Problem seen in discrete transfer function with varable z^-1, when calc ztrans of x(n)=n*u(n)
4 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
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 Commento
Walter Roberson
il 29 Set 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
Risposta accettata
Paul
il 29 Set 2022
Modificato: Paul
il 30 Set 2022
Code works exactly as advertised
u = @(n) heaviside(n) ; % change function name
syms n
x(n)=n*u(n)
X1=ztrans(x)
[num, den] = numden(X1)
As documented in sym2poly, it returns the polynomial in descending powers of the variable, in this case z
sym2poly(num)
[1 0] is the poly representation of z.
sym2poly(den)
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')
So we need two steps
X2 = tf(sym2poly(num),sym2poly(den),-1)
X2.Variable = 'z^-1'
9 Commenti
Paul
il 30 Set 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')
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')
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.
Più risposte (0)
Vedere anche
Categorie
Scopri di più su Arduino Hardware in Help Center e File Exchange
Prodotti
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!