[未答复] 二阶低通滤波推导的公式波特图与预期不符

zengxy3407 发表于 2022-3-29 22:43:11

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本帖最后由 zengxy3407 于 2022-3-29 23:00 编辑

求助各位路过的大神，我在推导二阶低通滤波的差分方程时碰到以下问题求解答：
大家都知道，二阶低通滤波器的标准传递函数如下：
$\frac{\omega_c^2}{s^2+2\xi\omega_cs+\omega_c^2}$

用反向差分法离散化，代入反向差分公式：
$s=\frac{1-z^{-1}}{T}$

然后简化，过程如下：
$G(z)=\frac{Y(z)}{X(z)}=\frac{\omega_c^2}{\left(\frac{1-z^{-1}}{T}\right)^2+2\xi\omega_c \frac{1-z^{-1}}{T} +\omega^2_c }$
$=\frac{\omega_c^2 T^2 }{(1-z^{-1})^2+2\xi\omega_c (1-z^{-1})T +\omega^2_c T^2}$
$=\frac{\omega_c^2 T^2 }{(1-2z^{-1}+z^{-2})+2\xi\omega_c (1-z^{-1})T +\omega^2_c T^2 }$
$=\frac{\omega_c^2 T^2}{z^{-2}-(2+2\xi\omega_c T)z^{-1}+1+2\xi\omega_cT +\omega^2_c T^2 }$

上下同乘以z^2:
$G(z)=\frac{\omega_c^2 T^2 z^2}{1-(2+2\xi\omega_c T)z+(1+2\xi\omega_cT +\omega^2_c T^2)z^2 }$

然后用MATLAB画bode图验证：

fs = 16000;
fc = 5000;
xi = 1/sqrt(2);
omgc = fc * 2 * pi;
T = 1/fs;
omgcT = omgc * T;
a2 = 1 + 2 * xi * omgcT + omgcT * omgcT;
a1 = -(2 + 2 * xi *omgcT);
a0 = 1;
b2 = omgcT * omgcT;
b1 = 0;
b0 = 0;
N = [b2 b1 b0]
D = [a2 a1 a0]
sys = tf(N, D, T)
bode(sys)
hold on
sysc = tf([omgc*omgc], [1 2*xi*omgc, omgc*omgc])
bode(sysc)
sysd = c2d(sysc, T)
bode(sysd)

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得到的结果跟MATLAB计算的不一样：

N =
3.8553 0 0
D =
7.6321 -4.7768 1.0000
sys =
3.855 z^2
-----------------------
7.632 z^2 - 4.777 z + 1
Sample time: 6.25e-05 seconds
Discrete-time transfer function.
sysc =
9.87e08
--------------------------
s^2 + 4.443e04 s + 9.87e08
Continuous-time transfer function.
sysd =
0.7094 z + 0.2623
------------------------
z^2 - 0.0905 z + 0.06224
Sample time: 6.25e-05 seconds
Discrete-time transfer function.

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得到的波特图也完全不对,连截止频率都差很多，这是为什么？

我还试了一下Matlab的zoh，得到的结果如下：

>> sysc = tf([1],[1,0])
sysc =
1
-
s
Continuous-time transfer function.
>> sysd = c2d(sysc, 0.1)
sysd =
0.1
-----
z - 1
Sample time: 0.1 seconds
Discrete-time transfer function.

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其实就是前向差分法， $s=\frac{z-1}{T}$ , 我用同样的方法试了也不对：
$G(z)=\frac{Y(z)}{X(z)}=\frac{\omega_c^2}{\left(\frac{z-1}{T}\right)^2+2\xi\omega_c \frac{z-1}{T} +\omega^2_c }$
$=\frac{\omega_c^2 T^2 }{(z-1)^2+2\xi\omega_cT (z-1) +\omega^2_c T^2}$
$=\frac{\omega_c^2 T^2 }{(1-2z+z^2)+2\xi\omega_cT (z-1) +\omega^2_c T^2 }$
$=\frac{\omega_c^2 T^2 }{z^2-(2-2\xi\omega_c T)z+1-2\xi\omega_cT +\omega^2_c T^2 }$