Neuromorphic computing: Worked examples — Hinglish deep dive

6.5.14 · D3 · Hardware › Advanced & Emerging Architectures › Neuromorphic computing

Yeh deep dive Neuromorphic computing ka maths drill karta hai jab tak koi bhi input aapko surprise na kare. Hum parent note ke do engines lete hain — Leaky Integrate-and-Fire (LIF) neuron aur Spike-Timing-Dependent Plasticity (STDP) — aur har possible case class ko cover karte hain.

Shuru karne se pehle, core tools ka ek reminder (sab parent note mein derive kiye gaye hain):

Yahan har symbol ka matlab: $R$ = leak resistance (ohms), $C$ = membrane capacitance (farads), $I$ = input current (amps), $V$ = membrane voltage, $V_{t h}$ = firing threshold, $τ_{m} = R C$ = membrane kitni jaldi charge/forget karta hai, $t_{p r e} / t_{p os t}$ = do connected neurons ke spike times, $A_{\pm}$ = maximum learning strength, $τ_{\pm}$ = timing kitni jaldi matter karna band kar deta hai. Agar koi bhi cheez shaky lagey, pehle parent ka RC derivation dobara padh lo.

The scenario matrix

Is topic ka har problem exactly inhi cells mein se ek mein aata hai. Neeche ke examples us cell ka label carry karte hain jo woh cover karte hain.

#	Case class	Kya khaas hai	Example
C1	Above threshold $R I > V_{t h}$	Neuron fire karta hai; $T$ finite hai	Ex 1
C2	Below threshold $R I < V_{t h}$	Log argument negative → kabhi fire nahi karta	Ex 2
C3	Exactly at threshold $R I = V_{t h}$	Degenerate: $V_{t h}$ sirf $t = \infty$ par reach karta hai	Ex 3
C4	Limiting: huge input $R I ≫ V_{t h}$	$T \to 0$ as $1/ I$ ; rate bina bound ke badhta hai (sirf refractory period cap karta hai)	Ex 4
C5	STDP potentiation $Δ t > 0$	Pre before post → strengthen	Ex 5
C6	STDP depression $Δ t < 0$	Pre after post → weaken	Ex 5
C7	STDP degenerate $Δ t = 0$	Simultaneous spikes → boundary	Ex 6
C8	Real-world word problem	Target rate ke liye $I$ choose karo	Ex 7
C9	Exam twist	Threshold current + rate just above it	Ex 8
C10	Sanity / energy	Silence = low power kyun (sparsity)	Ex 9

Jahan tak ho sake hum ek fixed "textbook neuron" use karte hain taaki numbers comparable rahein: $R = 10 M Ω = 1 0^{7} Ω, C = 1 nF = 1 0^{- 9} F, τ_{m} = R C = 10 ms, V_{t h} = 15 mV .$

Figure dekho: teen coloured curves same neuron hain jo teen alag currents se feed ho rahi hain. Orange curve (strong input) dashed threshold line ko cross karta hai; teal wala (medium) bas ustak pahunchta hai; plum wala (weak) line ke neeche hamesha ke liye flat ho jaata hai. Yahi teen shapes hain cells C1, C3, C2. Har example inhi curves mein se kisi ek ka ek point hai.

C1 — Above threshold: it fires

Worked example Example 1 — Time-to-spike aur rate (

$R I > V_{t h}$ ) Textbook neuron, constant input $I = 2 nA$ . Pata karo ki yeh fire karta hai ya nahi, first spike ka time, aur firing rate.

Forecast: abhi andaza lagao — kya yeh fire karta hai? ~10 ms se upar ya neeche spike ka time?

Steady-state voltage $R I$ . Kyun yeh step? Agar ceiling $R I$ , $V_{t h}$ se neeche hai toh neuron kabhi threshold reach nahi kar sakta — pehle hamesha check karo. $R I = (1 0^{7}) (2 \times 1 0^{- 9}) = 20 \times 1 0^{- 3} V = 20 mV .$ Kyunki $20 > 15$ , yeh fire karta hai. ✅
Time constant. Kyun? $τ_{m}$ decide karta hai charge-up kitni jaldi hoti hai. $τ_{m} = R C = 10 ms$ .
Time to first spike. Kyun? Boxed $T$ formula mein plug karo. $T = 10 ms \cdot ln \frac{20}{20 - 15} = 10 ms \cdot ln 4 \approx 13.86 ms .$
Firing rate. Kyun? Rate ek spike per $T$ hai (refractory time ignore karke). $f = 1/ T \approx 72.1 Hz$ .

Verify: $T$ wapas plug karo: $V (13.86 ms) = 20 (1 - e^{- 13.86/10}) = 20 (1 - 0.25) = 15 mV = V_{t h}$ . ✔ Units: $Ω \cdot A = V$ ✔, $Ω \cdot F = s$ ✔.

C2 — Below threshold: it never fires

C3 — Exactly at threshold: the degenerate boundary

Worked example Example 3 — Bilkul edge par (

$R I = V_{t h}$ ) $I$ aisa choose karo ki ceiling exactly threshold ke barabar ho.

Forecast: agar $R I$ exactly $V_{t h}$ par land kare, toh yeh kab (agar kabhi) spike karta hai?

Critical current solve karo. Kyun? Chahiye $R I = V_{t h} = 15 mV$ . $I_{cr i t} = \frac{V _{t h}}{R} = \frac{15 \times 1 0 ^{- 3}}{1 0 ^{7}} = 1.5 \times 1 0^{- 9} A = 1.5 nA .$
$T$ mein plug karo. Kyun? Limiting behaviour check karo. $T = τ_{m} ln \frac{15}{15 - 15} = τ_{m} ln (\frac{15}{0})$ . Argument $\to + \infty$ , toh $ln \to + \infty$ , toh $T \to \infty$ .
Interpret karo. Kyun? $V (t) = 15 (1 - e^{- t / τ_{m}})$ $15 mV$ ko approach karta hai lekin infinite-time limit mein hi reach karta hai. Toh neuron effectively finite time mein kabhi fire nahi karta — boundary "silent" side ki taraf belong karti hai.

Verify: $t = 100 ms = 10 τ_{m}$ par: $V = 15 (1 - e^{- 10}) = 15 (1 - 0.0000454) = 14.99932 mV$ — abhi bhi $15$ se neeche. Confirms $T = \infty$ . ✔ $1.5 nA$ se upar koi bhi current fire karta hai; us par ya neeche, nahi karta.

C4 — Limiting case: huge input

Worked example Example 4 — Bahut zyada current (

$R I ≫ V_{t h}$ ) Same neuron, $I = 100 nA$ .

Forecast: kya $T$ bina kisi limit ke shrink hota rehta hai jaise $I$ badhta hai, ya koi floor hai?

Ceiling. Kyun? $R I = (1 0^{7}) (100 \times 1 0^{- 9}) = 1 V = 1000 mV$ . $15 mV$ se bahut zyada upar.
Time to spike. Kyun? Plug in karo. $T = 10 ms \cdot ln \frac{1000}{1000 - 15} = 10 ms \cdot ln (1.01523) \approx 10 ms \times 0.015113 \approx 0.1511 ms .$
Limiting behaviour. Kyun? $R I ≫ V_{t h}$ ke liye, $ln \frac{R I}{R I - V _{t h}} \approx \frac{V _{t h}}{R I}$ (small-argument log). Toh $T \approx τ_{m} \frac{V _{t h}}{R I} = \frac{C V _{t h}}{I}$ . Yeh $1/ I$ ke saath shrink karta rehta hai, toh firing rate $f = 1/ T \approx I / (C V_{t h})$ $I \to \infty$ ke saath bina bound ke badhti hai — leak-imposed ceiling nahi hai. Sirf ek real neuron ka refractory period (har spike ke baad minimum dead time) rate ko cap karega. $T \approx \frac{C V _{t h}}{I} = \frac{( 1 0 ^{- 9} ) ( 15 \times 1 0 ^{- 3} )}{100 \times 1 0 ^{- 9}} = 1.5 \times 1 0^{- 4} s = 0.15 ms .$

Verify: Exact $0.1511 ms$ vs approximation $0.15 ms$ — ~1% ke andar match, aur approximation "capacitor pehle linearly charge hota hai" limit hai ( $V \approx I t / C$ ). $I$ double karo aur $T$ roughly half ho jaata hai — pure $1/ I$ scaling, koi floor nahi. ✔

C5 & C6 — STDP: strengthen vs weaken

Worked example Example 5 — Dono STDP branches ek saath

Learning params $A_{+} = 0.1$ , $A_{-} = 0.12$ , $τ_{+} = τ_{-} = 20 ms$ . (a) Pre spike at $5 ms$ , post at $8 ms$ . (b) Pre at $8 ms$ , post at $5 ms$ .

Forecast: kaun sa case synapse ko strengthen karega, aur kya unka magnitude equal hoga?

Case (a) sign. Kyun? $Δ t = t_{p os t} - t_{p r e}$ ka sign branch pick karta hai. $Δ t = 8 - 5 = + 3 ms > 0$ → potentiation (pre ne post ko cause karne mein help ki).
Case (a) magnitude. Kyun? Top branch use karo. $Δ w = 0.1 e^{- 3/20} = 0.1 \times 0.86071 = + 0.0861$ .
Case (b) sign. Kyun? $Δ t = 5 - 8 = - 3 ms < 0$ → depression (pre bahut late aaya).
Case (b) magnitude. Kyun? Bottom branch: $Δ w = - A_{-} e^{+ Δ t / τ_{-}} = - 0.12 e^{(- 3) /20} = - 0.12 \times 0.86071 = - 0.1033$ .

Verify: Same $∣Δ t ∣ = 3 ms$ , same exponential factor $0.86071$ , lekin magnitudes alag hain ( $0.0861$ vs $0.1033$ ) kyunki $A_{-} > A_{+}$ . STDP curve jaanbujhkar asymmetric hai — depression aksar zyada strong hoti hai taaki weights explode na karen. $Δ w$ ki units: dimensionless weight change. ✔

Figure $Δ w$ ko $Δ t$ ke against plot karta hai. Zero ke right (orange) = strengthen, timing loose hone ke saath decay karta hai; zero ke left (teal) = weaken. $Δ t = + 3$ par dot Ex 5a hai; $Δ t = - 3$ par dot Ex 5b hai — note karo ki left dip zyada gehri hai kyunki $A_{-} > A_{+}$ .

C7 — STDP degenerate: simultaneous spikes

Worked example Example 6 — Exactly saath mein (

$Δ t = 0$ ) Pre aur post dono $t = 7 ms$ par fire karte hain.

Forecast: strengthen, weaken, ya kuch nahi?

$Δ t$ compute karo. Kyun? $Δ t = 7 - 7 = 0$ — dono branches ke beech ki boundary.
Dono taraf se approach karo. Kyun? Rule $Δ t > 0$ aur $Δ t < 0$ ke liye defined hai lekin $0$ par nahi, toh limits check karo. Jaise $Δ t \to 0^{+}$ : $Δ w \to A_{+} e^{0} = + 0.1$ . Jaise $Δ t \to 0^{-}$ : $Δ w \to - A_{-} e^{0} = - 0.12$ .
Interpret karo. Kyun? Dono limits alag hain ( $+ 0.1$ vs $- 0.12$ ) — $Δ t = 0$ par jump discontinuity. Simultaneity causally ambiguous hai ("kisne kise cause kiya?"), toh real chips sirf $Δ w = 0$ set kar dete hain ya jo bhi spike pehle log hoi use use karte hain. Koi single "correct" value nahi hai; yeh ek modelling convention hai.

Verify: Limits differ karte hain → discontinuity confirmed. Yeh woh ek cell hai jahan formula ka genuinely koi defined output nahi hai, jo C3 ke lesson se match karta hai ki boundaries ko explicit handling chahiye. ✔

C8 — Real-world word problem

Worked example Example 7 — 100 Hz neuron design karo

Ek neuromorphic vision chip ko ek pixel-neuron chahiye jo steady light current ke neeche exactly $f = 100 Hz$ par fire kare. Textbook neuron use karke ( $τ_{m} = 10 ms$ , $R = 10 M Ω$ , $V_{t h} = 15 mV$ ), kaunsa constant input current $I$ chahiye?

Forecast: Ex 1 ke $2 nA$ se zyada ya kam jo $72 Hz$ deta tha?

Target period. Kyun? Rate $\to$ period: $T = 1/ f = 1/100 = 10 ms$ .
$T$ formula ko $I$ ke liye invert karo. Kyun? Hume $T$ pata hai, $I$ chahiye. $T = τ_{m} ln \frac{R I}{R I - V _{t h}}$ se shuru karo: $e^{T / τ_{m}} = \frac{R I}{R I - V _{t h}} \Rightarrow R I - V_{t h} = R I e^{- T / τ_{m}} \Rightarrow R I (1 - e^{- T / τ_{m}}) = V_{t h} .$ $I = \frac{V _{t h}}{R ( 1 - e ^{- T / τ_{m}} )}$
Numbers plug karo. Kyun? $T / τ_{m} = 10/10 = 1$ , toh $e^{- 1} = 0.36788$ , aur $1 - e^{- 1} = 0.63212$ . $I = \frac{15 \times 1 0 ^{- 3}}{1 0 ^{7} \times 0.63212} = \frac{15 \times 1 0 ^{- 3}}{6.3212 \times 1 0 ^{6}} = 2.373 \times 1 0^{- 9} A \approx 2.37 nA .$

Verify: $2.373 nA$ wapas feed karo: $R I = 23.73 mV$ ; $T = 10 ln \frac{23.73}{23.73 - 15} = 10 ln (2.7181) = 10 \times 1.0000 = 10.0 ms$ → $100 Hz$ . ✔ Aur $2.37 > 2 nA$ , forecast se match karta hai (tez firing ke liye zyada current chahiye). ✔

C9 — Exam twist: threshold current aur soft turn-on

Worked example Example 8 — Smallest spiking current, aur rate just above threshold par

Ek exam poochta hai: "Textbook neuron ke liye, koi bhi spiking produce karne wala sabse chhota current $I$ kya hai, aur us threshold current se $10%$ upar wala current kya firing rate deta hai?"

Forecast: threshold current par hi rate zero hai ya infinite-period?

Threshold current. Kyun? Spiking ke liye $R I > V_{t h}$ chahiye, toh infimum $R I = V_{t h}$ hai: $I_{0} = V_{t h} / R = 1.5 nA$ (yeh C3 boundary hai — exactly $I_{0}$ par, $T = \infty$ , rate $0$ ).
Dus percent upar. Kyun? $I = 1.1 I_{0} = 1.65 nA$ , toh $R I = 1.1 \times 15 = 16.5 mV$ .
Time to spike. Kyun? Plug in karo. $T = 10 ms \cdot ln \frac{16.5}{16.5 - 15} = 10 ln (11) = 10 \times 2.3979 = 23.98 ms$ .
Rate. $f = 1/ T = 1/0.02398 \approx 41.7 Hz$ .

Verify: Threshold ke just upar rate chhota lekin positive hai ( $41.7 Hz$ ), phir $I$ badhne ke saath steeply rise karta hai — LIF f–I curve ka classic "soft turn-on." Check: $V (23.98 ms) = 16.5 (1 - e^{- 2.398}) = 16.5 (1 - 0.0909) = 16.5 \times 0.9091 = 15.0 mV = V_{t h}$ . ✔

C10 — Sanity / energy check

Worked example Example 9 — Silence power kyun bachata hai

Maano ek neuromorphic layer mein $1000$ neurons hain jinhein sparse input feed ho rahi hai jahan sirf $30$ neurons threshold cross karte hain; baaki $970$ $R I < V_{t h}$ par baithe hain (Ex 2 situation). Agar har spike ki cost $E_{s p ik e} = 10 pJ$ hai aur idle neurons effectively $0$ cost karte hain, toh per timestep energy kya hai, aur yeh ek dense chip se kaise compare karta hai jo sabhi $1000$ ko "compute" karta hai?

Forecast: ek dense chip ke kaam ka roughly kitna fraction kiya jaata hai?

Active count. Kyun? Sirf firing neurons dynamic energy draw karte hain — yeh parent ke in-memory contrast se event-driven principle hai. Active $= 30$ .
Neuromorphic energy. Kyun? Sirf active spikes sum karo. $E = 30 \times 10 pJ = 300 pJ$ .
Dense-equivalent energy. Kyun? Ek dense accelerator har step par sabhi $1000$ units ke liye pay karta hai: $E_{d e n se} = 1000 \times 10 pJ = 10, 000 pJ$ .
Ratio. Kyun? $300/10000 = 0.03 = 3%$ .

Verify: Chip same layer ke liye $3%$ energy use karta hai kyunki $97%$ neurons sub-threshold the (C2) — C2 "never fires" case koi bug nahi hai, yeh poori energy story hai. Units: $pJ$ throughout, dimensionless ratio. ✔

Recall Self-test — har matrix cell ke liye ek

Kya $R I = V_{t h}$ wala neuron finite time mein kabhi fire karta hai? ::: Nahi — $T \to \infty$ ; boundary silent side par hai (dekho C3). LIF problem mein aap pehle hamesha kaunsi quantity compute karte ho, aur kyun? ::: $R I$ vs $V_{t h}$ — agar $R I \leq V_{t h}$ toh neuron kabhi fire nahi karta aur $T$ formula ka log negative/undefined ho jaata hai (C2). $R I ≫ V_{t h}$ ke liye, time-to-spike $I$ ke saath kaise scale karta hai? ::: $T \approx C V_{t h} / I$ , yaani $\propto 1/ I$ ; rate bina bound ke badhta hai, sirf refractory period cap karta hai (C4). Pre at 5 ms, post at 8 ms: strengthen ya weaken? ::: $Δ t = + 3 > 0$ → potentiation, strengthen (C5). STDP mein jab $Δ t = 0$ hota hai toh kya hota hai? ::: $+ A_{+}$ aur $- A_{-}$ ke beech jump discontinuity; convention-defined, usually $0$ (C7). Firing rate double karne ke liye $I$ (increase/decrease) karna padega? ::: Increase — zyada $R I$ se $T$ chhota hota hai (C8/C9).

See also: Neuromorphic computing (parent), Spiking Neural Networks (SNN), Hebbian learning, Memristors and ReRAM, RC circuits, Von Neumann architecture.