>>>>> "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

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

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.

Comments!!
 


> Ray

>>>>> "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 <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

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

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.

Ray

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.

Ray

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?).
-- 
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124

Tim Wescott wrote:
>
> More intuitively, if it never stops moving it has no final value and
you
> can't use the final value theorem.

It does not mean that an alternating sequence can not be considered to
be asymptotically converging to the average value. And this is what
seems to happend in this case. The function does not have to be
decaying to reach certain value finally. However, there is a problem I
see. I read in "Prokais .." Final value theorem is applied on one-sided
Z-transform. So when the sequence is two sided, is it correct to take
one-sided Z-transform and do the analysis.