Convolution with Fourier transform — Hinglish notes

4.7.20 · Maths › Partial Differential Equations

1. Convolution kya hai?

WHAT yeh karta hai: ek nayi function produce karta hai jo $f$ ka weighted blur hai, $g$ ko weight ki tarah use karke. WHY yeh matter karta hai: yeh frequency space mein multiplication ka real-space partner hai. HOW $g (x - u)$ padhein: $g$ ko flip karo (woh $- u$ ) aur use $x$ se shift karo.

2. Fourier transform convention (ek baar fix karo)

Hum use karte hain

f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \hat f(k)\, e^{ikx}\, dk.$$ > [!mistake] Constant convention par depend karta hai! > Ek galat-lekin-sahi-lagni wali trap: log "$\mathcal F\{f*g\} = \hat f\hat g$" yaad kar lete hain lekin **reverse** identity $\mathcal F\{fg\}=\frac{1}{2\pi}\hat f * \hat g$ mein ek $2\pi$ bhool jaate hain. *Kyun sahi lagta hai:* symmetry suggest karti hai ki dono clean hone chahiye. *Fix:* asymmetry isliye aati hai kyunki $\frac{1}{2\pi}$ sirf is convention mein **inverse** transform mein rehta hai. Track karo ki $2\pi$ kahan hai. --- ## 3. Convolution Theorem ki Derivation (scratch se) > [!formula] Convolution Theorem > $$\boxed{\;\mathcal F\{f * g\}(k) = \hat f(k)\,\hat g(k)\;}$$ **Derivation — har step justified:** $f*g$ ke transform ki definition se shuru karo: $$\mathcal F\{f*g\}(k) = \int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty} f(u)\,g(x-u)\,du\right] e^{-ikx}\,dx.$$ *Yeh step kyun?* Bas convolution ki definition ko FT definition mein plug karo. Integration ka order swap karo (Fubini — valid kyunki $f,g\in L^1$): $$= \int_{-\infty}^{\infty} f(u)\left[\int_{-\infty}^{\infty} g(x-u)\, e^{-ikx}\,dx\right] du.$$ *Kyun?* Double integral absolutely convergent hai, isliye order swap legal hai; hum andar wala $x$-integral isolate karna chahte hain. Substitute karo $v = x-u \Rightarrow x = v+u,\ dx=dv$: $$\int_{-\infty}^{\infty} g(v)\, e^{-ik(v+u)}\,dv = e^{-iku}\int_{-\infty}^{\infty} g(v)\,e^{-ikv}\,dv = e^{-iku}\,\hat g(k).$$ *Kyun?* Shift sirf ek phase $e^{-iku}$ factor out karta hai; bacha hua integral definition se **$\hat g(k)$ hi hai**. Wapas daalte hain: $$= \int_{-\infty}^{\infty} f(u)\, e^{-iku}\,\hat g(k)\,du = \hat g(k)\int_{-\infty}^{\infty} f(u)\,e^{-iku}\,du = \hat g(k)\,\hat f(k). \qquad\blacksquare$$ *Kyun?* $\hat g(k)$ $u$ par depend nahi karta, bahar nikaalo; bacha hua integral $\hat f(k)$ hai. **Iski gehri wajah:** FT shift operator ko diagonalize karta hai. $e^{-ikx}$ "shift by $u$" ke eigenfunctions hain jinka eigenvalue $e^{-iku}$ hai. Convolution poori tarah shifts se bana hai, toh eigenbasis mein woh plain multiplication ban jaata hai. ![[4.7.20-Convolution-with-Fourier-transform.png]] --- ## 4. Properties (aur har ek kyun hold karta hai) > [!formula] Convolution ka Algebra > - Commutative: $f*g = g*f$ — *$w=x-u$ substitute karo roles flip karne ke liye.* > - Associative: $(f*g)*h = f*(g*h)$ — *kyunki transforms sirf multiply karte hain, aur multiplication associative hai.* > - Distributive: $f*(g+h)=f*g+f*h$. > - Identity: $f * \delta = f$ — *kyunki $\hat\delta(k)=1$, toh $\hat f\cdot 1=\hat f$.* > [!intuition] $\delta$ identity kyun hai > Spectra ko $1$ se multiply karna kuch nahi badalta. Woh function jiska spectrum constantly $1$ hai woh Dirac delta hai. Toh $\delta$ se convolve karna sab kuch unchanged chhod deta hai — yeh "kuch nahi karna" wala blur hai. --- ## 5. Worked Examples > [!example] Example 1 — Do Gaussians ka Convolution > Maano $f(x)=e^{-ax^2},\ g(x)=e^{-bx^2}$. Theorem use karke $f*g$ nikalo. > > **Step 1.** Known transform: $\mathcal F\{e^{-ax^2}\}=\sqrt{\pi/a}\,e^{-k^2/4a}$. > *Kyun?* Standard Gaussian FT (exponent mein square complete karo). > > **Step 2.** Transforms multiply karo: > $$\widehat{f*g}=\sqrt{\tfrac{\pi}{a}}\sqrt{\tfrac{\pi}{b}}\,e^{-k^2/4a}e^{-k^2/4b}=\frac{\pi}{\sqrt{ab}}\,e^{-k^2\frac{a+b}{4ab}}.$$ > *Kyun?* Convolution theorem messy integral ko ek product mein badal deta hai. > > **Step 3.** Ise ek aur Gaussian $C e^{-cx^2}$ ke transform ki tarah pehchano jahan $\frac{1}{4c}=\frac{a+b}{4ab}\Rightarrow c=\frac{ab}{a+b}$. > > **Result:** $f*g = \sqrt{\dfrac{\pi}{a+b}}\; e^{-\frac{ab}{a+b}x^2}.$ > *Takeaway:* ek Gaussian convolved with ek Gaussian ek Gaussian hota hai — variances add hote hain (heat kernel yahi inherit karta hai). > [!example] Example 2 — Convolution se Heat Equation > $u_t = D u_{xx}$, $u(x,0)=f(x)$ solve karo. > > **Step 1.** $x$ mein FT: $\hat u_t = -Dk^2 \hat u$. *Kyun?* $\mathcal F\{u_{xx}\}=(ik)^2\hat u=-k^2\hat u$. > > **Step 2.** $t$ mein ODE solve karo: $\hat u(k,t)=\hat f(k)\,e^{-Dk^2 t}$. *Kyun?* $t$ mein linear first-order ODE. > > **Step 3.** Transforms ka product $\Rightarrow$ $x$ mein convolution: > $$u(x,t)=f * \mathcal F^{-1}\{e^{-Dk^2t}\} = f * G_t, \qquad G_t(x)=\frac{1}{\sqrt{4\pi Dt}}e^{-x^2/4Dt}.$$ > *Kyun?* $e^{-Dk^2t}$ $k$ mein ek Gaussian hai; uska inverse FT **heat kernel** $G_t$ hai. > > **Result:** $u(x,t)=\dfrac{1}{\sqrt{4\pi Dt}}\displaystyle\int_{-\infty}^{\infty} f(u)\,e^{-(x-u)^2/4Dt}\,du.$ Solution initial data hai jo ek spreading Gaussian se *blur* hoi hai. > [!example] Example 3 — Box ∗ Box = Triangle > Maano $f=g=\mathbb 1_{[-1,1]}$ (indicator). Tab $(f*g)(x)$ do unit-half-width boxes ka overlap length measure karta hai jo $x$ se shift hain: ek **triangle** width 4 ka, $x=0$ par peak $2$, $|x|\ge 2$ ke liye zero. *Kyun?* Overlap linearly shrink hota hai jab aap unhe slide karte ho. Isliye baar baar convolution spikes ko bumps mein smooth karta hai (aur ultimately Gaussians mein — CLT!). --- ## 6. Active Recall > [!recall]- Quick self-test (answers cover karo) > - Convolution integral state karo. → $\int f(u)g(x-u)du$. > - $\hat f\hat g$ real space mein kya banta hai? → $f*g$. > - $\delta$ convolution identity kyun hai? → $\hat\delta=1$. > - Heat kernel? → $\frac{1}{\sqrt{4\pi Dt}}e^{-x^2/4Dt}$. > [!recall]- Feynman: 12 saal ke bachche ko explain karo > Socho tumhare paas lamps ki ek row hai (function $f$) aur ek fuzzy lens hai (function $g$). Jab tum lens se dekhte ho, har lamp ki roshni spread hoti hai aur apne neighbors se blend hoti hai — woh blending **convolution** hai. Ab yahan magic hai: agar lamps dekhne ki jagah tum unke "musical notes" dekho (frequencies, Fourier transform), toh fuzzy lens bas har note ki volume ko ek fixed amount se upar ya neeche karta hai — woh simple **multiplication** hai. Toh lamp-world mein ek complicated blur note-world mein sirf knobs ghumana hai. Wapas jaane ke liye, tum "un-listen" karte ho aur knob-turning wapas blurring ban jaata hai. > [!mnemonic] Yaad rakho > **"Spectrum land mein multiply karo = back home blur ho jaao."** > Aur: **F**lip, **S**hift, **M**ultiply, **A**dd = convolution recipe (FSMA). --- ## 7. Flashcards #flashcards/maths Convolution $(f*g)(x)$ define karo. ::: $\int_{-\infty}^{\infty} f(u)g(x-u)\,du$ — $g$ ko flip karo, $x$ se shift karo, multiply karo, integrate karo. Convolution theorem state karo. ::: $\mathcal F\{f*g\} = \hat f \cdot \hat g$. Derivation ka key step jo phase factor produce karta hai. ::: $v=x-u$ substitute karne se $\int g(v)e^{-ik(v+u)}dv = e^{-iku}\hat g(k)$ milta hai. FT ke under convolution multiplication kyun ban jaata hai? ::: FT shift operator ko diagonalize karta hai; complex exponentials uske eigenfunctions hain, isliye shift se bane operations multiply karte hain. Convolution identity element kya hai aur kyun? ::: Dirac delta $\delta$, kyunki $\hat\delta(k)=1$ aur $1$ se multiply karna no-op hai. $e^{-ax^2}*e^{-bx^2}$ ke liye Gaussian ∗ Gaussian result. ::: $\sqrt{\pi/(a+b)}\,e^{-\frac{ab}{a+b}x^2}$ (ek Gaussian; precisions $c=ab/(a+b)$ ki tarah combine hote hain). Heat kernel $G_t(x)$. ::: $\frac{1}{\sqrt{4\pi Dt}}e^{-x^2/4Dt}$, $e^{-Dk^2t}$ ka inverse FT. Heat equation solution as a convolution. ::: $u(x,t)=f*G_t$, initial data ek spreading Gaussian se blur hota hai. Box ∗ box kya hota hai? ::: Ek triangle function (overlap length linearly decrease hoti hai). Reverse identity $\mathcal F\{fg\}$ (is convention ke saath). ::: $\frac{1}{2\pi}(\hat f * \hat g)$ — $2\pi$ note karo. --- ## 8. Connections - [[Fourier Transform]] — engine; differentiation → multiplication. - [[Heat Equation]] — data ko heat kernel se convolve karke solve hota hai. - [[Dirac Delta Function]] — convolution identity. - [[Gaussian Integral]] — Gaussian transforms ke liye chahiye. - [[Green's Functions]] — solution = source $*$ Green's function (same idea). - [[Central Limit Theorem]] — baar baar convolution → Gaussian. - [[Parseval's Theorem]] — same multiplication trick ka energy version. ## 🖼️ Concept Map ```mermaid flowchart TD PDE[PDEs heat wave Laplace] -->|FT turns d/dx into x| ALG[Algebraic equations] ALG -->|solution appears as| PROD[Product of transforms] PROD -->|inverse needs| CONV[Convolution] CONV -->|defined as| DEF[Integral of f u times g x-u] DEF -->|meaning| SMEAR[Smearing weighted blur] FTCONV[FT convention with 2pi in inverse] -->|fixes constant| THM[Convolution Theorem] DEF -->|plug into FT def| DERIV[Derivation via Fubini and shift] DERIV -->|substitute v equals x-u| PHASE[Shift factors phase e^-iku] PHASE -->|yields| THM[F of f star g equals fhat ghat] THM -->|deep reason| DIAG[FT diagonalizes shift operator] FTCONV -->|reverse identity| REV[F of fg equals 1 over 2pi fhat star ghat] ```