# Z-transform: Final Value theorem

Started by May 10, 2005
```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.

Now, In order to calculate, x(inf) = ?
I used final value theorem which is described as

x(inf) = lim (z-1)X(z) -- (2)
z->1

Then , X(inf) = 1/2; on substituting for X(z) by (1)

Query2: If the above is right, I am unable to understand how the
sequence would eventually become 1/2. What exactly does x(inf) mean
here?

"Final Value Theorem"

-Ajay

```
```"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).

--RY

> Now, In order to calculate, x(inf) = ?
> I used final value theorem which is described as
>
> x(inf) = lim (z-1)X(z) -- (2)
>          z->1

Since the ROC does not include z=1, you can't use this value
for z.
--
%  Randy Yates                  % "The dreamer, the unwoken fool -
%% Fuquay-Varina, NC            %  in dreams, no pain will kiss the brow..."
%%% 919-577-9882                %
```
```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).
>
> --RY
>
>
>
>>Now, In order to calculate, x(inf) = ?
>>I used final value theorem which is described as
>>
>>x(inf) = lim (z-1)X(z) -- (2)
>>         z->1
>
>
> Since the ROC does not include z=1, you can't use this value
> for z.

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

-------------------------------------------
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
```
```Tim Wescott <tim@seemywebsite.com> writes:

> 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).
> > --RY
>
> >
>
> >>Now, In order to calculate, x(inf) = ?
> >>I used final value theorem which is described as
> >>
> >>x(inf) = lim (z-1)X(z) -- (2)
> >>         z->1
> > Since the ROC does not include z=1, you can't use this value
>
> > for z.
>
> More intuitively, if it never stops moving it has no final value and
> you can't use the final value theorem.

Nice to see that the theory agrees with reality!
--
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124
```
```>>>>> "Randy" == Randy Yates <yates@ieee.org> writes:

>> Now, In order to calculate, x(inf) = ?
>> I used final value theorem which is described as
>>
>> x(inf) = lim (z-1)X(z) -- (2)
z-> 1

Randy> Since the ROC does not include z=1, you can't use this value
Randy> for z.

I think there's more to it than the ROC doesn't include 1.

Let X(z) = -1/z+1/2/z^2-1/3/z^3+...+(-1)^n/n/z^n+.... = log(1+1/z).  The
region of convergence is strictly less than 1, because the series
doesn't converge for z = -1 (and because log(1+1/z) is undefined at z
= -1).

But lim (z-1)X(z) as z -> 1 exists and is 0.  Which matches lim x(n)
as n -> infinity.  It helps, of course, that X(z) exists as z -> 1.

Ray

```
```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???

> --RY
>
>
> > Now, In order to calculate, x(inf) = ?
> > I used final value theorem which is described as
> >
> > x(inf) = lim (z-1)X(z) -- (2)
> >          z->1
>
> Since the ROC does not include z=1, you can't use this value
> for z.

I once again defer. Such a sequence has its DFT. So, ROC must include
ROC. No question of not having unit circle in ROC.

> --
> %  Randy Yates                  % "The dreamer, the unwoken fool -
> %% Fuquay-Varina, NC            %  in dreams, no pain will kiss the
brow..."
> %%% 919-577-9882                %
ELO

```
```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.

```
```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
```
```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
```