Z-transform: Final Value theorem

Started by May 10, 2005
Ajay wrote:
> Randy Yates wrote:
>
>>"Ajay" <mishraka@gmail.com> writes:
>>
>>
>>>Hi,
>>>
>>>I have encountered a strange conceptual problem while calculating
>
> value
>
>>>of sequence x(n) at Inf.
>>>
>>>the function x(n) is defined as
>>>x(n) = 1 when n is even
>>>       0 otherwise.
>>>
>>>I calculated z-transform of x(n) to be
>>>
>>>X(z) = 1/(z^2 -1); ROC : |z| = 1; -- (1)
>>>
>>>Query1 : Have I calculated the transform correctly? Transfrom seems
>
> to
>
>>>exist only on the unit circle.
>>
>>No. The ROC should be |z| < 1, and the z-transform should be
>
> 1/(1-z^{-2}).
>
>>(assuming the sequence is right-sided).
>>
>
> I think You meant "the sequence is left sided", how can a right sided
> sequence have ROC inside the circle?
>
> Anyway, here x(n) is two sided signal. It varies from -inf to inf.. so
> ROC has to be a ring. Any comment on this???

If n truly varies from -inf to inf, how can the transform exist?  It
doesn't converge anywhere, not even at z = 1.

Ray


Raymond Toy <raymond.toy@ericsson.com> writes:

> Randy Yates wrote:
> > Raymond Toy <raymond.toy@ericsson.com> writes:
> >
>
> >> But lim (z-1)X(z) as z -> 1 exists
>
> > Why? If the ROC for X(z) is |z| < 1, then how can
>
> > you say this limit exists? It seems to me you have
> > a domain error, i.e., the ROC is NOT |z| < 1 but
> > is a "bigger" set (when we say "ROC" do we not mean the *entire*
> > ROC?).
>
>
> If ROC truly means the entire region, then I have been a bit
> sloppy. Clearly in my example, the region of convergence is at least
> |z|<=1, except for the point -1.  (I think).  It seems, though, that
> in the context of z-transforms, the ROC is always taken to be a ring,
> with no accounting for special points on the ring boundary.

Yes, so to get back to your original point, why isn't there a
constraint in the application of the FVT that X(z) contain z = 1 in
its ROC?

Actually I read somewhere that the FVT is only applicable when the
poles of X(z) are inside the unit circle, but I didn't spend the time
to find out why. I suspect it will require knowing how the FVT is
derived. In the case of the X(z) provided by the OP, X(z) isn't even a
rational function, so maybe that's the problem?
--
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124

>>>>> "Randy" == Randy Yates <randy.yates@sonyericsson.com> writes:

Randy> Yes, so to get back to your original point, why isn't there a
Randy> constraint in the application of the FVT that X(z) contain z = 1 in
Randy> its ROC?

The fact that it's expressed as a limit is certainly a hint that X(z)
does not have to converge at z = 1.  If that were a requirement, we
wouldn't need the limit because everything would be well-defined.

And indeed, we can see this with x(n) = 1, n >= 0.  X(z) = 1/(1-1/z),
and X(z) does not exist at z = 1.  But limit (z-1)X(z) = 1 as z -> 1,
which is, of course, the final value of x(n).

Randy> Actually I read somewhere that the FVT is only applicable when the
Randy> poles of X(z) are inside the unit circle, but I didn't spend the time
Randy> to find out why. I suspect it will require knowing how the FVT is
Randy> derived. In the case of the X(z) provided by the OP, X(z) isn't even a

This is fairly straightforward.  For simplicity assume x(n) = 0 for n
< 0.  Then X(z) = sum x(n)/z^n and

(z-1)*X(z) = z*x(0) + (x(1)-x(0)) + sum (x(n+1)-x(n))/z^n.
n=0

So the issue is what is limit sum (x(n+1)-x(n))/z^n as z -> 1.  For
the appropriate conditions, I'd have to dig out my math books, but I
think you need uniform convergence to be able to interchange limit and
sum.  However, this is also a power series in 1/z, so something weaker
might be applicable.  Assuming that, we would get for the partial sum

M
sum (x(n+1)-x(n)) = x(M+1) - x(1)
n=1

which approaches limit x(n) - x(1), and

limit (z-1)*X(z) = limit x(n).

Ray

Raymond Toy wrote:
>
> If n truly varies from -inf to inf, how can the transform exist?  It
> doesn't converge anywhere, not even at z = 1.
>

FVT holds in case of one-sided Z-transform, X'(z). i.e

lim x(n) = lim (z-1)X'(z)
n->inf     z->1

As per convergence at z=1 is concerned, I think you answered own
question earlier in the thread. X'(z=1)needs not to be finite for lim
(z-1)X'(z) to exist. In other words, z=1 can be not present in ROC
still the limit can exist.

> Ray


Raymond Toy <raymond.toy@ericsson.com> writes:

> >>>>> "Randy" == Randy Yates <randy.yates@sonyericsson.com> writes:
>
>     Randy> Yes, so to get back to your original point, why isn't there a
>     Randy> constraint in the application of the FVT that X(z) contain z = 1 in
>     Randy> its ROC?
>
> The fact that it's expressed as a limit is certainly a hint that X(z)
> does not have to converge at z = 1.  If that were a requirement, we
> wouldn't need the limit because everything would be well-defined.

Yow! Ouch, that hurts. I gotta start refreshing my mind on this stuff.

> And indeed, we can see this with x(n) = 1, n >= 0.  X(z) = 1/(1-1/z),
> and X(z) does not exist at z = 1.  But limit (z-1)X(z) = 1 as z -> 1,
> which is, of course, the final value of x(n).

Yes. A beautifully simple example.

>     Randy> Actually I read somewhere that the FVT is only applicable when the
>     Randy> poles of X(z) are inside the unit circle, but I didn't spend the time
>     Randy> to find out why. I suspect it will require knowing how the FVT is
>     Randy> derived. In the case of the X(z) provided by the OP, X(z) isn't even a
>
> This is fairly straightforward.  For simplicity assume x(n) = 0 for n
> < 0.  Then X(z) = sum x(n)/z^n and
>
>   (z-1)*X(z) = z*x(0) + (x(1)-x(0)) + sum (x(n+1)-x(n))/z^n.
>                                       n=0

I believe this sum should be from n=1.

> So the issue is what is limit sum (x(n+1)-x(n))/z^n as z -> 1.  For
> the appropriate conditions, I'd have to dig out my math books, but I
> think you need uniform convergence to be able to interchange limit and
> sum.  However, this is also a power series in 1/z, so something weaker
> might be applicable.  Assuming that, we would get for the partial sum
>
>      M
>     sum (x(n+1)-x(n)) = x(M+1) - x(1)
>     n=1

I believe this sum should be from n=0, and you've dropped the z*x(0) term.
So it should be

x_M = x(0) + \sum_{n=0}^{M} (x(n+1) - x(n))

which would then make it

x_M = x(0) + x(M+1) - x(0)
= x(M+1).

Then as M --> \infty this gives you the final value of x(M).
Yes, that makes sense. But wherefore the "poles inside the
unit circle"? Is that part of making the sequence uniformly
convergent?
--
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124

>>>>> "Randy" == Randy Yates <randy.yates@sonyericsson.com> writes:

[corrections deleted]

Randy> Then as M --> \infty this gives you the final value of x(M).
Randy> Yes, that makes sense. But wherefore the "poles inside the
Randy> unit circle"? Is that part of making the sequence uniformly
Randy> convergent?

Don't know.  I forgot to look at my math books last night, so I don't
know the conditions that are really needed.  When I do look, perhaps
that will shed some light.

But certainly if you have poles, the series can't be uniformly
convergent in a neighborhood of the pole.  Uniform convergence is
usually on a closed neighborhood too, IIRC.

Ray