Questions tagged «math»

5
使用双线性变换产生的数学问题
因此,这与《食谱》有关,我大概在20年前尝试解决它,放弃了,并想起了未解决的问题。但这很直截了当,但我仍然陷入困境。 这是一个具有谐振频率和谐振的简单带通滤波器(BPF):Ω0Ω0\Omega_0QQQ H(s)=1QsΩ0(sΩ0)2+1QsΩ0+1H(s)=1QsΩ0(sΩ0)2+1QsΩ0+1 H(s) = \frac{\frac{1}{Q}\frac{s}{\Omega_0}}{\left(\frac{s}{\Omega_0}\right)^2 + \frac{1}{Q}\frac{s}{\Omega_0} + 1} 在共振频率 |H(jΩ)|≤H(jΩ0)=1|H(jΩ)|≤H(jΩ0)=1 |H(j\Omega)| \le H(j\Omega_0) = 1 并定义上下带限,以便 |H(jΩU)|2=∣∣H(jΩ02BW/2)∣∣2=12|H(jΩU)|2=|H(jΩ02BW/2)|2=12 |H(j\Omega_U)|^2 = \left|H\left(j\Omega_0 2^{BW/2} \right)\right|^2 = \tfrac12 |H(jΩL)|2=∣∣H(jΩ02−BW/2)∣∣2=12|H(jΩL)|2=|H(jΩ02−BW/2)|2=12 |H(j\Omega_L)|^2 = \left|H\left(j\Omega_0 2^{-BW/2} \right)\right|^2 = \tfrac12 我们称这些为“半功率带隙”。因为我们是音频,所以我们以八度为单位定义带宽,在模拟世界中,以八度为单位的与有关,如下所示:BWBWBWQQQ 1Q=2BW−12BW−−−−√=2sinh(ln(2)2BW)1Q=2BW−12BW=2sinh⁡(ln⁡(2)2BW) \frac{1}{Q} = \frac{2^{BW} - 1}{\sqrt{2^{BW}}} = 2 \sinh\left( \tfrac{\ln(2)}{2} BW \right) 我们正在使用双线性变换(具有预先扭曲的谐振频率),其映射为: sΩ0jΩΩ0←1tan(ω0/2)1−z−11+z−1←jtan(ω/2)tan(ω0/2)sΩ0←1tan⁡(ω0/2)1−z−11+z−1jΩΩ0←jtan⁡(ω/2)tan⁡(ω0/2)\begin{align} \frac{s}{\Omega_0} …
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