Different response to discrete time transfer function matlab and simulink

Hi,

If i use c2d in Matlab on a continuous function : Gz = c2d(tf([10],[1 5.5]),0.1), I get a function which produces a very similar results to the original. See result of Step(Gz) below. So far so good.

0.001461 z^2 + 0.005121 z + 0.00111
-----------------------------------
z^3 - 2.577 z^2 + 2.154 z - 0.5769

If I try to implement this discrete function in simulink the function behaves completely different. I'm not sure what I'm doing wrong here.

I use a discrete transfer function block, where the sample time is set to 0.1 num : [0.001461 0.005121 0.00111] denum : [1 -2.577 2.154 -0.5769]

input is a step of 1, also with a sample time of 0.1

simulink output is also attached.

Some pointers on what I'm doing wrong are much appreciated,

Thanks

M

Edit :

When i load the transfer function in to simulink with the LTI block, all seems to work correctly. So problem is I'm not implementing the discrete transfer function block correctly?