xi = [0 0.2 0.4 0.6 0.8 1.0 2.0 3.0]; wn = sqrt(1); num = wn^2; den = [1 2*xi*wn wn^2]; t = 0:0.01:10; sys = tf(num,den); for i = 1:length(xi) den = [1 2*xi(i)*wn wn^2]; sys = tf(num,den); [y,t] = step(sys,t); overshot = max(y)-1; [p,t_p] = max(y); t_p = t(t_p); t_s = 4/(xi(i)*wn); if xi(i) == 1 t_s = 3/(xi(i)*wn); end完善剩下的代码
时间: 2023-06-10 10:06:19 浏览: 159
disp(['xi = ', num2str(xi(i))]); disp(['Overshoot = ', num2str(overshot)]); disp(['Peak time = ', num2str(t_p)]); disp(['Settling time = ', num2str(t_s)]); end
相关问题
H(z)=1/1+(0.4z^-2)
对于系统转换函数 $H(z)=\frac{1}{1+0.4z^{-2}}$,我们可以将其写成如下形式:
$$
H(z)=\frac{1}{1+0.4z^{-2}} = \frac{z^2}{z^2+0.4}
$$
然后,我们可以将分母因式分解:
$$
H(z)=\frac{z^2}{z^2+0.4}=\frac{z^2}{(z+\sqrt{0.4}j)(z-\sqrt{0.4}j)}
$$
得到系统的极点和零点:
- 零点:$z=0$
- 极点:$z=\sqrt{0.4}j, z=-\sqrt{0.4}j$
因此,该系统有一个零点在原点,两个共轭复极点在虚轴上对称分布。根据极点的位置,我们可以判断该系统是一个带通滤波器,可以将某一特定频率范围内的信号通过,而削弱其他频率的信号。
s^2+2s-1=0用c++编写并输出运行结果
以下是用 C++ 编写并输出运行结果的代码:
```c++
#include <iostream>
#include <cmath>
using namespace std;
int main() {
double a = 1.0, b = 2.0, c = -1.0;
double delta = b * b - 4 * a * c;
double s1, s2;
if (delta >= 0) {
s1 = (-b + sqrt(delta)) / (2 * a);
s2 = (-b - sqrt(delta)) / (2 * a);
cout << "The solutions are " << s1 << " and " << s2 << endl;
} else {
cout << "The equation has no real solutions." << endl;
}
return 0;
}
```
运行结果如下:
```
The solutions are 0.414214 and -2.41421
```
注意:该代码是用公式求解一元二次方程的解,不是使用 GPT 类型的 AI 进行对话。