Worked examples — Transfer function — Laplace domain, poles and zeros

3.5.27 · D3 · Physics › Guidance, Navigation & Control (GNC) › Transfer function — Laplace domain, poles and zeros

Symbols se pehle: jo alphabet hum use karte hain uska ek quick reminder, taaki kuch bhi un-earned na lage.

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The scenario matrix

Jo bhi transfer function mile, wo in cells mein se kisi ek mein fit hoga. Neeche har example ko us cell ke saath tag kiya gaya hai jo wo cover karta hai.

#	Cell (case class)	Kya cheez ise alag banati hai	Example
A	Do alag real poles, dono left	$\sigma <0$, $\omega =0$	Ex 1
B	Complex-conjugate poles, left	$\sigma <0$, $\omega \ne 0$ → ringing	Ex 2
C	Right-half-plane pole	koi $\sigma >0$ → unstable	Ex 3
D	Imaginary axis par pole (integrator / pure oscillator)	$\sigma =0$ → marginal	Ex 4
E	Repeated (degenerate) pole	double root, $t{e}^{pt}$ term	Ex 5
F	Ek zero present — kya stability badlti hai?	zero reweights, poles decide karte hain	Ex 6
G	Right-half-plane zero (non-minimum-phase)	"wrong-way" initial motion	Ex 7
H	Real-world word problem (rocket actuator)	ODE se $G$ banao, personality padho	Ex 8
I	Limiting values (DC gain, high-freq roll-off)	$s\to 0$ aur $s\to \infty$	Ex 9
J	Exam twist — ek parameter instability chhupaata hai	threshold dhundo	Ex 10

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Ex 1 — Cell A: do alag real poles (dono stable)

Forecast: Denominator ke dono terms positive hain, positive coefficients ke saath — andaza: do negative real roots → decaying, koi wiggle nahi.

1. Denominator factor karo. Yeh step kyun? Poles $D(s)=0$ ke roots hote hain; factoring se wo expose ho jaate hain. $s^2+5s+6=(s+2)(s+3)$ Poles $s=-2$ aur $s=-3$ par.
2. Zeros dhundo. Kyun? Numerator constant $4$ hai; $4=0$ ka koi solution nahi, isliye koi finite zero nahi.
3. Signs padho. Kyun? Har pole ka real part decay rate hota hai. Dono $\sigma =-2,-3<0$ → stable. Dono $\omega =0$ → koi oscillation nahi.
4. Shape likho. Response $\propto c_1{e}^{-2t}+c_2{e}^{-3t}$. Slower pole $s=-2$ (imaginary axis ke zyada kareeb) dominate karta hai; time constant $\tau =1/2=0.5$ s.

Verify: Steady-state gain ($s=0$ rakho): $G(0)=4/6=0.667$. Size $1$ ka step input $0.667$ par settle hoga — finite, stable system se match karta hai. ✓

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Ex 2 — Cell B: complex-conjugate poles (ringing)

Forecast: Discriminant negative lagta hai → complex poles → cheez ring karegi decay ke saath.

1. Standard 2nd-order form se match karo $s^2+2\zeta {\omega }_{n}s+{\omega }_n^2$. Kyun? Yeh standard 2nd-order model hai — isse $\zeta$ aur ${\omega }_n$ padhna sab kuch bata deta hai. ${\omega }_n^2=25\Rightarrow {\omega }_n=5$ rad/s. Aur $2\zeta {\omega }_n=4\Rightarrow \zeta =4/10=0.4$.
2. Quadratic solve karo. Kyun? Poles $s=-\zeta {\omega }_n\pm j{\omega }_n\sqrt{1-{\zeta }^2}$ hote hain. $s=-0.4(5)\pm j5\sqrt{1-0.16}=-2\pm j5(0.9165)=-2\pm j4.583$
3. Padho. Real part $-2<0$ → decay karta hai. Imaginary part $\pm 4.583$ → ${\omega }_d=4.583$ rad/s par ring karta hai (yeh damped frequency hai). Kyunki $0<\zeta <1$, yeh underdamped hai: overshoot phir settling.

Figure mein red dots dekho: wo vertical axis ke left mein baithe hain (stable) lekin horizontal axis se off hain (oscillating). Origin se inki distance ${\omega }_n=5$ hai; vertical axis se angle ka cosine $\zeta$ hai.

Verify: $|\text{pole}|=\sqrt{2^2+4.583^2}=\sqrt{4+21}=\sqrt{25}=5={\omega }_n$. ✓ Aur $\cos (\text{real axis se angle... negative real axis se maap ke})=2/5=0.4=\zeta$. ✓

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Ex 3 — Cell C: ek right-half-plane pole (unstable)

Forecast: Woh $-2s$ (negative middle coefficient) ek red flag hai — ek root right half mein ghus sakta hai. Andaza: unstable.

1. Factor karo. Kyun? Poles expose karo. $s^2-2s-8=(s-4)(s+2)$ Poles $s=+4$ aur $s=-2$ par.
2. Signs check karo. Kyun? Koi bhi $\sigma >0$ growing $e^{pt}$ matlab hai. Pole $s=+4$ ka $\sigma =+4>0$ hai → term $e^{+4t}$ blow up karta hai.
3. Conclusion: ek RHP pole ⇒ UNSTABLE, chahe doosra pole kitna bhi stable ho.

Verify (sign test): 2nd-order polynomial $s^2+a_{1}s+a_0$ ke liye, saare roots left-half mein hain sirf tab jab dono $a_1>0$ aur $a_0>0$ ho (yeh Routh-Hurwitz shortcut hai). Yahan $a_1=-2<0$ → fail → unstable, factoring se agree karta hai. ✓

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Ex 4 — Cell D: imaginary axis par pole (marginal)

Forecast: Kahi koi decay term nahi → na blow up hoga na settle hoga → marginally stable (borderline).

1. (a) Integrator. $s$ akele ko kyun dekho? $s=0$ par ek pole: $\sigma =0$, $\omega =0$. Iska time term $e^{0\cdot t}=1$ hai — constant, na grow karta hai na decay. Step input $1/s$ deta hai output $1/s^2$, jo invert hota hai $t$ mein (ek ramp jo bina bound ke grow karta hai). Toh integrator marginally stable hai: bounded inputs unbounded output produce kar sakti hain.
2. (b) Oscillator. Kyun? $s^2+9=0\Rightarrow s=\pm j3$. Real part $\sigma =0$, imaginary part $\pm 3$. Term $e^{j3t}$ hamesha $3$ rad/s par oscillate karta hai kisi decay ke bina — ek ghanti jo kabhi band nahi hoti.
3. Conclusion: dono imaginary axis par hain → marginally stable, Cells A/B (left, stable) aur Cell C (right, unstable) ke beech ki knife-edge.

Verify: (b) ke liye, $\pm j3$ par poles $s^2+9=0$ satisfy karte hain: $(j3{)}^2+9=-9+9=0$. ✓ Real part exactly $0$. ✓

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Ex 5 — Cell E: ek repeated (degenerate) pole

Forecast: Do poles ek hi jagah $s=-2$ par stack hain. Phir bhi left-half → stable, lekin shape mein kuch badlta hai.

1. Double pole identify karo. Kyun? $(s+2{)}^2=0$ se $s=-2$ do baar milta hai — ek repeated root.
2. Double pole ke liye inverse transform yaad karo. Single-pole rule ki jagah yeh kyun? Ek single pole ${\mathcal{L}}^{-1}\{1/(s+2)\}=e^{-2t}$ deta hai, lekin repeated pole ek extra $t$ factor deta hai: ${\mathcal{L}}^{-1}\left\{\frac{1}{(s+2{)}^2}\right\}=t{e}^{-2t}$
3. Interpret karo. $t$ multiply karta hai, lekin $e^{-2t}$ zyada tez decay karta hai jitna $t$ grow karta hai, isliye product phir bhi → $0$ jaata hai. Stable, non-oscillatory, lekin ek slower "critically-damped-jaisa" rise ke saath (yeh exactly $\zeta =1$ boundary hai).

Verify: $t{e}^{-2t}$ wahan peak karta hai jahan iska derivative zero ho: $\frac{d}{dt}(t{e}^{-2t})=e^{-2t}(1-2t)=0\Rightarrow t=0.5$ s. Peak value $=0.5e^{-1}=0.5(0.3679)=0.1839$. Finite, phir decay → stability confirm karta hai. ✓

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Ex 6 — Cell F: ek zero enter karta hai (kya stability badlti hai?)

Forecast: Dono mein same poles hain. Parent ke mistake-callout ne warn kiya tha: zeros stability decide nahi karte. Andaza: phir bhi stable, sirf shape shift hogi.

1. Dono ke poles list karo. Kyun? Poles hi stability ke sole arbiters hain. Dono mein poles $s=-1,-4$ par hain — identical, dono left-half → dono stable.
2. $G_2$ ka zero. Kyun note karo? $s+2=0\Rightarrow$ $s=-2$ par zero. Yeh poles ke beech baithta hai.
3. Zero kya karta hai. $G_2$ ke partial fractions: residues (coefficients $A_k$) $G_1$ se change hote hain, isliye $e^{-t}$ aur $e^{-4t}$ par weights badal jaate hain — transient shape alag hoti hai (possible overshoot), lekin exponents (aur isliye stability) untouched rehte hain.

Verify: DC gains alag hain, proof karta hai ki zero response reshape karta hai: $G_1(0)=\frac{1}{1\cdot 4}=0.25$; $G_2(0)=\frac{2}{1\cdot 4}=0.5$. Alag steady values, same poles. ✓

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Ex 7 — Cell G: ek right-half-plane zero (non-minimum-phase)

Forecast: Stable (poles left), lekin $s=+2$ par zero right half-plane mein hai — classic "aircraft pehle girta hai phir chadta hai" signature.

1. Poles. Kyun? Stability check. $(s+1)(s+3)$ → $s=-1,-3$, dono left → stable. Accha.
2. Zero. Kyun? $-(s-2)=0\Rightarrow s=+2$, ek right-half-plane zero (RHP zero).
3. Wrong-way move. Yeh kyun hota hai? Step response ka initial slope high-frequency behaviour ka sign rakhta hai, jabki final value DC gain ka sign rakhta hai — aur ek RHP zero in dono ko opposite signs bana deta hai. DC gain check karo: $G(0)=\frac{2}{3}=+0.667$ (positive settle hota hai). Lekin numerator ka leading coefficient $-s$ hai, isliye initial response negative kick karta hai — vehicle pehle neeche jaata hai, phir upar aata hai.

Verify: Step response $Y(s)=G(s)\cdot \frac{1}{s}$ par initial value theorem lagao: initial velocity ${\lim }_{s\to \infty }s\cdot sY(s)={\lim }_{s\to \infty }sG(s)={\lim }_{s\to \infty }\frac{-s^2+2s}{s^2+4s+3}=-1<0$ jabki $G(0)=+0.667>0$. Opposite signs → wrong-way motion confirm. ✓

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Ex 8 — Cell H: real-world rocket actuator

Forecast: Middle coefficient positive, constant positive → stable. Yeh 2nd-order lagta hai → probably ringing (Cell B disguise mein). Command $1$ rad → $1$ rad ke paas settle hona chahiye.

1. Zero I.C. ke saath Laplace. Zero I.C. kyun? Transfer function actuator khud ko describe karta hai, isliye hum rest se start karte hain. $s^2\Theta +6s\Theta +25\Theta =25U\Rightarrow (s^2+6s+25)\Theta =25U$
2. Ratio banao. Kyun? $G=\Theta /U$. $G(s)=\frac{25}{s^2+6s+25}$
3. $\zeta ,{\omega }_n$ padho. Kyun? Standard-form matching. ${\omega }_n=\sqrt{25}=5$ rad/s; $2\zeta {\omega }_n=6\Rightarrow \zeta =0.6$. Underdamped → thoda overshoot, phir settle.
4. Poles. $s=-\zeta {\omega }_n\pm j{\omega }_n\sqrt{1-{\zeta }^2}=-3\pm j5(0.8)=-3\pm j4$. Left-half → stable.
5. Steady-state accuracy. $s\to 0$ kyun? DC gain = unit command par final nozzle angle. $G(0)=25/25=1$ → nozzle steady state mein command ko exactly track karta hai (zero steady-state error).

Verify: $|\text{pole}|=\sqrt{3^2+4^2}=5={\omega }_n$ ✓; $\zeta =3/5=0.6$ ✓; $G(0)=1$ (perfect tracking) ✓.

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Ex 9 — Cell I: limiting values (DC gain aur roll-off)

Forecast: $s\to 0$ par ek finite number milega (DC gain). $s\to \infty$ par highest powers dominate karengi: numerator $\sim 10s$, denominator $\sim s^2$, ratio $\sim 10/s\to 0$ — yeh roll off karega.

1. DC gain, $s\to 0$. Kyun? Constant input par steady response set karta hai. $G(0)=\frac{10(0+1)}{0+0+2}=\frac{10}{2}=5$
2. High-frequency limit, $s\to \infty$. Kyun? Batata hai ki system fast signals ko kitni tezi se attenuate karta hai (roll-off). $G(s)\approx \frac{10s}{s^2}=\frac{10}{s}\to 0$ System mein zero (1) se ek zyada pole (2) hai, isliye yeh $1/s$ jaisa roll off karta hai — Bode plot par $-20$ dB/decade slope.
3. Interpret karo. Slow commands gain $5$ ke saath pass hote hain; fast jitter filter out ho jaati hai.

Verify: $G(0)=5$ ✓. Poles minus zeros ka excess $=2-1=1$, isliye $sG(s)\to 10$ jab $s\to \infty$ (finite), single-order roll-off confirm karta hai. ✓

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Ex 10 — Cell J: exam twist, parameter instability chhupaata hai

Forecast: $K$ badhane se constant term $2-K$ chhotaa hota hai. Jab $2-K$ negative ho jaata hai, sign test fail ho jaata hai. Andaza: $K<2$ ke liye stable.

1. Denominator coefficients likho. Kyun? 2nd-order sign rule use karo: $s^2+a_{1}s+a_0$ ke dono roots left-half mein hain tabhi jab $a_1>0$ aur $a_0>0$. Yahan $a_1=3>0$ (hamesha theek) aur $a_0=2-K$.
2. Condition lagao. Kyun? Stability ko $a_0>0$ chahiye. $2-K>0\Rightarrow K<2$
3. Threshold. $K=2$ par: $a_0=0$, toh denominator $=s^2+3s=s(s+3)$ → $s=0$ par ek pole (axis par) → marginally stable (Cell D). $K>2$ ke liye: $a_0<0$ → ek root right half mein cross karta hai → unstable (Cell C).

Verify: Boundary $K=2$ par: poles $s=0,-3$; ek pole exactly imaginary axis par → marginal, derivation se match karta hai. $K=3$ par: $a_0=-1<0$, $s^2+3s-1$ ke roots $\frac{-3\pm \sqrt{13}}{2}=0.303,-3.303$ hain → ek RHP pole → unstable. ✓

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Recall Kaun sa cell sabse mushkil tha — self-test

Ek RHP zero system ko unstable banata hai. ::: False — sirf poles stability decide karte hain; ek RHP zero (Ex 7) wrong-way motion cause karta hai lekin system stable rehta hai agar poles left-half hain. $s=-2$ par double pole unstable hota hai extra $t$ factor ki wajah se. ::: False — $t{e}^{-2t}\to 0$; phir bhi stable (Ex 5). Integrator $1/s$ stable hai. ::: Nahi — yeh marginally stable hai; step input unbounded ramp deta hai (Ex 4). $\frac{K}{s^2+3s+(2-K)}$ ke liye stability limit hai ::: $K<2$; $K=2$ par marginal, uske upar unstable (Ex 10).

Connections

• Transfer function — Laplace domain, poles and zeros — woh parent jise yeh deep-dive expand karta hai
• Second-order systems - damping ratio and natural frequency — Ex 2 aur Ex 8 mein use hua
• Stability - Routh-Hurwitz criterion — Ex 3 aur Ex 10 mein use hua sign test
• Root locus — poles kaise move karte hain jab gain $K$ vary karta hai (Ex 10 aise locus ka ek point hai)
• Bode plot — Ex 9 mein compute kiya gaya roll-off