The iztrans(M) function computes the element-wise inverse Z transform of M.  The result, R, is formed as R[i,j] = iztrans(M[i,j], t, s).

f = iztrans(F) is the inverse Z transform of the scalar F with the default independent variable z.  If F is not a function of z, then F is  assumed to be a function of the independent variable returned by findsym(F,1). The default return is a function of n.

If F = F(n), then iztrans returns a function of k.

iztrans(F,k) makes F a function of the variable k instead of the default n.

iztrans(F,w,k) takes F to be a function of w instead of the default z. The summation is then with respect to k.

$\mathrm{with}\left(\mathrm{MTM}\right)\:$

$\mathrm{iztrans}\left(\frac{z}{z-2}\right)$

$2^n$

$\mathrm{iztrans}\left(\frac{n}{n-2}\right)$

$2^k$

$\mathrm{iztrans}\left(\frac{z}{z-2}\,s\right)$

$2^s$

$\mathrm{iztrans}\left(\frac{tz}{z-t\cdot 2}\,t\,s\right)$

$-\frac{{\left(\frac{1}{2}\right)}^s{z}^{s}z}{2}$

$M≔\mathrm{Matrix}\left(\left[\frac{z}{z-2}\,\frac{tz}{z-t\cdot 2}\right]\right)\:$

$\mathrm{iztrans}\left(M\right)$

$\left[\begin{array}{cc}{2}^n& 2^n{t}^{n}t\end{array}\right]$