英文:
Plotting response r(t) from R(s)
问题
对于系统函数 H(s)=s^2/(s^2+4) 和输入 E(s)=1/s^2。响应已计算为 R(s)=1/(s^2+4)。如何在时域绘制这个图形？
我尝试使用 lsim，但对我来说没有太多意义。我还尝试使用 step()，但我认为这是在输入是阶跃函数时使用的。
英文:
For system function H(s)=s^2/(s^2+4) and the input E(s)=1/s^2. The response was calculated to be R(s)=1/(s^2+4). How can I plot this in time domain?
I tried using lsim which didn't really make sense to me. I also tried using step() but I thought this is used when the input in a step function
答案1
得分: 1
以下是MATLAB代码的翻译：
使用以下MATLAB代码将为您提供系统的响应：
clear all; close all; clc; clf;
sys = tf([1 0 0],[1 0 4]);
t = 0:0.01:10;
u = t;
lsim(sys,u,t) % u,t define the input signal
您的输入是E(s)=1/s^2，其拉普拉斯反演等于t。这就是为什么在上面的代码中，我们有u = t。
以下是响应的绘图：
灰线是输入，即t。蓝色正弦线是系统的响应。它与您的响应兼容。响应是R(s)=1/(s^2+4)，其拉普拉斯反演是0.5*sin(2*t)，正如您在上面的图中看到的响应一样。
英文:
Using below MATLAB code will give you the response of system:
clear all; close all; clc; clf;
sys = tf([1 0 0],[1 0 4]);
t = 0:0.01:10;
u = t;
lsim(sys,u,t) % u,t define the input signal
Your input is E(s)=1/s^2 and its Laplace Inverse is equal to t. That's why in above code, we have u = t.
Below is the plot of response:
gray line is input which is t. The blue sinusoidal line is the response of system. It is compatible with your response. The response is R(s)=1/(s^2+4) and its Laplace Inverse is 0.5*sin(2*t), as you see in above plot as response.
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