双一次変換
s = 1 T ln  ( z ) = 2 T tanh − 1  z − 1 z + 1 = 2 T [ z − 1 z + 1 + 1 3 ( z − 1 z + 1 ) 3 + 1 5 ( z − 1 z + 1 ) 5 + 1 7 ( z − 1 z + 1 ) 7 + ⋯ ] ≈ 2 T z − 1 z + 1 ≈ 2 T 1 − z − 1 1 + z − 1 {\displaystyle {\begin{aligned}s&={\frac {1}{T}}\ln(z)\\&={\frac {2}{T}}\tanh ^{-1}{\frac {z-1}{z+1}}\\&={\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}}\\&\approx {\frac {2}{T}}{\frac {1-z^{-1}}{1+z^{-1}}}\end{aligned}}}