3.5.41 · Physics › Guidance, Navigation & Control (GNC)
Bode plot ek aisa tarika hai jisse hum dekh sakte hain ki ek system har possible frequency ki sine waves par kaise react karta hai — sab kuch ek hi graph mein.
KYA: Do stacked plots vs frequency ω (log axis par): ek gain ke liye (output amplitude kitna badhta/ghatta hai) aur ek phase ke liye (output kitna lag/lead karta hai).
KYUN: GNC mein hum controllers ko oscillating disturbances se drive karte hain (wind gusts, sensor noise, sloshing fuel). Hume pata hona chahiye ki loop kin frequencies par amplify karta hai, aur kitna delay karta hai — kyunki delay + gain hi instability ka kaaran banta hai.
KAISE: Hum transfer function G ( s ) ko imaginary axis s = j ω par evaluate karte hain, aur 20 log 10 ∣ G ( j ω ) ∣ (dB mein) aur ∠ G ( j ω ) (degrees mein) ko log 10 ω ke against plot karte hain.
Intuition Sine in → sine out
Ek linear time-invariant (LTI) system mein ek magic property hoti hai: usse frequency ω par ek pure sine do, aur steady state mein output bhi usi same frequency ki sine hogi — sirf amplitude scale hogi aur phase shift hogi. Aur kuch nahi badlega.
YEH KYUN hota hai (derive karo). Ek system lo jiska transfer function G ( s ) hai aur input u ( t ) = cos ω t = Re { e j ω t } hai. Complex exponential e j ω t kisi bhi LTI system ka eigenfunction hota hai:
y ( t ) = G ( j ω ) e j ω t
Complex number G ( j ω ) ko polar form mein likho:
G ( j ω ) = ∣ G ( j ω ) ∣ e j ϕ , ϕ = ∠ G ( j ω )
Tab
y ( t ) = ∣ G ( j ω ) ∣ e j ( ω t + ϕ ) ⇒ Re { y } = ∣ G ( j ω ) ∣ cos ( ω t + ϕ )
Definition Frequency response
Frequency response hai G ( j ω ) : transfer function ko s = j ω par evaluate kiya hua. Iska magnitude ∣ G ( j ω ) ∣ gain hai aur iska argument ∠ G ( j ω ) phase shift hai, dono ω ke functions hain.
Toh Bode plot literally complex-valued function G ( j ω ) ki ek picture hai — magnitude aur angle do alag graphs mein split ho jaate hain.
Intuition Products sums ban jaate hain
Real transfer functions simple factors ke products hote hain. Logarithms products ko sums mein badal dete hain — toh hum har factor ke Bode contributions ko aankhon se jod sakte hain. Yahi wajah hai ki Bode itna fast hai.
Derivation. Maano G = G 1 G 2 G 3 . Tab
∣ G ∣ = ∣ G 1 ∣∣ G 2 ∣∣ G 3 ∣
20 log 10 lo:
dB 20 log 10 ∣ G ∣ = 20 log 10 ∣ G 1 ∣ + 20 log 10 ∣ G 2 ∣ + 20 log 10 ∣ G 3 ∣
Aur phase seedha add hoti hai (product ka angle = angles ka sum):
∠ G = ∠ G 1 + ∠ G 2 + ∠ G 3
Definition Magnitude in decibels
∣ G ∣ dB ≡ 20 log 10 ∣ G ( j ω ) ∣
20 ka factor (10 nahi) isliye hai kyunki magnitude ek amplitude ratio hai; power ∝ amplitude2 , aur 10 log 10 ( amp 2 ) = 20 log 10 ( amp ) .
Log-frequency axis kyun: control systems kai decades span karte hain (0.01 rad/s slosh → 1000 rad/s structural modes). Log spacing sabko dikhata hai, aur — bonus — simple factors ke asymptotes straight lines ban jaate hain jinka slope dB/decade mein hota hai.
Koi bhi rational G ( j ω ) in factors mein decompose hoti hai. Yeh 4 seekh lo aur tum kuch bhi sketch kar sakte ho (80/20 core).
∣ K ∣ dB = 20 log 10 K (flat horizontal line) , ∠ K = 0 ∘
∣ G ∣ dB = 20 log 10 ω 1 = − 20 log 10 ω
Yeh ek straight line hai jiska slope − 20 dB/decade hai (ω mein har 10× par 20 dB drop).
∠ j ω 1 = ∠ e j 9 0 ∘ 1 = − 9 0 ∘ (constant)
∣ G ( j ω ) ∣ = 1 + ( ω / ω c ) 2 1
Asymptotes kyun:
Low freq ω ≪ ω c : term → 1 , toh ∣ G ∣ dB → 0 (flat).
High freq ω ≫ ω c : ∣ G ∣ ≈ ω c / ω , toh slope − 20 dB/decade.
ω = ω c par: ∣ G ∣ = 1/ 2 , yaani 20 log 10 ( 1/ 2 ) = − 3.01 dB — famous −3 dB corner .
Phase: ∠ G = − arctan ( ω / ω c ) : low par 0 ∘ , ω c par − 4 5 ∘ , high par − 9 0 ∘ .
Pole ka mirror image: slope +20 dB/decade corner ke upar, phase 0 ∘ → + 4 5 ∘ → + 9 0 ∘ tak badhti hai.
Worked example Example 1 — first-order lag
G ( s ) = 1 + s /5 10
Goal: Bode plot sketch karo; ω c = 5 rad/s, DC gain K = 10 .
DC gain in dB: 20 log 10 10 = 20 dB. Kyun? ω → 0 par ( 1 + s /5 ) term 1 ho jaata hai, toh gain = K .
Low-frequency asymptote: 20 dB par flat. Kyun? Pole ne apne corner ke neeche abhi "kick in" nahi kiya hai.
Corner ω = 5 par: magnitude = 20 − 3 = 17 dB. Kyun? Kisi bhi first-order corner par derived −3 dB dip.
5 rad/s ke upar: slope −20 dB/dec . Kyun? Pole ka high-frequency asymptote.
Phase: 0 ∘ (low) → − 4 5 ∘ at ω = 5 → − 9 0 ∘ (high). Kyun? ∠ = − arctan ( ω /5 ) .
Worked example Example 2 — pure double integrator
G ( s ) = s 2 1
Yeh ek rigid body ka force ke response mein position response hai (F = ma , twice-integrated).
∣ G ∣ dB = − 40 log 10 ω : slope −40 dB/dec har jagah. Kyun? Do integrators, har ek −20.
ω = 1 par 0 dB se guzarta hai. Kyun? log 10 1 = 0 ⇒ ∣ G ∣ = 1 .
Phase = − 18 0 ∘ constant. Kyun? Har 1/ ( j ω ) se − 9 0 ∘ milta hai; dono milake = − 18 0 ∘ .
GNC insight: ek plant jo − 18 0 ∘ phase par baitha hai wo instability ki edge par hai — controller ko crossover ke paas phase wapas add karni padegi (ek lead/PD term se).
Worked example Example 3 — stability margins padhna
Maano open-loop L ( j ω ) 0 dB par ω g c (gain-crossover) par cross karta hai aur −180° par ω p c (phase-crossover) par cross karta hai.
Phase margin = 18 0 ∘ + ∠ L ( j ω g c ) . Kyun? Kitne degrees ka extra lag phase ko −180° tak pahunchane se pehle absorb ho sakta hai us frequency par jahan gain unity hai (danger point).
Gain margin = − ∣ L ( j ω p c ) ∣ dB . Kyun? Kitne dB gain badha sakte ho jab tak magnitude −180° point par 0 dB tak nahi pahunch jaati.
Positive margins ⇒ stable, robust loop.
Common mistake "Magnitude ke liye
10 log 10 use karo."
Kyun sahi lagta hai: decibels famously power ke liye 10 log 10 ( P ) hote hain — yeh sach hai. Fix: Bode magnitude ek amplitude ratio hai, aur power ∝ amplitude². Toh 10 log 10 ( A 2 ) = 20 log 10 A . Gain ke liye hamesha 20 .
Common mistake "Corner woh jagah hai jahan phase = 0° ho."
Kyun sahi lagta hai: corner ek special "start" point jaisa lagta hai. Fix: corner ω c par ek single pole ki phase pehle se hi − 4 5 ∘ hai (halfway), aur magnitude − 3 dB hai, 0 nahi. Phase transition roughly ω c ke dono taraf ek decade tak phailti hai.
Common mistake "Slope −20 dB per octave hai."
Kyun sahi lagta hai: octaves (×2) aur decades (×10) ko confuse karna. Fix: standard Bode slopes per decade hote hain. (−20 dB/dec ≈ −6 dB/octave.)
Common mistake "Magnitude aur phase independent hain — main ek ko freely tune kar sakta hoon."
Kyun sahi lagta hai: yeh do alag graphs hain. Fix: minimum-phase systems ke liye yeh linked hain (Bode gain–phase relation): ek −20 dB/dec slope imply karta hai ≈ −90° phase. Tum steep roll-off nahi pa sakte bina phase lag chukaye — isliye aggressive filters loops ko destabilize kar dete hain.
Recall Pehle try karo, phir reveal karo
Hum G ko s = j ω par kyun evaluate karte hain? (sinusoidal steady-state; eigenfunction e j ω t )
Ek pole apne corner ke upar kaunsa slope add karta hai? (−20 dB/dec, −90°)
First-order corner par magnitude kya hoti hai? (−3 dB)
Phase margin define karo. (18 0 ∘ + ∠ L at gain crossover)
Recall Feynman: ek 12-saal ke bachhe ko explain karo
Socho ki tum ek bachhe ko swing par dhakka de rahe ho. Agar tum dheere dhakka do, swing aasani se follow karti hai (bada response). Agar tum bahut tez dhakka do, swing barely hilti hai (chota response) aur tumhare haath se peeche lag jaati hai. Bode plot bas ek chart hai jisme "swing kitni badi hai" aur "woh kitna peeche rehti hai" har ek pushing speed ke liye dikhaya jaata hai. Engineers ise use karte hain yeh ensure karne ke liye ki koi robot ya rocket khud hi hilna shuru na ho jaye jab koi cheez galat speed par use jilgila kar de.
"Pole neeche kheenchta hai: −20, −90." Ek P ole magnitude ko 20 dB/dec neeche P ush karta hai aur P hase ko 90° neeche. Ek Z ero ulta karta hai (upar +20, +90).
Bode plot kya dikhata hai? Frequency response G ( j ω ) ka magnitude (dB) aur phase (deg) vs log frequency.
s = j ω substitute kyun karte hain?Kyunki e j ω t LTI systems ka eigenfunction hai; sinusoid ka response ek scaled, phase-shifted sinusoid ∣ G ( j ω ) ∣ cos ( ω t + ∠ G ) hota hai.
Magnitude in dB ka formula? 20 log 10 ∣ G ( j ω ) ∣ ; factor 20 isliye kyunki power ∝ amplitude².
Ek pole corner ke upar kaunsa slope aur phase contribute karta hai? −20 dB/decade aur −90°.
Ek zero corner ke upar kaunsa slope aur phase deta hai? +20 dB/decade aur +90°.
First-order corner frequency par magnitude? −3 dB (exactly
20 log 10 ( 1/ 2 ) ).
First-order pole ke corner par phase? −45°.
Double integrator 1/ s 2 ka slope aur phase? −40 dB/decade, constant −180°.
Phase margin ki definition? 18 0 ∘ + ∠ L ( j ω g c ) gain-crossover (0 dB) frequency par.
Gain margin ki definition? Phase-crossover (−180°) frequency par 0 dB ke neeche dB mein: − ∣ L ( j ω p c ) ∣ d B .
Log frequency axis kyun? Kai decades span karta hai aur simple factors ke asymptotes ko dB/decade slopes wali straight lines banata hai.
Bode gain–phase relation (minimum phase)? Slope aur phase linked hain; jaise −20 dB/dec slope ≈ −90° phase imply karta hai.
Transfer function — woh G ( s ) jo hum s = j ω par evaluate karte hain.
Nyquist plot — wohi G ( j ω ) complex plane mein plot kiya; stability ke liye −1 ko encircle karta hai.
Stability margins — gain & phase margins seedhe Bode se padhe jaate hain.
Lead-lag compensator — crossover ke paas Bode plot ko shape karke design kiya jaata hai.
Low-pass filter — ek first-order pole; iska −3 dB corner iska cutoff hai.
Laplace transform — jahan se s , poles, aur zeros aate hain.
Loop shaping — GNC design method jo poori tarah Bode magnitude par driven hai.
Frequency response G of jw