pulsewidth and bandwidth
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From: Jitendra Rayala on 14 Jul 2010 01:28 On Jul 13, 5:53 am, dvsarwate <dvsarw...(a)gmail.com> wrote: > I don't understand what Jitendra Rayala <jray...(a)hotmail.com> wrote: > > In response to the OP's statement > > > > I know bandwidth = 1/pulsewidth. > > he writes > > > Strictly speaking, as far as I know, this is only true for Gaussian. > > If dt and df are the RMS width in time and frequency respectively, > > then the time-bandwidth product satisfies > > > dt*df >= 1/2 > > > ......Gaussian achieves the equality...... > > But if "Gaussian achieves the equality", then > dt*df = 1/2, not 1 as the OP wrote and to > which Jitendra seemed to agree when he > wrote "true for Gaussian". Perhaps Jitendra > will explain further which is his position, 1/2 or 1. Good point. It is 1/2 (or 1/4pi). When I said "true for Gaussian" what I actually had in mind was that the relation posted by OP holds only for a subset. It is the inequality that holds true in general. Jitendra > > Jitendra also correctly asserted that the > Gaussian pulse is neither time-limited nor > band-limited. *Any* consistent definition > of pulsewidth and bandwidth must take into > account the fact that in at least one of the > two domains, the support extends to infinity. > Definitions such as 99% energy containment > (or 92.4% energy containment as rb-j used) > etc always work and give finite measures > of pulsewidth and bandwidth, whereas some > signals have rms pulsewidth or rms bandwidth > equal to infinity. This corresponds to the notion > in probability theory that some random variables > e.g. Cauchy random variables, have infinite > variance, as has been noted with some surprise > in this newsgroup. > > --Dilip Sarwate
From: Greg Heath on 14 Jul 2010 05:27 On Jul 12, 8:26 pm, Jitendra Rayala <jray...(a)hotmail.com> wrote: > On Jul 7, 2:43 am, "Fan.Zhang" <zf624(a)n_o_s_p_a_m.sina.com> wrote: > > > Hi Experts, > > > I know bandwidth = 1/pulsewidth. > > Strictly speaking, as far as I know, this is only true for Gaussian. > If dt and df are the RMS width in time and frequency respectively, > then the time-bandwidth product satisfies > > dt*df >= 1/2 > > Gaussian achieves the equality but is neither time-limited nor band- > limited. > > > After reconsider the pulse shape and ISI, this question comes to my > > mind.But how it comes in this way? > > Comes from Cauchy-Schwarz >inequality:http://en.wikipedia.org/wiki/Cauchy%E2%80%>93Schwarz_inequality Physics fans: Doesn't this smell like the Heisenberg Uncertainty Principle? Greg
From: Greg Heath on 14 Jul 2010 07:29 On Jul 13, 8:53 am, dvsarwate <dvsarw...(a)gmail.com> wrote: > I don't understand what Jitendra Rayala <jray...(a)hotmail.com> wrote: > > In response to the OP's statement > > > > I know bandwidth = 1/pulsewidth. > > he writes > > > Strictly speaking, as far as I know, this is only true for Gaussian. > > If dt and df are the RMS width in time and frequency respectively, > > then the time-bandwidth product satisfies > > > dt*df >= 1/2 > > > ......Gaussian achieves the equality...... > > But if "Gaussian achieves the equality", then > dt*df = 1/2, not 1 as the OP wrote and to > which Jitendra seemed to agree when he > wrote "true for Gaussian". Perhaps Jitendra > will explain further which is his position, 1/2 or 1. > > Jitendra also correctly asserted that the > Gaussian pulse is neither time-limited nor > band-limited. *Any* consistent definition > of pulsewidth and bandwidth must take into > account the fact that in at least one of the > two domains, the support extends to infinity. > Definitions such as 99% energy containment > (or 92.4% energy containment as rb-j used) > etc always work and give finite measures > of pulsewidth and bandwidth, whereas some > signals have rms pulsewidth or rms bandwidth > equal to infinity. This corresponds to the notion > in probability theory that some random variables > e.g. Cauchy random variables, have infinite > variance, as has been noted with some surprise > in this newsgroup. This was probably the reason for using the following definition of half power bandwidth |X(BW/2)| = |X(0)|/sqrt(2) Some relevant formulae: http://en.wikipedia.org/wiki/... Fourier_transform#Tables_of_important_Fourier_transforms 104: x(t/a) <--> |a|*X(a*f) 201: rect(t/a) <--> |a|*sinc(a*f)) 206: exp(-(t/a)^2) <--> |a|*sqrt(pi)*exp(-pi*(a*f)^2) 207: exp(-|t/a|) <--> 2*|a|/(1 + (2*pi*a*f)^2) Hope this helps. Greg
From: dvsarwate on 14 Jul 2010 07:35 OK, I am confused even more. The OP said (without defining what is meant by pulsewidth or bandwidth) OP> bandwidth = 1/pulsewidth. Why? to which several people replied, and Robert Bristow-Johnson said (I think) that the relationship stated by the OP is true for Gaussian pulses if pulsewidth and bandwidth are defined in terms of 92.4% energy containment. Jitendra Rayala said in his second post that the OP's relationship holds for some pulses (but did not specify which ones he had in mind). He also said that when bandwidth and pulsewidth are taken to mean rms bandwidth df and rms pulsewidth dt, then the inequality JR1> dt*df >= 1/2 or perhaps JR2> dt*df >= 1/(4 pi) is applicable to *all* pulses (neither of which contradict the assertion that the product equals 1 for some pulses). For *most* pulses, the inequality would be strict: dt*df is *larger* than 1/2 or 1/(4 pi). But for a few select pulses (Gaussians?), it might be that dt*df *equals* 1/2 or 1/(4 pi). Does it? and which is the minimum value that is achieved: 1/2 or 1/(4 pi)? --Dilip Sarwate
From: Clay on 14 Jul 2010 10:28
On Jul 14, 5:27 am, Greg Heath <he...(a)alumni.brown.edu> wrote: > On Jul 12, 8:26 pm, Jitendra Rayala <jray...(a)hotmail.com> wrote: > > > > > > > On Jul 7, 2:43 am, "Fan.Zhang" <zf624(a)n_o_s_p_a_m.sina.com> wrote: > > > > Hi Experts, > > > > I know bandwidth = 1/pulsewidth. > > > Strictly speaking, as far as I know, this is only true for Gaussian. > > If dt and df are the RMS width in time and frequency respectively, > > then the time-bandwidth product satisfies > > > dt*df >= 1/2 > > > Gaussian achieves the equality but is neither time-limited nor band- > > limited. > > > > After reconsider the pulse shape and ISI, this question comes to my > > > mind.But how it comes in this way? > > > Comes from Cauchy-Schwarz >inequality:http://en.wikipedia.org/wiki/Cauchy%E2%80%>93Schwarz_inequality > > Physics fans: > > Doesn't this smell like the Heisenberg Uncertainty Principle? > > Greg- Hide quoted text - > > - Show quoted text - Yes, because position and momentum space representations in quantum mechanics are fourier transforms of each other. The non zero minimum uncertainlty arises since poistion and momentum are non commuting hermitian operators. Clay |