8/18/2019 Equations and Summary

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Previous semester’s work 

 The goal of my previous semester’s project is to simulate short term

plasticity in a synapse. The input is a current into the postsynaptic cell. The

output is the post-synaptic voltage.

 The post-synaptic neuron uses the integrate-and-re model, because actionpotentials were not the main focus. We only needed to know when an action

potential occurs, not how. o the e!uation for action potentials is"

 I (t )=V (t )− E leak 

 R  +C 

 dV (t )dT 

  (1)

Where  I  (t  )  is the input, either in the form of an injected current, or

synaptic current. #urrent injection is straightforward" the function $%t& in this

case is described by pulse length, and number of pulses.

When  I (t )  is the synaptic current, it is the result of action potentials in the

previous synapse, and needs to be calculated"

 I syn (t )=(V  ( t )− E syn) P s(t )(2)

Where  Esyn  is the synaptic voltage %a known parameter&, and  Ps(t  )  is the

synaptic conductance, measured as the probability of synaptic vesicle release.

 Ps(t  )  is determined as"

 Ps(t )= PmaxB ( y1( t )− y2(t )) (3)

Where  y1(t )  is the probability of vesicle opening, and  y2(t )  is the probability of 

vesicle closing. These two values are found iteratively"

 y1(t +1)= y

1 ( t )+0.01( y 1 (t )τ 

1)(4)

 y2 (t +1)= y2 ( t )+0.01(

 y 2 (t )τ 2 )(

5)

 The value ' is a normali(ation constant to ensure the ma)imum value of 

 Ps(t )  is *.

8/18/2019 Equations and Summary

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B=(( τ 2τ 1)τ riseτ 1 −( τ 2τ 

1)τ riseτ 2 )

−1

(6)

+nd rise is the rise time of the synapse, calculate as"

τ rise=  τ 

1τ 2

τ 1−τ 2(7)

inally, in order to simulate short term plasticity,  Ps(t  )  is modied by values

/ and , respectively synaptic depression, and synaptic facilitation. These are the

plasticity variables, found to be"

 D (t +1 )=d D(t )(8)

 F (t +1

)= F (t  )+ f (9

)

+fter the action potential, both / and recover e)ponentially back to * as

described by the following di0erential e!uation1"

τ  DdD (t )dt 

  =1− D ( t ) (10)

τ  F dF ( t )dt 

  =1− F (t )(11)

2ore details on the implementation of these e!uations, as well as graphs,are found in the report.