2a:null 2b:["$","title",null,{"children":"Tuning Curves | Steve Oney | Observable"}] 30:T60c,# (Optional) Analytically Computing LIF Responses Above, we computed the response of the LIF neuron by simulating it for multiple different inputs. However, we can also analytically predict its response without needing to simulate it. ${tex`v(t) = I(t) + (v(t-\Delta{}t) - I(t))e^{-\frac{\Delta{}t}{\tau_{rc}}}`} If we want to compute the time that it takes to get to the firing threshold ${tex`v_{th}`} under a constant input current ${tex`I`}, we first need to assume that the input current is high enough: ${tex`I > v_{th}`}. Then, we can solve for how long it takes to reach ${tex`v_{th}`} (which we will call ${tex`t_{th}`}) using the formula: ${tex`v_{th} = I + (0 - I)e^{-\frac{t_{th}}{\tau_{rc}}}`} Moving terms around, we get: ${tex`e^{-\frac{t_{th}}{\tau_{rc}}} = 1 - \frac{v_{th}}{I}`} giving us: ${tex`t_{th} = -\tau_{rc}\ln{\left(1 - \frac{v_{th}}{I}\right)}`} To compute the time that it takes to fire and recover, we need to add the refractory time (${tex`\tau_{ref}`}). This means the time that it takes to fire is: ${tex`\tau_{ref}-\tau_{rc}\ln{\left(1 - \frac{v_{th}}{I}\right)}`} and to get the firing rate ${tex`r(I)`} for a given input ${tex`I`}, we take its inverse: ${tex`r(I) = \frac{1}{\tau_{ref}-\tau_{rc}\ln{\left(1 - \frac{v_{th}}{I}\right)}}`} However, all of this only applies if ${tex`I > v_{th}`}, leaving us with the final equation: ${tex` r(I) = \begin{cases} 0, & \text{if } I \leq v_{th}\\ \frac{1}{\tau_{ref}-\tau_{rc}\ln{\left(1 - \frac{v_{th}}{I}\right)}}, & \text{otherwise} \end{cases} `}31:T56b,function plotSpikes(inp, runtime=20) { const TIME_STEP = 1e-3; const NUM_STEPS = runtime / TIME_STEP; // Simulate for runtime seconds const J = () => inp; const neuron = new LIF(0.02, 0, 0.2); const v_history = [ {t: 0, v: neuron.v} ]; let numSpikes = 0; for(let i = 0; i < NUM_STEPS; i++) { const j = J((i+1)*TIME_STEP); neuron.step(j, TIME_STEP); const out = neuron.output; if(out > 0) { numSpikes++; } v_history.push({t: (i+1)*TIME_STEP, v: neuron.v, j, out }); } const width = 1100; const height = 20; const padding = 5; const svg = DOM.svg(width + 2*padding, height + 2*padding); const spikeScaleX = d3.scaleLinear([0, v_history.length], [padding, padding+width]); v_history.forEach(({out}, t) => { if(out > 0) { const displayX = spikeScaleX(t); const locationIndicator = d3.select(svg).append("path") .attr("stroke", 'black') .attr("fill", "none") .attr("stroke-width", 0.5) .attr("d", `M${displayX},${padding}v${height}`); } }); // const plot = Plot.line(v_history, {x: "t", y: "out"}).plot(); return html`

${svg.outerHTML}

An input of ${inp} produces ${numSpikes} spike${numSpikes===1 ?'' : 's'} in ${runtime} seconds (${numSpikes / runtime} Hz)

`; }32:[] 2e:{"id":"af48ade400490b47","slug":"nmc-tuning-curves","title":"Tuning Curves","thumbnail":"7f40d74bd1a66e25ec844e49e8f3295ccb290526b3aa5a52a53d54f63feaac5d","access_level":"public","create_time":"$D2022-11-22T21:01:31.871Z","update_time":"$D2023-02-21T17:42:07.230Z","view_time":null,"first_public_time":"$D2023-01-24T15:21:54.957Z","fork":false,"stars":0,"starred":false,"owner":{"type":"team","id":"55b87b37da07de37","login":"soney","name":"Steve Oney","avatar_url":"https://avatars-next.observableusercontent.com/565926373d73cf99aa6425cf7c0d7ff79e41635d1a367894a5e3bce0299b18b9","tier":"starter_2024"},"version":482,"publish_version":null,"role":null,"authored":false,"authors":[{"type":"individual","id":"1d138075b46297f1","login":"soney","name":"Steve Oney","avatar_url":"https://avatars-next.observableusercontent.com/565926373d73cf99aa6425cf7c0d7ff79e41635d1a367894a5e3bce0299b18b9","tier":null}],"members":[{"type":"individual","id":"1c09ce4b7505d91c","login":"adanielescu","name":"Andreea Danielescu","avatar_url":"https://avatars-next.observableusercontent.com/5cd8a89ba2fda44883bc5dd98ba3f3c6cb6297d77c21b9ef0c0919c5ef608a83","tier":null,"role":"editor"},{"type":"individual","id":"79581e18d086caf4","login":"ericgallo","name":"","avatar_url":null,"tier":null,"role":"editor"},{"type":"individual","id":"93113dd5fb467076","login":"sharvari-bondre","name":"Sharvari Bondre","avatar_url":"https://avatars-next.observableusercontent.com/754b967a5cf1ce71589f615fed536b94bd68509f3412154d82572c4b6aa44ead","tier":null,"role":"editor"},{"type":"individual","id":"1d138075b46297f1","login":"soney","name":"Steve Oney","avatar_url":"https://avatars-next.observableusercontent.com/565926373d73cf99aa6425cf7c0d7ff79e41635d1a367894a5e3bce0299b18b9","tier":null,"role":"editor"}],"comments":[],"custom_thumbnail":false,"trashed":false,"subscribed":null,"forks":0,"collection":{"id":"ceab76d8d8a0acbc","slug":"neuromorphic-computing","title":"Neuromorphic Tutorials: Spiking Neural Networks (SNNs) from Scratch","owner":{"id":"55b87b37da07de37","login":"soney","name":"Steve Oney"}},"type":"notebook","body":{"title":"Tuning Curves","cells":[{"id":0,"value":"# Tuning Curves\n\nOne way to summarize the important aspects of what a neuron \"does\" is to summarize how it responds to a variety of inputs. We're going to first use the LIF model that we built in the last notebook:","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined"},{"id":5,"value":"class LIF {\n constructor(tau_rc=0.2, v_init=0, tau_ref=0.002, v_th=1){\n this.tau_rc = tau_rc;\n this.tau_ref = tau_ref;\n this.v = v_init;\n this.v_th = v_th;\n \n this.output = 0;\n this.refractory_time = 0;\n }\n\n step(I, t_step) {\n this.refractory_time -= t_step; // Subtract the amount of time thhat passed from our refractory time\n \n if(this.refractory_time < 0) { // If we aren't in our refractory period, update the voltage\n this.v = this.v * (1 - t_step/this.tau_rc) + (I * t_step / this.tau_rc);\n }\n\n if(this.v >= this.v_th) { // Fired\n this.refractory_time = this.tau_ref; // Set refractory time\n\n this.output = 1 / t_step; // Fire\n this.v = 0; // Reset potential\n } else {\n this.output = 0; // Don't fire\n }\n return this.output;\n }\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined"},{"id":8,"value":"So let's see what happens if we input a *constant* current. We're going to create a LIF neuron with parameters ${tex`\\tau_{rc} = 0.02`} and ${tex`\\tau_{ref} = 0.2`} and plot the spikes of our LIF neuron given a constant inputs of varying intensities.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined"},{"id":70,"value":"Let's see what happens when we call our function with an input *slightly* over 1A (`1.0000001`):","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined"},{"id":10,"value":"plotSpikes(1.0000001)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined"},{"id":24,"value":"We can see that our neuron fires *about* twice per second , making its firing rate ~2Hz. If we increase our input to 1.1A, we can see that it fires more frequently:","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined"},{"id":75,"value":"plotSpikes(1.1)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined"},{"id":77,"value":"Increase it some more, and it will fire even more frequently.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined"},{"id":79,"value":"plotSpikes(10)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined"},{"id":90,"value":"But we can only increase it so much, thanks to the refractory time. Note that if we increase our incoming current by a factor of 10 (100A), our spike frequency barely changes:","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined"},{"id":92,"value":"plotSpikes(100)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined"},{"id":121,"value":"However, if we don't input enough current, our neuron never spikes, as the potential leaks out faster than it can be added.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined"},{"id":123,"value":"plotSpikes(0.8)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined"},{"id":125,"value":"Let's see how our neuron reacts to different frequencies. ","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined"},{"id":127,"value":"function getSpikeCount(neuron, inp, runtime) {\n const TIME_STEP = 1e-3;\n const NUM_STEPS = runtime / TIME_STEP;\n\n let numSpikes = 0;\n for(let i = 0; i < NUM_STEPS; i++) {\n const output = neuron.step(inp, TIME_STEP);\n if(output > 0) {\n numSpikes++;\n }\n }\n return numSpikes;\n}","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined"},{"id":279,"value":"function getSpikeRate(neuron, inp, runtime=10) {\n return getSpikeCount(neuron, inp, runtime) / runtime;\n}","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined"},{"id":415,"value":"viewof tau_ref = Inputs.range([0, 2], {value: 0.2, step: 0.001, label: \"tau_ref\"})","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined"},{"id":417,"value":"viewof tau_rc = Inputs.range([0.1, 1], {value: 0.3, step: 0.001, label: \"tau_rc\"})","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined"},{"id":418,"value":"viewof v_th = Inputs.range([0.5, 1.2], {value: 1, step: 0.01, label: \"v_th\"})","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined"},{"id":129,"value":"{\n const [minI, maxI] = [0, 10.1];\n const stepSize = .01;\n\n const data = [];\n for(let i = minI; i<=maxI; i+=stepSize) {\n const neuron = new LIF(tau_rc, 0, tau_ref, v_th);\n data.push({i, hz: getSpikeRate(neuron, i)});\n }\n return Plot.line(data, {x: \"i\", y: \"hz\"}).plot();\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined"},{"id":213,"value":"So we can see that our neuron does not react to current until it reaches ${tex`v_{th}`} (`${v_th}`). At that point, it starts firing and its firing rate is proportional to the incoming current. The firing rate eventually flattens out, thanks to the refractory time (the period where the neuron can't fire).","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined"},{"id":470,"value":"# Summary\n\n- A tuning curve summarizes a neuron's response (in the form of a firing rate) to a variety of inputs\n- Neurons start firing at the threshold potential ${tex`v_{th}`} and the firing rate rises as the input increases (but rises at a decreasing rate)\n- As long as there is some refractory time, our neuron's firing rate will \"flatten\" out to a maximum value","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined"},{"id":475,"value":"---","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined"},{"id":216,"value":"$30","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined"},{"id":266,"value":"Let's plot out this result along with the one we computed above.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined"},{"id":455,"value":"viewof tau_ref2 = Inputs.range([0, 2], {value: 0.2, step: 0.001, label: \"tau_ref\"})","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined"},{"id":459,"value":"viewof tau_rc2 = Inputs.range([0.1, 1], {value: 0.3, step: 0.001, label: \"tau_rc\"})","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined"},{"id":461,"value":"viewof v_th2 = Inputs.range([0.5, 1.2], {value: 1, step: 0.01, label: \"v_th\"})","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined"},{"id":268,"value":"{\n const [minJ, maxJ] = [0, 2];\n const stepSize = .01;\n\n const data = [];\n for(let j = minJ; j<=maxJ; j+=stepSize) {\n const neuron = new LIF(tau_rc2, 0, tau_ref2, v_th2);\n const analyticalSolution = (j <= neuron.v_th) ? 0 : (1 / (neuron.tau_ref - neuron.tau_rc * Math.log(1 - neuron.v_th/j)));\n data.push({j, empirical: getSpikeRate(neuron, j), analytical: analyticalSolution});\n }\n const plot = Plot.plot({\n y: {\n grid: true,\n label: 'hz'\n },\n marks: [\n Plot.line(data, {x: \"j\", y: \"analytical\", stroke: \"red\", strokeWidth: 2}),\n Plot.line(data, {x: \"j\", y: \"empirical\", stroke: \"black\", strokeWidth: 1}),\n ]\n });\n return plot\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined"},{"id":315,"value":"We can see that the firing rate that we computed analytically matches what we observe in the simulation.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined"},{"id":468,"value":"---","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined"},{"id":478,"value":"# References\n\n- Chris Eliasmith's [Neural Engineering book](https://mitpress.mit.edu/9780262550604/neural-engineering/) contains the derivation used for analytically computing LIF responses\n- [The Nengo summer school lectures](https://www.youtube.com/watch?v=JCA2gOhN5oM&list=PLYLu6sY3jnoXTrkfA_WMn0nF_FGz-MaiJ&index=2) clearly describe how tuning curves work","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined"},{"id":53,"value":"$31","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined"}],"files":"$W32","stdlib":"1","titling":"auto"}} 2f:"$2e:body"