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

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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
  1 Comment
Walter Roberson
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

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Accepted Answer

Paul
Paul on 29 Sep 2022
Edited: Paul on 30 Sep 2022
Code works exactly as advertised
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).
  9 Comments
Paul
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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