On Jan 28, 5:36�pm, Randy Yates <ya...@ieee.org> wrote:> "Peter \"Timmy\" Timmins" <no.spam.thank....@invalid.invalid> writes: > > > "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message > >news:KfCdnWnUyezfueLUnZ2dnUVZ_ofinZ2d@centurytel.net... > > >> As you may know, the "z-domain" is for sampled data / systems much as the > >> s-domain is for continuous data / systems. > > > The z transform is for delayed signals, irrespective of whether or not > > they are sampled. > > Excuse me? Please show me a "t" in the z-transform. > > > z = e^(sT) is an analogue transformation. > > This is a mapping from C to C (C = the complex plane) - it is not a transform. > -- > % �Randy Yates � � � � � � � � �% "My Shangri-la has gone away, fading like > %% Fuquay-Varina, NC � � � � � �% �the Beatles on 'Hey Jude'" > %%% 919-577-9882 � � � � � � � �% � > %%%% <ya...@ieee.org> � � � � � % 'Shangri-La', *A New World Record*, ELOhttp://www.digitalsignallabs.com================================================================================ We can use Fourier analysis for True frequency domain analysis and find magnitude and phase response and Hilbert gives us information regarding phase however its z-transform and Laplace transform that dont really seem to be in some domain. So what is that domain Excuse me? Please show me a "t" in the z-transform.> z = e^(sT) is an analogue transformation.This is a mapping from C to C (C = the complex plane) - it is not a transform. If its not a transform then why call it a "Z-Transform" Also if I'm not moving off the topic, when Differential-PSK gives better result that coherent-PSK under Multipath fading environments (I observed this using scatter plots in Matlab using Rayleigh and Rician models) so why is it not a popular technique in modern cellular systems. Wouldn't 8-DPSK be better than pi/4 shifted QPSK? So we can leave out the carrier recovery circuit.
Using Z-transform with continous time signals
Started by ●January 27, 2009
Reply by ●January 28, 20092009-01-28
Reply by ●January 28, 20092009-01-28
Communications_engineer <communications_engineer@yahoo.com> writes:> On Jan 28, 5:36�pm, Randy Yates <ya...@ieee.org> wrote: >> "Peter \"Timmy\" Timmins" <no.spam.thank....@invalid.invalid> writes: >> >> > "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message >> >news:KfCdnWnUyezfueLUnZ2dnUVZ_ofinZ2d@centurytel.net... >> >> >> As you may know, the "z-domain" is for sampled data / systems much as the >> >> s-domain is for continuous data / systems. >> >> > The z transform is for delayed signals, irrespective of whether or not >> > they are sampled. >> >> Excuse me? Please show me a "t" in the z-transform. >> >> > z = e^(sT) is an analogue transformation. >> >> This is a mapping from C to C (C = the complex plane) - it is not a transform. >> -- >> % �Randy Yates � � � � � � � � �% "My Shangri-la has gone away, fading like >> %% Fuquay-Varina, NC � � � � � �% �the Beatles on 'Hey Jude'" >> %%% 919-577-9882 � � � � � � � �% � >> %%%% <ya...@ieee.org> � � � � � % 'Shangri-La', *A New World Record*, ELOhttp://www.digitalsignallabs.com > > ================================================================================ > > We can use Fourier analysis for True frequency domain analysis and > find magnitude and phase response and Hilbert gives us information > regarding phase however its z-transform and Laplace transform that > dont really seem to be in some domain. So what is that domainThis is not the topic or question for a usenet post. I've already given you an excellent reference on the subject; the most efficient use of your time (and ours) would be to check it out at your local library (or buy it) and read it. Or, perhaps even better, take a class in linear systems theory at your local college or university. It will take the better part of a semester to give you the foundation you need to understand the answers to the questions you're asking.> Excuse me? Please show me a "t" in the z-transform. > >> z = e^(sT) is an analogue transformation. > > This is a mapping from C to C (C = the complex plane) - it is not a > transform. > > If its not a transform then why call it a "Z-Transform"It isn't. That is, the mathematical relationship Timmins gave, z = e^(sT), isn't the z-transform. It's not a transform of any sort. -- % Randy Yates % "My Shangri-la has gone away, fading like %% Fuquay-Varina, NC % the Beatles on 'Hey Jude'" %%% 919-577-9882 % %%%% <yates@ieee.org> % 'Shangri-La', *A New World Record*, ELO http://www.digitalsignallabs.com
Reply by ●January 28, 20092009-01-28
On Jan 28, 6:36�am, Randy Yates <ya...@ieee.org> wrote:> "Peter \"Timmy\" Timmins" <no.spam.thank....@invalid.invalid> writes: > > > "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message > >news:KfCdnWnUyezfueLUnZ2dnUVZ_ofinZ2d@centurytel.net... > > >> As you may know, the "z-domain" is for sampled data / systems much as the > >> s-domain is for continuous data / systems. > > > The z transform is for delayed signals, irrespective of whether or not > > they are sampled. > > Excuse me? Please show me a "t" in the z-transform. > > > z = e^(sT) is an analogue transformation. > > This is a mapping from C to C (C = the complex plane) - it is not a transform. > -- > % �Randy Yates � � � � � � � � �% "My Shangri-la has gone away, fading like > %% Fuquay-Varina, NC � � � � � �% �the Beatles on 'Hey Jude'" > %%% 919-577-9882 � � � � � � � �% � > %%%% <ya...@ieee.org> � � � � � % 'Shangri-La', *A New World Record*, ELOhttp://www.digitalsignallabs.comLet's see ...... In order to take the inverse z-transform, I must take the contour integral in the z-plane. But, the integral (Ito integral aside) is defined for continuous functions ............. Also, isn't the the z-transform equal to X(z) = x(t) * SUM[ delta(t - nT) exp(sT)] ? Ummmmmm Maurice Givens
Reply by ●January 28, 20092009-01-28
On Jan 28, 12:23�pm, maury <maury...@core.com> wrote:> On Jan 28, 6:36�am, Randy Yates <ya...@ieee.org> wrote: > > > > > > > "Peter \"Timmy\" Timmins" <no.spam.thank....@invalid.invalid> writes: > > > > "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message > > >news:KfCdnWnUyezfueLUnZ2dnUVZ_ofinZ2d@centurytel.net... > > > >> As you may know, the "z-domain" is for sampled data / systems much as the > > >> s-domain is for continuous data / systems. > > > > The z transform is for delayed signals, irrespective of whether or not > > > they are sampled. > > > Excuse me? Please show me a "t" in the z-transform. > > > > z = e^(sT) is an analogue transformation. > > > This is a mapping from C to C (C = the complex plane) - it is not a transform. > > -- > > % �Randy Yates � � � � � � � � �% "My Shangri-la has gone away, fading like > > %% Fuquay-Varina, NC � � � � � �% �the Beatles on 'Hey Jude'" > > %%% 919-577-9882 � � � � � � � �% � > > %%%% <ya...@ieee.org> � � � � � % 'Shangri-La', *A New World Record*, ELOhttp://www.digitalsignallabs.com > > Let's see ...... In order to take the inverse z-transform, I must take > the contour integral in the z-plane. �But, the integral (Ito integral > aside) is defined for continuous functions ............. > > Also, isn't the �the z-transform equal to X(z) = x(t) * SUM[ delta(t - > nT) exp(sT)] ? �Ummmmmm > > Maurice Givens- Hide quoted text - > > - Show quoted text -Oops, the -n got away from me. It should be X(z) = x(t) * SUM[ delta(t - nT) exp(-nsT)] mia culpa Maurice Givens
Reply by ●January 28, 20092009-01-28
On Wed, 28 Jan 2009 10:32:33 -0800 (PST), maury <maury001@core.com> wrote:>mia culpaI think I caught her act once in Vegas... ;) Eric Jacobsen Minister of Algorithms Abineau Communications http://www.ericjacobsen.org Blog: http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php
Reply by ●January 28, 20092009-01-28
On Jan 28, 1:55�pm, Eric Jacobsen <eric.jacob...@ieee.org> wrote:> On Wed, 28 Jan 2009 10:32:33 -0800 (PST), maury <maury...@core.com> > wrote: > > >mia culpa > > I think I caught her act once in Vegas... > > ;) > > Eric Jacobsen > Minister of Algorithms > Abineau Communicationshttp://www.ericjacobsen.org > > Blog:http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.phpThanks Eric. I needed a laugh today! Maurice Givens
Reply by ●January 28, 20092009-01-28
Eric Jacobsen <eric.jacobsen@ieee.org> writes:> On Wed, 28 Jan 2009 10:32:33 -0800 (PST), maury <maury001@core.com> > wrote: > >>mia culpa > > I think I caught her act once in Vegas... > > ;)Wasn't she married to Woody Allen at one time? Oh no, wait... -- % Randy Yates % "How's life on earth? %% Fuquay-Varina, NC % ... What is it worth?" %%% 919-577-9882 % 'Mission (A World Record)', %%%% <yates@ieee.org> % *A New World Record*, ELO http://www.digitalsignallabs.com
Reply by ●January 28, 20092009-01-28
On Jan 28, 6:04�pm, Randy Yates <ya...@ieee.org> wrote:> Eric Jacobsen <eric.jacob...@ieee.org> writes: > > On Wed, 28 Jan 2009 10:32:33 -0800 (PST), maury <maury...@core.com> > > wrote: > > >>mia culpa > > > I think I caught her act once in Vegas... > > > ;) > > Wasn't she married to Woody Allen at one time? Oh no, wait... > -- > % �Randy Yates � � � � � � � � �% "How's life on earth? > %% Fuquay-Varina, NC � � � � � �% �... What is it worth?" > %%% 919-577-9882 � � � � � � � �% 'Mission (A World Record)', > %%%% <ya...@ieee.org> � � � � � % *A New World Record*, ELOhttp://www.digitalsignallabs.comYou're thinking of the one who invented the Farrow filter. Bob
Reply by ●January 29, 20092009-01-29
On Wed, 28 Jan 2009 07:36:39 -0500, Randy Yates wrote:> "Peter \"Timmy\" Timmins" <no.spam.thank.you@invalid.invalid> writes: > >> "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message >> news:KfCdnWnUyezfueLUnZ2dnUVZ_ofinZ2d@centurytel.net... >>> >>> As you may know, the "z-domain" is for sampled data / systems much as >>> the s-domain is for continuous data / systems. >> >> The z transform is for delayed signals, irrespective of whether or not >> they are sampled. > > Excuse me? Please show me a "t" in the z-transform.He's right. You _can_ use the z transform for analyzing systems that contain delays that are all integer multiples of some characteristic time. Not that there are many like that, but should you stumble across a system that's all done with equal-length bits of transmission line, you have the mathematical tools to analyze it. The 't' in the z-transform is the kT in z = e^(skT). So if your system is all in delays of T or greater, then your transfer function will all be in terms of e^(skT).> >> z = e^(sT) is an analogue transformation. > > This is a mapping from C to C (C = the complex plane) - it is not a > transform.OK, it's an analog mapping. It still shows how you can use the z transform to analyze continuous-time systems. -- http://www.wescottdesign.com
Reply by ●January 29, 20092009-01-29
On Jan 29, 12:31�am, Tim Wescott <t...@seemywebsite.com> wrote:> On Wed, 28 Jan 2009 07:36:39 -0500, Randy Yates wrote: > > "Peter \"Timmy\" Timmins" <no.spam.thank....@invalid.invalid> writes: > > >> "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message > >>news:KfCdnWnUyezfueLUnZ2dnUVZ_ofinZ2d@centurytel.net... > > >>> As you may know, the "z-domain" is for sampled data / systems much as > >>> the s-domain is for continuous data / systems. > > >> The z transform is for delayed signals, irrespective of whether or not > >> they are sampled. > > > Excuse me? Please show me a "t" in the z-transform. > > He's right. �You _can_ use the z transform for analyzing systems that > contain delays that are all integer multiples of some characteristic > time. �Not that there are many like that, but should you stumble across a > system that's all done with equal-length bits of transmission line, you > have the mathematical tools to analyze it. > > The 't' in the z-transform is the kT in z = e^(skT). �So if your system > is all in delays of T or greater, then your transfer function will all be > in terms of e^(skT). > > > > >> z = e^(sT) is an analogue transformation. > > > This is a mapping from C to C (C = the complex plane) - it is not a > > transform. > > OK, it's an analog mapping. �It still shows how you can use the z > transform to analyze continuous-time systems. > > --http://www.wescottdesign.comWhen Differential-PSK gives better result that coherent-PSK under Multipath fading environments (I observed this using scatter plots in Matlab using Rayleigh and Rician models and observed that when constellation was severely distortion the BER was much lower than what I got with PSK) so why is it not a popular technique in modern cellular systems. Is there some cost problem or some technical issue? Wouldn't 8-DPSK be better than pi/4 shifted QPSK? So we can leave out the carrier recovery circuit.
