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How would one go about calculating the Magnitude of...

$\mathcal{Z}$-transform = $$\frac{1}{1-z^{-2}}$$

I understand that z can be replaced with $${exp}^{j\omega}$$ and I am aware of the identities $${exp}^{j\omega}= {cos(\omega)+jsin(\omega)}$$.

$${exp}^{-j\omega}= {cos(\omega)-jsin(\omega)}$$.

I’m not 100% sure how to plug it all in though.

So I end up with... $$\frac{1}{1-z^{-2}}*\frac{z^{2}}{z^{2}}$$

To get $$\frac{z^{2}}{z^{2}-1}$$

Leading to $$\frac{exp^{2\omega}}{exp^{2\omega}-1}$$

And using eulers identity $$\frac{Cos(2\omega)+jsin(2\omega)}{Cos(2\omega)+jsin(2\omega)-1}$$

So I now have $$A(\mathcal{z})={cos(2\omega)+jsin(2\omega)}$$.

$$B(\mathcal{z})={cos(2\omega)+jsin(2\omega)-1}$$.

At this point I’m confused where to go. Do I just square and root? multiply by the complex conjugate? I’m just not sure. $$\frac{\sqrt{A(\mathcal{z})^{2}}}{\sqrt{B(\mathcal{z})^{2}}}$$

Any help or guidance would be amazing. This is my first Time doing this. Step by step would be greatly appreciated so that I could practice on more TF’s.

Thanks in advance to anyone who may help. Below is the response of the TF via an online calculator. I see the below as the response of a low-pass filter. However, I could be misreading it. enter image description here

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  • $\begingroup$ This is a pretty pointless exercise. Where does it come from? Replacing $z$ by $e^{j\omega}$ is meaningful if we want to compute the frequency response of a stable filter. However, the given transfer function has poles on the unit circle, hence there exists no frequency response. $\endgroup$
    – Matt L.
    Aug 22 at 11:13
  • $\begingroup$ @Matt L. This is a 3rd year uni question. That I’m working on. Sourced from a friend at uni. I’m not saying you’re wrong but I’ve put the transfer function in an online calculator and it gives a response that looks as if it is a low-pass filter. Will edit post to show response if possible. I would like to be able to calculate it myself using the information provided in the question (the TF). Any confirmation that Matt L is correct would be a great help. ( I did notice where the zero and poles were and was confused) $\endgroup$
    – New2Dsp
    Aug 22 at 11:42
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The transfer function

$$H(z)=\frac{1}{1-z^{-2}}\tag{1}$$

has two poles on the unit circle (at $z=\pm 1$). Consequently, the corresponding frequency response (i.e., the Fourier transform of the impulse response) doesn't exist as an ordinary function.$^1$

Nevertheless, it is possible to evaluate the magnitude of $H(z)$ on the unit circle $z=e^{j\omega}$ if we exclude the two points $z=1$ and $z=-1$. This is a basic exercise in complex numbers.

The following reformulation of $H(e^{j\omega})$ is not necessary, but it's an elegant way to immediately see its magnitude:

$$H(e^{j\omega})=\frac{1}{1-e^{-2j\omega}}=\frac{1}{e^{-j\omega}\left(e^{j\omega}-e^{-j\omega}\right)}=\frac{1}{e^{-j\omega}2j\sin(\omega)},\qquad 0<\omega<\pi\tag{2}$$

From $(2)$, the magnitude of $H(e^{j\omega})$ is

$$\left|H(e^{j\omega})\right|=\frac{1}{2\left|\sin(\omega)\right|},\qquad 0<\omega<\pi\tag{3}$$


$^1$ The Fourier transform of the impulse response corresponding to the given transfer function $(1)$ does exist as a distribution. It has Dirac delta impulses at $\omega=0$ and $\omega=\pi$.

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    $\begingroup$ The frequency responses of metastable and unstable filters may not "exist" in the system of thought that you are using, but they do exist and are meaningful if you put them into a closed-loop system that is stable. In that case, then yes, you can just compute $H(e^{j\omega})$ and get an input-output relationship that predicts what you will find in your closed loop system, across that component, if it is excited by a sine wave at $\omega$. $\endgroup$
    – TimWescott
    Aug 22 at 16:20
  • $\begingroup$ @TimWescott: I agree that marginally stable systems do have their applications. However, the existence of their frequency response (as an ordinary function) is not a matter of a "system of thought", it's just a mathematical question. And the answer is that it doesn't exist (again, as an ordinary function). The main point why I think that this exercise is pointless is that it appears to encourage students to blindly replace $z$ by $e^{j\omega}$ without understanding the meaning of the result. $\endgroup$
    – Matt L.
    Aug 23 at 10:05
  • $\begingroup$ The practical answer is that you can take a wholly unstable transfer function, calculate its frequency response as if it works, and use it in the classical Bode plot design technique to make a stable system. Then you can measure the "nonexistent" frequency response of that unstable thing. So I think you need to extend your mathematical constraints, somehow. Math should accurately reflect reality; it cannot make real physical things just disappear. $\endgroup$
    – TimWescott
    Aug 23 at 20:09

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