Z-transform of conjugated sequence ( a straightforward exercise)

[courteous]

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

Homework Statement

The conjugation property is expressed as x^*[n] \stackrel{Z}{\leftrightarrow} X^*(z^*)
This property follows in a straightforward maner from the definition of the z-transform, the details of which are left as an exercise.

Homework Equations

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

The Attempt at a Solution

Given a complex sequence, its z-transform is Z\{x[n]\} = \sum_{n=-\infty}^\infty (x_R[n] + jx_I[n]) z^{-n} = X_R(z) + jX_I(z) = X(z)

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

Now, how come I didn't get the z^*, as in X^*(z^*)?

 

[courteous]

What am I doing wrong in trying to show x^*[n] \stackrel{Z}{\leftrightarrow} X^*(z^*)?

 

[rude man]

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

Apply that to X*(z) = Ʃx*(n)z^-n.

 

[courteous]

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

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^*)

Is this correct? If so, what did I do wrong in the first attempt?

 

[rude man]

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 ...

 

[courteous]

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.

 

[rude man]

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.

 

[courteous]

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

So, 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)

Surely this is valid? :)

 

[rude man]

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

So, 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)

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.