On Wed, 02 Jul 2014 06:31:47 -0500, "kaz" <37480@dsprelated> wrote:>>Given an FIR filter of length N, I need to test if all the zeros are on >the unit circle or not, without factoring. Any ideas? >> >>Bob >> > >if you let Matlab do it for you (assuming Vlad is not irritated) then > >h = fir1(40,.1); -- example filter >h = round(h*2^15); >zplane(h,1);Hi, Bob doesn't need Vlad's permission to use Matlab. [-Rick-]
Coeff test
Started by ●July 2, 2014
Reply by ●July 3, 20142014-07-03
Reply by ●July 3, 20142014-07-03
Thanks for the suggestions. I'm actually trying to prove a theorom which is why I'm resistant to computational approaches. Bob
Reply by ●July 3, 20142014-07-03
On Thu, 3 Jul 2014 06:41:00 -0700 (PDT), radams2000@gmail.com wrote:>Thanks for the suggestions. I'm actually trying to prove a theorom which is why I'm resistant to computational approaches. > >BobYeah. Can't tell you how many times I've seen software "proofs" fall apart. We've seen a few here, too. I think my fave was proving a signal was moving faster than the speed of light with a software simulation. Eric Jacobsen Anchor Hill Communications http://www.anchorhill.com
Reply by ●July 3, 20142014-07-03
>Yeah. Can't tell you how many times I've seen software "proofs" fall >apart.Yes, analytic calculation seems more trustworthy... Let's start with A^2 - A^2 = A^2 - A^2 On the LHS, factor out A. On the RHS, use A^2-B^2 = (A+B)(A-B) gives A(A-A) = (A+A)(A-A) thus A = A+A :-) _____________________________ Posted through www.DSPRelated.com
Reply by ●July 3, 20142014-07-03
On 7/3/14 4:28 PM, mnentwig wrote:>> Yeah. Can't tell you how many times I've seen software "proofs" fall >> apart.well, if you have a finite number of cases in what you're trying to prove, maybe one can test each one with a computer program. http://en.wikipedia.org/wiki/Computer-assisted_proof> > Yes, analytic calculation seems more trustworthy... > > Let's start with > A^2 - A^2 = A^2 - A^2 > > On the LHS, factor out A. > On the RHS, use A^2-B^2 = (A+B)(A-B) > > gives >how do you get from here:> A(A-A) = (A+A)(A-A) > > thus >... to here?> A = A+A > > :-) > >http://mathforum.org/dr.math/faq/faq.false.proof.html -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by ●July 4, 20142014-07-04
>how do you get from here: > >> A(A-A) = (A+A)(A-A)why, cancel the common factor, of course. Chuck Norris can do it. _____________________________ Posted through www.DSPRelated.com
Reply by ●July 4, 20142014-07-04
>>how do you get from here: >> >>> A(A-A) = (A+A)(A-A) > >why, cancel the common factor, of course. >Chuck Norris can do it. > >_____________________________ >Posted through www.DSPRelated.com >It is divide by zero fun _____________________________ Posted through www.DSPRelated.com
Reply by ●July 4, 20142014-07-04
On Thu, 03 Jul 2014 16:53:04 -0400, robert bristow-johnson <rbj@audioimagination.com> wrote:>On 7/3/14 4:28 PM, mnentwig wrote: >>> Yeah. Can't tell you how many times I've seen software "proofs" fall >>> apart. > >well, if you have a finite number of cases in what you're trying to >prove, maybe one can test each one with a computer program. > >http://en.wikipedia.org/wiki/Computer-assisted_proof > >> >> Yes, analytic calculation seems more trustworthy... >> >> Let's start with >> A^2 - A^2 = A^2 - A^2 >> >> On the LHS, factor out A. >> On the RHS, use A^2-B^2 = (A+B)(A-B) >> >> gives >> > >how do you get from here: > >> A(A-A) = (A+A)(A-A) >> >> thus >> > >... to here? > >> A = A+A >>Hi Robert, Markus' problem is actually kind of interesting. It boils down to claiming: P*0 = Q*0 ['*' mean multiply] which is a true statement. Then he divides both sides of the equation by zero. But everyone correctly says, "It's illegal to divide by zero!" OK. I'm rather ignorant of number theory but I was thinking, "What's wrong, where's the danger, in dividing by zero? Why am I not allowed to divide a number by zero?" If I divide three by zero I obtain: 3/0 = infinity (1) If I divide five by zero I obtain: 5/0 = infinity (2) Here's my question: Does the "danger in dividing by zero" come from the fact that the 'infinity' in Eq. (1) does NOT equal the 'infinity' in Eq. (2)? Perhaps the danger is: we should NOT think of infinity as a number. [-Rick-]
Reply by ●July 4, 20142014-07-04
> >Here's my question: Does the "danger in >dividing by zero" come from the fact that >the 'infinity' in Eq. (1) does NOT equal the >'infinity' in Eq. (2)? > >Perhaps the danger is: we should NOT >think of infinity as a number. > >[-Rick-] > >If all mathematicians(and their DSP kids) go to church (or mosque...) for mass prayer they might be answered for all the mess of infinity and -infinity they have created. _____________________________ Posted through www.DSPRelated.com
Reply by ●July 4, 20142014-07-04
On Wednesday, July 2, 2014 10:37:25 PM UTC+12, radam...@gmail.com wrote:> Given an FIR filter of length N, I need to test if all the zeros are on the unit circle or not, without factoring. Any ideas? > > > > BobYou need ON or Inside/Outside? If the latter then a Schur Con (sure con!) test or Jury test. Spectral factorization will reflect all NMphase zeros back inside.
