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Does every function X(z) have an inverse Z transform?

Started by Unknown May 30, 2006
Hi,

Here is the formal question:

Is this statement true?

For every X(z) there exists a corresponding x(n), such that
Z{x(n)}=X(z)

Thank you.

andresmedina@hotmail.com wrote:
> Hi, > > Here is the formal question: > > Is this statement true? > > For every X(z) there exists a corresponding x(n), such that > Z{x(n)}=X(z) > > Thank you. >
Dunno, but here's some thoughts: I don't know (and I don't think there is) a way to go directly from X(z) to x(n). Obviously if X(z) is a rational fraction in z then there's a transform. If you can find a power series in 1/z such that X(z) = a_0 + a_1/z + a_2/z^2 + ... then the transform is simply x(n) = a_0, a_1, a_2, ... Mathematicians will want to argue about regions of convergence, and they may be right. By this hypothesis X(z) = e^{bz} implies that x(n) = 1, b, b^2/2, ... b^n/n!. If the series blows up then you have the z transform of something unstable, or you need to think in terms of the two-sided z transform and a non-causal x(n). Presumably you'd then have to split X(z) into a sum of X_nc(z) and X_c(z), where X_nc(z) could be found strictly in terms of a power series in z and X_c(z) could be found strictly in terms of a power series in 1/z. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/ "Applied Control Theory for Embedded Systems" came out in April. See details at http://www.wescottdesign.com/actfes/actfes.html
Tim Wescott wrote:
> andresmedina@hotmail.com wrote: > > > > Here is the formal question: > > > > Is this statement true? > > > > For every X(z) there exists a corresponding x(n), such that > > Z{x(n)}=X(z) > > > Dunno, but here's some thoughts: > > I don't know (and I don't think there is) a way to go directly from X(z) > to x(n).
i thought that this was: pi x[n] = 1/(2*pi) * integral{ X(e^(j*w)) * e^(j*w*n) dw} -pi where X(e^(j*w)) is X(z) with z = e^(j*w). this works for stable X(z) where all the poles are inside the unit circle. if you make the substitution of z = e^(j*w) and dz = j*e^(j*w) dw = j*z dw , then you get that circular line integral x[n] = 1/(2*pi*j) * integral{ X(z) * z^(n-1) dz} and the path for z can be expanded (or shrunk) from the unit circle to any simple closed curve that contains all of the poles. r b-j
Thank you for the replies.

OK, then my precise question would be:

Can you find a particular X(z) in which :
x[n] = 1/(2*pi*j) * integral{ X(z) * z^(n-1)  dz}

does not converge or does not exist, for every possible contour line?

On 30 May 2006 13:44:35 -0700, andresmedina@hotmail.com wrote:
> >Thank you for the replies. > >OK, then my precise question would be: > >Can you find a particular X(z) in which : >x[n] = 1/(2*pi*j) * integral{ X(z) * z^(n-1) dz} > >does not converge or does not exist, for every possible contour line? >
Hmm, the residue theorem comes to mind as a way to attack this problem. Might be worth a try: http://en.wikipedia.org/wiki/Cauchy_residue_theorem -- "The trouble with the world is that the stupid are cocksure and the intelligent are full of doubt" -- Bertrand Russell