# 雙線性轉換

## 離散時間估計

{\displaystyle {\begin{aligned}z&=e^{sT}\\&={\frac {e^{sT/2}}{e^{-sT/2}}}\\&\approx {\frac {1+sT/2}{1-sT/2}}\end{aligned}}}

{\displaystyle {\begin{aligned}s&={\frac {1}{T}}\ln(z)\\&={\frac {2}{T}}\left[{\frac {z-1}{z+1}}+{\frac {1}{3}}\left({\frac {z-1}{z+1}}\right)^{3}+{\frac {1}{5}}\left({\frac {z-1}{z+1}}\right)^{5}+{\frac {1}{7}}\left({\frac {z-1}{z+1}}\right)^{7}+\cdots \right]\\&\approx {\frac {2}{T}}{\frac {z-1}{z+1}}\\&={\frac {2}{T}}{\frac {1-z^{-1}}{1+z^{-1}}}\end{aligned}}}

${\displaystyle s\leftarrow {\frac {2}{T}}{\frac {z-1}{z+1}}.}$

${\displaystyle H_{d}(z)=H_{a}(s){\bigg |}_{s={\frac {2}{T}}{\frac {z-1}{z+1}}}=H_{a}\left({\frac {2}{T}}{\frac {z-1}{z+1}}\right).\ }$

## 例子

{\displaystyle {\begin{aligned}H_{a}(s)&={\frac {1/sC}{R+1/sC}}\\&={\frac {1}{1+RCs}}.\end{aligned}}}

 ${\displaystyle H_{d}(z)\ }$ ${\displaystyle =H_{a}\left({\frac {2}{T}}{\frac {z-1}{z+1}}\right)\ }$ ${\displaystyle ={\frac {1}{1+RC\left({\frac {2}{T}}{\frac {z-1}{z+1}}\right)}}\ }$ ${\displaystyle ={\frac {1+z}{(1-2RC/T)+(1+2RC/T)z}}\ }$ ${\displaystyle ={\frac {1+z^{-1}}{(1+2RC/T)+(1-2RC/T)z^{-1}}}.\ }$

## 一般的雙二階變換

${\displaystyle H_{a}(s)={\frac {b_{0}s^{2}+b_{1}s+b_{2}}{a_{0}s^{2}+a_{1}s+a_{2}}}={\frac {b_{0}+b_{1}s^{-1}+b_{2}s^{-2}}{a_{0}+a_{1}s^{-1}+a_{2}s^{-2}}}}$

${\displaystyle s\leftarrow K{\frac {1-z^{-1}}{1+z^{-1}}}}$

${\displaystyle H_{d}(z)={\frac {(b_{0}K^{2}+b_{1}K+b_{2})+(2b_{2}-2b_{0}K^{2})z^{-1}+(b_{0}K^{2}-b_{1}K+b_{2})z^{-2}}{(a_{0}K^{2}+a_{1}K+a_{2})+(2a_{2}-2a_{0}K^{2})z^{-1}+(a_{0}K^{2}-a_{1}K+a_{2})z^{-2}}}}$

${\displaystyle H_{d}(z)={\frac {{\frac {b_{0}K^{2}+b_{1}K+b_{2}}{a_{0}K^{2}+a_{1}K+a_{2}}}+{\frac {2b_{2}-2b_{0}K^{2}}{a_{0}K^{2}+a_{1}K+a_{2}}}z^{-1}+{\frac {b_{0}K^{2}-b_{1}K+b_{2}}{a_{0}K^{2}+a_{1}K+a_{2}}}z^{-2}}{1+{\frac {2a_{2}-2a_{0}K^{2}}{a_{0}K^{2}+a_{1}K+a_{2}}}z^{-1}+{\frac {a_{0}K^{2}-a_{1}K+a_{2}}{a_{0}K^{2}+a_{1}K+a_{2}}}z^{-2}}}.}$

{\displaystyle {\begin{aligned}y[n]={}&{\frac {b_{0}K^{2}+b_{1}K+b_{2}}{a_{0}K^{2}+a_{1}K+a_{2}}}\cdot x[n]+{\frac {2b_{2}-2b_{0}K^{2}}{a_{0}K^{2}+a_{1}K+a_{2}}}\cdot x[n-1]\\[8pt]&{}+{\frac {b_{0}K^{2}-b_{1}K+b_{2}}{a_{0}K^{2}+a_{1}K+a_{2}}}\cdot x[n-2]-{\frac {2a_{2}-2a_{0}K^{2}}{a_{0}K^{2}+a_{1}K+a_{2}}}\cdot y[n-1]\\[8pt]&{}-{\frac {a_{0}K^{2}-a_{1}K+a_{2}}{a_{0}K^{2}+a_{1}K+a_{2}}}\cdot y[n-2].\end{aligned}}}

## 頻率扭曲

${\displaystyle H_{d}(z)=H_{a}\left({\frac {2}{T}}{\frac {z-1}{z+1}}\right)\ }$
 ${\displaystyle H_{d}(e^{j\omega T})\ }$ ${\displaystyle =H_{a}\left({\frac {2}{T}}{\frac {e^{j\omega T}-1}{e^{j\omega T}+1}}\right)\ }$ ${\displaystyle =H_{a}\left({\frac {2}{T}}\cdot {\frac {e^{j\omega T/2}\left(e^{j\omega T/2}-e^{-j\omega T/2}\right)}{e^{j\omega T/2}\left(e^{j\omega T/2}+e^{-j\omega T/2}\right)}}\right)\ }$ ${\displaystyle =H_{a}\left({\frac {2}{T}}\cdot {\frac {\left(e^{j\omega T/2}-e^{-j\omega T/2}\right)}{\left(e^{j\omega T/2}+e^{-j\omega T/2}\right)}}\right)\ }$ ${\displaystyle =H_{a}\left(j{\frac {2}{T}}\cdot {\frac {\left(e^{j\omega T/2}-e^{-j\omega T/2}\right)/(2j)}{\left(e^{j\omega T/2}+e^{-j\omega T/2}\right)/2}}\right)\ }$ ${\displaystyle =H_{a}\left(j{\frac {2}{T}}\cdot {\frac {\sin(\omega T/2)}{\cos(\omega T/2)}}\right)\ }$ ${\displaystyle =H_{a}\left(j{\frac {2}{T}}\cdot \tan \left(\omega T/2\right)\right)\ }$

${\displaystyle \omega _{a}={\frac {2}{T}}\tan \left(\omega {\frac {T}{2}}\right)}$

${\displaystyle \omega ={\frac {2}{T}}\arctan \left(\omega _{a}{\frac {T}{2}}\right).}$

${\displaystyle -\infty <\omega _{a}<+\infty \ }$

${\displaystyle -{\frac {\pi }{T}}<\omega <+{\frac {\pi }{T}}.\ }$

${\displaystyle s\leftarrow {\frac {\omega _{0}}{\tan({\frac {\omega _{0}T}{2}})}}{\frac {z-1}{z+1}}.}$

## 參考資料

1. ^ Oppenheim, Alan. Discrete Time Signal Processing Third Edition. Upper Saddle River, NJ: Pearson Higher Education, Inc. 2010: 504. ISBN 978-0-13-198842-2.