Creating discrete-time model
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    Jasmina Zukorlic
 on 28 May 2018
  
    
    
    
    
    Commented: Star Strider
      
      
 on 29 May 2018
            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)
1 Comment
  Star Strider
      
      
 on 29 May 2018
				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.
Accepted Answer
  Abraham Boayue
      
 on 28 May 2018
        Use this line of code to get a negative exponent.
H = tf(P,Q,0.1,'variable','z^-1');
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