Exercises — Impulse response and transfer function (GNC connection)

KYA karte hain: dono sides ka Laplace transform lete hain, $y^{\prime }\to sY$ replace karte hain. Zero initial conditions kyun: exact rule hai $\mathcal{L}\{y^{\prime }\}=sY(s)-y(0)$ (yeh Laplace integral par integration by parts se aata hai). Yeh assume karke ki system $t=0$ se pehle rest mein tha, yaani $y(0)=0$, woh $-y(0)$ term gayab ho jaati hai aur derivative cleanly $sY$ par map ho jaata hai. Yeh essential hai: transfer function sirf zero-state (forced) response describe karta hai, isliye hum deliberately initial-condition terms drop karte hain. $3sY+6Y=U\Rightarrow (3s+6)Y=U\Rightarrow H(s)=\frac{Y}{U}=\frac{1}{3s+6}.$ Pole: denominator ko zero karo: $3s+6=0\Rightarrow s=-2$. Kyun matter karta hai: pole $s=-2$ par hai, real aur negative, toh impulse response $e^{-2t}$ ki tarah decay karta hai — stable (aur uska ROC $\Re (s)>-2$ imaginary axis contain karta hai).

KYA karte hain: Laplace transform invert karte hain. Hum standard pair $\mathcal{L}\{e^{-at}\}=\frac{1}{s+a}$ pehchante hain, yahan $a=5$ ke saath. $h(t)={\mathcal{L}}^{-1}\left\{\frac{1}{s+5}\right\}=e^{-5t},t\ge 0.$ Yeh pair kyun aur koi nahi: $H$ ka $s=-5$ par single real pole hai, aur single real pole hamesha ek single decaying exponential $e^{pt}$ mein invert hota hai jahan $p$ pole hai. Koi sine nahi, koi ramp nahi — woh complex ya repeated poles se aate hain. Causal (right-sided, $t\ge 0$) inverse choose karne se ROC $\Re (s)>-5$ fix ho jaata hai.

Step 1: $\mathcal{L}\{\delta (t)\}={\int }_0^{\infty }{e}^{-st}\delta (t)dt=e^0=1$ Dirac delta function ki sifting property se. Step 2: kyunki $Y=H\cdot U$ aur $U(s)=\mathcal{L}\{\delta \}=1$, $H(s)=\frac{Y(s)}{U(s)}=\frac{Y(s)}{1}=\frac{2}{s^2+1}.$ Kyun: delta feed karna exactly woh tarika hai jisse tum $H$ directly read off karte ho, kyunki $U(s)=1$ se $Y=H$ ho jaata hai.

Step 1 (Laplace world mein): $\mathcal{L}\{y^{\prime }\}=sY-y(0)$ use karte hain $y(0)=0$ ke saath, $sY+4Y=U\Rightarrow H(s)=\frac{1}{s+4}$. Step 2 (time par wapas): $h(t)={\mathcal{L}}^{-1}\{1/(s+4)\}=e^{-4t}$ for $t\ge 0$ (ROC $\Re (s)>-4$). Step 3 ($h(0^{+})$ check): $h(0^{+})=e^0=1$. ✓ Check kyun kaam karta hai: $t=0$ directly $h(t)=e^{-4t}$ mein dalo: $e^{-4\cdot 0}=1$. Zyada generally, first-order system $a_1{y}^{\prime }+a_{0}y=u$ ka $H(s)=\frac{1}{a_{1}s+a_0}$ hota hai, jiska impulse response $h(0^{+})=\frac{1}{a_1}$ se start hota hai — yahan $a_1=1$ ($y^{\prime }$ ka coefficient), toh $h(0^{+})=1$.

KYA karte hain: impulse response $h(t)=e^{-4t}$ ko input ke saath convolve karte hain, causal limits $0$ se $t$ use karte hue (dekho Convolution). Causal limits tab valid hain kyunki dono signals right-sided ROC mein rehte hain. $y(t)={\int }_0^{t}u(\tau )h(t-\tau )d\tau ={\int }_0^{t}1\cdot e^{-4(t-\tau )}d\tau .$ Step: $v=t-\tau$ substitute karo, $dv=-d\tau$; jaise $\tau :0\to t$, $v:t\to 0$: $y(t)={\int }_0^t{e}^{-4v}dv={\left[\frac{-e^{-4v}}{4}\right]}_0^t=\frac{1-e^{-4t}}{4}.$ Sanity check: jaise $t\to \infty$, $y\to \frac{1}{4}=H(0)=\frac{1}{0+4}$, DC gain. ✓

Step 1: $s^{2}Y+3sY+2Y=U\Rightarrow H(s)=\frac{1}{s^2+3s+2}=\frac{1}{(s+1)(s+2)}$. Step 2 (partial fractions): $A,B$ nikalo jahan $\frac{1}{(s+1)(s+2)}=\frac{A}{s+1}+\frac{B}{s+2}$. Cover-up: $A=\frac{1}{(-1+2)}=1$, $B=\frac{1}{(-2+1)}=-1$. Partial fractions kyun: har simple pole ek exponential mein invert hota hai, lekin sirf tab jab hum fraction ko single-pole pieces mein split kar lein. Step 3 (invert karo): $h(t)=e^{-t}-e^{-2t},t\ge 0.$

System stable hai agar aur sirf agar har pole ka $\Re (p)<0$ ho (imaginary axis ke strictly left mein). Figure teen cases ko complex plane par $\times$ marks ke roop mein plot karta hai taaki tum verdict dekh sako: green crosses (a) shaded left half-plane mein baithe hain, yellow crosses (b) bilkul vertical imaginary axis par baithe hain, aur red cross (c) $s=+2$ par forbidden right half-plane mein ghus jaata hai.

(a) poles $s=\frac{-2\pm \sqrt{4-20}}{2}=-1\pm 2j$. Real part $-1<0$ → stable (damped oscillation). (b) poles $s=\pm 2j$. Real part $=0$ → marginal (pure oscillation jo kabhi decay nahi karta). (c) $s^2-s-2=(s-2)(s+1)$, poles $s=2$ aur $s=-1$. Ek pole ka $\Re =+2>0$ → unstable ($e^{2t}$ ki tarah blow up karta hai).

KYA karte hain: standard form $\ddot{y}+2\zeta {\omega }_n\dot{y}+{\omega }_n^{2}y=u$ se match karo. Coefficients compare karo: ${\omega }_n^2=5\Rightarrow {\omega }_n=\sqrt{5}$; $2\zeta {\omega }_n=2\Rightarrow \zeta =\frac{1}{{\omega }_n}=\frac{1}{\sqrt{5}}\approx 0.447$. Kyunki $0<\zeta <1$ hai toh yeh underdamped hai. Damped frequency: ${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}=\sqrt{5}\cdot \sqrt{1-\frac{1}{5}}=\sqrt{5}\cdot \sqrt{\frac{4}{5}}=2.$ Impulse response (parent ka underdamped pair): $h(t)=\frac{1}{{\omega }_d}{e}^{-\zeta {\omega }_{n}t}\sin ({\omega }_{d}t)=\frac{1}{2}{e}^{-t}\sin (2t),t\ge 0.$ $e^{-t}$ kyun: $\zeta {\omega }_n=\frac{1}{\sqrt{5}}\cdot \sqrt{5}=1$, toh decay rate exactly $1$ hai — jo poles $-1\pm 2j$ ke real part se match karta hai jo 3.1(a) se hai. Figure yeh fingerprint dikhata hai: blue curve $h(t)$ hai jo ${\omega }_d=2$ par oscillate kar raha hai, yellow dashed envelope $\pm \frac{1}{2}{e}^{-t}$ se hugged hai jiska decay rate $1$ wahi hai jo pole ka real part hai. Ring khatam ho jaata hai, toh system stable hai — tum stability literally shrinking envelope se read kar sakte ho.

Poles: $s=\frac{-2b\pm \sqrt{4b^2-4}}{2}=-b\pm \sqrt{b^2-1}$. Real part stability govern karta hai.

• Agar $b>0$: $|b|<1$ ke liye poles $-b\pm j\sqrt{1-b^2}$ hain, real part $-b<0$ → stable. $b\ge 1$ ke liye, dono roots real aur negative hain (unka product $+1$, sum $-2b<0$) → stable.
• Agar $b=0$: poles $\pm j$, real part $0$ → marginal (undamped sinusoid, koi decay nahi).
• Agar $b<0$: real part $-b>0$ ban jaata hai → unstable. Conclusion: $b<0$ ke liye unstable; $b=0$ par marginal; saare $b>0$ ke liye stable. Damping term ki sign hi poori kahani hai.

Feedback kya karta hai: $u=-ky$ ko $\dot{y}=u$ mein substitute karo: $\dot{y}+ky=0\Rightarrow \text{closed-loop pole at }s=-k.$ (Forced response ke liye, reference $r$ se closed-loop transfer function $\frac{H_p}{1+k{H}_p}=\frac{1/s}{1+k/s}=\frac{1}{s+k}$ hai.) Dekho Feedback control systems. Pole choose karo: hum $s=-3$ chahte hain, toh $k=3$ set karo. Closed-loop impulse response: $H_{cl}(s)=\frac{1}{s+3}\Rightarrow h_{cl}(t)=e^{-3t}$, $t\ge 0$. Yeh punchline kyun hai: ek open-loop marginal system (pole at $0$) pure feedback gain $k>0$ choose karke stabilise ho jaata hai; bada $k$ pole ko aur left push karta hai (faster decay) — exactly woh GNC idea jo parent ke double-integrator remark mein hai.

Yahan $\theta$ woh output hai jise hum zero par steer karna chahte hain, $\dot{\theta }$ uski rate (angular velocity) hai, aur $u$ commanded torque hai. Step 1 (loop close karo): $\ddot{\theta }=u=-k_p\theta -k_d\dot{\theta }$ se milta hai $\ddot{\theta }+k_d\dot{\theta }+k_p\theta =0.$ Characteristic polynomial: $s^2+k_{d}s+k_p=0$. Step 2 (target poles): hum $-2\pm 2j$ roots chahte hain. Un roots ke saath polynomial banao: $(s+2-2j)(s+2+2j)=(s+2{)}^2+2^2=s^2+4s+8.$ Step 3 (coefficients match karo): $k_d=4$, $k_p=8$. P alone kyun nahi, PD kyun: pure $k_p$ se $s^2+k_p=0$ milta hai, poles $\pm j\sqrt{k_p}$ — abhi bhi marginal (undamped oscillation). Derivative term $k_d\dot{\theta }$ damping supply karta hai ($s^1$ coefficient), poles ko imaginary axis se hataata hai stable left half-plane mein. Position feedback akele damp nahi kar sakta; rate feedback bhi chahiye.

(a) Transfer function & poles. $H(s)=\frac{1}{s^2+2s+2}=\frac{1}{(s+1{)}^2+1}.$ Poles: $s=\frac{-2\pm \sqrt{4-8}}{2}=-1\pm j$. Dono ka $\Re =-1<0$. Yahan $\zeta {\omega }_n=1$ aur ${\omega }_d=1$ ($(s+1{)}^2+1$ se), toh ${\omega }_n=\sqrt{2}$, $\zeta =1/\sqrt{2}\approx 0.707$.

(b) Impulse response. Pair $\mathcal{L}\{e^{-at}\sin bt\}=\frac{b}{(s+a{)}^2+b^2}$ se match karo $a=1,b=1$ ke saath: $H(s)=\frac{1}{(s+1{)}^2+1}=\frac{1}{1}\cdot \frac{1}{(s+1{)}^2+1}\Rightarrow h(t)=e^{-t}\sin t,t\ge 0.$ (Kyunki $b=1$ hai, koi division nahi chahiye.)

(c) Laplace se step response. $U(s)=\mathcal{L}\{1\}=1/s$ ke saath: $Y(s)=H(s)U(s)=\frac{1}{s(s^2+2s+2)}.$ Partial fractions $\frac{1}{s(s^2+2s+2)}=\frac{A}{s}+\frac{Bs+C}{s^2+2s+2}$. Multiply out: $1=A(s^2+2s+2)+(Bs+C)s$.

• $s=0$: $1=2A\Rightarrow A=\frac{1}{2}$.
• $s^2$ coeff: $0=A+B\Rightarrow B=-\frac{1}{2}$.
• $s^1$ coeff: $0=2A+C\Rightarrow C=-1$. Toh $Y(s)=\frac{1/2}{s}-\frac{\frac{1}{2}s+1}{s^2+2s+2}$. Second term ko $(s+1{)}^2+1$ ke upar rewrite karo: $\frac{1}{2}s+1=\frac{1}{2}(s+1)+\frac{1}{2}$, toh $Y(s)=\frac{1/2}{s}-\frac{1}{2}\cdot \frac{s+1}{(s+1{)}^2+1}-\frac{1}{2}\cdot \frac{1}{(s+1{)}^2+1}.$ Term by term invert karo ($\mathcal{L}\{e^{-t}\cos t\}=\frac{s+1}{(s+1{)}^2+1}$ aur $\mathcal{L}\{e^{-t}\sin t\}=\frac{1}{(s+1{)}^2+1}$ use karke): $y_{\text{step}}(t)=\frac{1}{2}-\frac{1}{2}{e}^{-t}\cos t-\frac{1}{2}{e}^{-t}\sin t,t\ge 0.$

(d) Final value & verdict. Jaise $t\to \infty$, $e^{-t}$ terms gayab ho jaate hain: $y_{\text{step}}(\infty )=\frac{1}{2}=H(0)=\frac{1}{0+0+2}.✓$ $-1\pm j$ par poles ka negative real part hai → stable, ek underdamped decay DC gain $1/2$ ki taraf. (Stability imaginary axis ko ROC ke andar rakhti hai, toh $H(0)$ ek valid number hai jo read kiya ja sakta hai.)

Closed-loop char. poly: Ex 4.2 se, $s^2+k_{d}s+k_p$. $s^2+2\zeta {\omega }_{n}s+{\omega }_n^2$ se match karo:

• ${\omega }_n^2=9\Rightarrow k_p=9$.
• $2\zeta {\omega }_n=2(1)(3)=6\Rightarrow k_d=6$. Poles: $s^2+6s+9=(s+3{)}^2=0$ → repeated pole $s=-3$ (critically damped). Impulse response: $H_{cl}(s)=\frac{1}{(s+3{)}^2}$, aur pair $\mathcal{L}\{t{e}^{-at}\}=\frac{1}{(s+a{)}^2}$ use karke $a=3$ ke saath: $h_{cl}(t)=t{e}^{-3t},t\ge 0.$ Critically damped kyun: $\zeta =1$ overshoot ($\zeta <1$, ringing) aur sluggishness ($\zeta >1$) ke beech boundary hai; response target tak as fast as possible pahunchta hai bina oscillate kiye — settling manoeuvre ke liye ideal. Repeated real pole $t{e}^{-3t}$ shape produce karta hai: rise karta hai, peak karta hai, phir decay karta hai.

1.1 $H=\frac{1}{3s+6}$, pole $-2$ ::: 1.2 $h=e^{-5t}$ ::: 1.3 $H=\frac{2}{s^2+1}$ 2.1 $h=e^{-4t}$ ::: 2.2 $y=\frac{1-e^{-4t}}{4}$ ::: 2.3 $h=e^{-t}-e^{-2t}$ 3.1 stable / marginal / unstable ::: 3.2 $\zeta =1/\sqrt{5},{\omega }_n=\sqrt{5},{\omega }_d=2,h=\frac{1}{2}{e}^{-t}\sin 2t$ ::: 3.3 unstable $b<0$, marginal $b=0$ 4.1 $k=3,h=e^{-3t}$ ::: 4.2 $k_d=4,k_p=8$ 5.1 $h=e^{-t}\sin t$, step $\to \frac{1}{2}$, stable ::: 5.2 $k_p=9,k_d=6$, pole $-3$ (double), $h=t{e}^{-3t}$