Inverse Z-transform with negative Z-exponents

Question

Michael 2015-2-14

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https://ww2.mathworks.cn/matlabcentral/answers/178458-inverse-z-transform-with-negative-z-exponents

评论： OTHMAN nesrine 2015-7-3

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)'?

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Answer 1

Michael 2015-2-14

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在 MATLAB Online 中打开

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];

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Answer 2

Azzi Abdelmalek 2015-2-14

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https://ww2.mathworks.cn/matlabcentral/answers/178458-inverse-z-transform-with-negative-z-exponents#answer_167989

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num=[1 0 0 0];
den=conv([-1 1],[-0.2 1])
g=tf(num,den,1,'variable','z^-1')

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Michael 2015-2-14

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 2015-7-3