Creating discrete-time model
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Hello, can someone please tell me what am I doing wrong in writing this expression in MATLAB:

This is the result I'm obtaining: H =
                  8 z^5 - 5 z^4 - 4 z^3 + z^2 + 3 z - 2
  ----------------------------------------------------------------------
  0.0648 z^6 + 0.1134 z^5 - 0.6184 z^4 + 1.436 z^3 - 1.7 z^2 + 1.6 z - 1
And here is my code:
Nd=[-8 5 4 -1 -3 2];
Dd=[-0.0648 -0.1134 0.6184 -1.436 1.7 -1.6 1 ];
P=Nd;Q=Dd;
H = tf(P,Q,0.1)
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  Star Strider
      
      
 2018년 5월 29일
				If you want to code the transfer function in the image you posted, you need to enter the coefficients in the correct order. Here, that means using fliplr (since I do not want to re-type the vectors):
Nd = [-8 5 4 0 -1 -3 2];
Dd = [-0.0648 -0.1134 0.6184 -1.436 1.7 -1.6 1 ];
P = fliplr(Nd);
Q = fliplr(Dd);
H = tf(P,Q,0.1,'variable','z^-1')
H =
                     2 - 3 z^-1 - z^-2 + 4 z^-4 + 5 z^-5 - 8 z^-6
    ------------------------------------------------------------------------------
    1 - 1.6 z^-1 + 1.7 z^-2 - 1.436 z^-3 + 0.6184 z^-4 - 0.1134 z^-5 - 0.0648 z^-6
Sample time: 0.1 seconds
Discrete-time transfer function.
채택된 답변
  Abraham Boayue
      
 2018년 5월 28일
        Use this line of code to get a negative exponent.
H = tf(P,Q,0.1,'variable','z^-1');
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