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Z-transform of conjugated sequence ( a straightforward exercise)

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

Homework Statement


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.


Homework Equations


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


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:
 
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Answers and Replies

  • #2
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What am I doing wrong in trying to show [itex]x^*[n] \stackrel{Z}{\leftrightarrow} X^*(z^*)[/itex]?
 
  • #3
rude man
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Fact: x*y = (xy*)* for any complex numbers x and y.

Apply that to X*(z) = Ʃx*(n)z-n.
 
  • #4
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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?
 
  • #5
rude man
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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 ...
 
  • #6
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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.
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.
 
  • #7
rude man
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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.
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.
 
  • #8
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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? :)
 
  • #9
rude man
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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? :)
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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