# Problem seen in discrete transfer function with varable z^-1, when calc ztrans of x(n)=n*u(n)

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Commented: Paul on 30 Sep 2022
Hi dears,
Who knows why X1 and X2 are not the same?
X2 should be (z^-1)/(1-z^-1)^2 or (z^-1)/(1 - 2 z^-1 + z^-2)
Thanks
sympref('HeavisideAtOrigin', 1); % by default u(0)=0.5 so we set U(0)=1
u = @(n) heaviside(n) ; % change function name
u0=u(0)
syms n
x(n)=n*u(n)
X1=ztrans(x)
[num, den] = numden(X1);
X2 = tf(sym2poly(num), sym2poly(den),-1, 'variable', 'z^-1')
X2_var=X2.variable Walter Roberson on 29 Sep 2022
My tests show it is related to specifying the variable. If you let the variable default to 'z' then you get a z in the numerator

Paul on 29 Sep 2022
Edited: Paul on 30 Sep 2022
u = @(n) heaviside(n) ; % change function name
syms n
x(n)=n*u(n)
x(n) = X1=ztrans(x)
X1 = [num, den] = numden(X1)
num =
z
den = As documented in sym2poly, it returns the polynomial in descending powers of the variable, in this case z
sym2poly(num)
ans = 1×2
1 0
[1 0] is the poly representation of z.
sym2poly(den)
ans = 1×3
1 -2 1
Here, we are telling tf that sym2poly(num) is the poly representation with variable z^-1. But wrt to z^-1, [1 0] = 1 + 0*z^-1 = 1, which is exactly what we get.
X2 = tf(sym2poly(num), sym2poly(den),-1, 'variable', 'z^-1')
X2 = 1 ----------------- 1 - 2 z^-1 + z^-2 Sample time: unspecified Discrete-time transfer function.
So we need two steps
X2 = tf(sym2poly(num),sym2poly(den),-1)
X2 = z ------------- z^2 - 2 z + 1 Sample time: unspecified Discrete-time transfer function.
X2.Variable = 'z^-1'
X2 = z^-1 ----------------- 1 - 2 z^-1 + z^-2 Sample time: unspecified Discrete-time transfer function.
Finally, be very careful using heaviside. The default value of heaviside(0) is 1/2, which is (almost?) never what you want for discrete-time problems. It didn't matter here becasue u(0) = 0. Use sympref to control the value of heaviside(0).
Paul on 30 Sep 2022
They only need to be the same size if that's what the problem requires. For an example of when it's not required
H(z) = (1 + z^-1) / (1 + 2*z^-1 + 3*z^-2)
H = tf([1 1],[1 2 3],-1,'Variable','z^-1')
H = 1 + z^-1 ------------------- 1 + 2 z^-1 + 3 z^-2 Sample time: unspecified Discrete-time transfer function.
You can, of course, zero-pad the numerator if you wish (zero-pad to the right for z^-1)
H = tf([1 1 0],[1 2 3],-1,'Variable','z^-1')
H = 1 + z^-1 ------------------- 1 + 2 z^-1 + 3 z^-2 Sample time: unspecified Discrete-time transfer function.
but you're not obligated to do so. The only reuqirement is that num and den represent the system for the Variable that's being used.

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