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Signed-off-by: David Rotermund <54365609+davrot@users.noreply.github.com>
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@ -7,4 +7,4 @@ $$\tau_e \frac{dA_e(t)}{dt} = -A_e(t) + g_e\left( \frac{I(t)}{A_i(t)+\sigma} \ri
$$\tau_i \frac{dA_i(t)}{dt} = -A_i(t) + g_i\left( I(t) \right)$$ $$\tau_i \frac{dA_i(t)}{dt} = -A_i(t) + g_i\left( I(t) \right)$$
Here, $g_X$ are gain functions for $x\in\{e,i\}$ with $g_X(I) = m_X(I-\theta_X)$ for $I>\theta_X$, and $0$ otherwise, while $A_e$ and $A_i$ could be interpreted as internal activations. Here, $g_X$ are gain functions for $x\in\{e,i\}$ with $g_X(I) = m_X(I-\theta_X)$ for $I>\theta_X$, and $0$ otherwise, while $A_e$ and $A_i$ could be interpreted as internal activations.
Default parameters: $\tau_e=10$ ms, $\tau_i=40$ ms, $\theta_{\{e,i\}}=0$, $m_e=m_i=1$ (nA)${}^{-1}$, $I=1$ nA, $\sigma=0.25$. From the activation $A_e$, an output rate can be derived via $r(t) = r_0 A_e(t)$ with, let's say, $r_0=100$ Hz. Default parameters: $\tau_e=10$ ms, $\tau_i=40$ ms, $\theta_{\{e,i\}}=0$, $m_e=m_i=1 nA^{-1}$, $I=1$ nA, $\sigma=0.25$. From the activation $A_e$, an output rate can be derived via $r(t) = r_0 A_e(t)$ with, let's say, $r_0=100$ Hz.