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
Reply by scc28●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 Tim Wescott●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 Randy Yates●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 scc28●February 12, 20082008-02-12
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