Frequency separation by an excitatory-inhibitory network

Abstract

We consider a situation in which individual features of the input are represented in the neural system by different frequencies of periodic firings. Thus, if two of the features are presented concurrently, the input to the system will consist of a superposition of two periodic trains. In this paper we present an algorithm that is capable of extracting the individual features from the composite signal by separating the signal into periodic spike trains with different frequencies. We show that the algorithm can be implemented in a biophysically based excitatory-inhibitory network model. The frequency separation process works over a range of frequencies determined by time constants of the model’s intrinsic variables. It does not rely on a “resonance” phenomenon and is not tuned to a discrete set of frequencies. The frequency separation is still reliable when the timing of incoming spikes is noisy.

Appendix

Equations for an I cell

$$ C_m \frac{d V}{d t}=-I_{\rm ion}(V)-I_{\rm syn}+I_{\rm input}. $$

The intrinsic current I _ion consists of the leak, sodium, and potassium currents, and the bias current I:

$$\begin{array}{rll} I_{\rm ion}&=&g_L(V-V_L)+g_{Na}(m_\infty(V))^3h(V-V_{Na})\\&&+\,g_K(1-h)^4(V-V_K)-I,\\ \frac{dh}{dt}&=&(h_\infty(V)-h)/\tau_h(V), \end{array}$$

where

$$\begin{array}{rll} m_\infty(V)&=&1/(1+\exp(-(V+37)/7)),\\ h_\infty(V)&=&1/(1+\exp((V+41)/4)),\\ \tau_h(V) &=& 0.69/(\alpha_h(V)+\beta_h(V)),\\ \alpha_h(V) &=& 0.128\exp(-(46+V)/18),\\ \beta_h(V) &=& 4/(1+\exp(-(23+V)/5)). \end{array}$$

The parameters are given in Table 1.

Equations for an E cell

$$ C_m \frac{d V}{d t}=-I_{\rm ion}(V)-I_{\rm syn}+I_{\rm input}. $$

The intrinsic current I _ion consists of the leak, sodium, and potassium currents, and the T-type current—outward current activated by hyperpolarization:

$$\begin{array}{rll} I_{\rm ion}&=&g_L(V-V_L)+g_{Na}(m_\infty(V))^3h(V-V_{Na})\\ &&+\,g_K(1-h)^4(V-V_K)+g_T(m_{\infty,T})^2h_TV,\\ \frac{dh}{dt}&=&(h_\infty(V)-h)/\tau_h(V),\\ \frac{dh_T}{dt}&=&(h_{\infty,T}(V)-h)/\tau_{h,T}(V), \end{array}$$

where

$$\begin{array}{rll} m_\infty(V)&=&1/(1+\exp(-(V+37)/7)),\\ m_{\infty,T}(V)&=&1/(1+\exp(-(V+60)/6.2)),\\ h_\infty(V)&=&1/(1+\exp((v+41)/4)),\\ h_{\infty,T}(V)&=&1/(1+\exp((v+84)/4)),\\ \tau_h(V) &=& 0.83/(\alpha_h(V)+\beta_h(V)),\\ \tau_{h,T}(V) &=& 28-\exp((V+25)/10.5),\\ \alpha_h(V) &=& 0.128*\exp(-(46+V)/18),\\ \beta_h(V) &=& 4/(1+\exp(-(23+V)/5)). \end{array}$$

External input to the E cell is equal to zero (I _input = 0). Parameter values are given in Table 1.