DSP / Control System / Z-Transform Question

Suppose I have a system modeled as y’'=u. Taking the Laplace Transform with zero initial conditions yields G(s) = 1/(s^2).

No biggie, I am interested in the discrete time transfer function so I look up a Z-Transform table and get:

If, G(s) = 1/(s^2), then G(z) = Tz/((z-1)^2)

However, using the state space representation one may find the DT transfer function (assuming A,B,C,D matrices are DT) using:

G(z) = C*(sI - A)^-1*B + D

These yield two totally different transfer functions and accordingly step responses. Using Matlab I get:

k_sys=tf(1,[1 0 0]) %define 1/(s^2)
t_sys_d=c2d(k_sys,1) %convert to D.T. transfer function

0.5 z + 0.5

z^2 - 2 z + 1

Which is NOT the Z transform from the table, but does match the
G(z) = C*(sI - A)^-1*B + D.

Can anyone explain why the Z-Transform tables do not yield the same discrete time transfer function?

Sources: Readings | Dynamic Systems and Control | Electrical Engineering and Computer Science | MIT OpenCourseWare (page 5 for D.T. Transfer Function Equation from D.T. State Space Representation)

Laplace and Z Transforms (Z Transform Table)

Thinking you are finding/comparing the difference between continuous and discrete time.
Z transform for discrete

laplace transform for continuous.

I think understand that piece. I just don’t understand why the discrete time transfer functions given in the tables do not match the transfer function given by G(z) = C*(sI - A)^-1*B + D.

In continuous, G(s) = 1/(s^2) yields G(z) = Tz/((z-1)^2) using the tables

Mapping G(s) = 1/(s^2) into the continuous state space, then discrete state space, then using the G(z) = C*(sI - A)^-1*B + D yields a
G(z) =
0.5 z + 0.5

z^2 - 2 z + 1

and my arithmetic is confirmed with Matlab so I am sure I’m doing it right. I just don’t understand why two ostensibly equal methods are yielding different results.