参考链接：一阶RC低通数字滤波器算法实现
一阶RC无源低通滤波器电路图如下图所示
从时域角度可以列写出输出电压与输入电压的关系：
\[\left\{ {\begin{array}{*{20}{c}}
{{V_{in}} - {V_{out}} = R \cdot i}\\
{i = C \cdot \frac{{d{V_{out}}}}{{dt}}}
\end{array}} \right.
\]
下式带入上式得到：
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RC \cdot \frac{{d{V_{out}}}}{{dt}} + {V_{out}} = {V_{in}}
\]
采用后（前）向差分，对上式进行离散化：
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RC \cdot \frac{{{V_o}\left( k \right) - {V_o}\left( {k - 1} \right)}}{{{T_s}}} + {V_o}\left( k \right) = {V_i}\left( k \right)
\]
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RC \cdot \frac{{{V_o}\left( {k + 1} \right) - {V_o}\left( k \right)}}{{{T_s}}} + {V_o}\left( k \right) = {V_i}\left( k \right)
\]
化简得到：
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{V_o}\left( k \right) = \frac{{RC}}{{RC + {T_s}}} \cdot {V_o}\left( {k - 1} \right) + \frac{{{T_s}}}{{RC + {T_s}}} \cdot {V_i}\left( k \right)
\]
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{V_o}\left( {k + 1} \right) = \left( {1 - \frac{{{T_s}}}{{RC}}} \right) \cdot {V_o}\left( k \right) + \frac{{{T_s}}}{{RC}} \cdot {V_i}\left( k \right) = {V_o}\left( k \right) + \frac{{{T_s}}}{{RC}}\left[ {{V_i}\left( k \right) - {V_o}\left( k \right)} \right]
\]
其中：
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RC = \frac{1}{{2\pi {f_{cut}}}}
\]
1. 已知截止频率，求系数
将RC表达式带入输入输出表达式中，得到下式：
\[{V_o}\left( k \right) = \frac{1}{{1 + 2\pi {f_{cut}} \cdot {T_s}}} \cdot {V_o}\left( {k - 1} \right) + \frac{{2\pi {f_{cut}} \cdot {T_s}}}{{1 + 2\pi {f_{cut}} \cdot {T_s}}} \cdot {V_i}\left( k \right)
\]
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{V_o}\left( k \right) = \left( {1 - 2\pi {f_{cut}} \cdot {T_s}} \right) \cdot {V_o}\left( {k - 1} \right) + 2\pi {f_{cut}} \cdot {T_s} \cdot {V_i}\left( k \right)
\]
2. 已知系数，求截止频率
按照上式，所以两个系数表达式分别为：
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\left\{ {\begin{array}{*{20}{c}}
{co{f_1} = \frac{1}{{1 + 2\pi {f_{cut}} \cdot {T_s}}}}\\
{co{f_2} = \frac{{2\pi {f_{cut}} \cdot {T_s}}}{{1 + 2\pi {f_{cut}} \cdot {T_s}}}}
\end{array}} \right.
\]
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\left\{ {\begin{array}{*{20}{c}}
{co{f_1} = 1 - 2\pi {f_{cut}} \cdot {T_s}}\\
{co{f_2} = 2\pi {f_{cut}} \cdot {T_s}}
\end{array}} \right.
\]
若采样频率为Ts，所以截止频率为
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{f_{cut}} = \frac{{1 - co{f_1}}}{{2\pi {T_s} \cdot co{f_1}}} = \frac{{co{f_2}}}{{2\pi {T_s} \cdot \left( {1 - co{f_2}} \right)}}
\]
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{f_{cut}} = \frac{{1 - co{f_1}}}{{2\pi {T_s}_1}} = \frac{{co{f_2}}}{{2\pi {T_s}}}
\]