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

• courteous
In summary: Clearly not. Andf I don't mind admitting I don't know why it isn't. This math is - pardon the pun - complex.In summary, the z-transform of a conjugated sequence is Z\{x^*[n]\} = \sum_{n=-\infty}^\infty (x_R[n] + jx_I[n]) z^{-n}.

#### 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 $X^*(z^*)$?

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

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

Apply that to X*(z) = Ʃx*(n)z-n.

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?

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 ...

rude man said:
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.

courteous said:
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.

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? :)

courteous said:
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.

## 1. What is the Z-transform of a conjugated sequence?

The Z-transform of a conjugated sequence is a mathematical operation that converts a discrete-time signal into a complex-valued function of a complex variable. It is used to analyze and manipulate discrete-time signals in the field of signal processing and control systems.

## 2. How is the Z-transform of a conjugated sequence calculated?

The Z-transform of a conjugated sequence is calculated using the formula: X(z) = Σ x(n) * z^(-n), where x(n) is the discrete-time signal and z is the complex variable.

## 3. What is the significance of the Z-transform of a conjugated sequence?

The Z-transform of a conjugated sequence is significant because it allows for the representation of discrete-time signals in the frequency domain, making it easier to analyze and manipulate them. It also provides a way to transfer properties of continuous-time signals to discrete-time signals.

## 4. How is the Z-transform of a conjugated sequence used in practical applications?

The Z-transform of a conjugated sequence is used in various practical applications such as digital filter design, system analysis and control, and spectral analysis of signals. It is also used in the design of digital communication systems and signal reconstruction from sampled data.

## 5. What are some common properties of the Z-transform of a conjugated sequence?

Some common properties of the Z-transform of a conjugated sequence include linearity, time-shifting, convolution, and frequency shifting. These properties can be used to simplify calculations and manipulate signals in the frequency domain.