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So I am taking a signal processing course in EE and my professor is an Engineer who really likes math however his book which we use for the class falls in the dreadful purgatory of math books in my opinion: too "rigorous" to be intuitive and way too abridged and takes leaps which make it impossible to consider rigorous.

It leaves me scratching my head on the chapter on the relationship between z-transforms, Fourier transforms, DTFTs and DFTs.

Here are some extracts from the book:

Comparing \$V(z)\$ (implicitly meaning the z transform of a sequence say v[n]) with the Laplace transform \$V_s(s)\$ we note that the two transforms are realted by a simple change of variables. In particular letting \$z = e^{Ts}\$ we have:

\$V(z)|_{z = e^{Ts}} = V(e^{Ts}) = \sum\limits_{n = - \infty }^{\infty}v_c(nT)e^{-nTs} = V_s(s)\$ (where \$V_s\$ is the Laplace transform of the ideally sampled function-with --fs = 1/T--\$v_c\$ which itself has a Laplace transform)

Now this is where he starts losing me:

We note that the transformation \$z = e^{Ts}\$ transforms the axis \$s = j\omega\$ into the unit circle : \$z = e^{j\omega t} = e^{j\Omega}\$ where \$\Omega \triangleq \omega T\$ which is the relation between the discrete-time domain angular frequency \$\Omega\$ in radians and the angular frequency of the continuous-time domain frequency \$\omega\$ in radians/s. The vertical line \$s = \sigma_0 + j\omega\$ in the s plane is transformed into a circle \$z = e^{\sigma_0 T } e^{jT\omega}\$ in the z-plane. In fact a pole at \$s = \alpha +j\beta\$ is transformed into a pole \$z = e^{(\alpha + j\beta)T}\$ of radius \$r = e^{\alpha T}\$ and angle \$\Omega = \beta T\$ in the z plane.

That last paragraph doesn't mean much to me and I am having a hard time with the following concepts which seem pretty important:

  1. What does a discrete-time domain angular frequency stand for?
  2. What does he mean by the z plane in that context
  3. In the end if we consider transforms as morphism (I absolutely have not enough serious math background to consider morphisms for functional spaces but have some intuition into the concept form isomorphims in abstract LA), does he mean that the substitution \$z = e^{j\omega T}\$ is a morphism from "a sort of functional space resulting from the application of the z transform to sequences " to "a functional space consisting of complex functions defined on the circle?" (sorry if that last bit was cringy for some of you, just trying to get a grasp, thanks)
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    \$\begingroup\$ Hang on; I'm fixing the MathJax... :) \$\endgroup\$ – bitsmack Dec 8 '15 at 23:44
  • \$\begingroup\$ Thanks, I just remembered we use backslashes around here for math mode. And I'm still postponing learning regular expressions \$\endgroup\$ – SolipsistElvis Dec 8 '15 at 23:45
  • \$\begingroup\$ No problem! Your MathJaX skills are impressive :) Did I break anything? \$\endgroup\$ – bitsmack Dec 8 '15 at 23:53
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From the last highlighted paragraph:

  1. It's perhaps easier to relate a unit delay in the z and s domains; thus \$e^{-sT}\$ delays a continuous signal by \$\small T\$ secs, and \$z^{-1}\$ is the unit delay function in the z-domain, which delays a discrete signal by one sampling increment. Hence \$e^{-sT}\leftrightarrow z^{-1}\$, or \$e^{sT}\leftrightarrow z^{1}=z\$ is often a more convenient form.

  2. To go from the s-domain to the frequency domain we use \$s \rightarrow j\omega\$. Using the equivalence in 1., above we may go from the z-domain to frequency domain by: \$\small z\rightarrow e^{st} \rightarrow e^{j\omega T} = cos(\omega T) +j\:sin(\omega T)\$. Now, \$\small \omega T\$ is an angle, which is proportional to \$\small \omega\$, and it's convenient to denote this angle, \$\small \Omega\$, and think of \$\small e^{j\Omega}=cos\:\Omega\:+\:j\:sin\:\Omega\$ as a vector that rotates counter-clockwise from zero radians as frequency increases from \$\small\omega=0\$ rad/sec.

  3. Now consider an s-plane root, \$\small s=-\alpha\$. This would transform to a z-plane root: \$\small z \rightarrow e^{-\alpha T}\$, which is real, positive and less than unity in magnitude.

  4. Moving on, a complex root in the s-plane: \$\small s=\:-\alpha\:+ j\:\omega\$ transforms to \$\small z \rightarrow e^{-\alpha T\:+ j\:\omega T}=\: e^{-\alpha T}\:\small (cos\:\Omega\:+\:j\:sin\Omega)\$, which is a counter-clockwise rotating vector with radius: \$\small e^{-\alpha T}\$. The conjugate root would rotate clockwise with the same radius. Note that the radius is \$\small \lt 1\$, i.e. the vectors are inside the unit circle.

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  • \$\begingroup\$ Thanks, jeez, no wonder I was confused, he refers to big Omega as both a frequency and an angle. Obviously by just looking at the units it is an angle. May I ask, I never really took much time to understand the z domain as opposed to frequency or laplace. For example the variable jw in a Laplace TF Vout/Vin, omega woudl be the frequency. But how can one interpret the axis of the z domaine? And if Omega is an angle, where does the "concept of frequency go"? \$\endgroup\$ – SolipsistElvis Dec 9 '15 at 0:56
  • \$\begingroup\$ because T is a constant (=sampling increment), it's easier to write \$\Omega=\omega T\$ and treat \$\Omega\$ as the variable frequency. The vector wraps itself around in a never ending circle as \$\Omega\$ increases without bound! Hence aliasing... \$\endgroup\$ – Chu Dec 9 '15 at 1:08
  • \$\begingroup\$ It is important to realize that the z-domain is different than the s-domain, and that they "visualize" different things. Yes, in the s-plane, your imaginary axis corresponds to the frequency domain, but in the z-domain that is no longer true. Instead, the correspondence is that it is "on" that circle. \$\endgroup\$ – Pål-Kristian Engstad Dec 9 '15 at 1:09
  • \$\begingroup\$ So could we say something like : in the context of the discrete fourier transform, the z plane resulting from the map \$z \rightarrow e^{-sT} \$ as the quotient ring \$ s \, mod(2\pi j \omega) \$? Actually I just realized that is nonsense but could you expand a little bit on the correspondance in the z domain and how to have some sort of intuition into it? Or does it require some like serious math \$\endgroup\$ – SolipsistElvis Dec 9 '15 at 1:23
  • \$\begingroup\$ \$\small z\rightarrow e^{-j\omega T}=cos\Omega+j\:sin\Omega\$ maps to the unit circle. Note that if \$\alpha\lt 0\$ the real root is in the right half s-plane, i.e. unstable, and the z-plane vector magnitudes are >1, i.e. outside the unit circle. The unit circle is the stability boundary in the z-plane. \$\endgroup\$ – Chu Dec 9 '15 at 1:30

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