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

1. Sep 26, 2012

### courteous

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

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

3. 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 $X^*(z^*)$?

Last edited: Sep 26, 2012
2. Sep 28, 2012

### courteous

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

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

3. Sep 28, 2012

### rude man

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.

4. Sep 29, 2012

### courteous

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

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?

5. Sep 29, 2012

### rude man

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 ...

6. Sep 29, 2012

### courteous

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.

7. Sep 29, 2012

### rude man

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.

8. Sep 29, 2012

### courteous

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

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? :)

9. Sep 29, 2012

### rude man

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.