Exercises — Convolution theorem — proof, applications

Test: convolution mein dono arguments ==milakar $t$ bante hain== ($\tau +(t-\tau )=t$), aur upper limit $t$ honi chahiye.

• (a) Arguments hain $\tau$ aur $t-\tau$ — ye milakar $t$ bante hain. Limits $0$ se $t$ hain. ✅ Ye convolution hai jisme $f(\tau )={\tau }^2$ (matlab $f(t)=t^2$) aur $g(t-\tau )=t-\tau$ (matlab $g(t)=t$).
• (b) Dono factors mein $\tau$, $\tau$ hai — arguments $\tau$ aur $\tau$ hain, ye milakar $t$ nahi bante. ❌ Ye sirf ek ordinary integral hai, convolution nahi.
• (c) Upper limit $\infty$ hai, $t$ nahi. ❌ Ye woh causal convolution nahi jo humne define ki (ye ek alag cheez hai). Answer: sirf (a); $f(t)=t^2,g(t)=t$.

Product split karo. $\frac{1}{s^3(s-2)}=\underset{F(s)}{\underbrace{\frac{1}{s^3}}}\cdot \underset{G(s)}{\underbrace{\frac{1}{s-2}}}$. Standard pairs yaad karo (Laplace Transform — definition and properties se): ${\mathcal{L}}^{-1}\left\{\frac{1}{s^3}\right\}=\frac{t^2}{2}=f(t),{\mathcal{L}}^{-1}\left\{\frac{1}{s-2}\right\}=e^{2t}=g(t).$ $t^2/2$ kyun? Kyunki $\mathcal{L}\{t^n\}=n!/s^{n+1}$, isliye $\mathcal{L}\{t^2\}=2/s^3$, aur hume $2$ se divide karna padta hai. Theorem apply karo: ${\mathcal{L}}^{-1}\left\{\frac{1}{s^3(s-2)}\right\}={\int }_0^t\frac{{\tau }^2}{2}{e}^{2(t-\tau )}d\tau .$ Ho gaya — ye maangi gayi convolution form hai.

Direct. Yahaan $f(t)=1,g(t)=1$: $(1\ast 1)(t)={\int }_0^{t}1\cdot 1d\tau =[\tau {]}_0^t=t.$ Theorem check. $\mathcal{L}\{1\}=\frac{1}{s}$, toh $s$-world mein product hai $\frac{1}{s}\cdot \frac{1}{s}=\frac{1}{s^2}$, aur ${\mathcal{L}}^{-1}\{1/s^2\}=t$. ✅ Same answer. Ye sabse clean proof hai ki convolution ≠ pointwise product: $1\cdot 1=1$ lekin $1\ast 1=t$.

$f(t)=t,g(t)=e^t$. Commutativity ki wajah se ($f\ast g=g\ast f$) jise integrate karna aasaan ho use flip ke saath rakho. $f(\tau )=\tau$ rakho: $(t\ast e^t)(t)={\int }_0^t\tau e^{t-\tau }d\tau =e^t{\int }_0^t\tau e^{-\tau }d\tau .$ $e^t$ bahar kyun nikala? Kyunki $e^{t-\tau }=e^t{e}^{-\tau }$ aur running clock $\tau$ ke liye $e^t$ constant hai. Integration by parts karo ($u=\tau ,dv=e^{-\tau }d\tau ,v=-e^{-\tau }$): ${\int }_0^t\tau e^{-\tau }d\tau =[-\tau e^{-\tau }{]}_0^t+{\int }_0^t{e}^{-\tau }d\tau =-t{e}^{-t}+[-e^{-\tau }{]}_0^t=-t{e}^{-t}-e^{-t}+1.$ $e^t$ se multiply karo: $(t\ast e^t)(t)=e^t(1-e^{-t}-t{e}^{-t})=\boxed{e^t-1-t}.$ Theorem se sanity check: $\mathcal{L}\{t\}\mathcal{L}\{e^t\}=\frac{1}{s^2}\cdot \frac{1}{s-1}$, aur $\mathcal{L}\{e^t-1-t\}=\frac{1}{s-1}-\frac{1}{s}-\frac{1}{s^2}$. Common denominator $s^2(s-1)$: numerator $s^2-s(s-1)-(s-1)=s^2-s^2+s-s+1=1$. Toh ye $\frac{1}{s^2(s-1)}$ ke barabar hai. ✅

Split: $\frac{1}{s}\cdot \frac{1}{s^2+4}$. Inverses: ${\mathcal{L}}^{-1}\{1/s\}=1$ aur ${\mathcal{L}}^{-1}\{1/(s^2+4)\}=\frac{1}{2}\sin 2t$ (kyunki $\mathcal{L}\{\sin at\}=a/(s^2+a^2)$, yahaan $a=2$ hai, isliye $2$ se divide karo). ${\mathcal{L}}^{-1}={\int }_0^{t}1\cdot \frac{1}{2}\sin (2(t-\tau ))d\tau =\frac{1}{2}{\int }_0^t\sin (2t-2\tau )d\tau .$ $w=2t-2\tau ,dw=-2d\tau$ rakho; jab $\tau :0\to t$, toh $w:2t\to 0$: $=\frac{1}{2}\cdot \frac{-1}{2}[-\cos w]...\Rightarrow \frac{1}{2}[\frac{\cos (2t-2\tau )}{2}{]}_0^t=\frac{1}{4}(\cos 0-\cos 2t)=\boxed{\frac{1}{4}(1-\cos 2t)}.$ Check: $\mathcal{L}\{\frac{1}{4}(1-\cos 2t)\}=\frac{1}{4}(\frac{1}{s}-\frac{s}{s^2+4})=\frac{1}{4}\cdot \frac{(s^2+4)-s^2}{s(s^2+4)}=\frac{1}{4}\cdot \frac{4}{s(s^2+4)}=\frac{1}{s(s^2+4)}$. ✅

Laplace lo, $\mathcal{L}\{y^{\prime \prime }\}=s^{2}Y-sy(0)-y^{\prime }(0)=s^{2}Y$ use karke (dono zero hain): $s^{2}Y+4Y=F(s)\Rightarrow Y=\underset{\text{transfer function }H(s)}{\underbrace{\frac{1}{s^2+4}}}F(s).$ $Y$ do $s$-functions ka product hai, isliye uska inverse ek convolution hai: $y(t)=h(t)\ast f(t),h(t)={\mathcal{L}}^{-1}\{H(s)\}=\frac{1}{2}\sin 2t.$ Ye kyun beautiful hai: $h(t)=\frac{1}{2}\sin 2t$ system ka unit impulse ke liye response hai. Koi bhi input $f$ produce karta hai: $\boxed{y(t)=\frac{1}{2}{\int }_0^t\sin (2(t-\tau ))f(\tau )d\tau }$ — output wo input hai jo impulse response se smeared ho gayi. Yahi "smearing" hai convolution.

Convolution pehchano. ${\int }_0^t(t-\tau )y(\tau )d\tau =(t\ast y)(t)$ kyunki arguments $t-\tau$ aur $\tau$ milakar $t$ bante hain. (Dekho Volterra Integral Equations.) Dono sides ka Laplace lo. $\mathcal{L}\{e^t\}=\frac{1}{s-1}$, $\mathcal{L}\{t\}=\frac{1}{s^2}$, aur convolution → product: $Y=\frac{1}{s-1}+\frac{1}{s^2}Y.$ Algebraically solve karo (yahi poora point hai — integral algebra ban gaya): $Y\left(1-\frac{1}{s^2}\right)=\frac{1}{s-1}\Rightarrow Y\cdot \frac{s^2-1}{s^2}=\frac{1}{s-1}\Rightarrow Y=\frac{s^2}{(s-1)(s^2-1)}.$ Note karo $s^2-1=(s-1)(s+1)$, toh $Y=\frac{s^2}{(s-1{)}^2(s+1)}$. Partial fractions (dekho Inverse Laplace Transform — partial fractions): $\frac{s^2}{(s-1{)}^2(s+1)}=\frac{A}{s-1}+\frac{B}{(s-1{)}^2}+\frac{C}{s+1}.$

• $s=1$: $B(1+1)=1\Rightarrow B=\frac{1}{2}$.
• $s=-1$: $C(-1-1{)}^2=1\Rightarrow C\cdot 4=1\Rightarrow C=\frac{1}{4}$.
• $s^2$ coefficients compare karo: $A+C=1\Rightarrow A=1-\frac{1}{4}=\frac{3}{4}$. $Y=\frac{3/4}{s-1}+\frac{1/2}{(s-1{)}^2}+\frac{1/4}{s+1}.$ Term by term invert karo: $\boxed{y(t)=\frac{3}{4}{e}^t+\frac{1}{2}t{e}^t+\frac{1}{4}{e}^{-t}}.$ Spot check at $t=0$: $y(0)=\frac{3}{4}+0+\frac{1}{4}=1=e^0$. Aur RHS at $t=0$ hai $e^0+0=1$. ✅

$y(0)=0,y^{\prime }(0)=1$ ke saath Laplace: $\mathcal{L}\{y^{\prime \prime }\}=s^{2}Y-s\cdot 0-1=s^{2}Y-1$. $s^{2}Y-1+Y=\frac{1}{s^2+1}\Rightarrow (s^2+1)Y=1+\frac{1}{s^2+1}.$ $Y=\underset{\text{initial velocity se}}{\underbrace{\frac{1}{s^2+1}}}+\underset{\text{driven part}}{\underbrace{\frac{1}{(s^2+1{)}^2}}}.$ Pehla term $\sin t$ mein invert hota hai. Doosra term product hai $\frac{1}{s^2+1}\cdot \frac{1}{s^2+1}$, toh ye $\sin t\ast \sin t$ hai. Parent note se (App 2): $\sin t\ast \sin t=\frac{1}{2}(\sin t-t\cos t).$ Jodo: $\boxed{y(t)=\sin t+\frac{1}{2}\sin t-\frac{1}{2}t\cos t=\frac{3}{2}\sin t-\frac{1}{2}t\cos t}.$ IVs check karo: $y(0)=0$ ✅. $y^{\prime }(t)=\frac{3}{2}\cos t-\frac{1}{2}\cos t+\frac{1}{2}t\sin t=\cos t+\frac{1}{2}t\sin t$, toh $y^{\prime }(0)=1$ ✅.

Tarika 1 — convolution. Split karo $\frac{s}{(s^2+1{)}^2}=\frac{s}{s^2+1}\cdot \frac{1}{s^2+1}$. Inverses: ${\mathcal{L}}^{-1}\{s/(s^2+1)\}=\cos t$, ${\mathcal{L}}^{-1}\{1/(s^2+1)\}=\sin t$. Toh hume $\cos t\ast \sin t$ chahiye: $\cos t\ast \sin t={\int }_0^t\cos \tau \sin (t-\tau )d\tau .$ Product-to-sum: $\cos \tau \sin (t-\tau )=\frac{1}{2}[\sin t-\sin (2\tau -t)]$ ($\sin X\cos Y=\frac{1}{2}[\sin (X+Y)+\sin (X-Y)]$ use karke, $X=t-\tau ,Y=\tau$ ke saath). $=\frac{1}{2}{\int }_0^t\sin td\tau -\frac{1}{2}{\int }_0^t\sin (2\tau -t)d\tau =\frac{1}{2}t\sin t-\frac{1}{2}[\frac{-\cos (2\tau -t)}{2}{]}_0^t.$ Doosra bracket: $-\frac{1}{2}(-\frac{1}{2})[\cos t-\cos (-t)]=\frac{1}{4}[\cos t-\cos t]=0$. $\Rightarrow \cos t\ast \sin t=\frac{1}{2}t\sin t.$ Tarika 2 — known transform. $\mathcal{L}\{t\sin t\}=\frac{2s}{(s^2+1{)}^2}$ (table / $\mathcal{L}\{\cos t\}$ differentiate karne ki trick). Isliye ${\mathcal{L}}^{-1}\{s/(s^2+1{)}^2\}=\frac{1}{2}t\sin t$. Agree karte hain: dono dete hain $\boxed{\frac{1}{2}t\sin t}$. ✅

General $\omega >0$. Zero IVs ke saath Laplace: $(s^2+{\omega }^2)Y=F(s)$, toh $Y=\frac{1}{s^2+{\omega }^2}F(s).$ Ab ${\mathcal{L}}^{-1}\{1/(s^2+{\omega }^2)\}=\frac{1}{\omega }\sin \omega t$ (kyunki $\mathcal{L}\{\sin \omega t\}=\omega /(s^2+{\omega }^2)$, toh $\omega$ se divide karo). Ye impulse response $h(t)$ hai. Product $H(s)F(s)$ convolution mein invert hota hai: $y(t)=h\ast f=\frac{1}{\omega }{\int }_0^t\sin (\omega (t-\tau ))f(\tau )d\tau .■$ Degenerate $\omega \to 0$. Equation ban jaati hai $y^{\prime \prime }=f(t)$. Limit ko kernel ke andar lo: jab $\omega \to 0$, $\frac{\sin (\omega (t-\tau ))}{\omega }\to (t-\tau )$ (kyunki chhote $x$ ke liye $\sin x\approx x$). Toh $y(t)\to {\int }_0^t(t-\tau )f(\tau )d\tau .$ Interpretation / check: $y^{\prime \prime }=f$ zero IVs ke saath matlab $f$ ko twice integrate karo; aur sach mein ${\int }_0^t(t-\tau )f(\tau )d\tau$ exactly double antiderivative hai (ek baar differentiate karo ⇒ ${\int }_0^{t}f(\tau )d\tau$, do baar ⇒ $f(t)$). ✅ Kernel gracefully degrade karta hai — koi blow-up nahi.

$y(t)={\int }_0^t{e}^{-(t-\tau )}\cdot 1d\tau =e^{-t}{\int }_0^t{e}^{\tau }d\tau =e^{-t}(e^t-1)=\boxed{1-e^{-t}}.$ $t=0$ ke paas: $1-e^{-t}\approx t$ (linearly $0$ se utha — system abhi "fill" nahi hua). $y(0)=0$: koi bhi input accumulate hone se pehle output nahi. Jab $t\to \infty$: $e^{-t}\to 0$, toh $y\to 1$. Output input level par saturate ho jaata hai — classic first-order "charging" curve. Ye frequency-domain cousin ka step response hai. Theorem se cross-check: $H(s)=\frac{1}{s+1}$, $F(s)=\frac{1}{s}$, product $\frac{1}{s(s+1)}=\frac{1}{s}-\frac{1}{s+1}$, inverse $1-e^{-t}$. ✅

$t>ϵ$ ke liye: $(f\ast g_{ϵ})(t)={\int }_0^{t}f(t-\tau )g_{ϵ}(\tau )d\tau =\frac{1}{ϵ}{\int }_0^{ϵ}f(t-\tau )d\tau .$ Sirf $[0,ϵ]$ kyun? Kyunki $\tau >ϵ$ ke liye $g_{ϵ}(\tau )=0$ hai — kernel ek patla, ooncha pulse hai jiska area $1$ hai. Woh last integral $t$ par khatam hone wali $ϵ$ width ki window par $f$ ka average hai. Jab $ϵ\to 0$, ek continuous function ka ek shrinking window par average point value par converge karta hai: ${\lim }_{ϵ\to 0}(f\ast g_{ϵ})(t)=f(t).$ Interpretation: unit-area pulse convolution ka identity hai — isse convolve karne par $f$ wapas milti hai. Yahi Dirac delta ka role hai: $f\ast \delta =f$. L3/L5.1 mein har impulse-response derivation secretly isi par rely karti hai.