Worked examples — Photonic and optical interconnects

Before we start, every quantity we use is defined in the parent Photonic and optical interconnects. We only need five formulas, restated here so nothing is used unearned:

Here $\Delta \varphi$ (read "delta phi") is the extra phase — how far one arm's light wave is shifted relative to the other, measured in radians where $\pi$ radians $=180^{\circ }$ = half a wavelength of shift.

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The scenario matrix

Every worked example below is tagged with the cell it covers.

#	Cell class	What makes it special	Example
A	Copper $L^2$ scaling	double the length, watch delay quadruple	Ex 1
B	Copper vs photonic crossover	find the length where light wins	Ex 2
C	Degenerate input ($L\to 0$)	zero-length wire, what do formulas say?	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 ($\Delta \varphi =\pi /2$)	the 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 quadruples

1. Convert units so seconds come out clean. $r=0.2\Omega /\text{mm}$, $c_{\ell }=0.2\text{pF/mm}=0.2\times 10^{-12}\text{F/mm}$. Why this step? $\Omega \times \text{F}=\text{seconds}$, so keeping both "per mm" and multiplying by $L^2$ in mm gives an answer in seconds directly.
2. Apply $\tau =r{c}_{\ell }{L}^2$ at $L=5$. ${\tau }_5=(0.2)(0.2\times 10^{-12})(5^2)=0.04\times 10^{-12}\times 25=1.0\times 10^{-12}\text{s}=1\text{ps}$. Why this step? Direct substitution — this is the whole point of the $L^2$ law.
3. Apply at $L=10$. ${\tau }_{10}=(0.2)(0.2\times 10^{-12})(10^2)=0.04\times 10^{-12}\times 100=4.0\times 10^{-12}\text{s}=4\text{ps}$. Why this step? Same formula, only $L$ changed from 5 to 10.

Verify: ratio ${\tau }_{10}/{\tau }_5=4/1=4$. Doubling $L$ quadrupled delay ($2^2=4$), exactly as the $L^2$ law predicts. Units: $\Omega \cdot \text{F}=\text{s}$. ✓ (If you forecast "$2\times$" you fell for the linear-intuition trap.)

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Example 2 — Cell B: where does light overtake copper?

Figure s01: horizontal axis is wire length $L$ in mm, vertical axis is time in ps. The black straight line is the photon transit $t_{ph}=13.33L$ (linear). The red parabola is copper delay ${\tau }_{cu}=0.04L^2$. They meet at the marked crossover point $L=333$ mm — left of it copper is faster, right of it light wins.

1. Write both times as functions of $L$ (in mm). Copper: ${\tau }_{cu}=0.04L^2$ ps. Photon: $t_{ph}=\frac{nL}{c_0}$ with $n=4$ and $c_0=3\times 10^8\text{m/s}$. Why this step? To find a crossover you need both curves as functions of the same variable in the same units.
2. Convert $c_0$ into mm/ps, once, cleanly. Multiply top and bottom to change metres→mm and seconds→ps: $c_0=3\times 10^8\frac{\text{m}}{\text{s}}\times \frac{10^3\text{mm}}{1\text{m}}\times \frac{1\text{s}}{10^{12}\text{ps}}=3\times 10^{-1}\frac{\text{mm}}{\text{ps}}=0.3\frac{\text{mm}}{\text{ps}}.$ Why this step? With $c_0$ in mm/ps and $L$ in mm, the transit $t_{ph}=nL/c_0$ comes out directly in ps — no leftover unit mismatch.
3. Write the photon time in ps. $t_{ph}=\frac{4L}{0.3\text{mm/ps}}=13.33L$ ps (with $L$ in mm). Check with $L=1$ mm: $t_{ph}=4/0.3=13.33$ ps. ✓ Why this step? Now both curves are in ps, ready to equate.
4. Set equal and solve. $0.04L^2=13.33L\Rightarrow L=13.33/0.04=333\text{mm}$. Why this step? Divide both sides by $L$ (valid for $L\ne 0$); the crossover length is where the parabola catches the line — the red dot in the figure.

Verify: at $L=333$ mm, copper $\tau =0.04\times 333^2\approx 4435$ ps; photon $t=13.33\times 333\approx 4439$ ps. Equal to rounding. ✓ Below $\sim 333$ mm the photon is actually slower per metre (index $n=4$!) — this is exactly the latency myth the parent warns about. Photonics wins on bandwidth/energy, not raw short-haul latency.

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Example 3 — Cell C: the degenerate zero-length wire

1. Take limits. ${\tau }_{cu}=r{c}_{\ell }{L}^2\to 0$ and $t_{ph}=nL/c_0\to 0$ as $L\to 0$. Why this step? A zero-length wire has no delay in either model — a sanity floor every model must pass.
2. Compare the rate of approach. Near $L=0$, ${\tau }_{cu}\propto L^2$ shrinks faster than $t_{ph}\propto L$. Why this step? For very short wires copper's quadratic term is tiny, which is why on-chip short hops still use copper — photonics only pays off past the crossover (Ex 2).

Verify: plug $L=0$: ${\tau }_{cu}=r{c}_{\ell }\cdot 0=0$, $t_{ph}=n\cdot 0/c_0=0$. Both exactly $0$. ✓ Degenerate case is consistent (no negative or infinite nonsense). The physical caveat: real links still have fixed overheads (modulator, TIA latency) the $L$-formulas ignore — so the true zero-length time is not zero, it's the fixed E-O-E overhead. See Transimpedance Amplifier (TIA).

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

Figure s02: three phasor diagrams side by side. In each, the black arrow is arm 1 (fixed, pointing right) and the red arrow is arm 2 rotated by $\Delta \varphi$; the dashed arrow is their sum. Left ($\Delta \varphi =0$) sums to full length → $P=1$. Middle ($\Delta \varphi =\pi /2$) → $P=0.5$. Right ($\Delta \varphi =\pi$) the red arrow points back and the sum collapses to zero → $P=0$.

1. State D, $\Delta \varphi =0$. $P_{\text{out}}=1\cdot {\cos }^2(0/2)=1\cdot {\cos }^{2}0=1\cdot 1=1\text{mW}$. Why this step? At zero phase difference the two arm-waves add crest-to-crest (constructive interference) — the two phasors in the figure point the same way, full length. Bit 1.
2. State E, $\Delta \varphi =\pi$. $P_{\text{out}}=1\cdot {\cos }^2(\pi /2)=1\cdot 0^2=0\text{mW}$. Why this step? A $\pi$ shift means one wave is a full half-wavelength behind — crest meets trough, they cancel (destructive interference). The phasors point opposite, sum to zero. Bit 0.

Verify: ${\cos }^{2}0=1$ ✓ and ${\cos }^2(\pi /2)=0$ ✓. Full power → "1", zero power → "0". The voltage that produces exactly $\pi$ of phase is called $V_{\pi }$ — the modulator's switching voltage. Energy conservation check: the "missing" power at $\Delta \varphi =\pi$ isn't destroyed, it exits the other (complementary) output port of the interferometer.

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Example 5 — Cell F: the tricky half-power midpoint

1. Substitute $\Delta \varphi =\pi /2$. $P_{\text{out}}=1\cdot {\cos }^2(\pi /4)$. Why this step? Half of $\pi /2$ is $\pi /4$; that's the argument of the cosine.
2. Evaluate. $\cos (\pi /4)=\frac{1}{\sqrt{2}}$, so ${\cos }^2(\pi /4)=\frac{1}{2}$. Thus $P_{\text{out}}=0.5\text{mW}$. Why this step? Exactly half power — the quadrature point.
3. Why special? At $\Delta \varphi =\pi /2$ the ${\cos }^2$ curve is steepest (its slope is maximal), so a small voltage wiggle gives the biggest, most linear power swing. Why this step? Linearity matters for analog optical links where you want output $\propto$ input, not a hard on/off.

Verify: ${\cos }^2(\pi /4)=(1/\sqrt{2}{)}^2=1/2$ ✓ → $0.5\text{mW}$, exactly halfway between the $1\text{mW}$ and $0\text{mW}$ of Ex 4. Slope of ${\cos }^2(x/2)$ is $-\frac{1}{2}\sin (x)$; at $x=\pi /2$ that is $-\frac{1}{2}$, the largest magnitude on the interval — confirms "steepest". ✓

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Example 6 — Cell G: ordinary WDM throughput

1. Apply $B_{\text{total}}=N\times B$. $B_{\text{total}}=8\times 50=400\text{Gb/s}$. Why this step? Independent colors don't share capacity, so throughput simply adds — see Micro-ring resonators, one ring per color.
2. Express in Tb/s. $400\text{Gb/s}=0.4\text{Tb/s}$. Why this step? $1\text{Tb/s}=1000\text{Gb/s}$; just moving the decimal for readability.

Verify: $8\times 50=400$ ✓. A single 50 Gb/s copper lane would carry only $50\text{Gb/s}$, so WDM here is an $8\times$ density win on one waveguide. ✓

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Example 7 — Cell H: the finite-band limiting case

1. Count how many spacings fit. $N_{\max }=\frac{\Delta f}{\Delta f_{\text{ch}}}=\frac{4.4\times 10^{12}}{100\times 10^9}=44$. Why this step? Each channel needs its own guard-band-inclusive slot of $100\text{GHz}$; the band holds only so many slots. This is the hard ceiling the "unlimited bandwidth" mistake ignores.
2. Aggregate at that ceiling. $B_{\text{total}}=N_{\max }\times B=44\times 25=1100\text{Gb/s}=1.1\text{Tb/s}$. Why this step? Even at the physical max, $N$ is finite, so throughput is finite.

Verify: $4.4\times 10^{12}/(100\times 10^9)=44$ ✓, and $44\times 25=1100$ ✓. This disproves "crank $N$ to infinity": spectrum is finite and spacing is bounded below by crosstalk. Tighter spacing (e.g. 50 GHz) would double $N$ but costs sharper filters and more thermal tuning power. ✓

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

1. Apply $E_{\text{bit}}=P_{\text{total}}/B_{\text{total}}$. $E_{\text{bit}}=\frac{2\text{W}}{1.6\times 10^{12}\text{bit/s}}$. Why this step? Watts are Joules/second, bit/s is bits/second; dividing cancels seconds → Joules/bit.
2. Compute. $E_{\text{bit}}=1.25\times 10^{-12}\text{J/bit}=1.25\text{pJ/bit}$. Why this step? $2/1.6=1.25$, and $10^0/10^{12}=10^{-12}$, i.e. picojoules.

Verify: $2/(1.6\times 10^{12})=1.25\times 10^{-12}$ ✓ $=1.25\text{pJ/bit}$. This is just above the sub-pJ target ($<1\text{pJ/bit}$) — competitive with, and roughly distance-independent unlike, high-speed copper's several-pJ/bit that rises with length. See Energy per bit as an efficiency metric. ✓

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

1. (a) Apply $I=R_{\lambda }P$. $I=0.8\text{A/W}\times 0.5\times 10^{-3}\text{W}=4\times 10^{-4}\text{A}=400\mu \text{A}$. Why this step? A photodiode is a current source, not a voltage source — this is the parent's mistake #4. The tiny current then needs a Transimpedance Amplifier (TIA) to become a usable voltage.
2. (b) Compute the photonic transit for 10 cm. $t_{ph}=\frac{nL}{c_0}=\frac{4\times 0.1\text{m}}{3\times 10^8\text{m/s}}=1.33\times 10^{-9}\text{s}=1.33\text{ns}$. Why this step? Light in silicon crawls at $c_0/4$; we need its transit time to compare against copper.
3. (b) Compute the copper signal delay for 10 cm. $t_{cu}=\frac{L}{v_{cu}}=\frac{0.1\text{m}}{2\times 10^8\text{m/s}}=0.5\times 10^{-9}\text{s}=0.5\text{ns}$. Why this step? A copper trace's electrical signal travels near half the speed of light — faster per centimetre than $c_0/4$.
4. (b) Compare and rebut. $t_{cu}=0.5\text{ns}<t_{ph}=1.33\text{ns}$, so copper is lower latency over this short 10 cm hop. Why this step? Confirms the parent's latency myth: at short lengths copper is lower latency. Photonics wins on bandwidth density and energy, not short-haul latency.

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 ✓; $0.5\text{ns}<1.33\text{ns}$ ✓. The student loses the latency argument. ✓

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Active recall

Recall Which

$\Delta \varphi$ gives full power, and which gives 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". Why is $N$ not unlimited in WDM? ::: Finite optical band ($\sim$4.4 THz C-band) divided by minimum crosstalk-safe channel spacing gives a hard ceiling on $N$. A photodiode outputs current or voltage? ::: Current, $I=R_{\lambda }P$; a TIA converts it to voltage. Over 10 cm of silicon waveguide, is photonic latency lower than copper? ::: No — light at $c_0/4$ is slower per cm than a copper signal; photonics wins on bandwidth/energy, not short-haul latency.