Shows how points on the S-Plane map on to points on the Z-Plane. This provides a visual interpretation of the relationship between the Laplace and Z transforms
this is an amazing visualization, thank you so much, not only for this video but the whole series on Z-transform. It was eye-opening. Developed my own little version of Z-surface visualization in Python to nail it down in my head. I absolutely love this visualization of how S-space is just z-space when sampling frequency goes to infinity, quite bamboozled by it, making some nice connections in my brain.
Good question. One way to picture this is to play the video backwards from the end up to time 1 min 55 sec! After this point (in reverse time) you'd have to use your imagination and picture that the yellow 'circle' points would continue to rotate past the -1+0j and become negative imaginary terms (aliased negative frequencies). Similarly, the red 'circle' points would rotate past the -1+0j and become positive imaginary terms (aliased positive frequencies). Eventually the green radial lines would also become associated with aliased negative frequencies - and if the sampling frequency continued to reduce eventually all the points that are associated with aliased negative frequencies would become aliased positive frequencies (and so on ).
You're correct (kind of!). The signal z^-n does grow exponentially for values of z that lie inside the unit circle. However, the z-transform multiplies the signal z^-n by the signal we are interested in analysing for stability e.g. x[n] (and then sums the resulting multiplied terms). If x[n] is growing exponentially and z^-n is also growing (which is the case for all values of z within the unit circle) then the sum of the products of x[n] and z[n] will never converge within the unit circle. On the other hand when x[n] is decaying and z^-n is growing then there will a set of values of z that lie within the unit circle for which the product of x[n] and he signal z^-n will converge. I suspect it's quite difficult to interpret these previous sentences - if so you could take a look at page 83 of www.researchgate.net/publication/370660126_The_z-transform_A_practical_overview or take the following video ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-dEJp46SFgV4.html. Hopefully, one of these could provide some insight.
@@ddorran I read the paper that you have linked, however I didnt understand one thing, you say that the region of convergence lies outside the circle with radius equal to that of the pole of the system. In your example you used -0.5. My question is why is it that you called -0.5 the pole? Wasn't that the impulse response? Also all of this implies that the region of stability/convergence is > Pole value circle, not that region of stability is
@@AbhinavRao-te9co "My question is why is it that you called -0.5 the pole?" The answer is because H(z) goes toward infinity when z= -0.5 (poles exist at values of z for which H(z) goes toward infinity). "Also all of this implies that the region of stability/convergence is > Pole value circle, not that region of stability is
@@ddorran "Having said that, regions outside the unit circle are associated with instability because any pole outside the unit circle will cause the system to be unstable (since the unit circle will not lie in the region of convergence)." Is there any proof for this or should i just learn this as a given?
@@AbhinavRao-te9co I don't have a formal proof to hand but there must be one out there! There are a couple of aspects to look at: A formal proof that the region of convergence lies outside the pole furthest from the origin (or proof that the region of convergence cannot contain poles, is another way of saying this); A formal proof that an unstable system contains poles that lie outside the unit circle. If both of these can be proven then it follows that the unit circle will not lie in the region of convergence, for unstable causal systems.
