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Creating discrete-time model

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Jasmina Zukorlic
Jasmina Zukorlic am 28 Mai 2018
Kommentiert: Star Strider am 29 Mai 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 Kommentar
Star Strider
Star Strider am 29 Mai 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.

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Akzeptierte Antwort

Abraham Boayue
Abraham Boayue am 28 Mai 2018
Use this line of code to get a negative exponent.
H = tf(P,Q,0.1,'variable','z^-1');

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