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→Appendix A: Derivation of a temporal filter for short-term depression
</math>
</math>
The aim is to derive a filter<math>\chi</math>that relates the output synaptic current $I$ to the input rate $R$.
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 $R$ 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 $R_1 \rho(t)$ of the firing rate $R(t)$ around a constant rate $R_0$, 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,
<math>
<math>
</math>
</math>
目的是导出一个过滤器 $\chi$，它将输出突触电流 $I$ 与输入速率 $R$ 联系起来。请注意，由于输入速率 $R$ 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 $R_0$ 附近的发射率 $R(t)$ 的小扰动 $R_1 \rho(t)$，即
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 $R$ produce small perturbations in the variable $x$ around its steady state value $x_0>0$ :
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>
<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 \, .
\label{eq:appA_x01}
\label{eq:appA_x01}
</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 $x(t)$ around the steady-state value $x_0$ by approximating the product
We can now linearize the dynamics of $x(t)$ around the steady-state value $x_0$ by approximating the product
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>
</math>
</math>
在方程式中的位置。 \ref{eq:appA_rx} 我们删除了二阶项 $x_1 R_1\rho$，因为我们假设 $R_1\ll R_0$ 和 $|x_1|\ll x_0$。堵塞方程式。 \ref{eq:appA_rx} 转化为等式。 \ref{eq:appA_x} 产生
[数学]\displaystyle{ \begin{eqnarray} {\frac<nowiki>{{\rm d} x}{{\rm d}t}} = \frac{1-x}{\tau_{d}}</nowiki> - U R_0 x - U x_0 R + U x_0 R_0\,.\label{eq:appA_xlin} \end{eqnarray} }[/math]
We now take the Fourier transform at both sides of Eq. \ref{eq:appA_xlin}
We now take the Fourier transform at both sides of Eq. \ref{eq:appA_xlin}
<math>
<math>
where from Eq. \ref{eq:appA_x01} we used $U R_0 \tau_{d}=1/x_0 - 1$.
where from Eq. \ref{eq:appA_x01} we used $U R_0 \tau_{d}=1/x_0 - 1$.
<nowiki>我们现在对等式两边进行傅里叶变换。 \ref{eq:appA_xlin} [数学]\displaystyle{ \begin{eqnarray} j\omega \tau_{d} \widehat{x} = -\widehat{x} - U R_0 \tau_{d} \widehat{x } - U x_0 \tau_{d}\widehat{R} + (1+ U R_0 \tau_{d} x_0) \delta(\omega) \label{eq:appA_xhat0} \end{eqnarray} }[/math]我们定义了傅里叶变换对 [math]\displaystyle{ \begin{eqnarray} \widehat{x}(\omega) := \int \!{\rm d}{t}\, x(t) \exp( -j\omega t ) \quad; \quad x(t) = \frac{1}{2\pi}\int \!{\rm d}\omega\, \widehat{x}(\omega) \exp(j\omega t) \label{ eq:appA_ft} \end{eqnarray} }[/math] 和 $j=\sqrt{-1}$ 是虚数单位。求解方程。 \ref{eq:appA_xhat0} 对于变量 $\widehat{x}$，我们找到 [math]\displaystyle{ \begin{eqnarray} \widehat{x} = -\frac{U\tau_{d}x_0}{ 1/x_0 + j \omega \tau_{d}} \widehat{R} + x_0 (2-x_0) \delta(\omega) \label{eq:appA_xhat} \end{eqnarray} }[/math] 从哪里方程。 \ref{eq:appA_x01} 我们使用了 $U R_0 \tau_{d}=1/x_0 - 1$。</nowiki>
Next, we plug Eq. （11）into Eq. （8） to linearize the dynamics of the synaptic current
<math>
\begin{eqnarray}
I &=& \tau_{s}AU (R_0x+x_0R-x_0R_0)\\
&=& I_0 \left( \frac{x}{x_0}+ \frac{R}{R_0}-1\right) \label{eq:appA_Ilin}
\end{eqnarray}
</math>
around the steady-state value <math>
I_0 = \tau_{s}AU x_0 R_0
</math>.
接下来，我们方程式（11）插入方程式（8）线性化突触电流的动态性
<math>
<math>
\end{eqnarray}
\end{eqnarray}
</math>
</math>
围绕静态值 <math>
I_0 = \tau_{s}AU x_0 R_0
</math>.
By taking the Fourier transform at both sides of Eq. \ref{eq:appA_Ilin}, using Eq. \ref{eq:appA_xhat}, we obtain
By taking the Fourier transform at both sides of Eq. （16）, using Eq. （15）, we obtain
<math>
<math>
\begin{eqnarray}
\begin{eqnarray}
</math>
</math>
To interpret the result, we plug into Eq. \ref{eq:appA_Ihat} the Fourier transform $\widehat{R}=R_0\delta(\omega)+R_1 \widehat{\rho}$,
通过对等式（16）两边进行傅里叶变换。使用等式（15），我们得到 <math>
\begin{eqnarray}
\widehat{I} &=& I_0 \frac{\widehat{x}}{x_0} + I_0 \frac{\widehat{R}}{R_0} - I_0 \delta(\omega) \\
&=& \frac{I_0}{R_0} \widehat{\chi} \widehat{R} + I_0(1-x_0) \delta(\omega)
\label{eq:appA_Ihat}
\end{eqnarray}
</math>，其中我们定义了过滤器<math>
\begin{eqnarray}
\widehat{\chi}(\omega) := 1- \frac{1/x_0 -1}{1/x_0 + j\omega \tau_{d}} = \frac{1+(\tau_{d}\omega)^2x_0+j\omega\tau_{d}(1-x_0)}{1/x_0+(\tau_{d}\omega)^2 x_0}\,.
\label{eq:appA_chihat}
\end{eqnarray}
</math>
To interpret the result, we plug into Eq.（17）the Fourier transform<math>
\widehat{R}=R_0\delta(\omega)+R_1 \widehat{\rho}
</math>,
which yields
which yields
</math>
</math>
Finally, the inverse Fourier transform of Eq. \ref{eq:appA_Ihat_final} reads
为了解释结果，我们插入方程式。 （17）傅里叶变换<math>
\widehat{R}=R_0\delta(\omega)+R_1 \widehat{\rho}
</math>，产生
<math>
\begin{eqnarray}
\widehat{I}(\omega) = I_0 \delta(\omega) + \frac{I_0 R_1}{R_0} \widehat{\chi}(\omega) \widehat{\rho}(\omega)\,.
\label{eq:appA_Ihat_final}
\end{eqnarray}
</math>
Finally, the inverse Fourier transform of Eq.（19）reads
<math>
<math>
\begin{eqnarray}
\begin{eqnarray}
</math>
</math>
<nowiki>最后，等式的傅里叶逆变换。 \ref{eq:appA_Ihat_final} 读取 [math]\displaystyle{ \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] with [math]\displaystyle{ \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\lt 0 \end{cases}\,. \label{eq:appA_chi_final} \end{eqnarray} }[/math]</nowiki>
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>
\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>
<nowiki>通过对等式两边进行傅里叶变换。 \ref{eq:appA_Ilin}，使用等式。 \ref{eq:appA_xhat}，我们得到 [math]\displaystyle{ \begin{eqnarray} \widehat{I} &=& I_0 \frac{\widehat{x}}{x_0} + I_0 \frac{\widehat{ R}}{R_0} - I_0 \delta(\omega) \\ &=& \frac{I_0}{R_0} \widehat{\chi} \widehat{R} + I_0(1-x_0) \delta(\omega ) \label{eq:appA_Ihat} \end{eqnarray} }[/math] 我们定义了过滤器 [math]\displaystyle{ \begin{eqnarray} \widehat{\chi}(\omega) := 1- \frac {1/x_0 -1}{1/x_0 + j\omega \tau_{d}} = \frac{1+(\tau_{d}\omega)^2x_0+j\omega\tau_{d}(1- x_0)}{1/x_0+(\tau_{d}\omega)^2 x_0}\,. \label{eq:appA_chihat} \end{eqnarray} }[/math]</nowiki>
Therefore the output current <math>
I
</math>is the sum of the steady-state current <math>
I_0
</math> and the filtered perturbation <math>
\frac{I_0 R_1}{R_0} \int {\rm d}\tau \, \chi(\tau) \rho(t-\tau)
</math> where <math>
\chi
</math>is the filter we are interested in.
因此，输出电流<math>
I
</math> 是稳态电流 <math>
I_0
</math>和滤波后的扰动之和，其中<math>
\chi
</math>是过滤器。
==参考文献References==
==参考文献References==