Translation to alpha synapse equation

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1

Hello,

I want to ask basic question how to translate the equation of alpha function.
When we think alpha function below;
image ![image|365x63]

Then differentiate g(t) with respect to t,
image

Then, define x = t/tau, resulted dx/dt=1/tau.

However, it looks incorrect since the result doesn’t match the example of Brian2 website.

How can we introduce the equation used in Brian implement??

Thank you very much for your kindness for me in advance!

2

Hi. This looks almost correct to me, but in the last step you should rather get
\frac{dg}{dt} = \frac{1}{\tau}\left( e^\frac{-t}{\tau} - g\right). Then, by setting x = e^\frac{-t}{\tau} you get

\frac{dg}{dt} = \frac{1}{\tau}\left(x - g\right)\\ \frac{dx}{dt} = -\frac{x}{\tau}

which are the equations we give in the Brian documentation.

3

Hi Maecel,

Thank you so much for your answer! That’s exactly what I want to know!

May I ask a related question?
When we think about biexponential synapse equation, can we do the same strategy? Are there some reference describing the process of biexponential synapse equation?

4

Yes, the same approach should also work for biexponential synapses. We don’t have the process detailed anywhere, but the result is at the bottom of the same documentation page I linked earlier.

5

Hi Marcel,

Thank you so much for your teaching and kindness!!

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