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Creating discrete-time model

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Jasmina Zukorlic
Jasmina Zukorlic 2018 年 5 月 28 日
コメント済み: Star Strider 2018 年 5 月 29 日
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 件のコメント
Star Strider
Star Strider 2018 年 5 月 29 日
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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採用された回答

Abraham Boayue
Abraham Boayue 2018 年 5 月 28 日
Use this line of code to get a negative exponent.
H = tf(P,Q,0.1,'variable','z^-1');

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