Difference in sample time of discrete model and input

Question

Jatin Sharma 2020-5-5

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https://ww2.mathworks.cn/matlabcentral/answers/523062-difference-in-sample-time-of-discrete-model-and-input

回答： Abhishek Kumar 2020-12-13

I was going through the algorithms on how MATLAB evaluates convolution in CT system. Example code given on MATLAB is

w2 = 62.83^2;
h = tf(w2,[1 2 w2]);
dt = 0.016;
ts = 0:dt:5;
us = (rem(ts,1) >= 0.5);
hd = c2d(h,dt);
lsim(hd,us)

In the above example sample time of input and TF are the same. I have removed ts from lsim because it was making no difference in this case. So in the output I got 313 samples.

w2 = 62.83^2;
h = tf(w2,[1 2 w2]);
dt = 0.016;
ts = 0:0.02:5;
us = (rem(ts,1) >= 0.5);
hd = c2d(h,dt);
lsim(hd,us)

In the above code sample time of input and TF are different. If i put Ts then it gave an error but when I removed it worked fine. So in the output I had 250 samples. Some of the starting samples matched also.

So my question is how the matlab calculates output in 2nd case ??

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Answer 1

Abhishek Kumar 2020-12-13

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https://ww2.mathworks.cn/matlabcentral/answers/523062-difference-in-sample-time-of-discrete-model-and-input#answer_574425

在 MATLAB Online 中打开

Hi Jatin, given the two codes you have used, the continuous time transfer function is same in both the cases, the differnce is in length of number of samples

>> h1
h1 =
 
        3948
  ----------------
  s^2 + 2 s + 3948
 
Continuous-time transfer function.
>>dt1 = 0.016;
>>ts1 = 0:dt1:5;
>> length(ts1)
ans =
   313

>> h2
h2 =
 
        3948
  ----------------
  s^2 + 2 s + 3948
 
Continuous-time transfer function.
>>dt2 = 0.016;
>>ts2 = 0:0.02:5;
>> length(ts2)
ans =
   251