Creating discrete-time model
Ältere Kommentare anzeigen
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
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.
Akzeptierte Antwort
Weitere Antworten (0)
Kategorien
Mehr zu MATLAB finden Sie in Hilfe-Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!