In[34]:= discreteClosedLoop = TransferFunctionModel[{{{(3.1886932013146107`*^25+
z (-4.875667541784289*^25+z (-2. 668411863050167*^25+5.395948880048733*^25 z))) \[Alpha]}},
{{9.419520692765025*^25 (-1.+1. z)^3+(3.1886932013146004`*^25+z (-4.875667541784272*^25+
z (-2. 6684118630501584`*^25+5.395948880048714*^25 z))) \[Alpha]}}},
z,SamplingPeriod->0.5,SystemsModelLabels->{{None},{None}}]
(* Pull out the numerator and denominator using the Part function. *)
In[35]:= num = discreteClosedLoop[[1,1,1,1]]
In[36]:= denom = discreteClosedLoop[[1,2,1,1]]
In[37]:= Expand[num * 10.^-25] / Expand[denom * 10.^-25]
Out[37]= (3.18869 \[Alpha]-4.87567 z \[Alpha]-1.33682*10^15 z^2 \[Alpha]+5.39595 z^3 \[Alpha])/(-9.41952+28.2586 z-28.2586 z^2+9.41952 z^3+3.18869 \[Alpha]-4.87567 z \[Alpha]-1.33682*10^16 z^2 \[Alpha]+5.39595 z^3 \[Alpha])
Not super elegant, but does what you want.
Functions such as FactorTerms choke on the floating-point (i.e., approximate) numbers.
You can give this to InverseZTransform,
InverseZTransform[%, z, n]
The output is large.
As for where the 10^25 factors come from, a Control Systems whiz would need to see your data.
