I should know the answer but I am ashamed not to. :-( z^-1 z X(z) = ------------------------ = -------------------- 1 - 2^0.5 * z^-1 + z^-2 z^2 - 2^0.5 * z + 1 If I only look at the first form, I anwer yes. However, I cannot see any pole at z=0 in the second form. It has a well defined limit of 0 as z->0. What is your answer? Thank you very much. Jean Castonguay �lectrocommande Pascal
Has this X(z) a pole at z=0 or not?
Started by ●January 7, 2005
Reply by ●January 7, 20052005-01-07
"Jean Castonguay" <jcastong@riq.qc.ca> writes:> I should know the answer but I am ashamed not to. :-( > > z^-1 z > X(z) = ------------------------ = -------------------- > 1 - 2^0.5 * z^-1 + z^-2 z^2 - 2^0.5 * z + 1 > > If I only look at the first form, I anwer yes. > > However, I cannot see any pole at z=0 in the second form. It has a > well defined limit of 0 as z->0. > > What is your answer?Hi Jean, If you define a "pole" as a place where the denominator of a rational function goes to zero, then you cannot say the first form has a pole a z=0 since, as you have deduced, it is undefined at z=0. -- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124
Reply by ●January 7, 20052005-01-07
>>>>> "Jean" == Jean Castonguay <jcastong@riq.qc.ca> writes:Jean> I should know the answer but I am ashamed not to. :-( Jean> z^-1 z Jean> X(z) = ------------------------ = -------------------- Jean> 1 - 2^0.5 * z^-1 + z^-2 z^2 - 2^0.5 * z + 1 Jean> If I only look at the first form, I anwer yes. Jean> However, I cannot see any pole at z=0 in the second form. It has a Jean> well defined limit of 0 as z->0. Jean> What is your answer? I would say the first form has a (gratutious) removable singularity (pole) at z = 0. Much like like how sin(z)/z has a gratuitous removable singularity at z = 0. Ray
Reply by ●January 7, 20052005-01-07
"Jean Castonguay" <jcastong@riq.qc.ca> wrote in message news:IbxCuxQzw2Ms-pn2-RuOLoy5J8UbI@localhost...>I should know the answer but I am ashamed not to. :-( > > z^-1 z > X(z) = ------------------------ = -------------------- > 1 - 2^0.5 * z^-1 + z^-2 z^2 - 2^0.5 * z + 1 > > If I only look at the first form, I anwer yes. > > However, I cannot see any pole at z=0 in the second form. It has a > well defined limit of 0 as z->0. >Answer to what? Is there is a pole at z=0? z^-1 X1(z) = ------------------------ 1 - 2^0.5 * z^-1 + z^-2 z X2(z) = ------------------------ = z^2*X1(z) z^2 - 2^0.5 * z + 1 Well, they are the same rational functions with one being in the z plane and the other maybe being in the z^-1 plane (Note the symmetry of coefficients in the denominator). So, one is the reciprocal of the other if they are taken in the same plane - with the unit circle being the same in both cases. The first form X1, in the z^-1 plane has a zero at z^-1=0 and has poles at z^-1 at +0.707+j.707 and +0.707-j0.707. The second form X2, in the z plane has the same zero at z=0 and has poles at z=+0.707+j.707 and +0.707-j0.707. Now, if you want to take the first form X1 in the z plane as we did X2, instead of in the z^-1 plane, then it has a zero at z=infinity and poles still at +0.707+j.707 and +0.707-j0.707 because they are on the unit circle and don't change with the transformation. When dealing with poles and zeros at infinity and at zero, you have to be a bit careful. If we just stay in the z plane we see that X1 has a zero at infinity. We note that X1(z) has been multiplied by z^2 to get X2(z) which introduces a double zero at z=0 and eliminates the zero at infinity. The symmetry of the demoninator doesn't give us a clue. Try it with zeros that aren't on the unit circle: z^-1 X1(z^-1) = ------------------ z^-2 - z^-1 + 0.5 substituting z for z^-1: z X1(z) = ------------------ z^2 - z^1 + 0.5 So, X1(z^-1) has the same singularities in the z^-1 plane as X1(z) has in the z plane. multiplying the original expression for X1(z^-1) by z^2: z X1(z^-1) = ------------------ 0.5*z^2 - z + 1 Since z X2(z) = --------------- 0.5*z^2 - z + 1 Then X2(z)= X1(z^-1) X2(z) and X1(z^-1) have a zero at z=0 and poles at z=1 +/- j; that is, in the z plane. X1(z^-1) has a zero at z^-1=infinity and poles at z^-1 = 0.5 +/- j*0.5; that is, in the z^-1 plane, which is consisent with the former statement about the z plane. X1(z) has a zero at z=0 and poles at z=0.5 +/- j*0.5 as above. The use of the z^-1 plane instead of the z plane is a not so common variation amongst authors of texts. Kaiser in an early paper: "Some Practical Considerations in the Realization of Linear Digital Filters" plots poles and zeros in the z^-1 plane. Most authors work with the z plane. Obviously, for stability, the poles have to be inside the unit circle in the z plane and outside the unit circle in the z^-1 plane. Note that the zero plots for linear phase FIR filters are the same in either plane due to radial reciprocal symmetry. Does that help? Fred
Reply by ●January 7, 20052005-01-07
What is '2�.5' intended to mean? I assume there's a font problem. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●January 7, 20052005-01-07
What is '2�.5' intended to mean? I assume there's a font problem. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●January 7, 20052005-01-07
Jerry Avins wrote:> What is '2�.5' intended to mean? I assume there's a font problem. > > JerryIf you use Mozilla and you're seeing something that looks like "two degrees dot five" then it's because Fred gave you 2 ^ 0 . 5 without the spaces -- Mozilla automagically turns number uparrow number into an exponent. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Reply by ●January 7, 20052005-01-07
Tim Wescott wrote:> Jerry Avins wrote: > >> What is '2�.5' intended to mean? I assume there's a font problem. >> >> Jerry > > > If you use Mozilla and you're seeing something that looks like "two > degrees dot five" then it's because Fred gave you 2 ^ 0 . 5 without the > spaces -- Mozilla automagically turns number uparrow number into an > exponent.That didn't occur to me because I also see z^-1 etc. Thanks fir the word. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●January 7, 20052005-01-07
"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message news:saKdnSaRDYrqn0LcRVn-qQ@centurytel.net...> > Answer to what? Is there is a pole at z=0? > > z^-1 > X1(z) = ------------------------ > 1 - 2^0.5 * z^-1 + z^-2 > > z > X2(z) = ------------------------ = z^2*X1(z) > z^2 - 2^0.5 * z + 1Hey Fred, Last time I did algebra, multiplying the numerator and the denominator by the same (non-zero) thing left you with what you started with. So, z X2(z) = ------------------------ = X1(z) z^2 - 2^0.5 * z + 1 Yeah? The anomaly occurs when z=0, because you've got a whole multiply by 0/0 thing going on. Cheers, Syms.
Reply by ●January 13, 20052005-01-13
On Fri, 7 Jan 2005 21:43:20 UTC, "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote: Thank you, Mr. Marshall. You gave me a lot of information that I have yet to master.> > > > The use of the z^-1 plane instead of the z plane is a not so common > variation amongst authors of texts. Kaiser in an early paper: "Some > Practical Considerations in the Realization of Linear Digital Filters" plots > poles and zeros in the z^-1 plane. Most authors work with the z plane.Would you know where I can find the paper? When did the Z-transform �won� over other transforms? I have a French-language book dated 1970 that deals with a gamma-transform which is Z-transform with an opposite sign for the exponent.> > Obviously, for stability, the poles have to be inside the unit circle in the > z plane and outside the unit circle in the z^-1 plane. Note that the zero > plots for linear phase FIR filters are the same in either plane due to > radial reciprocal symmetry. > > Does that help? >Yes of course. -- Jean Castonguay �lectrocommande Pascal
