Hi, I have two questions which seem somewhat basic but confuse me: 1. We know that a sequence has to be "summable" to have Fourier Transform (ie. DTFT converges). Is this the same as saying that a "bounded" sequence will have Fourier Transform? 2. Do all sequences have z-transform? In other words, can a sequence have z-transform with no ROC (region of convergence)? Thanks in advance. regards, scc28
Two DSP theoretical questions
Started by ●February 12, 2008
Reply by ●February 12, 20082008-02-12
"scc28" <stanleyche@hotmail.com> writes:> [...] > 2. Do all sequences have z-transform? In other words, can a sequence have > z-transform with no ROC (region of convergence)?This one right off is a resounding NO. If there is no place in the complex plane in which the z-transform of the sequence converges, then the z-transform of the sequence exists NOWHERE. That is tantamount to saying that the sequence doesn't have a z-transform. -- % Randy Yates % "She has an IQ of 1001, she has a jumpsuit %% Fuquay-Varina, NC % on, and she's also a telephone." %%% 919-577-9882 % %%%% <yates@ieee.org> % 'Yours Truly, 2095', *Time*, ELO http://www.digitalsignallabs.com
Reply by ●February 12, 20082008-02-12
On Tue, 12 Feb 2008 17:06:13 -0600, scc28 wrote:> Hi, I have two questions which seem somewhat basic but confuse me: > > 1. We know that a sequence has to be "summable" to have Fourier > Transform (ie. DTFT converges). Is this the same as saying that a > "bounded" sequence will have Fourier Transform?No. If x_n = 1 for all n from 0 to infinity, then the sum of all x_n is infinite, even though x_n is pretty strongly bounded.> 2. Do all sequences have z-transform? In other words, can a sequence > have z-transform with no ROC (region of convergence)? >Randy's answer: nuh uh. The z transform is the result of a sum, if the sum doesn't converge for any z then it's 'answer' is meaningless. -- Tim Wescott Control systems and communications consulting http://www.wescottdesign.com Need to learn how to apply control theory in your embedded system? "Applied Control Theory for Embedded Systems" by Tim Wescott Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html
Reply by ●February 14, 20082008-02-14
Hi thanks for the helpful insight. I guess I need to rephrase my 2nd question. If a z transform of a sequence does not have any ROC, can it still be written in a closed mathematical expression? thanks in advance>On Tue, 12 Feb 2008 17:06:13 -0600, scc28 wrote: > >> Hi, I have two questions which seem somewhat basic but confuse me: >> >> 1. We know that a sequence has to be "summable" to have Fourier >> Transform (ie. DTFT converges). Is this the same as saying that a >> "bounded" sequence will have Fourier Transform? > >No. If x_n = 1 for all n from 0 to infinity, then the sum of all x_n is>infinite, even though x_n is pretty strongly bounded. > >> 2. Do all sequences have z-transform? In other words, can a sequence >> have z-transform with no ROC (region of convergence)? >> >Randy's answer: nuh uh. The z transform is the result of a sum, if the >sum doesn't converge for any z then it's 'answer' is meaningless. > >-- >Tim Wescott >Control systems and communications consulting >http://www.wescottdesign.com > >Need to learn how to apply control theory in your embedded system? >"Applied Control Theory for Embedded Systems" by Tim Wescott >Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html >
Reply by ●February 15, 20082008-02-15
On Thu, 14 Feb 2008 12:28:06 -0600, scc28 wrote: (top posting fixed)> >>On Tue, 12 Feb 2008 17:06:13 -0600, scc28 wrote: >> >>> Hi, I have two questions which seem somewhat basic but confuse me: >>> >>> 1. We know that a sequence has to be "summable" to have Fourier >>> Transform (ie. DTFT converges). Is this the same as saying that a >>> "bounded" sequence will have Fourier Transform? >> >>No. If x_n = 1 for all n from 0 to infinity, then the sum of all x_n is > >>infinite, even though x_n is pretty strongly bounded. >> >>> 2. Do all sequences have z-transform? In other words, can a sequence >>> have z-transform with no ROC (region of convergence)? >>> >>Randy's answer: nuh uh. The z transform is the result of a sum, if the >>sum doesn't converge for any z then it's 'answer' is meaningless. >> > Hi thanks for the helpful insight. I guess I need to rephrase my 2nd > question. If a z transform of a sequence does not have any ROC, can it > still be written in a closed mathematical expression? > > thanks in advance >Yes: X(z) = e^(j * theta(z)) * infinity That's assuming you're good enough to solve an unbounded summation for the phase angle, of course. -- Tim Wescott Control systems and communications consulting http://www.wescottdesign.com Need to learn how to apply control theory in your embedded system? "Applied Control Theory for Embedded Systems" by Tim Wescott Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html
