Tobin Fricke's Lab Notebook (![[info]](http://l-stat.livejournal.com/img/userinfo.gif){kind=small} nibot_lab) wrote,@ 2009-06-03 18:33:00 |
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z transform
Suppose
Then take the Laplace transform:
The z-transform is related to the Laplace transform via:
Solve that for z:
where
Now, left to my own devices, I would try to expand this equation about z=0, hoping to get a power series in
Question: Can we use this scheme to get a finite-difference approximation to the derivative?
Trivia: Wikipedia tells me that the z-transform was invented in part by Lotfi Zadeh, whom I remember passing in the hallways of Soda hall at Berkeley!
Suppose
y(t) = (d/dt) x(t)Then take the Laplace transform:
Y(s) = s X(s)The z-transform is related to the Laplace transform via:
z = exp(T s)Solve that for z:
s = (1/T) ln zwhere
|z| = 1.Now, left to my own devices, I would try to expand this equation about z=0, hoping to get a power series in
z^(-1), where, of course, z^(-1) is the delay operator... So we should get a finite-difference approximation to our original differential equation, right? But it fails immediately because the logarithm has a pole at zero.Question: Can we use this scheme to get a finite-difference approximation to the derivative?
Trivia: Wikipedia tells me that the z-transform was invented in part by Lotfi Zadeh, whom I remember passing in the hallways of Soda hall at Berkeley!
