Document for Impedance-class

[eacheon]

It is highly interesting. But for a beginner I am having a hard time understanding some of the formula, say this one
http://www.neuron.yale.edu/neuron/stati ... #Impedance

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...
yields the linearized impedance matrix [(jwc - di/dv)v = i0 ] * exp(jwt) 
...

Is there a better typeset version of these paragraphs as pdf available? I know I should have picked up a EE textbook, but it is better if i have a clearer version of your formula here so it would be faster for me to learn.

[ted]

Is there a better typeset version of these paragraphs as pdf available?

No, but anyone who wants to create one is welcome to. The Programmer's Reference
documentation is generated automatically from plain ASCII files with just enough "markup"
to implement a very minimalist style.

I see one small typographical error near the end of the introductory paragraphs on the
Impedance class: in this line

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in the line
    	(jwc + gnabar*m^3*h)   gnabar*3*m^2*h*(v-ena)   gnavar*m^3*(v-ena)
    	                                            this v ^
    	                                            should be a b, like so:
    	(jwc + gnabar*m^3*h)   gnabar*3*m^2*h*(v-ena)   gnabar*m^3*(v-ena)
    	                                                   ^

[eacheon]

I see one small typographical error near the end of the introductory paragraphs on the
Impedance class: in this line

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in the line
    	(jwc + gnabar*m^3*h)   gnabar*3*m^2*h*(v-ena)   gnavar*m^3*(v-ena)
    	                                            this v ^
    	                                            should be a b, like so:
    	(jwc + gnabar*m^3*h)   gnabar*3*m^2*h*(v-ena)   gnabar*m^3*(v-ena)
    	                                                   ^

Knowing this is nice, Ted, and thanks for the prompt answer. I guess the question that I am stuck on may be something very apparent for many:

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Specifically, the cable equation, c*dv/dt = i(v), where the d2v/dx2 compartmental terms are in i, yields the linearized impedance matrix [(jwc - di/dv)v = i0 ] * exp(jwt) ...

what does "v = i0" mean in the above solution? How does the c*dv/dt = i(v) lead to this solution?

[ted]

Sorry about the delayed reply. I lost track of this thread while traveling.

what does "v = i0" mean in the above solution?

It's not "v = i0". It's
(jwc - di/dv)v = i0

How does the c*dv/dt = i(v) lead to this solution?

It's not a solution. It is an equation that is the equivalent, in the frequency domain, of the
time-domain equation
c*dv/dt = i(v)
where the d2v/dx2 compartmental terms are contained in i. It is derived by applying the
Fourier transformation to the time-domain formulation of the cable equation.