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Biologically Inspired Intelligent Systems

Biologically Inspired Intelligent Systems. Lecture 4 Dr. Roger S. Gaborski. Developing a Neuron Model. Develop physical model of neuron Want to write an algorithm for the operation of a neuron Steps: Develop mechanical model of neuron Develop electrical model of mechanical neuron

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Biologically Inspired Intelligent Systems

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  1. Biologically Inspired Intelligent Systems Lecture 4 Dr. Roger S. Gaborski

  2. Developing a Neuron Model • Develop physical model of neuron • Want to write an algorithm for the operation of a neuron • Steps: • Develop mechanical model of neuron • Develop electrical model of mechanical neuron • Develop mathematical model of electrical model

  3. Quick Review – derivatives and integrals • Derivatives – ‘rate of change’ • At a specified point, the slope of the tangent line to the function at that point • Integrals – ‘area under the curve’ between points a and b

  4. A Quick Introduction to Differential Equations • Reference Textbooks • Spikes, decisions and actions Dynamical Foundation of Neuroscience Author: Hugh R. Wilson • Dynamic Modeling of Environmental Systems Author: M. Deaton and J. Winebrake • Internet sources

  5. A Quick Introduction to Differential Equations • Exponential function and its derivative • Describes processes of growth and decay • Assume a colony of bacteria has a population of F at t=0 • Assume fraction k of population reproduces per unit time • Increase in population is described by a differential equation: • dF/dt = kF • Rate of change of the population is proportional to the population F

  6. Why are Differential Equations Important in BIIS? • DE describe the change in neural responses between the present time t and a time t+dt • Describes change in terms of physiologically relevant variables at time t • Order of DE is defined as the highest derivative in the equation • Simplest DE: first order, linear with constant coefficients

  7. Fundamental First Order Equation(from Wilson, p13) • dx/dt = -(1/τ)x • The rate of change of function x(t) as a function of time is equal to a constant times the function itself • Units of τ and t are both milliseconds • We need to find a function that will satisfy the DE • In this example, the DE is solved by substituting the exponential fcn for x(t)

  8. Exponential Function Solution • x(t) = Aeαt • Derivative of x(t) is dx/dt = αA eαt • Substitute into DE: dx/dt = -(1/τ)x Aα eαt = -(1/τ) Aeαt Solve for α by canceling common terms yields α = -(1/τ)

  9. Solution, continued • The solution is simply x(t) = Ae(-t/τ) To determine the value of A we need to know an initial condition, x0, a value for x(t) at t=0 x(t) = x0 e-t/τ

  10. MATLABHow fast will a population die out? >> t=[0:99]; >> tau=10; >> x = exp(-1*t/tau); >> figure, plot([0:99],x) >> xlabel('time') >> ylabel('function value')

  11. What if I change the constant tau?

  12. CircuitsResistors and Capacitors • Using resistors and capacitors to model neurons • Simple circuits • Kirchoff’s Circuit Laws

  13. Kirchoff’s Circuit Laws • Kirchoff’s Current Law: • The current entering a node equals the current leaving the node • Kirchoff’s Voltage Law • The sum of the voltages around a loop equal zero

  14. Kirchoff’s Current Law • Conservation of charge: • The charge entering a node equals the charge leaving the node • IB + IC + ID = IA • (IB + IC + ID) – IA =0 • Notation: • Current flowing into a node is negative • Current flowing out of a node is positive http://www.physics.uoguelph.ca

  15. Kirchoff’s Voltage Law • Battery • Rectangles 1 and 2 are resistors. • Voltage drop around the loop is zero • Each component has a current going through it and a voltage across it • How are the voltages related?

  16. Voltage Law + V1 - + V2 - The sum of voltages around the loop equals zero: VB – V1 –V2 = 0 Vb = V1 + V2 +VB-

  17. Kirchoff’s Laws • http://www.youtube.com/watch?v=nEfd-jMWzsU&feature=related • http://www.youtube.com/watch?v=gNQwjdQHOfU&NR=1

  18. Question • If VB = 1.0 volts • R1 = 18 ohms • R2 = 2 ohms • What is V1? And V2?

  19. Ohm’s Law • V = IR • R = 18+2 = 20 • I = V/R = 1/20 = .05 • V1 = IR1 = .05 * 18 = .9 volts • V2 = IR2 = .05 * 2 = .1 volts

  20. Membrane Equation • Examples taken from Dr. David Heeger’s handout (http://www.cns.nyu.edu/~david/handouts/membrane.pdf) • ASSIGNMENT: Read Heeger’s paper Summarize the ideas presented in the paper, including the mathematical development Email your report by Tuesday, Sept 21th

  21. Passive Neural Membrane Model

  22. What’s a Capacitor? • Two conducting plates separated by an insulator

  23. Membrane Model • A resistor and capacitor can be used to model a passive membrane • Cellular membranes are insulators  Capacitor • Sodium-Potassium pump (keeps electrical potential inside the cell below outside potential level  Battery • Current leaks through the membrane Resistor

  24. Mechanical ‘Balloon Model’

  25. Balloon Analogy • Leaky water balloon connected to a water pump • Pump off- balloon is empty • Pump turned on- balloon begins to fill with water • Initially, balloon expands rapidly (force of incoming water greater than elastic force of unexpanded balloon) • Elastic force increases as balloon fills • Balloon fills more slowly

  26. Balloon reaches equilibrium: • Water flowing into balloon from pump equals water flowing out of the holes • Water pressure inside balloon equals pressure exerted by the pump

  27. Balloon – Electrical Relationships • Water volume  electrical charge (units: coulombs) • Rate of water flow into and out of balloon  flow of electrical charge electrical current (units: amp = coul/sec) • Size and number of holes in balloon  resistance (units: ohms), electrical conductance is reciprocal of resistances (siemems = 1/ohms) • Water pressure  electrical potential (units: volts, 1 volts moves 1 amp of current through a 1 siemem conductor)

  28. Dynamic Equilibrium • When the balloon is fully inflated water flowing into and out of the balloon is equal • Neuron at rest • Some current is leaking out of cell • Leakage is balanced by current from sodium-potassium pump • Net current flow into and out of the cell is zero

  29. Net Current in/out of Balloon • Depends on the difference between the pressure in the balloon and the pressure exerted by the pump • When equal, net current is zero and balloon in equilibrium

  30. Balloon Behavior • Express behavior in terms of: • Water volume • Water current (flow) • Water pressure inside of balloon • Size and number of holes • Elasticity of balloon • How does the balloon example relate to the RC electrical circuit model?

  31. Ohm’s Law • V = IR (voltage = current x resistance) • I = V/R = V* g (where g is conductance) • For our model: Ig = g(Vm – E) Where: • Ig is current through the resistor • Vm is the membrane potential (voltage between the inside and outside of the membrane – modeled as voltage drop across the resistor and battery)

  32. A Balloon Stores Water / A Capacitor Stores Charge • The balloon stores water • ‘capacitance’ of the balloon is the volume of water stored per unit of applied pressure. • The capacitance of the balloon depends on the elasticity of the balloon • For a given water pressure an elastic balloon will store more water than a stiff (less elastic) balloon

  33. A Balloon Stores Water / A Capacitor Stores Charge • Electrical capacitor stores charge • The capacitance is the charge per unit applied voltage: C = q/Vm or q = CVm • C is capacitance • q is charge • Vm is the membrane potential

  34. Water / Current Flow • When the pump is first turned on the balloon expands quickly, then more slowly as it finally approaches equilibrium • Water pressure also changes over time – first it is zero, then equals pressure exerted by the pump

  35. For Electrical Model • Voltage across the capacitor and current through the capacitor changes with time • When switch first closed, current is large through the capacitor than decreases as the capacitor is charged • Voltage across the capacitor is first zero (before switch is closed), then increases rapidly at first, at equilibrium the voltage across the capacitor equals the voltage of the battery

  36. Current through Capacitor Ic = dq = d (CVm) = C dVm dt dt dt Ic is the current through the capacitor

  37. Net current in and out of Balloon • Ic net current into the balloon in terms of capacitance and pressure change: • Ic = C dVm/ dt • Ig net current out of the balloon in terms of conductance (g) and pressure difference (Vm – E): • Ig = g(Vm – E)

  38. Net Current In = Net Current Out • Net current in plus net current out = 0 • C dVm/ dt + g(Vm – E) = 0 • First order, ordinary differential equation • Describes full behavior of RC circuit

  39. How does Vm change with time after the switch is closed? • This is the solution to the differential equation • Assume the switch closes at t = 0 • SOLUTION: Vm(t) = u(t) E [1-e-(t/τ)] • Where: τ = C/g = RC • u(t) is zero for t<0 and one for t>0

  40. Verify Solution • Take derivative of solution: dVm/dt = δ(t) E [ 1-e-(t/τ)] + u(t) E (1/τ)e-(t/τ) δ(t) = 1 when t=0, but when t = 0 bracket term is zero, so first term is zero Then: dVm/dt = u(t) E (1/τ)e-(t/τ) = (-g/C )(Vm – u(t) E) For t>0, C dVm/dt = -g(Vm-E)

  41. Current Injection

  42. Current Clamp • Used to inject current into the circuit • Inject current can vary with time • The current can be either positive or negative • Three branches in circuit, each with a current into or out of neuron • Sum of the three currents must be zero

  43. Sum of currents • C dVm/dt + g(Vm-E)-Iinject = 0 • This is a differential equation • Iinject is negative because it flows into the neuron, the other two currents flow out of the neuron • How does the membrane potential Vm vary over time in response to any time-varying injection current? (solution to differential equation)

  44. Solution to DE • Consider a step current • First, assume injection current = 0 long enough so that membrane potential is at rest, Vm(0) = E • Assume current stepped up to some fixed value • The membrane potential will reach a new steady state value after a period of time, Vm(∞) = Vs

  45. Steady State Condition • What is Vs and what is the time-course of the membrane potential as it changes from E to Vs? • In the steady state condition dVm/dt =0 that is, the membrane potential is not changing so its 1st derivative = 0 • Therefore, g(Vs –E) – Iinject = 0 Vs = E + (1/g)Iinject

  46. Time-course • The membrane potential changes exponentially starting with E to final value Vs • Vm(t) = u(t)Vs[1-e-(t/τ)] +u(t)E[e-(t/τ)] • At t=0, current turned on, first term in bracket is 0 and Vm = E • When t>>0, Vm(∞) = Vs • When current turned off, Vm returns to resting potential

  47. Current Steps (different current step sizes) Steady state membrane potential Double conductance, Steady state value reduced Current: on at t=0, off at t=150msec

  48. Post Synaptic Potential (PSP) First action potential Second action potential Several action potentials Gerstner and Kistler Spiking Neuron Models. Single Neurons, Populations, PlasticityCambridge University Press, 2002

  49. If action potentials are not too dense (don’t occur too closely together) the PSPs linearly add together • If the PSP reaches the threshold an action potential is fired by the post-synaptic neuron and the membrane potential returns to its resting state • If the action potentials occur too close together the summation linearity breaks down. When the threshold is reached an action potential is fired, but the neuron enters a stage of hyper-polarization below the resting state.

  50. Models – resistors, capacitors and batteries Switch Battery Resistor V: voltage C: capacitance Vc: voltage across capacitance R: resistance VR: voltage across resistor i: current Capacitor

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