot hai. Pink dashed lines asymptotes hain: corner ke neeche ek flat line $0$ dB pe aur uske upar ek $- 20$ dB/dec wali line. Yellow dot corner $ω = 20$ pe true curve ko mark karta hai, jo asymptotes ke crossing point se $3$ dB neeche hai.

DC value. Jab $ω \to 0$ , $G \to 1/1 = 1 \Rightarrow 20 lo g_{10} 1 = 0$ dB. Yeh step kyun? Corner ke neeche $(1 + s /20)$ term $\approx 1$ hai, isliye pole ne abhi "kick in" nahi kiya.
At the corner $ω = 20$ : $∣ G ∣ = 1/ 1 + 1^{2} = 1/ 2$ . $20 lo g_{10} (1/ 2) = - 3.01 dB$ Yeh step kyun? Yeh universal −3 dB corner hai jo parent note mein derive kiya gaya hai.
High frequency $ω ≫ 20$ : $∣ G ∣ \approx 20/ ω$ , slope −20 dB/dec (upper pink dashed line). Yeh step kyun? $s /20$ term dominate karta hai, isliye magnitude $1/ ω$ ki tarah girta hai.
Phase: $∠ G = - arctan (ω /20)$ : $0^{\circ}$ at DC, $- 4 5^{\circ}$ at $ω = 20$ , $\to - 9 0^{\circ}$ high. Yeh step kyun? $1/ (1 + j ω /20)$ ka angle denominator ke angle ka minus hota hai.

Verify: at $ω = 200$ (one decade above corner), $∣ G ∣_{dB} = 20 lo g_{10} (1/ 1 + 100) = - 20.04$ dB — corner se lagbhag $20$ dB neeche, jo −20 dB/dec slope confirm karta hai. ✓

Pink dashed asymptotes figure mein dekho: woh exactly corner pe milte hain, aur true curve (blue) us meeting point se $3$ dB neeche sagging karti hai.

Example 3 — Cell C: single real zero (lead)

Worked example

$G (s) = 1 + s /2$ , corner $ω_{c} = 2$ rad/s Forecast: ek zero pole ka mirror hota hai. Corner ke upar slope aur $ω = 2$ pe phase guess karo.

DC value. $ω \to 0 \Rightarrow G \to 1 \Rightarrow 0$ dB. Kyun? Pole ke corner se neeche jaisa hi — term $\approx 1$ hoti hai.
At the corner $ω = 2$ : $∣ G ∣ = 1 + 1 = 2 \Rightarrow + 3.01$ dB. Yeh step kyun? Zero apne corner pe $+ 3$ dB add karta hai — pole ke versus sign flip ho jaata hai.
High frequency: slope +20 dB/dec. Kyun? $∣1 + j ω /2∣ \approx ω /2$ linearly badhta hai.
Phase: $∠ G = + arctan (ω /2)$ : $0^{\circ} \to + 4 5^{\circ}$ at $ω = 2 \to + 9 0^{\circ}$ high. Yeh step kyun? Zero numerator mein hai, isliye uska angle add (positive) hota hai, subtract nahi.

Verify: at $ω = 20$ (one decade up), $∣ G ∣_{dB} = 20 lo g_{10} 1 + 100 = + 20.04$ dB — $20$ dB upar, +20 dB/dec confirm karta hai. ✓ Ex 2 se compare karo: identical magnitude size, opposite sign. Yeh pole↔zero mirror hai. Yeh exactly Lead-lag compensator ka ingredient hai.

Example 4 — Cell D (degenerate): pole at the origin

Worked example

$G (s) = \frac{50}{s}$ — gain ke saath ek integrator Forecast: yahan koi corner nahi hai. Curve $0$ dB kahan cross karta hai, aur har frequency pe phase kya hai?

Figure mein magnitude (dB, blue) versus $ω$ (log axis, x-labelled "frequency w") plot hai. Line ek single straight ramp hai slope $- 20$ dB/dec ke saath bina kisi flat section ke — yeh degenerate signature hai. Yellow dot mark karta hai jahan yeh $0$ dB cross karta hai $ω = 50$ pe; pink arrow low-frequency end par point karta hai jahan magnitude bina limit ke badhti rehti hai.

Magnitude. $∣ G (j ω) ∣ = 50/ ω$ , isliye $∣ G ∣_{dB} = 20 lo g_{10} 50 - 20 lo g_{10} ω = 33.98 - 20 lo g_{10} ω$ Yeh step kyun? $lo g$ quotient ko difference mein badal deta hai; $50$ poori $- 20$ dB/dec line ko upar shift karta hai.
Crossover (0 dB). $∣ G ∣_{dB} = 0$ set karo: $20 lo g_{10} ω = 33.98 \Rightarrow ω = 50$ rad/s. Yeh step kyun? $∣ G ∣ = 1$ tab hoga jab $50/ ω = 1$ , yaani $ω = 50$ — gain numerator ke barabar hai.
Phase. $∠ (50/ j ω) = ∠50 - ∠ (j ω) = 0^{\circ} - 9 0^{\circ} = - 9 0^{\circ}$ , sabhi $ω$ ke liye. Yeh step kyun? $j ω$ (angle $+ 9 0^{\circ}$ ) se divide karne par har cheez se $9 0^{\circ}$ subtract hota hai — ek flat phase line, integrator ka degenerate hallmark.

Verify: at $ω = 50$ : $∣ G ∣_{dB} = 33.98 - 20 lo g_{10} 50 = 0.00$ dB. ✓ Aur $ω = 5$ deta hai $33.98 - 13.98 = 20$ dB — das guna lower frequency, $20$ dB higher, origin case ke through slope confirm karta hai.

Degenerate feature: Ex 2/3 ke unlike yahan koi flat low-frequency section nahi hai — slope $ω \to 0$ tak poori tarah hold karta hai jahan magnitude blow up ho jaati hai (ek Low-pass filter pole eventually flat ho jaata, ek origin pole kabhi nahi hota).

Example 5 — Cell E: repeated (high-order) factor

Worked example

$G (s) = \frac{1}{s ^{2}}$ — double integrator Forecast: do integrators stacked. Slope aur phase guess karo.

Magnitude. $∣ G ∣ = 1/ ω^{2} \Rightarrow ∣ G ∣_{dB} = - 40 lo g_{10} ω$ : slope −40 dB/dec. Yeh step kyun? $1/ (j ω)$ ke do factors mein se har ek $- 20$ dB/dec contribute karta hai; logs add hote hain, isliye $- 20 - 20 = - 40$ .
Crossover. $∣ G ∣ = 1$ tab hota hai jab $ω^{2} = 1 \Rightarrow ω = 1$ rad/s $\Rightarrow$ $0$ dB pe $ω = 1$ se pass karta hai. Yeh step kyun? $lo g_{10} 1 = 0$ .
Phase. $∠ (1/ (j ω)^{2}) = - 9 0^{\circ} - 9 0^{\circ} = - 18 0^{\circ}$ , constant. Yeh step kyun? Har pole $- 9 0^{\circ}$ deta hai; woh add ho jaate hain.

Verify: at $ω = 10$ : $∣ G ∣_{dB} = - 40 lo g_{10} 10 = - 40$ dB — crossover se ek decade upar, $40$ dB neeche. ✓

GNC insight: $- 18 0^{\circ}$ pe baithna matlab hai zero phase margin agar gain crossover yahan land kare — loop knife-edge pe hai. Ek lead term ko crossover ke paas phase wapas inject karna hoga.

Example 6 — Cell F: pole aur zero (lead–lag shape)

Worked example

$G (s) = \frac{1 + s /1}{1 + s /100}$ — zero at 1, pole at 100 Forecast: do corners. Magnitude $ω = 1$ aur $ω = 100$ ke beech mein kya karta hai?

Figure mein do y-axes hain: left (blue) magnitude dB mein hai, right (pink) phase degrees mein hai, dono $ω$ ke against log x-axis pe. Blue curve $0$ dB pe flat hai, do yellow corner lines ( $ω = 1$ aur $ω = 100$ ) ke beech $+ 20$ dB/dec pe ramp karta hai, phir $+ 40$ dB pe flat ho jaata hai. Pink curve ek phase bump tak uthti hai aur wapas girti hai — pink dot $ω = 10$ pe uski peak mark karta hai.

$ω = 1$ se neeche: dono terms $\approx 1 \Rightarrow 0$ dB, flat. Kyun? Abhi koi corner active nahi hua.
1 aur 100 ke beech: zero active hai (+20 dB/dec) lekin pole abhi nahi — net slope +20 dB/dec. Yeh step kyun? Sirf zero ka corner pass hua hai, isliye sirf uski slope add hoti hai.
$ω = 100$ se upar: ab pole bhi activate ho jaata hai: $+ 20 - 20 = 0$ dB/dec — phir flat, higher level pe. Yeh step kyun? Slopes add hote hain; zero aur pole dono corners ke upar cancel ho jaate hain.
High-frequency level. Jab $ω \to \infty$ , $G \to (s /1) / (s /100) = 100 \Rightarrow 20 lo g_{10} 100 = 40$ dB. Yeh step kyun? $s$ terms top aur bottom dominate karte hain; unka ratio corner ratio $100/1$ hai.
Phase. $∠ G = arctan (ω /1) - arctan (ω /100)$ : $ω \approx 10$ ke paas $+ 9 0^{\circ}$ ki taraf uthta hai, phir $0^{\circ}$ ki taraf wapas girata hai — ek phase bump. (Yahan zero ka corner $1$ rad/s hai, isliye uska arctan argument $ω /1$ hai, yaani $ω$ khud.) Yeh step kyun? Zero phase pehle add karta hai; pole use baad mein cancel karta hai; peak geometric mean $1 \cdot 100 = 10$ pe lie karti hai.

Verify: geometric mean $ω = 10$ pe: $∠ G = arctan (10/1) - arctan (10/100) = 84.2 9^{\circ} - 5.7 1^{\circ} = 78.5 8^{\circ}$ . ✓ Yeh positive phase bump exactly wahi hai jo ek lead compensator provide karta hai — dekho Loop shaping.

Example 7 — Cell G (sign case): negative gain

Worked example

$G (s) = \frac{- 10}{1 + s /8}$ Forecast: minus sign kaunse graph ko change karta hai — magnitude ya phase?

Magnitude. $∣ - 10∣ = 10$ ; sign absolute value ke neeche disappear ho jaata hai. dB ko uske do parts mein split karo: $∣ G ∣_{dB} = flat gain term = 20 dB 20 lo g_{10} 10 - frequency-dependent pole term 20 lo g_{10} 1 + (ω /8)^{2}$ Pehla term constant $20$ dB hai (flat region). Doosra term corner ke neeche $0$ hai, $ω = 8$ pe $3$ dB hai, aur uske upar $20 lo g_{10} (ω /8)$ ki tarah badhta hai (the −20 dB/dec roll-off). Yeh positive-gain version se identical hai. Yeh step kyun? Magnitude $∣ G ∣$ hai; $- 1$ se multiply karne par koi length nahi badalti. Constant term ko $ω$ -dependent term se alag karne par saaf ho jaata hai ki sign sirf phase ko touch karta hai.
Phase. $∠ (- 10) = 18 0^{\circ}$ (ya equivalently $- 18 0^{\circ}$ ). Isliye $∠ G = - 18 0^{\circ} - arctan (ω /8)$ Yeh step kyun? $- 10$ ek negative real number hai — yeh negative real axis ke taraf point karta hai, angle $18 0^{\circ}$ . Pole ka lag phir uske upar subtract karta hai.

Verify: at $ω = 8$ : $∠ G = - 18 0^{\circ} - 4 5^{\circ} = - 22 5^{\circ}$ (equivalently $+ 13 5^{\circ}$ ). ✓ Poora phase curve positive-gain case se $\pm 18 0^{\circ}$ shift hua hai — ek subtlety jo Nyquist plot pe stability conclusions ko flip kar deti hai.

Example 8 — Cell H: pure limiting behaviour (bina calculator ke)

Worked example

$G (s) = \frac{100 ( 1 + s /10 )}{s ( 1 + s /1000 )}$ — sirf do ends padho Forecast: koi bhi arithmetic karne se pehle low-frequency slope aur high-frequency slope guess karo.

Jab $ω \to 0$ . Sirf lowest-order terms rakho: numerator $\to 100$ , denominator $\to s = j ω$ . Isliye $G \to 100/ (j ω)$ : slope −20 dB/dec, phase −90°. Yeh step kyun? Zero frequency ke paas, har $(1 + s / corner)$ collapse hokar $1$ ho jaata hai; sirf origin pole bachta hai.
Jab $ω \to \infty$ . Sirf highest-order terms rakho: numerator $\to 100 s /10 = 10 s$ , denominator $\to s \cdot s /1000 = s^{2} /1000$ . Isliye $G \to 10 s \cdot 1000/ s^{2} = 10000/ s$ : slope −20 dB/dec, phase −90°. Yeh step kyun? Bahut bade $ω$ pe, har $(1 + s / corner) \to s / corner$ ; $s$ ki net powers count karo: numerator mein $s^{1}$ hai, denominator mein $s^{2}$ hai → net $s^{- 1}$ .

Verify (slopes ka count): poles at origin aur 1000 (do, har ek −20), zero at 10 (ek, +20): high $ω$ pe net order $= - 2 + 1 = - 1 \Rightarrow - 20$ dB/dec. ✓ Dono ends ek bhi logarithm evaluate kiye bina match karte hain — limiting-value cell ka yahi point hai.

Example 9 — Cell I: real-world word problem

Worked example Reaction-wheel attitude loop

Ek spacecraft ka attitude torque pe double integrator ki tarah respond karta hai (angle = force twice-integrated), aur ek controller ek lead term add karta hai. Open loop hai $L (s) = \frac{20 ( 1 + s /2 )}{s ^{2}}$ Gain-crossover frequency $ω_{g c}$ (jahan $∣ L ∣ = 1$ ) aur wahan phase margin find karo. Forecast: zero ke bina, plant $- 18 0^{\circ}$ pe baitha hota hai (Ex 5). Kya zero use bachata hai?

$∣ L ∣ = 1$ set up karo. $∣ L ∣ = \frac{20 1 + ( ω /2 ) ^{2}}{ω ^{2}} = 1$ . Yeh step kyun? Gain crossover unit magnitude se define hoti hai ( $0$ dB).
Numerically solve karo. Yeh deta hai $ω_{g c} \approx 4.61$ rad/s. Yeh step kyun? Koi clean closed form nahi; yeh ek single-variable root hai, ek baar solve karo.
Wahan phase. $∠ L = arctan (ω /2) - 18 0^{\circ}$ . At $ω = 4.61$ : $arctan (2.305) = 66.5 5^{\circ}$ , isliye $∠ L = 66.55 - 180 = - 113.4 5^{\circ}$ . Yeh step kyun? Double integrator $- 18 0^{\circ}$ contribute karta hai; zero $+ 66. 6^{\circ}$ wapas add karta hai.
Phase margin $= 18 0^{\circ} + ∠ L = 180 - 113.45 = 66.5 5^{\circ}$ . Yeh step kyun? Parent note se definition: $- 18 0^{\circ}$ se pehle kitna lag bacha hai.

Verify: plug $ω = 4.61$ back: $∣ L ∣ = 20 1 + (2.305)^{2} / (4.61)^{2} = 20 (2.512) /21.25 \approx 1.00$ . ✓ Phase margin $+ 66. 6^{\circ}$ healthy hai — lead zero ne us plant ko bachaya jiska akele $0^{\circ}$ margin tha. Dekho Stability margins.

Example 10 — Cell J: exam twist, dono crossovers & dono margins

Worked example

$L (s) = \frac{40}{s ( 1 + s /4 ) ( 1 + s /20 )}$ Forecast: guess karo ki phase margin ya gain margin binding hogi.

Phase crossover $ω_{p c}$ : $∠ L = - 18 0^{\circ}$ solve karo. $∠ L = - 9 0^{\circ} - arctan (ω /4) - arctan (ω /20) = - 18 0^{\circ}$ Isliye $arctan (ω /4) + arctan (ω /20) = 9 0^{\circ}$ , jo deta hai $ω_{p c} = 4 \cdot 20 = 8.944$ rad/s. Yeh step kyun? $9 0^{\circ}$ tak sum karne wale do arctans matlab hai unke arguments reciprocals hain: $tan α$ aur $tan (9 0^{\circ} - α) = 1/ tan α$ multiply karke $1$ dete hain. Yahan $(ω /4) (ω /20) = 1 \Rightarrow ω^{2} = 80 \Rightarrow ω = 80$ , do corner frequencies $4$ aur $20$ ka geometric mean. Isliye do lags ka $- 18 0^{\circ}$ phase point hamesha unke corners ke geometric mean pe hota hai (plus origin pole se $- 9 0^{\circ}$ hamen exactly $- 18 0^{\circ}$ tak le jaata hai).
Gain margin $= - ∣ L (j ω_{p c}) ∣_{dB}$ . $∣ L (j 8.944) ∣ = \frac{40}{8.944 1 + ( 8.944/4 ) ^{2} 1 + ( 8.944/20 ) ^{2}}$ $= 40/ (8.944 \cdot 2.449 \cdot 1.096) = 1.667$ , isliye $∣ L ∣_{dB} = 20 lo g_{10} 1.667 = 4.44$ dB, jo deta hai gain margin $= - 4.44$ dB. Yeh step kyun? $ω_{p c}$ pe magnitude $0$ dB se zyada hai, isliye gain badhana danger nahi — loop already safe point ke baad hai. Negative gain margin matlab unstable.
Gain crossover $ω_{g c}$ : $∣ L ∣ = 1$ numerically solve karo $\Rightarrow ω_{g c} \approx 12.9$ rad/s. Yeh step kyun? Gain crossover wahan hai jahan $∣ L ∣ = 1$ ; teen poles ke roll off ke saath, hum single magnitude equation numerically solve karte hain.
Phase margin $= 18 0^{\circ} + ∠ L (j ω_{g c})$ . At $ω = 12.9$ : $∠ L = - 90 - arctan (3.225) - arctan (0.645) = - 90 - 72.79 - 32.83 = - 195. 6^{\circ}$ , isliye phase margin $= 180 - 195.6 = - 15. 6^{\circ}$ (negative → unstable). Yeh step kyun? Dono margins agree karte hain: yeh loop unstable hai aur iski gain kam karni hogi (ya lead add karna hoga).

Verify: $ω_{p c} = 80 = 8.944$ . ✓ $∣ L (j ω_{p c}) ∣ = 1.667$ . ✓ Dono margins negative — consistent stability verdict. Fix: $40$ ko reduce karo taaki gain crossover $ω_{p c}$ se neeche aaye, positive margins restore ho.

Example 11 — Cell K: underdamped second-order (complex poles)

Worked example

$G (s) = \frac{ω _{n}^{2}}{s ^{2} + 2 ζ ω _{n} s + ω _{n}^{2}}$ with $ω_{n} = 10$ rad/s, $ζ = 0.1$ Naye symbols (yahan banaye gaye): $ω_{n}$ natural frequency hai — jahan do poles "chahte" hain oscillate karna; $ζ$ (zeta) damping ratio hai — oscillation kitni strongly suppress hoti hai ( $ζ = 1$ critically damped, $ζ \to 0$ pure ringing). Jab $ζ < 1$ do poles ek complex-conjugate pair hote hain, aur magnitude curve ek clean corner ke bajaay ek resonant peak badhata hai. Forecast: $ζ = 0.1$ (bahut light damping) ke saath, guess karo peak kitna tall hai aur kahan hai.

Figure mein magnitude (dB, blue, left) aur phase (deg, pink, right) $ω$ ke against log axis pe hain. Notice karo ki blue curve sirf $ω_{n} = 10$ pe nahi bend karti — yeh ek sharp yellow-dotted peak tak spike karti hai $ω_{n}$ ke paas, phir $- 40$ dB/dec pe roll off karti hai. Pink phase $ω_{n}$ pe $- 9 0^{\circ}$ se through drop karti hai aur $- 18 0^{\circ}$ tak swing karti hai, aur drop steep hai kyunki damping choti hai.

Low frequency $ω ≪ ω_{n}$ : $G \to ω_{n}^{2} / ω_{n}^{2} = 1 \Rightarrow 0$ dB, flat. Yeh step kyun? $s^{2}$ aur $2 ζ ω_{n} s$ terms $ω_{n}$ se neeche negligible hain; sirf $ω_{n}^{2} / ω_{n}^{2}$ bachta hai.
High frequency $ω ≫ ω_{n}$ : $G \to ω_{n}^{2} / (- ω^{2})$ , slope −40 dB/dec, phase −180°. Yeh step kyun? Do poles saath roll off karte hain — double integrator jaisa, lekin shift hua hua.
Peak. Evaluate at $ω = ω_{n}$ : denominator $= (j ω_{n})^{2} + 2 ζ ω_{n} (j ω_{n}) + ω_{n}^{2} = - ω_{n}^{2} + 2 ζ ω_{n}^{2} j + ω_{n}^{2} = 2 ζ ω_{n}^{2} j$ . Isliye $∣ G (j ω_{n}) ∣ = \frac{ω _{n}^{2}}{2 ζ ω _{n}^{2}} = \frac{1}{2 ζ} = \frac{1}{0.2} = 5 \Rightarrow 20 lo g_{10} 5 = 13.98 dB$ Yeh step kyun? Denominator ke real parts exactly $ω_{n}$ pe cancel ho jaate hain, sirf tiny imaginary damping term $2 ζ ω_{n}^{2}$ bachta hai — ek chota denominator matlab ek badi magnitude. Yahi resonant peak hai. Damping jitni halki, peak utni tall: peak $\approx 1/ (2 ζ)$ .
Phase. At $ω_{n}$ denominator purely imaginary hai ( $+ j$ ), isliye $∠ G = - 9 0^{\circ}$ . Yeh $0^{\circ} \to - 9 0^{\circ} \to - 18 0^{\circ}$ sweep karta hai, aur sweep chote $ζ$ ke liye sharp hoti hai. Yeh step kyun? Complex poles $ω_{n}$ ke aas-paas ek narrow band mein phase ko $- 18 0^{\circ}$ se rotate karte hain.

Verify: peak $= 1/ (2 \cdot 0.1) = 5$ , yaani $20 lo g_{10} 5 = 13.98$ dB. ✓ (True peak $ω_{r} = ω_{n} 1 - 2 ζ^{2} \approx 9.9$ rad/s pe hai, essentially $ω_{n}$ pe chote $ζ$ ke liye.) GNC insight: ek lightly damped structural mode (fuel slosh, flexible boom) exactly isi spike ki tarah dikhta hai — agar wahan loop gain $0$ dB se zyada hai, toh loop resonance ko unstable drive kar sakta hai. Loop shaping use notch out kar deta hai.

Recall

Recall Kaun si cell kaun si hai?

Negative gain phase (by ±180°) change karta hai, magnitude nahi ::: magnitude $∣ G ∣$ hai, sign se unaffected Origin pe pole ka koi corner nahi hota aur constant −90° phase hoti hai ::: yeh $ω \to 0$ pe kabhi flat nahi hota Do corners ke beech, low zero aur high pole wale case mein slope +20 dB/dec hai, phir pole ke upar flat ::: slopes add hote hain Do arctans tab $90°$ mein sum karte hain jab ::: unke arguments ka product $1$ ho (geometric-mean frequency) Negative gain margin matlab ::: loop unstable hai — gain already −180° pe safe level se zyada hai Underdamped pole pair peak karta hai height pe ::: approximately $1/ (2 ζ)$ $ω_{n}$ ke paas