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Complex integral for z-transform causality

  1. Oct 2, 2015 #1
    This relates to z-transform causality, but I'll try to phrase it as a complex analysis question. Suppose I have a function ##X(z)## whose poles are all inside the unit circle, and which has the property
    [tex]\lim_{|z|\to\infty} \frac{X(z)}{z} = 0[/tex]
    Is that sufficient to guarantee that
    [tex]\frac{1}{2\pi i}\oint_C X(z) z^{n-1} dz = 0 [/tex]
    for ##n < 0## (where ##C## is the unit circle)?

    My thinking is that the residue at infinity is
    [tex] \operatorname{Res}_{\infty}\left[X(z) z^{n-1} \right] = \lim_{|z| \to\infty} X(z) z^n = 0[/tex]
    for ## n < 0 ## because of the stated property of ##X(z)##. Then, since we've specified that there are no other poles outside the unit circle, the integral would have to be proportional to this residue, and thus it would have to be zero. Is that correct?

    Alternatively, suppose I know that ##X(z)## has at least as many poles as zeros, and again all the poles are inside the unit circle. Is that enough to guarantee that the above integral will be zero for ##n<0##? I know it's true for rational ##X(z)## but I'm not quite sure how to prove it in general.
     
  2. jcsd
  3. Oct 5, 2015 #2

    Svein

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    You also assume that ##X(z)## is meromorphic - right? Then [itex]\frac{1}{2\pi i}\oint_{C}X(z)dz=\sum_{\lvert z\rvert<1} Res(X(z)) [/itex] regardless of how ##X(z)## behaves outside ##C##.
    Now, since ##\frac{X(z)}{z}_{z\rightarrow \infty}\rightarrow 0##, ##X(z)## must have at least as many poles as zeros (the zeros need not be inside ##C##, however). Furthermore, there are three possibilities for ##z=0##:
    • ##X## has neither a pole nor a zero at ##z=0##. Then [itex]\frac{X(z)}{z} [/itex] has a simple pole at ##z=0## and that residue is added to the other residues for ##X##.
    • ##X## has a simple pole at ##z=0##. Then that residue is already part of the integral.
    • ##X## has a zero at ##z=0##. Then sooner or later ##\frac{X(z)}{z^{n}}## will have a simple pole at ##z=0## and that residue is added to the other residues for ##X##.
    After that simple pole is done, we are left with ##\frac{K}{z^{n}}##, n>1, at ##z=0##. But [itex] \oint_{C}\frac{K}{z^{n}}dz=0[/itex] for n>1, so it has no effect on the sum of the other residues.
     
    Last edited: Oct 5, 2015
  4. Oct 5, 2015 #3
    Yes, I can assume ##X(z)## is meromorphic.

    I'm a little confused by your analysis... unless I'm misunderstanding, it seems to neither prove nor disprove my assertion. Yes, the integral will be equal to the residues inside ##C##, but I'm effectively arguing that those residues must add up to zero in the specified situation.

    A related theorem is the inside-outside theorem (http://mathworld.wolfram.com/Inside-OutsideTheorem.html). In that case, if (#poles)>(#zeros+1), then the residues inside the curve plus the residues outside the curve must add to zero. So if all the poles are inside ##C##, then the residues must add to zero. That's exactly what I want, but the statement on that page is restricted to rational functions. I'm wondering if it can be generalized, either by insisting that ##X(z)/z \to 0## as ##|z| \to \infty##, or by insisting that (#poles)>(#zeros+1).

    So maybe another way to ask my question is this: is there a way to generalize the inside-outside theorem to non-rational functions? Perhaps you can say that the residues inside equal the residues outside, but you have to include residues at infinity?

    Apologies if I'm just being slow here. I'm an electrical engineer, so proving things isn't my strong suit.
     
  5. Oct 5, 2015 #4

    Svein

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    I do not agree. Let us take an example: [itex]X(z)=\frac{(z+2)}{(z-\frac{i}{2})(z+\frac{1}{2})} [/itex]. Here all poles are inside ##C## and there are two poles and one zero.
    Now the basic integral [itex]\frac{1}{2\pi i}\oint_{C}X(z)dz = R_{z=\frac{i}{2}}+R_{z=-\frac{1}{2}}=\frac{4+i}{1+i}+\frac{-3}{1+i}=1 [/itex] - which is not zero.
     
  6. Oct 5, 2015 #5
    That example doesn't satisfy the required condition (#poles)>(#zeros)+1. The key to the whole thing is that the integrand needs to go to zero faster than ##1/z##.

    Interestingly, though, the inside-outside theorem still seems to apply to your example if you include poles at infinity. The residue at infinity is
    [tex] \lim_{z\to 0} \frac{-1}{z} X\left(\frac{1}{z}\right) = \lim_{z\to 0} \frac{-1 - 2z}{(1 - iz/2)(1+z/2)} = -1 [/tex]
    So we could have calculated ##\oint_C X(z) d z## by adding the negative of all the residues outside ##C##, which in this case is just the residue at infinity.

    Edit: In the last post I said
    I realize that wasn't clear at all. What I should have said was "I'm wondering whether the inside-outside theorem can be generalized, either by insisting that ##zX(z) \to 0## as ##z\to 0##, or by insisting that (#poles of ##X##)>[(#zeros of ##X##)+1], where ##X## is the integrand in the inside-outside theorem, and not from my original question."
     
    Last edited: Oct 5, 2015
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