ztrans command changing sample period

I have a question about MATLAB's ztrans function:

When I evaluate the command:

ztrans( exp(-t) )

I get the result:

ans =
z/(z - 1/exp(1))

Is there any way to change the sampling period MATLAB uses to calculate the z-transform? Right now it seems as if its using a default period of 1. Is it possible to have the function return a result such as:

z/(z - 1/exp(1 * T))

Answer 1

[Paulo Silva]

easy question and easy answer

syms t T
ztrans(exp(-t*T))

ans =
z/(z - 1/exp(T))

[Walter Roberson]

Scooped! :)

[Paul Prechtl]

but here T plays the role of the inverse time constant and not the sampling time

[Paul]

Even though ztrans yielded the desired result in this case, it's a bit misleading. Basically, ztrans is treating t as if its the integer independent variable of the sampled signal, even though that's not what t really represents. I think it's clearer to show in a few steps

syms x t
x = exp(-t)

x = 

syms n integer
syms T
xn = subs(x,t,n*T)

xn = 

syms z
X = ztrans(xn,n,z)

X = 

Or, using symfun objects

x(t) = exp(-t)

x(t) = 

X(z) = ztrans(x(n*T),n,z)

X(z) = 

Answer 2

[Walter Roberson]

ztrans is not using a period: it is doing an infinite summation, as documented.

I do not have access to the Symbolic Math toolbox at the moment, but it would be interesting to

iztrans(z/(z - 1/exp(1 * T)),z)

and see what you would have to transform to get that result.

[Paulo Silva]

iztrans(z/(z - 1/exp(T)),z)

ans=

((1/exp(T))^z*exp(T) - exp(T)*kroneckerDelta(z, 0))/exp(T) + kroneckerDelta(z, 0)

iztrans(z/(z - 1/exp(1)),z)

ans =

(6627126856707895/18014398509481984)^z

[Paulo Silva]

I made a mistake, it's

iztrans(z/(z - 1/exp(1)))

ans =

(6627126856707895/18014398509481984)^n

It's something close to 0.3679^n