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

# Z-transform: Final Value theorem

Started by ●May 10, 2005

Reply by ●May 11, 20052005-05-11

Reply by ●May 11, 20052005-05-11

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

Reply by ●May 11, 20052005-05-11

>>>>> "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

Reply by ●May 12, 20052005-05-12

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

Reply by ●May 12, 20052005-05-12

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=0I 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=1I 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

Reply by ●May 12, 20052005-05-12

>>>>> "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