Hi All, Here is a doubt regarding Z-Transform with respect to poles and zeros. Here is the problem example: x(n) = a^n * u(n) X(z) = (n=0 to n=infinity)SUM x(n) * z^ -n; X(z) = 1 + az^-1+(az^-1)^2 +(az^-1)^3+.....+infinity; .....(eq 1) X(z) = 1/(1-az^-1); provided |az^-1|<1 => |z| >|a| ......(eq 2) now X(z) can be written as X(z) = z/(z-a). ......(eq 3) Now the question is what are the poles and zeros location of X(z)???????? DSP Books says Zero at z = 0 Pole at z = a from (eq 3) but how can z=0 be substituted in the equation for X(z) = 1/(1-az^-1) as 0 < |a| which is not in ROC and it makes the original X(z) calculated using (eq 1) to go to infinity. But ROC plots in DSP books show a zero at z=0, Are the DSP Texts stating wrong? or is my understanding wrong?

# Z-Transform(poles, Zeros and ROC)

Started by ●December 26, 2006

Reply by ●December 26, 20062006-12-26

PrasadBC(CISC Tech) skrev:> Hi All,> now X(z) can be written as X(z) = z/(z-a). ......(eq 3) > > Now the question is what are the poles and zeros location of > X(z)???????? > > DSP Books says Zero at z = 0 Pole at z = a from (eq 3) > > but how can z=0 be substituted in the equation for X(z) = 1/(1-az^-1) > as 0 < |a| which is not in ROC and it makes the original X(z) > calculated using (eq 1) to go to infinity.Because z^-1 means 1/z, and multiplying X(x) by z/z removes all 1/z terms from the expression. Rune

Reply by ●December 26, 20062006-12-26

On Tue, 26 Dec 2006 12:46:31 -0000, Rune Allnor <allnor@tele.ntnu.no> = wrote:> > PrasadBC(CISC Tech) skrev: >> Hi All, > >> now X(z) can be written as X(z) =3D z/(z-a). ......(eq 3=)>> >> Now the question is what are the poles and zeros location of >> X(z)???????? >> >> DSP Books says Zero at z =3D 0 Pole at z =3D a from (eq 3) >> >> but how can z=3D0 be substituted in the equation for X(z) =3D 1/(1-az=^-1)>> as 0 < |a| which is not in ROC and it makes the original X(z)=>> calculated using (eq 1) to go to infinity. > > Because z^-1 means 1/z, and multiplying X(x) by z/z removes > all 1/z terms from the expression.That doesn't really address the question though, which was (as I = understand it) "How can convergent points exist outside the ROC?", as th= e = term "ROC" would imply "the set of values of z which cause X(z) to = converge". I guess the answer is that singularities can exist outside t= he = ROC, at which convergence still occurs, i.e. the definition of ROC is = actually "a subset of the values of z which cause X(z) to converge". On a related note, if we look at the expanded summation form of X(z) (as= = given by the OP): X(z) =3D 1 + a^1.z^-1 + a^2.z^-2 + a^3.z^-3 + ... (Eq.4) it would appear that at z=3D0, this should be equal to 1, and hence isn'= t a = zero! This problem seems to be caused by considering the general form o= f = the Z-transform: inf X(z) =3D SUM x[n].z^-n (Eq.5) n=3D0 What is the value at z=3D0? This requires evaluating the summation term= = when n=3D0, which is x[0].0^0, which would appear to be undefined. We c= an = get round this problem by re-defining the Z-transform as: inf X(z) =3D x[0] + SUM x[n].z^-n (Eq.6) n=3D1 which in turn leads to (Eq.4). But this definition would seem to preclu= de = the existence of zeros at z=3D0 (unless x[0]=3D0). In summary, I'm thoroughly confused! -- = Oli

Reply by ●December 26, 20062006-12-26

On 26 Dec 2006 04:46:31 -0800, "Rune Allnor" <allnor@tele.ntnu.no> wrote:>Because z^-1 means 1/z, and multiplying X(x) by z/z removes >all 1/z terms from the expression.I think that you misunderstood the OP's question (either that, or I'm reading more into it than is actually there). It looks to me like he is addressing exactly the same issue, using exactly the same example, that I did in this thread: http://groups.google.com/group/comp.dsp/browse_thread/thread/de6ea7b9c51ca3ec/517a6e6a88507733 Basically, it is this: How can a transfer function have a zero at a location outside its region of convergence? Does it make any sense at all to discuss behavior outside the ROC? -- Greg Berchin

Reply by ●December 26, 20062006-12-26

Oli Charlesworth skrev:> That doesn't really address the question though, which was (as I > understand it) "How can convergent points exist outside the ROC?",If that's the OP's question, I didn't catch it.> as the > term "ROC" would imply "the set of values of z which cause X(z) to > converge". I guess the answer is that singularities can exist outside the > ROC, at which convergence still occurs, i.e. the definition of ROC is > actually "a subset of the values of z which cause X(z) to converge".The function X(z) = z/(z-a) has exactly one point where it does not converge. So pedantically speaking, I suppose the ROC can be defined as "everywhere except z=a". I think the usual terminology in context of the z transform is that the ROC denotes a region 0 <= b < |z| < c <= inf where no poles exist.> In summary, I'm thoroughly confused!Me too. After posting my first response, I thought along these lines: Let P be a polynomial in x, P(x) = a_n x^n + ... + a_1 x + 1 Multiply by x^-n/x^-n = 1 and re-order coefficients: P'(x) = x^-n +a_1 x^(n-1) + ... + a_n Since we only multiplied by 1, P'(x)=P(x). However, P'(0) = infinite because of the 1/x terms, while P(0) = 1. Where is the flaw? And no, this is not a troll. I don't see what cancels the effect of the 1/x terms for x = 0. Rune

Reply by ●December 26, 20062006-12-26

On Tue, 26 Dec 2006 13:29:27 -0000, Rune Allnor <allnor@tele.ntnu.no> = wrote:> > Oli Charlesworth skrev: > >> That doesn't really address the question though, which was (as I >> understand it) "How can convergent points exist outside the ROC?", > > If that's the OP's question, I didn't catch it. > >> as the >> term "ROC" would imply "the set of values of z which cause X(z) to >> converge". I guess the answer is that singularities can exist outsid=e =>> the >> ROC, at which convergence still occurs, i.e. the definition of ROC is=>> actually "a subset of the values of z which cause X(z) to converge". > > The function X(z) =3D z/(z-a) has exactly one point where it does > not converge. So pedantically speaking, I suppose the ROC > can be defined as "everywhere except z=3Da".I think you are confusing two things, "convergence", and "singularities"= . The function X(z) =3D z/(z-a) does indeed have a singularity at z=3Da, b= ut it = is not meaningful to talk about "convergence" of X(z). The term "convergence" (and therefore "ROC") can only apply to an infini= te = summation; in our case it is meaningful to talk about the convergence of= : X(z) =3D SUM x[n].z^-n (Eq.7) which has a clearly-defined ROC. If we only consider values within that= = ROC, it is valid to specify X(z) as: X(z) =3D z/(z-a) (Eq.8) However, (Eq.8) is *not* the same as (Eq.7) outside the ROC. Specifying a Z-transform on its own doesn't uniquely define the original= = sequence. However, specifying a Z-transform *and* the ROC does uniquely= = specify x[n].> I think the usual terminology in context of the z transform is > that the ROC denotes a region 0 <=3D b < |z| < c <=3D inf where > no poles exist. > >> In summary, I'm thoroughly confused! > > Me too. After posting my first response, I thought along these > lines: > > Let P be a polynomial in x, > > P(x) =3D a_n x^n + ... + a_1 x + 1 > > Multiply by x^-n/x^-n =3D 1 and re-order coefficients: > > P'(x) =3D x^-n +a_1 x^(n-1) + ... + a_n > > Since we only multiplied by 1, P'(x)=3DP(x). > > However, P'(0) =3D infinite because of the 1/x terms, > while P(0) =3D 1. > > Where is the flaw?The flaw is that you multiplied by x^-n, not (x^-n/x^-n)! -- = Oli

Reply by ●December 26, 20062006-12-26

Oli Charlesworth skrev:> On Tue, 26 Dec 2006 13:29:27 -0000, Rune Allnor <allnor@tele.ntnu.no> > wrote: > > > > > Oli Charlesworth skrev: > > > >> That doesn't really address the question though, which was (as I > >> understand it) "How can convergent points exist outside the ROC?", > > > > If that's the OP's question, I didn't catch it. > > > >> as the > >> term "ROC" would imply "the set of values of z which cause X(z) to > >> converge". I guess the answer is that singularities can exist outside > >> the > >> ROC, at which convergence still occurs, i.e. the definition of ROC is > >> actually "a subset of the values of z which cause X(z) to converge". > > > > The function X(z) = z/(z-a) has exactly one point where it does > > not converge. So pedantically speaking, I suppose the ROC > > can be defined as "everywhere except z=a". > > I think you are confusing two things, "convergence", and "singularities". > > The function X(z) = z/(z-a) does indeed have a singularity at z=a, but it > is not meaningful to talk about "convergence" of X(z). > > The term "convergence" (and therefore "ROC") can only apply to an infinite > summation; in our case it is meaningful to talk about the convergence of: > > X(z) = SUM x[n].z^-n (Eq.7) > > which has a clearly-defined ROC. If we only consider values within that > ROC, it is valid to specify X(z) as: > > X(z) = z/(z-a) (Eq.8) > > However, (Eq.8) is *not* the same as (Eq.7) outside the ROC. > > Specifying a Z-transform on its own doesn't uniquely define the original > sequence. However, specifying a Z-transform *and* the ROC does uniquely > specify x[n].The Z transform is defined as the sum from negative infinite to positive infinite. The Z transform (7) is well-behaved everywhere except at z=a, for a suitable choise of summation limits. It is guaranteed not to converge for z = a, regardless of summation limits. Since the ZT is linear, the ZT of one function can be separated as the sum of ZTs two functions. Since the sign of the exponent n is crucial for the discussion, we can make an ARBITRARY choise to separate x[n] to one series for negative n and another for non-negative n: ZT{x[n]} = ZT{x1[n]} + ZT{x2[n]} (1) where ZT{x1[n]}^{-} = sum_{-inf}^-1 x[n] z^-n (2) ZT{x2[n]}^{+} = sum{0}^{+inf} x[n] z^-n (3) Exploiting the linearity of the ZT like this and employing a divide-and-conquer strategy, one can see that it converges everywhere except at the poles. The convention is to define as the ROC the region where the ZT converges under the summation limits as in (3). This choise is CONVENIENT since we usually want to discuss causal sequences. Rune

Reply by ●December 26, 20062006-12-26

On Tue, 26 Dec 2006 14:11:13 -0000, Rune Allnor <allnor@tele.ntnu.no> = wrote:> > Oli Charlesworth skrev: >> On Tue, 26 Dec 2006 13:29:27 -0000, Rune Allnor <allnor@tele.ntnu.no>=>> wrote: >> > >> > The function X(z) =3D z/(z-a) has exactly one point where it does >> > not converge. So pedantically speaking, I suppose the ROC >> > can be defined as "everywhere except z=3Da". >> >> I think you are confusing two things, "convergence", and =>> "singularities". >> >> The function X(z) =3D z/(z-a) does indeed have a singularity at z=3Da=, but =>> it >> is not meaningful to talk about "convergence" of X(z). >> >> The term "convergence" (and therefore "ROC") can only apply to an =>> infinite >> summation; in our case it is meaningful to talk about the convergence==>> of: >> >> X(z) =3D SUM x[n].z^-n (Eq.7) >> >> which has a clearly-defined ROC. If we only consider values within t=hat>> ROC, it is valid to specify X(z) as: >> >> X(z) =3D z/(z-a) (Eq.8) >> >> However, (Eq.8) is *not* the same as (Eq.7) outside the ROC. >> >> Specifying a Z-transform on its own doesn't uniquely define the origi=nal>> sequence. However, specifying a Z-transform *and* the ROC does uniqu=ely>> specify x[n]. > > The Z transform is defined as the sum from negative infinite > to positive infinite. The Z transform (7) is well-behaved everywhere > except at z=3Da, for a suitable choise of summation limits.This is untrue. The Z-transform of a^n.u[n] is: X(z) =3D SUM a^n.z^-n and is not well-behaved (in the sense of convergence) for *any* value of= = |z|<=3D|a|, not just z=3Da. (This is in contrast to the expression z/(z-a), which is not well-behave= d = only at z=3Da. Hence it is a different function, which is why we must = specify the ROC when discussing z/(z-a)).> It is > guaranteed not to converge for z =3D a, regardless of summation > limits.This, however, is true (with the proviso that we choose the upper limit = as = +inf; for any other upper limit, it is a finite sum, and will always = "converge").> Since the ZT is linear, the ZT of one function can be separated > as the sum of ZTs two functions. Since the sign of the exponent n > is crucial for the discussion, we can make an ARBITRARY > choise to separate x[n] to one series for negative n and > another for non-negative n: > > ZT{x[n]} =3D ZT{x1[n]} + ZT{x2[n]} (1) > > where > > ZT{x1[n]}^{-} =3D sum_{-inf}^-1 x[n] z^-n (2) > ZT{x2[n]}^{+} =3D sum{0}^{+inf} x[n] z^-n (3) > > Exploiting the linearity of the ZT like this and employing a > divide-and-conquer strategy, one can see that it converges > everywhere except at the poles.Again, the Z-transform does not converge *anywhere* outside the ROC, i.e= . = in our example, it doesn't converge for any |z|<=3D|a| (except possibly = at = the singular values specified as the zeros; I'm not too sure on that one= !). -- = Oli

Reply by ●December 26, 20062006-12-26

Oli Charlesworth skrev:> Again, the Z-transform does not converge *anywhere* outside the ROC, i.e. > in our example, it doesn't converge for any |z|<=|a| (except possibly at > the singular values specified as the zeros; I'm not too sure on that one!).So what is your point? That anything that happens inside the ROC has to be discarded? If that's the case, all one would need was the outermost pair of poles, and whatever zeros might be present in the ROC. Rune

Reply by ●December 26, 20062006-12-26

On Tue, 26 Dec 2006 14:41:57 -0000, Rune Allnor <allnor@tele.ntnu.no> = wrote:> > Oli Charlesworth skrev: > >> Again, the Z-transform does not converge *anywhere* outside the ROC, ==>> i.e. >> in our example, it doesn't converge for any |z|<=3D|a| (except possib=ly at>> the singular values specified as the zeros; I'm not too sure on that ==>> one!). > > So what is your point? That anything that happens inside the ROC > has to be discarded? If that's the case, all one would need was the > outermost pair of poles, and whatever zeros might be present > in the ROC.I'm not sure I had a point! (and I'm not sure what you mean by "discarde= d") However, to analyse stability of a causal system (causality requires tha= t = the ROC include infinity), it is true that all you need to know are the = = locations of the outermost poles, as this places a bound on the ROC (the= = ROC cannot by definition include poles). If this ROC doesn't include th= e = unit circle, then the system is unstable. This is the formal explanation for why poles *must* be inside the unit = circle for a causal system to be stable. -- = Oli