Bode plot — magnitude and phase vs frequency: Question bank

3.5.41 · D5 · Physics › Guidance, Navigation & Control (GNC) › Bode plot — magnitude aur phase vs frequency

Definition Symbols

$s$ , $j$ , aur $G (s)$ — pehle yeh padho

$j$ imaginary unit hai: woh number jiske liye $j^{2} = - 1$ (engineers $j$ likhte hain, $i$ nahi, taaki electric current se clash na ho). $j$ se multiply karna matlab "kisi quantity ko complex plane mein $9 0^{\circ}$ rotate karna," aur exactly isi tarah yeh ek phase shift encode karta hai.
$s$ Laplace variable hai (dekho Laplace transform): ek complex number $s = σ + j ω$ . Iska imaginary part wahi frequency $ω$ hai jo hum horizontal axis par plot karte hain. $σ = 0$ set karna (yani $s = j ω$ ) matlab "imaginary axis par seedha upar chalna," har steady-state sine frequency ko sweep karna.
$G (s)$ transfer function hai (dekho Transfer function): plant ke liye output aur input ka ratio, $s$ ki function ke roop mein likha hua. Ise $s = j ω$ par evaluate karne se yeh frequency response $G (j ω)$ ban jaata hai, jiska size aur angle do Bode curves ban jaate hain.

Figure — Bode plot — magnitude and phase vs frequency

True or false — justify karo

Neeche har prompt ek poora statement hai. True ya false decide karo, phir reveal karne se pehle ek saas mein bolo kyon.

Bode plot kisi bhi system ke liye kaam karta hai, linear ho ya nahi.

False — sirf linear time-invariant (LTI) systems mein "sine in → same-frequency sine out" property hoti hai; ek nonlinear system nayi frequencies create kar sakta hai, isliye har

ω

ke liye ek single gain-and-phase pair ka matlab hi nahi banta.

Ek first-order pole apni corner frequency par magnitude ko exactly 3 dB giraa deta hai.

True —

ω = ω_{c}

par hume

∣ G ∣ = 1/ 2

milta hai, aur

20 lo g_{10} (1/ 2) = - 3.01

dB, jo defining "−3 dB corner" hai.

Corner frequency par ek single pole ki phase

0^{\circ}

hoti hai, kyunki wahi se pole "shuru" hota hai.

False — corner par phase already

- 4 5^{\circ}

hoti hai (apne

0^{\circ} \to - 9 0^{\circ}

swing ke beech mein); corner ek magnitude landmark hai, phase-starting line nahi.

- 20

dB/decade ka standard Bode slope

- 20

dB/octave ke barabar hai.

False — ek decade

\times 10

hai aur ek octave

\times 2

;

- 20

dB/decade sirf lagbhag

- 6

dB/octave hai, isliye dono numbers

lo g_{10} 2 \approx 0.3

ke factor se alag hain.

Do transfer functions ko multiply karna matlab unke Bode magnitude curves ko dB mein add karna.

True —

∣ G_{1} G_{2} ∣ = ∣ G_{1} ∣∣ G_{2} ∣

, aur

20 lo g_{10}

lene se woh product ek sum ban jaata hai; phase bhi add hoti hai. Yahi poori wajah hai ki Bode sketching itni fast hai.

Ek pure integrator

1/ s

ki magnitude plant ke DC gain par depend karti hai.

False —

1/ s

ki magnitude

- 20 lo g_{10} ω

hai bina kisi free constant ke; yeh hamesha

ω = 1

rad/s par

0

dB se guzarti hai aur kisi bhi cheez se nirbhata nahi

- 20

dB/dec slope ke saath.

Agar ek plant exactly

- 18 0^{\circ}

phase par hai jahan uska gain

1

(

0

dB) hai, to closed loop aaram se stable hai.

False — yeh exactly danger point hai jo upar define kiya gaya hai: zero phase margin matlab loop marginally unstable hai aur koi bhi extra lag ise sustained oscillation mein dhakael deta hai.

Ek minimum-phase system ke liye, tum magnitude curve ko freely shape kar sakte ho bina phase affect kiye.

False — Bode gain–phase relation inhe bind karta hai: ek

- 20

dB/dec slope force karta hai lagbhag

- 9 0^{\circ}

phase, isliye steep roll-off hamesha phase lag ka kharch uthaata hai.

Ek first-order zero

(1 + s / ω_{c})

pole ka mirror image hai:

+ 20

dB/dec aur phase

+ 9 0^{\circ}

tak utha rahi hai.

True — ek zero wahan multiply karta hai jahan ek pole divide karta hai, isliye asymptote mein har sign palat jaata hai; exactly isi tarah ek lead network phase wapas khareedt a hai.

∣ G ∣_{dB} = 20 lo g_{10} ∣ G ∣

mein jo factor

20

hai woh isliye hai kyunki decibels ke liye hamesha

20

use karte hain.

False — power decibels

10 lo g_{10}

use karte hain;

20

isliye aata hai kyunki gain ek amplitude ratio hai aur power

\propto

amplitude

^{2}

, isliye

10 lo g_{10} (A^{2}) = 20 lo g_{10} A

Galti dhundo

Har line mein ek flawed claim hai. Ek sentence mein galti batao.

"Ek double integrator

1/ s^{2}

ka slope

- 20

dB/decade hai kyunki yeh abhi bhi sirf ek integrator hai."

Galti yeh hai: do integrators mein se har ek

- 20

contribute karta hai, isliye slope

- 40

dB/decade hai aur phase

- 9 0^{\circ}

nahi balki

- 18 0^{\circ}

hai.

"Kyunki phase ek alag graph hai,

9 0^{\circ}

phase lag badalta hai ki system us frequency par ek signal ko kitna amplify karta hai."

Galti yeh hai: ek fixed frequency par sirf magnitude curve amplification set karti hai — phase shift sirf bataata hai ki output kitna lag karta hai, na ki yeh kitna bada hai, isliye phase us frequency par amplification nahi badalta (halanki gain aur phase milkar loop stability decide karte hain).

"Gain margin us frequency par padha jaata hai jahan phase

0^{\circ}

hai."

Galti yeh hai: gain margin ==phase-crossover

ω_{p c}

== par padha jaata hai jahan phase

= - 18 0^{\circ}

hoti hai (danger-point phase),

0^{\circ}

nahi.

"Phase margin nikalene ke liye, gain crossover par phase ko

18 0^{\circ}

se minus karo."

Galti yeh hai: yeh

18 0^{\circ} + ∠ L (j ω_{g c})

hai jo gain-crossover

ω_{g c}

par evaluate hota hai — tum (negative) phase ko add karte ho, kyunki wahan phase typically

- 13 5^{\circ}

jaisi hoti hai, jo ek positive margin deta hai.

"Control loop ke andar sharper cutoff wala low-pass filter hamesha safer choice hai."

Galti yeh hai: ek sharper low-pass zyaada tezi se roll off karta hai, jo gain–phase relation ke anusaar crossover ke paas zyaada phase lag dump kar deta hai — aksar loop ko stabilize karne ki jagah destabilize kar deta hai.

"Bode plot par horizontal axis linear scale par frequency hai isliye tum

ω

directly padh sakte ho."

Galti yeh hai: frequency axis logarithmic hai (

lo g_{10} ω

ω

rad/s mein); yahi wajah hai ki simple factors kaafi decades mein straight-line asymptotes ke roop mein dikhte hain.

"Ek constant gain

K

low-frequency asymptote ko

20 lo g_{10} K

dB per decade tilts karta hai."

Galti yeh hai: ek constant gain

20 lo g_{10} K

dB ka ek flat horizontal shift hai jisme zero slope aur

0^{\circ}

phase hai — yeh level badalta hai, tilt nahi.

1/ (1 + s / ω_{c})

ki corner frequency

s = ω_{c}

hai, isliye yeh real axis par hai."

Galti yeh hai: corner ek frequency value

ω = ω_{c}

rad/s hai jo

s = j ω

par evaluate hoti hai; hum imaginary axis par upar chalte hain, aur corner wahan hai jahan

ω

pole ki magnitude ke barabar ho.

Why questions

Har ek ka jawab mechanism ke ek ya do sentences mein do. Definitions box se yaad karo ki $L (j ω)$ loop transfer function (plant × controller × sensor) hai, aur $s = j ω$ imaginary axis par upar chalta hai.

Hum

G (s)

ko specifically

s = j ω

par kyun evaluate karte hain, kisi aur point par kyun nahi?

Kyunki

e^{j ω t}

ek LTI system ka eigenfunction hai, isliye

s = j ω

(imaginary axis,

σ = 0

) exactly woh steady-state gain aur phase extract karta hai jo frequency

ω

(rad/s mein) ki ek pure sine par apply hoti hai.

Magnitude ka logarithm lene se hand-sketching kyun possible ho jaati hai?

Logarithms Transfer function ke factors ke products ko sums mein convert karte hain, isliye har pole aur zero ek additive straight-line piece contribute karta hai jise tum aankhon se stack kar sakte ho.

Pole ki phase transition corner ke har taraf roughly ek decade tak kyun spread hoti hai?

Kyunki

∠ = - arctan (ω / ω_{c})

dheere dheere badlata hai — yeh corner se ek decade neeche

0^{\circ}

ke paas hai aur ek decade upar

- 9 0^{\circ}

ke paas, isliye swing smooth hai, ek jump nahi.

GNC engineers delay ki parwah kyun karte hain, sirf gain ki nahi?

Kyunki delay (phase lag) aur loop gain milkar negative feedback ko positive feedback mein badal dete hain: kaafi lag ek correcting signal ko ek reinforcing signal mein palat deta hai, oscillation create karta hai.

Ek lead compensator

- 18 0^{\circ}

ke paas baith e hue loop ki madad kyun karta hai?

Ek lead network ka zero crossover ke paas positive phase add karta hai, phase margin restore karta hai taaki corrective action us se pehle pahunche jab tak loop response danger point tak lag na ho jaaye.

Stability judge karne ke liye Nyquist plot kabhi kabhi Bode se zyaada preferred kyun hota hai?

Bode margins assume karte hain ki danger point

L (j ω)

ka ek single crossover hai, lekin un loops ke liye jo

- 18 0^{\circ}

ko kai baar cross karte hain, Nyquist encirclement count woh rigorous stability verdict deta hai jo alag margins misread kar sakte hain.

Loop gain badhane se dono margins kyun shrink hote hain?

L

ki poori magnitude curve ko utha ne se gain crossover

ω_{g c}

higher frequency par push ho jaata hai jahan phase zyaada negative hai (kam phase margin) aur level

- 18 0^{\circ}

point par

0

dB ke zyaada kareeb aa jaata hai (kam gain margin).

Edge cases

Woh degenerate aur limiting scenarios jo topic chhupa leta hai.

ω \to 0^{+}

aur

ω \to \infty

par

1/ s

ki phase ka kya hota hai?

Kuch nahi badalta — integrator ki phase har frequency par constant

- 9 0^{\circ}

hai; sirf iska magnitude

+ \infty

ω = 1

rad/s par

0

dB tak sweep karta hai.

Exactly DC (

ω = 0

) par ek first-order pole ki magnitude kya hai?

Yeh exactly

1

hai (yani DC-gain level se

0

dB upar), kyunki jab

ω = 0

hota hai tab

(1 + j ω / ω_{c})

term

1

ke barabar hoti hai, isliye pole abhi "kick in" nahi kiya hai.

Agar do poles ek hi corner frequency

ω_{c}

share karein to kya hoga?

Unka contribution simply double ho jaata hai: slope corner ke upar

- 40

dB/dec ho jaati hai aur phase

- 18 0^{\circ}

ki taraf jaati hai, kyunki har pole apna

- 20

dB/dec aur

- 9 0^{\circ}

add karta hai.

Ek pure time-delay element

e^{- s τ}

do Bode curves ke saath kya karta hai?

Iska magnitude

∣ e^{- j ω τ} ∣ = 1

hai (=

0

dB, flat, har frequency par unchanged), lekin iska phase

- ω τ

radians hai — ek lag jo

ω

badhne ke saath unbounded badhti jaati hai. Isliye ek pure delay gain mein kuch nahi leta lekin phase margin ko tezi se tezi se khaata hai, ek classic Loop shaping trap jo straight-line Bode sketch (jisme koi corner nahi hota jaanne ko) poori tarah chhupa leta hai.

Agar ek transfer function mein right-half-plane (non-minimum-phase) zero ho, to kya "sharper slope means more lag" intuition phir bhi sahi rehti hai?

Nahi — ek non-minimum-phase zero magnitude slope add karta hai ek normal zero ki tarah lekin phase subtract karta hai ek pole ki tarah, isliye woh gain–phase link jo minimum-phase systems ke liye hold karta hai toot jaata hai.

Ek right-half-plane (unstable) pole Bode phase ko ek ordinary pole ki tulna mein kya karta hai?

Ek RHP pole same

- 20

dB/dec magnitude roll-off rakhta hai lekin iska phase ulte direction mein jaata hai — yeh

- 9 0^{\circ}

ki jagah

+ 9 0^{\circ}

contribute karta hai — isliye magnitude normal lagti hai jabki phase waise drift karti hai jo minimum-phase reader expect karta hai uske ulta, ek classic trap non-minimum-phase analysis mein.

Ek resonant (lightly damped) complex-pole pair ke paas apni natural frequency ke kareeb Bode magnitude kya karta hai?

Gentle

- 3

dB corner ki jagah yeh ek sharp peak banata hai jo

0

dB se kaafi upar uthta hai aur jiska height damping girane ke saath badhta hai; straight-line asymptote yeh miss kar leta hai, isliye Loop shaping ko explicitly peak account karni padti hai.

Agar phase kisi bhi finite frequency par

- 18 0^{\circ}

tak kabhi nahi pahunchti to gain margin kya hai?

Gain margin infinite hai — bina kisi phase-crossover

ω_{p c}

ke koi gain level nahi hai jo us mechanism se loop ko unstable kare, isliye margin unbounded hai (halanki doosre criteria phir bhi bind kar sakte hain).

Agar gain-crossover frequency

ω_{g c}

ko wahan push kiya jaaye jahan phase exactly

- 18 0^{\circ}

ho to phase margin ka kya hoga?

Phase margin

0^{\circ}

ho jaati hai, danger point —

ω_{g c}

ω_{p c}

se coincide karta hai aur loop ek steady oscillation sustain karta hai, stable aur unstable closed loop ke beech ki boundary.

Recall Ek-line self-check

Upar ke answers cover karo aur unme se teen ko scratch se re-derive karo — ek "true/false," ek "spot the error," aur ek "edge case" chuno. Agar koi bhi reason sirf haan/na ke roop mein nikla, to parent note ka woh section dobara dekho jiska woh hissa hai.