Exercises — Bode plot — magnitude and phase vs frequency

HUM KYA KARTE HAIN: zero phase ke saath flat line ka matlab hai koi poles, koi zeros, koi integrators nahi — sirf ek constant gain $K$. KYUN: poles/zeros/integrators sab curve ko modhte hain; sirf ek plain number $K$ use flat rakhta hai $0^{\circ }$ phase ke saath. dB ko ratio mein convert karo. $|K{|}_{\text{dB}}=20{\log }_{10}K$ se: $26=20{\log }_{10}K\Rightarrow {\log }_{10}K=1.3\Rightarrow K=10^{1.3}\approx \mathrm{19.95.}$ Answer: $G(s)=K\approx 20$ (lagbhag 20 ka pure gain).

KYA/KYUN: slope $-20$ dB/dec har jagah (koi corner nahi, koi flattening nahi) plus constant $-90^{\circ }$ phase single integrator $1/s$ ki fingerprint hai. Level check karo: $|1/(j\omega ){|}_{\text{dB}}=-20{\log }_{10}\omega$, jo $\omega =1$ par $0$ dB hai. ✓ Answer: $G(s)=\frac{1}{s}$.

KYA/KYUN: "flat, phir neeche $-20$ dB/dec, phase $0^{\circ }\to -90^{\circ }$" ek single first-order pole hai (ek basic Low-pass filter). Bend ki location corner ${\omega }_c$ hai. Answer: $G(s)=\frac{1}{1+s/10}$, corner ${\omega }_c=10$ rad/s. ${\omega }_c$ par magnitude $-3$ dB hai aur phase $-45^{\circ }$ hai.

(a) DC gain. $\omega \to 0$ par $(1+s/20)$ term $1$ ho jaata hai, toh $|G|=50$. $20{\log }_{10}50=20(1.699)=33.98\text{ dB}\approx 34\text{ dB}.$ (b) Corner $\omega =20$ par. Pole wahaan exactly $-3$ dB contribute karta hai, toh $33.98-3.01=30.97\text{ dB}\approx 31\text{ dB}.$ (c) $\omega =200$ par (corner se ek decade upar). Asymptote par corner ke upar har decade $20$ dB girta hai. $\omega =20$ se $\omega =200$ tak ek decade hai $\Rightarrow$ flat level se $-20$ dB: $33.98-20=13.98\text{ dB}\approx 14\text{ dB}.$

KYA: har factor ki phase add karo (product ka angle = angles ka sum).

• Constant $100$: $0^{\circ }$.
• Integrator $1/s$: $-90^{\circ }$ (hamesha).
• Zero $(1+j\omega /2)$ $\omega =2$ par: $\arctan (2/2)=\arctan (1)=+45^{\circ }$. $\angle G(j2)=0-90^{\circ }+45^{\circ }=-45^{\circ }.$

(a) $20{\log }_{10}(0.5)=20(-0.301)=-6.02$ dB $\approx -6$ dB. (Amplitude ka aadha hona $=-6$ dB — yaad karne layak.) (b) $-40=20{\log }_{10}r\Rightarrow {\log }_{10}r=-2\Rightarrow r=10^{-2}=0.01$.

Corners: integrator ka koi corner nahi hai (uska $-20$ dB/dec $\omega \to 0$ se shuru hota hai); do poles $\omega =10$ aur $\omega =100$ par corner karte hain. Starting slope: akela integrator sabse kam frequencies par $-20$ dB/dec deta hai. Har pole apna corner cross hone ke baad $-20$ dB/dec add karta hai.

Neeche wali figure padho (alt-text: log-frequency Bode magnitude, blue exact curve with dashed orange straight-line asymptotes; do red dots corners ko mark karte hain 10 aur 100 rad/s par). Blue curve true magnitude hai; dashed orange lines hamare haath se khainche asymptotes hain. Left se right follow karo: line dheere dheere girti hai ($-20$ dB/dec) pehle red dot $\omega =10$ par tak, jahaan woh tez hoke $-40$ dB/dec ho jaati hai; doosre red dot $\omega =100$ par woh phir $-60$ dB/dec pe kink karti hai. Har red dot ek pole ka "switch on" hona hai. Dhyaan do ki blue curve har kink ke thoda neeche baithti hai — woh $-3$ dB rounding hai jo straight lines ignore karti hain.

KYA: asymptote levels ko band by band chalte huye dekho jab tak woh $0$ dB tak pahunche. $\omega =10$ se neeche asymptote integrator line hai $|G{|}_{\text{dB}}=20{\log }_{10}(1000/\omega )$.

• $\omega =10$ par: $20{\log }_{10}(1000/10)=20{\log }_{10}100=40$ dB. Abhi bhi $0$ se upar.
• $10$ aur $100$ ke beech slope $-40$ dB/dec hai. Solve karo kahan $0$ hit karta hai: $0=40-40({\log }_{10}\omega -{\log }_{10}10)\Rightarrow {\log }_{10}\omega -1=1\Rightarrow {\log }_{10}\omega =2\Rightarrow \omega =100.$ Yeh exactly agla corner par land karta hai. Answer: ${\omega }_{gc}\approx 100$ rad/s (asymptote estimate). Upar wali figure mein, yahi woh jagah hai jahan dashed orange line grey $0$ dB horizontal se milti hai.

KYA: phase margin $=180^{\circ }+\angle L(j{\omega }_{gc})$. $\omega =100$ par phases ka sum karo:

• Integrator: $-90^{\circ }$.
• Pole @10: $-\arctan (100/10)=-\arctan (10)=-84.3^{\circ }$.
• Pole @100: $-\arctan (100/100)=-\arctan (1)=-45^{\circ }$. $\angle L(j100)=-90-84.3-45=-219.3^{\circ }.$ $\text{PM}=180^{\circ }+(-219.3^{\circ })=-39.3^{\circ }.$ Negative phase margin ⇒ closed loop UNSTABLE hai. Is plant ko madad chahiye (L5 dekho). Margins ki poori details Stability margins mein hain.

KYA/KYUN: ek first-order low-pass ${\omega }_c$ ke upar $-20$ dB/dec se roll off karta hai. Hum chahte hain $600$ rad/s par $20$ dB attenuation. $20$ dB at $-20$ dB/dec = corner se ek decade upar. Toh corner ko noise se ek decade neeche hona chahiye: ${\omega }_c=\frac{600}{10}=60\text{ rad/s}.$ Signal band check: $\omega =6$ rad/s par, corner se ratio $6/60=0.1$ hai, toh $|F(j6)|=\frac{1}{\sqrt{1+0.1^2}}\Rightarrow |F{|}_{\text{dB}}=20{\log }_{10}\frac{1}{\sqrt{1+0.1^2}}=-20{\log }_{10}\sqrt{1+0.1^2}.$ (KYUN minus sign aata hai: ${\log }_{10}(1/x)=-{\log }_{10}x$ — reciprocal ka log, log ka negative hota hai. Kyunki $F$ ek square root ka $1$ over hai, uska dB us square root ke dB ka minus hai. Yahi rule hai jo pole ke high-frequency asymptote slope ko neeche karta hai.) Numerically $\sqrt{1+0.01}=1.004988$, toh $|F{|}_{\text{dB}}=-20{\log }_{10}(1.004988)=-0.0432$ dB — essentially zero loss. ✓ Requirement ($\le 3$ dB) easily meet ho gayi. Answer: $F(s)=\frac{1}{1+s/60}$, corner ${\omega }_c=60$ rad/s.

KYA/KYUN: ek Lead-lag compensator lead mode mein phase inject karta hai low/negative phase margin ko bachane ke liye. Maan lo $\alpha =z/p$ (jisme $0<\alpha <1$). Tab $\sin {\varphi }_{\max }=\frac{p-z}{p+z}=\frac{1-\alpha }{1+\alpha }.$ ${\varphi }_{\max }=45^{\circ }$ set karo, toh $\sin 45^{\circ }=\frac{1}{\sqrt{2}}=0.7071$: $\frac{1-\alpha }{1+\alpha }=0.7071\Rightarrow 1-\alpha =0.7071(1+\alpha )\Rightarrow 0.2929=1.7071\alpha \Rightarrow \alpha =\mathrm{0.1716.}$ ${\omega }_m=\sqrt{zp}=100$ aur $z=\alpha p$ ke saath: $\sqrt{\alpha p\cdot p}=100\Rightarrow p\sqrt{\alpha }=100$: $p=\frac{100}{\sqrt{0.1716}}=\frac{100}{0.4142}=241.4\text{ rad/s},z=\alpha p=0.1716\times 241.4=41.4\text{ rad/s}.$ Answer: $z\approx 41.4$ rad/s, $p\approx 241.4$ rad/s. Neeche wali figure padho (alt-text: log-frequency phase plot, green curve rising from 0 to a peak then falling; ek red dot +45° peak ko mark karta hai omega_m = 100 par; dashed blue aur orange verticals zero ko 41.4 aur pole ko 241.4 rad/s par mark karte hain). Green curve compensator ki phase hai. Woh blue zero line ke baad $41.4$ rad/s par chadhna shuru karti hai, red dot ($+45^{\circ }$) par exactly ${\omega }_m=100$ par peak karti hai — do dashed lines ke geometric middle mein — phir orange pole line ke baad $241.4$ rad/s par dheere neeche aati hai. Yeh bump woh "phase hai jo hum loop ko crossover par wapas dete hain."

KYA/KYUN: cascading transfer functions ko multiply karta hai, toh phases add hote hain. Lead apna $+45^{\circ }$ exactly $\omega =100$ par add karta hai. Solution 3.3 se, $\angle L_0(j100)=-219.3^{\circ }$. Lead ki phase ${\omega }_m=100$ par add karo, jo design se exactly $+45^{\circ }$ hai: $\angle L(j100)=-219.3^{\circ }+45^{\circ }=-174.3^{\circ }.$ $\text{PM}=180^{\circ }+(-174.3^{\circ })=+5.7^{\circ }.$ Loop ab (barely) stable hai. Ek single $45^{\circ }$ lead ne $-39^{\circ }$ margin ko $+6^{\circ }$ par palat diya — ek real compensator aur gain trim aur zyaada lead add karega, lekin mechanism exactly yahi hai: gain $0$ dB cross kare wahaan phase wapas khareedna. Yeh Loop shaping ek move mein hai.

KYA/KYUN: Transfer functions jo minimum-phase hain, unke liye slope aur phase ek saath locked hain: har $-20$ dB/dec slope $\Rightarrow$ lagbhag $-90^{\circ }$ phase. $-40$ dB/dec slope unme se do hai: $\text{phase}\approx 2\times (-90^{\circ })=-180^{\circ }.$ Ek-line reason: steep roll-off hi extra phase lag hai — yeh ek minimum-phase system ke liye same physical fact hai, toh aggressive filter aur bada phase lag alag nahi ho sakte. Woh lag exactly woh cheez hai jo tumhari phase margin khaata hai aur loops ko destabilise karta hai (Nyquist plot aur Stability margins dekho).

Step 1 — crossover $\omega =1$ par force karo. $K=1$ ke saath, $L(j\omega )=\frac{1+j\omega /{\omega }_z}{(j\omega {)}^2}.$ $\omega =1$ par: $|L|=\frac{|1+j/{\omega }_z|}{1}=\sqrt{1+1/{\omega }_z^2}$. Crossover ke liye ($|L|=1$) hume yeh $=1$ chahiye, jo ${\omega }_z\to \infty$ force karega; baaki standard trick yeh hai ki zero ko crossover par rakho ${\omega }_z={\omega }_{gc}=1$ taaki wahaan phase maximise ho, aur chhoti magnitude bump accept karo. ${\omega }_z=1$ lo. Step 2 — $\omega =1$ par phase.

• Double integrator $1/(j\omega {)}^2$: $-180^{\circ }$.
• Zero $(1+j\omega /1)$ $\omega =1$ par: $+\arctan (1)=+45^{\circ }$. $\angle L(j1)=-180^{\circ }+45^{\circ }=-135^{\circ }.$ $\text{PM}=180^{\circ }+(-135^{\circ })=+45^{\circ }\ge 30^{\circ }.✓$ Answer: ${\omega }_z=1$ rad/s se PM $=45^{\circ }$ milta hai. PD zero ne phase ko deadly $-180^{\circ }$ (marginal) se khींchकर comfortable $-135^{\circ }$ tak pahunchaya — exactly wahi "phase wapas add karo" ka lesson Example 2 se.

Ek first-order low-pass corner ko ek noise tone se kitne decades neeche hona chahiye $20$ dB rejection ke liye? ::: Ek decade. Ek $45^{\circ }$ lead ke liye $z/p$ kya hona chahiye? ::: $\alpha =\frac{1-\sin 45^{\circ }}{1+\sin 45^{\circ }}\approx 0.172$. Ek $-40$ dB/dec minimum-phase slope kya phase imply karta hai? ::: Lagbhag $-180^{\circ }$. Negative phase margin ka matlab closed loop… hai? ::: Unstable. Ek $1/s^2$ plant ke liye crossover par PD zero add karne se kya phase margin milta hai? ::: $+45^{\circ }$.