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I'm new on the forum and hence not familiar with formal writing of a post so before going in on this post I wanted to make this clear.

I'm an second year engineering student with a small knowledge of stability, discrete systems, continuous systems, laplace-transform, fourier-transform etcetera. This year I'm taking control system classes and we have given a system of the following discrete signal: $u_{0}, u_{1}+\frac{u_{0}}{2}, u_{2}+\frac{u_{1}}{2},+\frac{u_{0}}{4},...,u_{k}+\frac{u_{k-1}}{2}+...+\frac{u_{0}}{2^k}$

It is asked to design a filter system to reconstruct the initial signal. Therefore, we need to find the z-transform of the given signal. That's at least my plan. From the z-transform can other properties of the system be found: stability, poles, zeros, block-scheme, impulse response, ...

But there is my problem: I have rewritten the given signal like this: $u[k]=\sum_{i=0}^{k}\frac{u_{i}}{2^{k-i}}=\frac{1}{2^{k}}\sum_{i=0}^{k}{u_{i}2^{i}}$. But now I don't have a clue on how to derive the z-transformation. I've tried some other things like rewriting the formula, however it did not work in finding the z-transform.

Rewriting the formula

What can I do?

Extra question: If the z-transform is found of this equation, can this be considered as the transfer function of the filter system?

EDIT: Due to a good suggestion in the comment section, I have written out a small block diagram and derived the transfer function, but I still have trouble finding a way to design a filter from the given question because the coefficients of $u$ are not further defined. Here I give my new insights:

newly found block diagram

Many Thanks in advance.

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  • $\begingroup$ Note that you are not obliged to use the $z$ transform to get the inverse filter equation... Homework constraint? $\endgroup$
    – Damien
    Commented Mar 5, 2022 at 11:51
  • $\begingroup$ No, it is not a homework constraint. It was just intuitively in my mind better to use a z transform. Due to your suggestion I have written out a block diagram (I updated my question with my new notations), but I have no idea how to derive a filter from this diagram because the given u-constants are not further defined. $\endgroup$ Commented Mar 5, 2022 at 12:26

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Let us define the $Z$ transform by

$$ U(z) = \sum_{n=0}^{\infty}u_n z^{-n} $$

$$ V(z) = \sum_{n=0}^{\infty}v_n z^{-n} $$

Here we implicitly assume that $u_n=0, v_n=0$ for $n<0$ (causal signals).

From the relation $u_{0}, u_{1}+\frac{u_{0}}{2}, u_{2}+\frac{u_{1}}{2},+\frac{u_{0}}{4},...,u_{k}+\frac{u_{k-1}}{2}+...+\frac{u_{0}}{2^k}$, we can directly derive

$$ v_k=u_k+ \frac{1}{2}v_{k-1}$$

$$ u_k = v_k - \frac{1}{2}v_{k-1} $$

and then that

$$U(z)=\left(1-\frac{z^{-1}}{2}\right)V(z)$$

$$V(z)=\left(\frac{1}{1-\frac{z^{-1}}{2}}\right)U(z)$$

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    $\begingroup$ I guess the there is a typo in the final expressions. That should be $z^{-1}$ instead of $z$, no? $\endgroup$
    – KBS
    Commented Mar 5, 2022 at 16:02
  • $\begingroup$ @KBS Corrected. Thank you. $\endgroup$
    – Damien
    Commented Mar 5, 2022 at 16:55
  • $\begingroup$ No worries. Here to help. $\endgroup$
    – KBS
    Commented Mar 5, 2022 at 16:58

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