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I want to study how the Z transform changes with the sampling rate T, in a closed loop system.

The command I'm using to do this is:

InverseZTransform[(-1 - T + z + E^T (1 + (-1 + T) z))/
                  (-T +  E^T (1 + (-2 + T) z + z^2)), z, n]

However, it takes a while to do so, and the result that appears is insanely huge and impractical.

For substitutions of T = 1, T = 0, etc, the inverse transform is very very easily found. I was wondering if someone could give any suggestions as to whats going on, because I really don't want to have to solve it by hand.

Also, for some reason, Apart[y[z] , z] also doesn't work. (y[z] is the function to invert).

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At least in this case, it is better to work with the most generic form of your rational function, and then plug-in the desired coefficients later. Witness the following:

FullSimplify[InverseZTransform[(g + f z)/(c + b z + a z^2), z, n], n ∈ Integers]
   (2^(-1 - n) (Sqrt[b^2 - 4 a c] ((-b + Sqrt[b^2 - 4 a c])/a)^n g + Sqrt[b^2 - 4 a c]
    (-((b + Sqrt[b^2 - 4 a c])/a))^n g + (-((b + Sqrt[b^2 - 4 a c])/a))^n (2 c f - b g) +
    ((-b + Sqrt[b^2 - 4 a c])/a)^n (-2 c f + b g)) (-1 + UnitStep[-n]))/
   (c Sqrt[b^2 - 4 a c])

which in itself is already complicated.

From there, make the necessary substitutions to get what you want:

Simplify[% /. {g -> E^T - T - 1, f -> 1 + E^T (T - 1),
               c -> E^T - T, b -> E^T (T - 2), a -> E^T}]

whose output I shall omit. The result is still complicated, but not as much as the one obtained from directly evaluating InverseZTransform[] on the OP's expression.

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How about creating a function?

f = Function[{T, n}, InverseZTransform[(-1 - T + z + Exp[T] (1 + (-1 + T) z))/(-T + 
    Exp[T] (1 + (-2 + T) z + z^2)), z, n]][T, n];

Then

Plot[f /. {T -> x, n -> 3}, {x, 0, 5}] 

enter image description here

Simplify[f /. {T -> x, n -> 3}]

-1 + 6 x + E^(-2 x) x - 5 x^2 + x^3 + E^-x (1 - 6 x + 3 x^2)

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