Creating discrete-time model

조회 수: 2 (최근 30일)
Jasmina Zukorlic
Jasmina Zukorlic 2018년 5월 28일
댓글: Star Strider 2018년 5월 29일
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)
  댓글 수: 1
Star Strider
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.

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채택된 답변

Abraham Boayue
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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