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? Please help me understand this engimatic result obtained after applying "Final Value Theorem" -Ajay

# Z-transform: Final Value theorem

Started by ●May 10, 2005

Reply by ●May 10, 20052005-05-10

"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->1Since 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 % %%%% <yates@ieee.org> % 'Eldorado Overture', *Eldorado*, ELO http://home.earthlink.net/~yatescr

Reply by ●May 10, 20052005-05-10

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

Reply by ●May 10, 20052005-05-10

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

Reply by ●May 10, 20052005-05-10

>>>>> "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

Reply by ●May 11, 20052005-05-11

Randy Yates wrote:> "Ajay" <mishraka@gmail.com> writes: > > > Hi, > > > > I have encountered a strange conceptual problem while calculatingvalue> > 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 seemsto> > exist only on the unit circle. > > No. The ROC should be |z| < 1, and the z-transform should be1/(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. Please comment!!> -- > % Randy Yates % "The dreamer, the unwoken fool - > %% Fuquay-Varina, NC % in dreams, no pain will kiss thebrow..."> %%% 919-577-9882 % > %%%% <yates@ieee.org> % 'Eldorado Overture', *Eldorado*,ELO> http://home.earthlink.net/~yatescr

Reply by ●May 11, 20052005-05-11

Tim Wescott wrote:> > More intuitively, if it never stops moving it has no final value andyou> 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.

Reply by ●May 11, 20052005-05-11

Raymond Toy <raymond.toy@ericsson.com> writes:> But lim (z-1)X(z) as z -> 1 existsWhy? 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

Reply by ●May 11, 20052005-05-11

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

Reply by ●May 11, 20052005-05-11

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