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

• Sep 28th 2012, 01:47 AM
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 $\displaystyle x^*[n] \stackrel{Z}{\leftrightarrow} X^*(z^*)$.
This property follows in a straightforward maner from the definition of the $\displaystyle z$-transform, the details of which are left as an exercise.

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

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

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

Now, how come I didn't get the $\displaystyle z^*$, as in $\displaystyle X^*(z^*)$? :confused:

PS. I apologize beforehand for copying verbatim from physicsforums.
• Sep 28th 2012, 08:28 PM
chiro
Re: Z-transform of a conjugated sequence ("a straightforward" exercise)
Hey courteous.

Did you try conjugating the z first and then doing the other one after?
• Sep 28th 2012, 10:31 PM
courteous
Re: Z-transform of a conjugated sequence ("a straightforward" exercise)
I was hinted at this fact: $\displaystyle x^*y = (xy^*)^*$. This then leads to

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

But, what went wrong in my first attempt? (Wondering) Surely $\displaystyle X^*(z^*) \neq X^*(z)$?
• Sep 28th 2012, 10:55 PM
chiro
Re: Z-transform of a conjugated sequence ("a straightforward" exercise)
I'm wondering if you apply the conjugation operation if you need to do it to the whole thing and not just your basis coefficients.
• Sep 28th 2012, 11:11 PM
courteous
Re: Z-transform of a conjugated sequence ("a straightforward" exercise)
Chiro, do you mean, if the following step is correct?
Quote:

Originally Posted by courteous
$\displaystyle \sum_{n=-\infty}^\infty \left(x[n](z^*)^{-n}\right)^* = \left(\sum_{n=-\infty}^\infty x[n](z^*)^{-n}\right)^*$

I think it is, since $\displaystyle (x+y)^* = (x^*+y^*)$? So, is my 2nd attempt valid ... and what went wrong with the 1st attempt (in the OP)?
• Sep 28th 2012, 11:13 PM
chiro
Re: Z-transform of a conjugated sequence ("a straightforward" exercise)
No what I mean is that if you conjugate the x[n], you also have to conjugate the z as well.
• Sep 28th 2012, 11:24 PM
courteous
Re: Z-transform of a conjugated sequence ("a straightforward" exercise)
So, if one computes the z-transform of a conjugated sequence $\displaystyle Z\{x^*[n]\}$, one should also conjugate the $\displaystyle z$ in the very definition of the z-transform?

Like so $\displaystyle Z\{x^*[n]\} = \sum_{n=-\infty}^\infty x^*[n](z^*)^{-n}$ ?
• Sep 28th 2012, 11:38 PM
chiro
Re: Z-transform of a conjugated sequence ("a straightforward" exercise)
I'm not sure and I'm not going to tell you something I don't know, but this might be something to check out.

One thing I do know is that in integral transforms (like the continuous versions of these transforms like the Fourier Transform), these transforms deal with a standard definition of a particular variable, and if you want to conjugate the actual variable, then you are going to have to make sure that conjugation is applied every-where else down the line.

Now if if you are doing an inversion (which it looks like), it means that if you have done a transform with respect to a conjugated variable, then the inverse must also be done with respect to a conjugated variable and not the non-conjugated variable because if you don't do that, the two are not equivalent.

If you want to prove this yourself, what you should do is start from the forward transform and then look at the inverse and what happens when you use the conjugate for the forward but then disregard for the inverse: if what I'm saying is right, then you should get a contradiction and that will show you mathematically why you need to conjugate your variable.

The easiest way to think about this is that you have your inverse transform say F^(-1) and your normal transform F where F^(-1)(F(z*)) = z* and F^(-1)(F(z)) = z but in the context of functions and not a single variable.
• Sep 29th 2012, 12:55 AM
courteous
Re: Z-transform of a conjugated sequence ("a straightforward" exercise)
Well, I (very) likely know even (much) less about this (or mathematics in general). But, surely one needn't change (forward) Z-transform definition between inputs, say, $\displaystyle x[n]=\{..., 1-j2, 5+j, ...\}$ and $\displaystyle x^*[n]=\{..., 1+j2, 5-j, ...\}$?

Also, as far as I can tell, there is no inverse Z-transform involved.

Chiro, in short: is there anything wrong with post #3 (and how is it different from the 1st attempt in post #1)?
• Sep 29th 2012, 01:50 AM
chiro
Re: Z-transform of a conjugated sequence ("a straightforward" exercise)
There is an inverse:

Z-transform - Wikipedia, the free encyclopedia

Also look at the wiki and look at the very definition: it means that I said is right. You need to consider what X(z) in comparison to X(z*) where you need to replace z with z*.

So if you are proving results related to z*, you need to look at X(z*) as opposed to X(z) just like there is a difference between X(a) and X(b) for any appropriate a and b.
• Sep 29th 2012, 06:30 AM
courteous
Re: Z-transform of a conjugated sequence ("a straightforward" exercise)
Quote:

Originally Posted by chiro

I know that the inverse Z-transform exists, but (as far as I can tell) it (or its properties) isn't called for in this problem.

Quote:

Originally Posted by chiro
Also look at the wiki and look at the very definition: it means that I said is right. You need to consider what X(z) in comparison to X(z*) where you need to replace z with z*.

I take it you mean definition of a forward Z-transform? You're correct: this example from Harvey Mudd (under "Conjugation") proceeds in the manner you suggested (by substituting z with z*). Also, post #3 seems correct, since Wikipedia has the exact same proof under Z-transform properties, complex conjugation.