Filter löschen
Filter löschen

Inverse Z-transform with negative Z-exponents

14 Ansichten (letzte 30 Tage)
Michael
Michael am 14 Feb. 2015
Kommentiert: OTHMAN nesrine am 3 Jul. 2015
I am faced with a Z-transform problem for school, and I already know the code to handle most of the problem using matrices for the numerator and denominator. My trouble is that the problem uses negative-exponents for the Zs.
x(z) = (z^-3)/((1 - z^-1)(1 - 0.2z^-1))
If this problem had only positive exponents, but the same coefficients, I would approach inputing this particular problem like this:
MATLAB code
syms z
%x(z) = (z^3)/((-z^1 + 1)(-0.2z^1 + 1))
num=[1 0 0 0];
den=conv([-1 1],[-0.2 1]);
Any thoughts for using a similar input style with negative exponents? Would it really be as simple as something like 'fun(-1)'?

Akzeptierte Antwort

Michael
Michael am 14 Feb. 2015
Well, I might be a moron. I just realized that the entire problem can be simplified before you even involve Matlab.
x(z) = (z^-3)/((1-z^-1)(1-0.2z^-1))
But it can be simplified as:
x(z) = 1/(z^3 -1.2z^2 + 0.2z)
Which would make the Matlab code for inputting the equation:
syms z
num = [0 0 0 1];
den = [1 -1.2 0.2 0];

Weitere Antworten (1)

Azzi Abdelmalek
Azzi Abdelmalek am 14 Feb. 2015
num=[1 0 0 0];
den=conv([-1 1],[-0.2 1])
g=tf(num,den,1,'variable','z^-1')
  2 Kommentare
Michael
Michael am 14 Feb. 2015
Doesn't seem so. The third coefficient in the denominator seems to be wrong, going by hand-convolution ("1" instead of "-0.2") - and the numerator is simply a 1 instead of z^-3.
But it is a lot closer than I had been getting.
OTHMAN nesrine
OTHMAN nesrine am 3 Jul. 2015
try with function 'filt' , see help

Melden Sie sich an, um zu kommentieren.

Produkte

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by