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

If f = f(z), then ztrans returns a function of w.

By definition, $F\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{f\left(n\right)}{z^n}$ where the summation above proceeds with respect to z.

ztrans(f,w) makes f a function of the variable w instead of the default z.

ztrans(f,k,w) takes f to be a function of k instead of the default n. The summation is then with respect to w.

The ztrans(M) function computes the element-wise Z-transform of M.  The result, R, is formed as R[i,j] = ztrans(M[i,j], n,z).

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

$\mathrm{ztrans}\left(\sin \left(\frac{\mathrm{\pi }}{2}n\right)\right)$

$\frac{z}{z^2+1}$

$\mathrm{ztrans}\left(\sin \left(\frac{\mathrm{\pi }}{2}z\right)\right)$

$\frac{w}{w^2+1}$

$\mathrm{ztrans}\left(\sin \left(\frac{\mathrm{\pi }}{2}n\right)\,s\right)$

$\frac{s}{s^2+1}$

$\mathrm{ztrans}\left(k\sin \left(\frac{\mathrm{\pi }}{2}n\right)\,k\,s\right)$

$\frac{\sin \left(\frac{\mathrm{\pi }n}{2}\right)s}{{\left(s-1\right)}^2}$

$M≔\mathrm{Matrix}\left(\left[\sin \left(\frac{\mathrm{\pi }}{2}n\right)\,k\sin \left(\frac{\mathrm{\pi }}{2}n\right)\right]\right)\:$

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

$\left[\begin{array}{cc}\frac{z}{z^2+1}& \frac{kz}{z^2+1}\end{array}\right]$