Question

How this equation:
I_{s}=\sum_{b=1}^{Nu}w_{b}\int_{-\infty }^{t}d\tau K_{s}(t-\tau)u_{b}(\tau)
differentiates to this:
\tau_{s}\frac{\mathrm{d} I_{s} }{\mathrm{d} t}=-I_{s}+\sum_{b=1}^{Nu}w_{b}u_{b}
given that Ks(t)=exp(-tau/t)/tau

Answer 1

\[I_{s}=\sum_{b=1}^{Nu}w_{b}\int\limits_{-\infty }^{t}d\tau K_{s}(t-\tau)u_{b}(\tau)\]

\[\tau_{s}\frac{\mathrm{d} I_{s} }{\mathrm{d} t}=-I_{s}+\sum_{b=1}^{Nu}w_{b}u_{b}\]

Answer 2

Let me just rewrite this question so it's easier to read:
\[\large I_s=\sum_{b=1}^\nu w_b\int_{-\infty}^t\frac{\exp\left[\frac{-\tau}{t-\tau}\right]}{\tau}u_b(\tau)~d\tau\]
And you're asking how the derivative (with respect to t, apparently) is
\[\large \tau_s\frac{dI_s}{dt}=-I_s+\sum_{b=1}^\nu w_bu_b\]
Am we to assume \(\large u_b\) is the Heaviside step function?

Answer 3

ub() is convolution of pb() with a boxcar filter and pb() is a sum of dirac functions
i don't know what is Heaviside step function, sry!

Answer 4

I don't remember much about convolution, and I've never heard of the boxcar filter you mention. Sorry I can't help!

By the way, the Heaviside step function is
\[u_t(x)=u(x-t)=\begin{cases}1&\text{for }x\ge t\\0&\text{for }x<t\end{cases}\]
You might know it by some other name.

Answer 5

u(t) is the firing rate of a neuron and is continous in time,i think you can assume that it's just an arbitrary function of time!
About Heaviside,yes it is similar to Delta dirac function
btw, boxcar filter is a simple filter which is 1 in [a,b] and zero otherwise

Answer 6

what is \(K_s(t) \) ?

Answer 7

@phi, it's the given as exponential function in the integral.
\[\large K_s(t)=\frac{\exp\left[\frac{-\tau}{t}\right]}{\tau}\]

Answer 8

Do we know anything about \( w_b(t)\) ?

in the meantime, I think we can use this
http://mathmistakes.info/facts/CalculusFacts/learn/doi/doi.html

Answer 9

w is constant