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: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-241j-dynamic-systems-and-control-spring-2011/readings/MIT6_241JS11_chap25.pdf (page 5 for D.T. Transfer Function Equation from D.T. State Space Representation)
http://lpsa.swarthmore.edu/LaplaceZTable/LaplaceZFuncTable.html (Z Transform Table)