Worked examples — Photonic and optical interconnects

6.5.15 · D3 · Hardware › Advanced & Emerging Architectures › Photonic and optical interconnects

Shuru karne se pehle, hum jo bhi quantities use karte hain wo parent Photonic and optical interconnects mein define hain. Hume sirf paanch formulas chahiye, yahan restate kiye hain taaki kuch bhi bina samjhe use na ho:

Yahan $\Delta \varphi$ (padho "delta phi") extra phase hai — ek arm ki light wave kितna shift hui doosre ke relative, radians mein measured, jahan $\pi$ radians $=180^{\circ }$ = half a wavelength of shift.

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Scenario matrix

Neeche har worked example us cell ke saath tagged hai jo wo cover karta hai.

#	Cell class	Kya special hai isme	Example
A	Copper $L^2$ scaling	length double karo, delay quadruple hoti dekho	Ex 1
B	Copper vs photonic crossover	wo length dhundho jahan light jeet jaati hai	Ex 2
C	Degenerate input ($L\to 0$)	zero-length wire, formulas kya kehte hain?	Ex 3
D	MZM "1" state ($\Delta \varphi =0$)	full brightness, bit = 1	Ex 4
E	MZM "0" state ($\Delta \varphi =\pi$)	perfect cancellation, bit = 0	Ex 4
F	MZM half-power ($\Delta \varphi =\pi /2$)	tricky midpoint	Ex 5
G	WDM ordinary	plain $N\times B$ throughput	Ex 6
H	WDM limiting case	$N$ bounded by finite optical band	Ex 7
I	Energy/bit word problem	real datacenter budget	Ex 8
J	Exam twist / trap	photodetector current, latency myth	Ex 9

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Example 1 — Cell A: copper delay quadruple hoti hai

1. Units convert karo taaki seconds clean nikal sakein. $r=0.2\Omega /\text{mm}$, $c_{\ell }=0.2\text{pF/mm}=0.2\times 10^{-12}\text{F/mm}$. Yeh step kyun? $\Omega \times \text{F}=\text{seconds}$, isliye dono "per mm" rakhke $L^2$ ko mm mein multiply karne se answer directly seconds mein aata hai.
2. $\tau =r{c}_{\ell }{L}^2$ apply karo $L=5$ par. ${\tau }_5=(0.2)(0.2\times 10^{-12})(5^2)=0.04\times 10^{-12}\times 25=1.0\times 10^{-12}\text{s}=1\text{ps}$. Yeh step kyun? Direct substitution — yahi to $L^2$ law ka poora point hai.
3. $L=10$ par apply karo. ${\tau }_{10}=(0.2)(0.2\times 10^{-12})(10^2)=0.04\times 10^{-12}\times 100=4.0\times 10^{-12}\text{s}=4\text{ps}$. Yeh step kyun? Same formula, sirf $L$ 5 se 10 hua.

Verify: ratio ${\tau }_{10}/{\tau }_5=4/1=4$. $L$ double karne se delay quadruple ho gayi ($2^2=4$), bilkul waisa jैसा $L^2$ law predict karta hai. Units: $\Omega \cdot \text{F}=\text{s}$. ✓ (Agar tumne "$2\times$" forecast kiya tha, tum linear-intuition trap mein fase.)

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Example 2 — Cell B: light copper ko kahan overtake karti hai?

Figure s01: horizontal axis wire length $L$ in mm hai, vertical axis time in ps hai. Black straight line photon transit $t_{ph}=13.33L$ hai (linear). Red parabola copper delay ${\tau }_{cu}=0.04L^2$ hai. Yeh marked crossover point $L=333$ mm par milte hain — iske left mein copper faster hai, right mein light jeet jaati hai.

1. Dono times ko $L$ ke function ke roop mein likho (mm mein). Copper: ${\tau }_{cu}=0.04L^2$ ps. Photon: $t_{ph}=\frac{nL}{c_0}$ jahan $n=4$ aur $c_0=3\times 10^8\text{m/s}$. Yeh step kyun? Crossover dhundhne ke liye dono curves ko same variable mein same units mein chahiye.
2. $c_0$ ko mm/ps mein ek baar, cleanly convert karo. Metres→mm aur seconds→ps badalne ke liye upar-neeche multiply karo: $c_0=3\times 10^8\frac{\text{m}}{\text{s}}\times \frac{10^3\text{mm}}{1\text{m}}\times \frac{1\text{s}}{10^{12}\text{ps}}=3\times 10^{-1}\frac{\text{mm}}{\text{ps}}=0.3\frac{\text{mm}}{\text{ps}}.$ Yeh step kyun? Jab $c_0$ mm/ps mein ho aur $L$ mm mein, to transit $t_{ph}=nL/c_0$ directly ps mein aata hai — koi leftover unit mismatch nahi.
3. Photon time ps mein likho. $t_{ph}=\frac{4L}{0.3\text{mm/ps}}=13.33L$ ps ($L$ mm mein). $L=1$ mm se check karo: $t_{ph}=4/0.3=13.33$ ps. ✓ Yeh step kyun? Ab dono curves ps mein hain, equate karne ke liye ready.
4. Equal rakho aur solve karo. $0.04L^2=13.33L\Rightarrow L=13.33/0.04=333\text{mm}$. Yeh step kyun? Dono sides ko $L$ se divide karo (valid for $L\ne 0$); crossover length wahan hai jahan parabola line ko pakad leti hai — figure mein red dot.

Verify: $L=333$ mm par, copper $\tau =0.04\times 333^2\approx 4435$ ps; photon $t=13.33\times 333\approx 4439$ ps. Rounding tak equal. ✓ $\sim 333$ mm se neeche photon actually slower per metre hai (index $n=4$ ki wajah se!) — yahi exactly latency myth hai jo parent warn karta hai. Photonics bandwidth/energy par jeetता hai, raw short-haul latency par nahi.

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Example 3 — Cell C: degenerate zero-length wire

1. Limits lo. ${\tau }_{cu}=r{c}_{\ell }{L}^2\to 0$ aur $t_{ph}=nL/c_0\to 0$ jab $L\to 0$. Yeh step kyun? Zero-length wire ka koi bhi model mein zero delay hona chahiye — ek sanity floor jo har model ko pass karna chahiye.
2. Approach ki rate compare karo. $L=0$ ke paas, ${\tau }_{cu}\propto L^2$ shrink hota hai $t_{ph}\propto L$ se faster. Yeh step kyun? Bahut chhoti wires ke liye copper ka quadratic term bahut tiny hota hai, isliye on-chip short hops abhi bhi copper use karte hain — photonics tabhi pay off karta hai jab crossover ke baad (Ex 2).

Verify: $L=0$ plug karo: ${\tau }_{cu}=r{c}_{\ell }\cdot 0=0$, $t_{ph}=n\cdot 0/c_0=0$. Dono exactly $0$. ✓ Degenerate case consistent hai (koi negative ya infinite nonsense nahi). Physical caveat: real links mein abhi bhi fixed overheads hote hain (modulator, TIA latency) jo $L$-formulas ignore karte hain — isliye true zero-length time zero nahi hota, wo fixed E-O-E overhead hota hai. Dekho Transimpedance Amplifier (TIA).

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Example 4 — Cells D & E: MZM full-on aur full-off states

Figure s02: teen phasor diagrams side by side. Har ek mein, black arrow arm 1 hai (fixed, right point karta hua) aur red arrow arm 2 hai jo $\Delta \varphi$ se rotated hai; dashed arrow unka sum hai. Left ($\Delta \varphi =0$) full length tak sum hota hai → $P=1$. Middle ($\Delta \varphi =\pi /2$) → $P=0.5$. Right ($\Delta \varphi =\pi$) red arrow wapas point karta hai aur sum zero ho jaata hai → $P=0$.

1. State D, $\Delta \varphi =0$. $P_{\text{out}}=1\cdot {\cos }^2(0/2)=1\cdot {\cos }^{2}0=1\cdot 1=1\text{mW}$. Yeh step kyun? Zero phase difference par dono arm-waves crest-to-crest add hoti hain (constructive interference) — figure mein dono phasors same direction mein point karte hain, full length. Bit 1.
2. State E, $\Delta \varphi =\pi$. $P_{\text{out}}=1\cdot {\cos }^2(\pi /2)=1\cdot 0^2=0\text{mW}$. Yeh step kyun? $\pi$ shift ka matlab ek wave full half-wavelength peeche hai — crest meets trough, cancel ho jaate hain (destructive interference). Phasors opposite point karte hain, sum zero. Bit 0.

Verify: ${\cos }^{2}0=1$ ✓ aur ${\cos }^2(\pi /2)=0$ ✓. Full power → "1", zero power → "0". Exactly $\pi$ phase produce karne wala voltage $V_{\pi }$ kehlaata hai — modulator ka switching voltage. Energy conservation check: $\Delta \varphi =\pi$ par "missing" power destroy nahi hoti, wo interferometer ke doosre (complementary) output port se nikalta hai.

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Example 5 — Cell F: tricky half-power midpoint

1. $\Delta \varphi =\pi /2$ substitute karo. $P_{\text{out}}=1\cdot {\cos }^2(\pi /4)$. Yeh step kyun? $\pi /2$ ka half $\pi /4$ hai; yahi cosine ka argument hai.
2. Evaluate karo. $\cos (\pi /4)=\frac{1}{\sqrt{2}}$, isliye ${\cos }^2(\pi /4)=\frac{1}{2}$. Thus $P_{\text{out}}=0.5\text{mW}$. Yeh step kyun? Exactly half power — quadrature point.
3. Special kyun hai? $\Delta \varphi =\pi /2$ par ${\cos }^2$ curve steepest hoti hai (iska slope maximum hota hai), isliye ek chhoti si voltage wiggle se sabse bada, sabse linear power swing milta hai. Yeh step kyun? Linearity matter karti hai analog optical links mein jahan aap output $\propto$ input chahte ho, na ki hard on/off.

Verify: ${\cos }^2(\pi /4)=(1/\sqrt{2}{)}^2=1/2$ ✓ → $0.5\text{mW}$, exactly Ex 4 ke $1\text{mW}$ aur $0\text{mW}$ ke beech mein. ${\cos }^2(x/2)$ ka slope $-\frac{1}{2}\sin (x)$ hai; $x=\pi /2$ par wo $-\frac{1}{2}$ hai, interval par largest magnitude — "steepest" confirm karta hai. ✓

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Example 6 — Cell G: ordinary WDM throughput

1. $B_{\text{total}}=N\times B$ apply karo. $B_{\text{total}}=8\times 50=400\text{Gb/s}$. Yeh step kyun? Independent colors capacity share nahi karte, isliye throughput simply add hoti hai — dekho Micro-ring resonators, ek ring per color.
2. Tb/s mein express karo. $400\text{Gb/s}=0.4\text{Tb/s}$. Yeh step kyun? $1\text{Tb/s}=1000\text{Gb/s}$; readability ke liye sirf decimal shift kiya.

Verify: $8\times 50=400$ ✓. Ek single 50 Gb/s copper lane sirf $50\text{Gb/s}$ carry karta, isliye WDM yahan ek waveguide par $8\times$ density win hai. ✓

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Example 7 — Cell H: finite-band limiting case

1. Count karo kitne spacings fit hote hain. $N_{\max }=\frac{\Delta f}{\Delta f_{\text{ch}}}=\frac{4.4\times 10^{12}}{100\times 10^9}=44$. Yeh step kyun? Har channel ko apna guard-band-inclusive slot of $100\text{GHz}$ chahiye; band mein sirf itne hi slots fit hote hain. Yahi wo hard ceiling hai jo "unlimited bandwidth" wali galti ignore karti hai.
2. Us ceiling par aggregate karo. $B_{\text{total}}=N_{\max }\times B=44\times 25=1100\text{Gb/s}=1.1\text{Tb/s}$. Yeh step kyun? Physical max par bhi $N$ finite hai, isliye throughput finite hai.

Verify: $4.4\times 10^{12}/(100\times 10^9)=44$ ✓, aur $44\times 25=1100$ ✓. Yeh disprove karta hai "$N$ ko infinity tak crank karo": spectrum finite hai aur spacing crosstalk se neeche bounded hai. Tighter spacing (e.g. 50 GHz) $N$ double kar dega lekin sharper filters aur zyada thermal tuning power chahiye. ✓

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Example 8 — Cell I: energy-per-bit word problem

1. $E_{\text{bit}}=P_{\text{total}}/B_{\text{total}}$ apply karo. $E_{\text{bit}}=\frac{2\text{W}}{1.6\times 10^{12}\text{bit/s}}$. Yeh step kyun? Watts Joules/second hain, bit/s bits/second hai; divide karne par seconds cancel ho jaate hain → Joules/bit.
2. Compute karo. $E_{\text{bit}}=1.25\times 10^{-12}\text{J/bit}=1.25\text{pJ/bit}$. Yeh step kyun? $2/1.6=1.25$, aur $10^0/10^{12}=10^{-12}$, yani picojoules.

Verify: $2/(1.6\times 10^{12})=1.25\times 10^{-12}$ ✓ $=1.25\text{pJ/bit}$. Yeh sub-pJ target ($<1\text{pJ/bit}$) se thoda upar hai — high-speed copper ke several-pJ/bit ke saath competitive hai, jo length ke saath badhti hai, unlike photonics jo roughly distance-independent hai. Dekho Energy per bit as an efficiency metric. ✓

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Example 9 — Cell J: exam twist (photodetector current + latency trap)

1. (a) $I=R_{\lambda }P$ apply karo. $I=0.8\text{A/W}\times 0.5\times 10^{-3}\text{W}=4\times 10^{-4}\text{A}=400\mu \text{A}$. Yeh step kyun? Ek photodiode current source hai, voltage source nahi — yahi parent ka mistake #4 hai. Isko phir ek Transimpedance Amplifier (TIA) chahiye usable voltage banane ke liye.
2. (b) 10 cm ke liye photonic transit compute karo. $t_{ph}=\frac{nL}{c_0}=\frac{4\times 0.1\text{m}}{3\times 10^8\text{m/s}}=1.33\times 10^{-9}\text{s}=1.33\text{ns}$. Yeh step kyun? Silicon mein light $c_0/4$ par crawl karti hai; copper se compare karne ke liye uska transit time chahiye.
3. (b) 10 cm ke liye copper signal delay compute karo. $t_{cu}=\frac{L}{v_{cu}}=\frac{0.1\text{m}}{2\times 10^8\text{m/s}}=0.5\times 10^{-9}\text{s}=0.5\text{ns}$. Yeh step kyun? Ek copper trace ka electrical signal light ki speed ke kaafi kareeb speed se travel karta hai — per centimetre $c_0/4$ se faster.
4. (b) Compare karo aur rebut karo. $t_{cu}=0.5\text{ns}<t_{ph}=1.33\text{ns}$, isliye is chote 10 cm hop par copper lower latency hai. Yeh step kyun? Parent ka latency myth confirm hota hai: chhoti lengths par copper lower latency hoti hai. Photonics bandwidth density aur energy par jeetata hai, short-haul latency par nahi.

Verify: (a) $0.8\times 5\times 10^{-4}=4\times 10^{-4}\text{A}=400\mu \text{A}$ ✓. (b) photon $4\times 0.1/(3\times 10^8)=1.33\times 10^{-9}$ s ✓; copper $0.1/(2\times 10^8)=0.5\times 10^{-9}$ s ✓; $0.5\text{ns}<1.33\text{ns}$ ✓. Student latency argument haar jaata hai. ✓

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Active recall

Recall Kaun sa

$\Delta \varphi$ full power deta hai, aur kaun sa zero? $\Delta \varphi =0\Rightarrow {\cos }^{2}0=1\Rightarrow$ full power (bit 1). $\Delta \varphi =\pi \Rightarrow {\cos }^2(\pi /2)=0\Rightarrow$ zero power (bit 0). ::: $0$ → "1", $\pi$ → "0". WDM mein $N$ unlimited kyun nahi hota? ::: Finite optical band ($\sim$4.4 THz C-band) ko minimum crosstalk-safe channel spacing se divide karne par $N$ ki ek hard ceiling milti hai. Ek photodiode current output karta hai ya voltage? ::: Current, $I=R_{\lambda }P$; ek TIA use voltage mein convert karta hai. Silicon waveguide ke 10 cm par, kya photonic latency copper se kam hai? ::: Nahi — $c_0/4$ par light per cm copper signal se slow hoti hai; photonics bandwidth/energy par jeetata hai, short-haul latency par nahi.