Photonic and optical interconnects: Visual walkthrough

6.5.15 · D2 · Hardware › Advanced & Emerging Architectures › Photonic and optical interconnects

This page zooms into one machine from the parent topic: the Mach-Zehnder Modulator.

Step 1 — Light wave asal mein kya hoti hai (rotating arrow)

Figure — Photonic and optical interconnects

Step 2 — Light ko do arms mein split karna

Formula Do arm-arrows — power se split hote hain, field se nahi

Ek lossless coupler ko energy conserve karni hogi: andar jaane wali power barabar hogi bahar aane wali power ke. Power (field) $^{2}$ ke hisaab se hoti hai, toh agar andar aane wala field $E_{0}$ hai (power $\propto E_{0}^{2}$ ) aur yeh do equal halves mein divide ho, toh har arm power carry karta hai $\propto \frac{1}{2} E_{0}^{2}$ . Square undo karo field paane ke liye: $E_{arm} = \frac{E _{0}}{2}$

$E_{0}$ — incoming arrow ka field (uski power $\propto E_{0}^{2}$ ).
Denominator mein $2$ — yahi cheez hai jo har arm ko exactly half power carry karaati hai: $(E_{0} / 2)^{2} = \frac{1}{2} E_{0}^{2}$ . ✔

$2$ kyun aur $2$ kyun nahi? Power ko half mein split karna, field ko half mein split karna nahi hai. Field ko half karne se har arm ko power ka sirf ek quarter milega ( $(\frac{1}{2} E_{0})^{2} = \frac{1}{4} E_{0}^{2}$ ), toh naively "half the field" total energy ka adha waste kar deta hai — ek lossless coupler ke liye impossible. Honest normalization hai $1/ 2$ .

Step 3 — Ek arm ko twist karna: phase shift $Δ ϕ$ (aur bias $ϕ_{0}$ )

Formula Recombine karne se thoda pehle dono arms

$E_{1} = \frac{E _{0}}{2}, E_{2} = \frac{E _{0}}{2} e^{i (ϕ_{0} + Δ ϕ)}$

$E_{1}$ — arm 1, untouched, reference angle $0$ pe arrow.
$ϕ_{0}$ — fixed DC bias jo hum ek baar tune karte hain.
$Δ ϕ$ — fast data twist.
$e^{i (ϕ_{0} + Δ ϕ)}$ — arm 2 ka arrow, total angle se aage spin kiya hua.
Dono abhi bhi length $E_{0} / 2$ ke hain — phase shift arrow ko rotate karta hai, chhota nahi karta.

Core derivation ko clean rakhne ke liye, hum bias ko ek symbol mein absorb karte hain: maano $φ \equiv ϕ_{0} + Δ ϕ$ total arm-to-arm phase hai. Hum formula $φ$ mein derive karte hain, phir end mein dono dials read karte hain.

$e^{i φ}$ kyun, "amplitude mein $φ$ add karo" kyun nahi? Phase shift ek rotation hai, stretch nahi. $e^{i φ}$ se multiply karna exactly "is arrow ko $φ$ se rotate karo aur uski length rakho." Yahi pure reason hai ki hum Step 1 mein arrow notation kyun adopt kiya.

Step 4 — Recombining: do arrows ko tip-to-tail add karna

Step 5 — Field se power tak: arrow ki length ko square karna

Formula Sum ko square karna, term by term

$∣ E_{o u t} ∣^{2} = \frac{1}{2} E_{0}^{2} 1 + e^{i φ}^{2}$ (prefactor hai $(E_{0} / 2)^{2} = \frac{1}{2} E_{0}^{2}$ ). Ab $1 + e^{i φ}$ ka length-squared: arrow ko uske conjugate se multiply karo: $1 + e^{i φ}^{2} = (1 + e^{i φ}) (1 + e^{- i φ}) = 1 + e^{i φ} + e^{- i φ} + 1$

Do $1$ 's — har unit arrow ka length-squared 1 hai.
$e^{i φ} + e^{- i φ}$ — cross terms; Euler ke rule se ek arrow plus uska mirror $2 cos φ$ deta hai (sideways parts cancel, forward parts add).

$\Rightarrow 1 + e^{i φ}^{2} = 2 + 2 cos φ = 2 (1 + cos φ)$ Toh power hai $∣ E_{o u t} ∣^{2} = \frac{1}{2} E_{0}^{2} \cdot 2 (1 + cos φ) = E_{0}^{2} (1 + cos φ) \cdot \frac{1}{1} / ...$ Dhyan raho — constants ko cleanly collect karo: $∣ E_{o u t} ∣^{2} = \frac{1}{2} E_{0}^{2} \cdot 2 (1 + cos φ) = E_{0}^{2} (1 + cos φ) \cdot \frac{1}{2} \cdot 2 = \frac{1}{2} E_{0}^{2} \cdot 2 (1 + cos φ) .$ Cleanly: $∣ E_{o u t} ∣^{2} = E_{0}^{2} \frac{1 + cos φ}{1} \cdot \frac{1}{2} \cdot 2$ . Confusion se bachne ke liye hum exact tidy line use karte hain jo next step mein use hogi: $∣ E_{o u t} ∣^{2} = \frac{1}{2} E_{0}^{2} (1 + cos φ) \cdot 2 = E_{0}^{2} (1 + cos φ) .$

Step 6 — Clean form: half-angle identity

Formula Half-angle cleanup

Tidy result $∣ E_{o u t} ∣^{2} = E_{0}^{2} (1 + cos φ)$ se start karo aur $1 + cos θ = 2 cos^{2} (θ /2)$ use karo: $∣ E_{o u t} ∣^{2} = E_{0}^{2} \cdot 2 cos^{2} (\frac{φ}{2})$ Full input power input arrow squared hai, $P_{in} \propto E_{0}^{2}$ ; ek symmetric ideal MZM $P_{in}$ ko single bright port pe bhejta hai, toh standard normalization ke saath output ratio hai $P_{o u t} = P_{in} cos^{2} (\frac{φ}{2}) = P_{in} cos^{2} (\frac{ϕ _{0} + Δ ϕ}{2})$

$P_{in}$ — woh full power jo enter ki.
$cos^{2} (φ /2)$ — transfer function: 1 (bright) se 0 (dark) tak ek dial.
$φ = ϕ_{0} + Δ ϕ$ — bias $ϕ_{0}$ choose karta hai ki tum hill pe kahan baithe ho; data $Δ ϕ$ tumhe uske saath move karta hai.

Half-angle kyun? Kyunki yeh poori interference ko ek non-negative squared term mein package karta hai — power kabhi negative nahi hoti, aur $cos^{2}$ kabhi negative nahi hota. Yeh form on/off behaviour ko ek glance mein obvious banata hai.

Step 7 — Har case: curve ko sab phases ke liye padhna

$φ$	arm-2 arrow point karta hai	arrows...	$cos^{2} (φ /2)$	$P_{o u t}$	matlab
$0$	arm 1 ke saath same	fully reinforce karte hain	$1$	$P_{in}$	bit 1 (bright)
$π /2$	90° aage	partial add	$cos^{2} (45°) = \frac{1}{2}$	$\frac{1}{2} P_{in}$	half-bright (quadrature)
$π$	opposite	exactly cancel	$0$	$0$	bit 0 (dark)
$3 π /2$	270° aage	phir partial add	$\frac{1}{2}$	$\frac{1}{2} P_{in}$	half-bright
$2 π$	full turn, start pe wapas	fully reinforce	$1$	$P_{in}$	phir bright (periodic)

One-picture summary

Recall Feynman retelling — walkthrough kisi dost ko explain karo

Light ko ek chhoti arrow ki tarah imagine karo jo clock face pe spin kar rahi hai; arrow jitni lambi hogi, light utni hi bright hogi. Hum incoming arrow ko do channels mein split karte hain — lekin yahan ek subtlety hai: hum energy ko half mein split karte hain, aur kyunki energy length-squared hai, har arm ka arrow $1/ 2$ jitna lamba hota hai (karib 0.71), aadha nahi. Ek channel pe hum ek knob (ek voltage) push karte hain jo woh arrow aur $Δ ϕ$ angle aage spin kara deta hai; ek slow background dial $ϕ_{0}$ bhi hoti hai jo hum ek baar ek tiny heater se set karte hain yeh decide karne ke liye ki hill pe hum kahan se start karte hain. Phir hum dono arrows wapas saath laate hain aur unhe tip-to-tail lagate hain. Agar twisted arrow apne partner ke same direction mein point kare, dono ek lambe arrow mein stack ho jaate hain: bright, ek 1. Agar woh exact opposite direction mein point kare, cancel ho jaate hain aur kuch nahi bachta: dark, ek 0. Detector arrows spinning nahi dekh sakta (bahut fast — 200 trillion times per second), toh woh sirf average energy feel karta hai, jo arrow ki length squared hai — aur "ek arrow plus ek twisted arrow" pe woh squaring karna exactly woh jagah hai jahan se $cos^{2}$ aata hai. Knob ko ek full circle se aage ghoomao aur tum wapas bright pe loop ho jaate ho, toh device ko actually sirf twist ko "same direction" se "opposite direction" tak swing karna hota hai ek bit on aur off flip karne ke liye.

Recall Self-test

Lossless 50/50 split ke baad har arm mein field? ::: $E_{0} / 2$ (har arm power ka half carry karta hai, toh field $E_{0} / 2$ hai, $E_{0} /2$ nahi). Imaginary unit $i$ ka geometric matlab kya hai? ::: 90° turn; $i^{2} = - 1$ se define hota hai. Kisi arrow pe $e^{i φ}$ se multiply karna kya karta hai? ::: Use angle $φ$ se rotate karta hai bina uski length change kiye. Conjugate $\overset{z}{ˉ}$ kya karta hai, aur ise kyun use karte hain? ::: Arrow ko mirror karta hai ( $e^{i θ} \to e^{- i θ}$ ); $z \overset{z}{ˉ} = ∣ z ∣^{2}$ real length-squared deta hai, yaani power. $ϕ_{0}$ kya hai aur ise kaise set karte hain? ::: Static DC bias phase; ek heater se ek baar set kiya jaata hai device ko $cos^{2}$ hill pe chosen point pe park karne ke liye (jaise $π /2$ quadrature). Kaun sa total phase $φ$ "0" deta hai? ::: $φ = π$ (arrows opposite, cancel ho jaate hain). $P_{o u t}$ $φ$ mein periodic kyun hai? ::: Phase ek angle hai; $2 π$ add karne se arrow same jagah wapas aa jaata hai, toh output repeat karta hai.

Dekho bhi: Micro-ring resonators (WDM ke liye wavelength-selective cousin), Co-packaged optics, aur Energy per bit as an efficiency metric.