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第204行： − 我们考虑方程式中基于速率的动态。方程式（3）用于抑郁症主导的突触 <math>u^+ \approx U</math>和比抑郁症动力学快得多的突触反应 (<math>\tau_s \ll \tau_d</math>):+ 第215行：
第215行： + + 第221行：
第223行： − 目的是导出一个过滤器 <math>\chi</math>，它将输出突触电流 <math>I</math>与输入速率 <math>R</math> 联系起来。请注意，由于输入速率 <math>R</math>以乘法方式进入方程，因此输入-输出传递函数是非线性的。然而，线性滤波器可以通过考虑在恒定速率 <math>R_0</math>附近的发射率<math>R(t)</math>的小扰动 <math>R_1 \rho(t)</math>，即 + − − <math> − R(t):=R_0 + R_1 \rho (t)\, \quad\text{with}\quad R_0,R_1>0 \quad\text{and}\quad R_1\ll R_0 \, . − \label{eq:appA_pert} − − We assume that such small perturbations in <math>R</math>produce small perturbations in the variable<math>x</math>around its steady state value<math>x_0>0</math>:+ 第234行：
第231行： − + − We can now linearize the dynamics of $x(t)$ around the steady-state value $x_0$ by approximating the product + − − 第361行：
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第364行： + + + + − \begin{eqnarray} − \chi(t)=\delta(t) - \frac{1/x_0-1}{\tau_{d}} \begin{cases} \displaystyle {\exp\left(-\frac{t}{x_0\tau_{d}}\right)} & \text{for}\quad t\ge0 \\ 0 & \text{for}\quad t<0 \end{cases}\,. − \label{eq:appA_chi_final} − \end{eqnarray} − </math> − − 最后，等式（19）的傅里叶逆变换读取 <math> − \begin{eqnarray} − I(t) = I_0 + \frac{I_0 R_1}{R_0} \int {\rm d}\tau \, \chi(\tau) \rho(t-\tau) − \label{eq:appA_I_final} − \end{eqnarray} − </math>以及<math>
→Appendix A: Derivation of a temporal filter for short-term depression
We consider the rate-based dynamics in Eq. （3）for depression-dominated synapses (<math>u^+ \approx U</math>) and for synaptic responses that are much faster than the depression dynamics (<math>\tau_s \ll \tau_d</math>):
We consider the rate-based dynamics in Eq. （3）for depression-dominated synapses (<math>u^+ \approx U</math>) and for synaptic responses that are much faster than the depression dynamics (<math>\tau_s \ll \tau_d</math>):
我们考虑方程式中基于速率的动态。等式（3）用于抑郁症主导的突触 <math>u^+ \approx U</math>和比抑郁症动力学快得多的突触反应 (<math>\tau_s \ll \tau_d</math>):
<math>
<math>
The aim is to derive a filter<math>\chi</math>that relates the output synaptic current <math>I</math>to the input rate<math>R</math>.
The aim is to derive a filter<math>\chi</math>that relates the output synaptic current <math>I</math>to the input rate<math>R</math>.
Note that because the input rate<math>R</math>enters the equations in a multiplicative fashion the input-output transfer function is non linear. Yet a linear filter can be derived by considering small perturbations <math>R_1 \rho(t)</math>of the firing rate <math>R(t)</math>around a constant rate <math>R_0</math>, that is,
Note that because the input rate<math>R</math>enters the equations in a multiplicative fashion the input-output transfer function is non linear. Yet a linear filter can be derived by considering small perturbations <math>R_1 \rho(t)</math>of the firing rate <math>R(t)</math>around a constant rate <math>R_0</math>, that is,
目的是导出一个过滤器 <math>\chi</math>，它将输出突触电流 <math>I</math>与输入速率 <math>R</math> 联系起来。请注意，由于输入速率 <math>R</math>以乘法方式进入方程，因此输入-输出传递函数是非线性的。然而，线性滤波器可以通过考虑在恒定速率 <math>R_0</math>附近的发射率<math>R(t)</math>的小扰动 <math>R_1 \rho(t)</math>，即
<math>
<math>
</math>
</math>
We assume that such small perturbations in <math>R</math>produce small perturbations in the variable<math>x</math>around its steady state value<math>x_0>0</math>
</math>
我们假设 <math>R</math>中的这种小扰动会在变量<math>x</math>中围绕其稳态值<math>x_0>0</math>产生小的扰动:
<math>
<math>
x(t) = x_0 + x_1(t)\quad\text{with}\quad x_0 = \frac{1}{1+UR_0\tau_{d}} \quad\text{and}\quad |x_1(t)| \ll x_0 \, .
x(t) = x_0 + x_1(t)\quad\text{with}\quad x_0 = \frac{1}{1+UR_0\tau_{d}} \quad\text{and}\quad |x_1(t)| \ll x_0 \, .
</math>
</math>
<nowiki>我们假设 $R$ 中的这种小扰动会在变量 $x$ 中围绕其稳态值 $x_0>0$ 产生小的扰动： [math]\displaystyle{ x(t) = x_0 + x_1(t)\quad\ text{with}\quad x_0 = \frac{1}{1+UR_0\tau_{d}} \quad\text{and}\quad |x_1(t)| \ll x_0 \, . \label{eq:appA_x01} }[/math]</nowiki>
We can now linearize the dynamics of <math>x(t)</math> around the steady-state value <math>x_0</math>by approximating the product
我们现在可以通过近似乘积将 <math>x(t)</math>的动态线性化为围绕稳态值<math>x_0</math>
我们现在可以通过近似乘积将 $x(t)$ 的动态线性化为围绕稳态值 $x_0$
<math>
<math>
Finally, the inverse Fourier transform of Eq.（19）reads
Finally, the inverse Fourier transform of Eq.（19）reads
最后，等式（19）的傅里叶逆变换读取
<math>
<math>
\begin{eqnarray}
\begin{eqnarray}
\end{eqnarray}
\end{eqnarray}
</math>
</math>
with
with
以及
<math>
<math>
\begin{eqnarray}
\begin{eqnarray}
\chi(t)=\delta(t) - \frac{1/x_0-1}{\tau_{d}} \begin{cases} \displaystyle {\exp\left(-\frac{t}{x_0\tau_{d}}\right)} & \text{for}\quad t\ge0 \\ 0 & \text{for}\quad t<0 \end{cases}\,.
\chi(t)=\delta(t) - \frac{1/x_0-1}{\tau_{d}} \begin{cases} \displaystyle {\exp\left(-\frac{t}{x_0\tau_{d}}\right)} & \text{for}\quad t\ge0 \\ 0 & \text{for}\quad t<0 \end{cases}\,.