不好意思，这几天比较忙，希望你还在关注。
我用matlab测试了一下，他的公式有些问题，也有可能我程序哪里看错了。
表中的结果应该是近似过的
(ra//rb)*c*(1+A) 为第一级低通滤波
c*rc/(1+A) 为第二级高通滤波
中心频率就是(high+low)/2
分析和前面的帖子一样。
你把下面的程序贴到matlab里，运行一下，看Mid_Freq_Hz就可以了
C(1) = 0.1e-6;
C(2) = 0.1e-6;
C(3) = 0.047e-6;
C(4) = 0.0022e-6;
C(5) = 0.0022e-6;
Ra(1) = 11e3;
Ra(2) = 2.7e3;
Ra(3) = 1.5e3;
Ra(4) = 7.5e3;
Ra(5) = 2e3;
Rb(1) = 27e3;
Rb(2) = 6.3e3;
Rb(3) = 3.3e3;
Rb(4) = 18e3;
Rb(5) = 4.3e3;
Rc(1) = 91e3;
Rc(2) = 22e3;
Rc(3) = 11e3;
Rc(4) = 63e3;
Rc(5) = 15e3;
A(1) = 4.1;
A(2) = 4.1;
A(3) = 3.7;
A(4) = 4.2;
A(5) = 4.2;
Low_pass = zeros(5,1);
High_pass = zeros(5,1);
Mid_Freq = zeros(5,1);
Mid_Freq_Hz = zeros(5,1);
i=1;
while i<=5
ra = Ra(i);
rb = Rb(i);
rc = Rc(i);
c = C(i);
a = A(i);
Low_pass(i) = ((ra*rb)/(ra+rb))*c*(1+a);
High_pass(i) = c*rc/(1+a);
Mid_Freq(i) = (High_pass(i)+Low_pass(i))/2;
%Mid_Freq(i) = c*(ra*rb/(ra+rb))*rc;
Mid_Freq_Hz(i) = 1/(Mid_Freq(i)*6.28);
i = i+1;
end
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