Help on derivative of a function in discrete time

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Hello all Please can you help me out calculating derivatives of discrete time functions in matlab r(k+1) from r(k)
Given a discrete signal r(k)=0.5*sin*(pi*k*ts) for k=1:1=100
I found in one book that they used an extrapolation method as follows:
dr(k)=(r(k)-r_1)/ts;
dr_1=(r_1-r_2)/ts;
r1(k)=2*r(k)-r_1;
dr1(k)=2*dr(k)-dr_1;
to calculate first and second derivatives. i didn't understand this method
I tried rather to use a simple formula:
r_1=0;
r_1(k)=r(k)-r_1/ts;
r_1=r(k);
I don't know if this is correct ?
Thank you so much for your precious help
Best regards

回答(1 个)

Walter Roberson
Walter Roberson 2015-6-14
You change the same variable in all three lines. You have
r_1=0;
r_1(k)=r(k)-r_1/ts;
r_1=r(k);
You set all of r_1 in the first line. In the second line you overwrite only part of r_1 . But in the third line you overwrite all of it, using a value that was not changed in the previous lines. The effect is going to be as if you had only one line
r_1 = r(k);
I cannot really advise on the correctness of the method you showed from the book as I do not know what r1 is intended to represent that is different from dr. Also we would need to know how r_1 and r_2 are initialized. My suspicion is that the book might also show them changing, possibly like
r_2 = r_1;
r_1 = r(k);
so that the variables represent the two previous values.
  3 个评论
lady bird
lady bird 2015-6-15

Thank you Walter Roberson for your answer

in fact, i did a mistake my code was rather as follows:

    r_1=0;
    r1(k)=r(k)-r_1/ts;
    r_1=r(k); 

where r1(k) is the first derivative, and i dont know if this is correct

In the other hand, yes i just checked in that book , there is an update of the variables as u just stated

   r_2 = r_1;
   r_1 = r(k);

r_1 and r_2 are initialized to zero at first

Walter Roberson
Walter Roberson 2015-6-16
They are using dr() as the first derivative, it appears, so I do not understand why they would also have r1.
The r_1 and r_2 make sense to me now.

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