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Z-transform of conjugated sequence ( a straightforward exercise)

  1. Sep 26, 2012 #1
    Z-transform of a conjugated sequence ("a straightforward" exercise)

    1. The problem statement, all variables and given/known data
    The conjugation property is expressed as [tex]x^*[n] \stackrel{Z}{\leftrightarrow} X^*(z^*)[/tex]
    This property follows in a straightforward maner from the definition of the [itex]z[/itex]-transform, the details of which are left as an exercise.


    2. Relevant equations
    Z-transform definition: [tex]X(z)=\sum_{n=-\infty}^\infty x[n]z^{-n}[/tex]


    3. The attempt at a solution
    Given a complex sequence, its z-transform is [tex]Z\{x[n]\} = \sum_{n=-\infty}^\infty (x_R[n] + jx_I[n]) z^{-n} = X_R(z) + jX_I(z) = X(z)[/tex]

    Hence, the z-transform of a conjugated sequence [tex]Z\{x^*[n]\} = \sum_{n=-\infty}^\infty (x_R[n] - jx_I[n]) z^{-n} = X_R(z) - jX_I(z) = X^*(z)[/tex]

    Now, how come I didn't get the [itex]z^*[/itex], as in [itex]X^*(z^*)[/itex]? :confused:
     
    Last edited: Sep 26, 2012
  2. jcsd
  3. Sep 28, 2012 #2
    Re: Z-transform of conjugated sequence ("a straightforward" exercise)

    What am I doing wrong in trying to show [itex]x^*[n] \stackrel{Z}{\leftrightarrow} X^*(z^*)[/itex]?
     
  4. Sep 28, 2012 #3

    rude man

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    Re: Z-transform of conjugated sequence ("a straightforward" exercise)

    Fact: x*y = (xy*)* for any complex numbers x and y.

    Apply that to X*(z) = Ʃx*(n)z-n.
     
  5. Sep 29, 2012 #4
    Re: Z-transform of conjugated sequence ("a straightforward" exercise)

    rude man, thank you for the nudge in the right direction:

    [tex]Z\{x^*[n]\} = \sum_{n=-\infty}^\infty x^*[n]z^{-n} = \sum_{n=-\infty}^\infty \left(x[n](z^*)^{-n}\right)^* = \left(\sum_{n=-\infty}^\infty x[n](z^*)^{-n}\right)^* = \left(X(z^*)\right)^* = X^*(z^*)[/tex]

    Is this correct? If so, what did I do wrong in the first attempt?
     
  6. Sep 29, 2012 #5

    rude man

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    Re: Z-transform of conjugated sequence ("a straightforward" exercise)

    Yes, that's right.

    As for what's wrong with your first try: unfortunately I never had to deal with complex signals so I need to bone up on this stuff more myself. Sorry this is the case.

    Maybe this will help: Z(Re{x(n)}) = (1/2){X(z) + X*(z*)} and
    Z(Im{x(n)}) = (1/2j){X(z) - X*(z*)}.

    So that Z(Re{x(n)} + jZ(Im{x(n)}) = X(z) as required.

    Stay tuned, I hope to do more later ...
     
  7. Sep 29, 2012 #6
    Re: Z-transform of conjugated sequence ("a straightforward" exercise)

    That's like cat chasing its tail. :) And it also doesn't help shedding light on my 1st, erroneous attempt. Hope you'll find something out; I'll most surely post as well if I find the error.
     
  8. Sep 29, 2012 #7

    rude man

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    Re: Z-transform of conjugated sequence ("a straightforward" exercise)

    Point is, you have written Z(Re{x(n)}) as though it has no complex part, but it does, unless x(n) is itself real. Same goes for Z(Im{x(n)}) where you just stuck a j in front of another z-transformed function as though it too were real.

    Thanks for getting back if you do find the error. I would really like to know.
     
  9. Sep 29, 2012 #8
    Re: Z-transform of conjugated sequence ("a straightforward" exercise)

    You mean those "R" and "I" indices (as in [itex]X_R[/itex] and [itex]X_I[/itex])? Say [itex]x[n]=\{..., 1-j2, 5+j, ...\}[/itex]. Then, [itex]x_R[n]=\{..., 1, 5, ...\}[/itex], [itex]x_I[n]=\{..., -2, 1, ...\}[/itex], and so [itex]x[n] = x_R[n] + jx_I[n][/itex] ... also, [itex]x^*[n]=\{..., 1+j2, 5-j, ...\}[/itex].

    So, [tex]Z\{x[n]\} = \sum_{n=-\infty}^\infty (x_R[n] + jx_I[n]) z^{-n} = \sum_{n=-\infty}^\infty x_R[n] z^{-n} + \sum_{n=-\infty}^\infty jx_I[n] z^{-n} = \sum_{n=-\infty}^\infty x_R[n] z^{-n} + j\sum_{n=-\infty}^\infty x_I[n] z^{-n} = X_R(z) + jX_I(z) = X(z)[/tex]

    Surely this is valid? :)
     
  10. Sep 29, 2012 #9

    rude man

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    Re: Z-transform of conjugated sequence ("a straightforward" exercise)

    Apparently not. Andf I don't mind admitting I don't know why it isn't. This math is - pardon the pun - complex.

    From what I suggested before:

    Z{x*[n]} = Z{Re(x[n])} - jZ{Im(x[n])} which I know you agree with.

    = (1/2){X(z) + X*(z*)} - (j/2j){X(z) - X*(z*)} which I got from a table of Z transforms.
    =X*(z*) QED

    I'm afraid I've exhausted my reservoir of knowledge in this situation. Again, my apologies.
     
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