Difference in sample time of discrete model and input

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 ??

Answer 1

[Abhishek Kumar]

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

hence, some of the coefficients are similar you can go through the documentation of functions used for better understanding..