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

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
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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