Prev: Floating point DSPsNext: MATLAB indexing issue From: robert bristow-johnson on 12 Jul 2010 13:31 On Jul 12, 4:05 am, Greg Heath wrote:> On Jul 8, 3:09 pm, robert bristow-johnson > wrote: > > > On Jul 7, 5:43 am, "Fan.Zhang" wrote: > > > > Hi Experts, > > > > I know bandwidth = 1/pulsewidth. > > > well, they *are* proportional to the reciprocal of each other.  but > > the constant of proportionality is dependent on the definitions of > > bandwidth or pulsewidth. > > > > After reconsider the pulse shape and ISI, this question comes to my > > > mind.But how it comes in this way? > > > the inverse-proportionality comes from the fundamental time and > > frequency scaling property of the Fourier Transform. > > For example, if you use an FT table and consider > the power spectra (proportional to the square of > the FT magnitude) of the gaussian time pulse > x1 = exp(-(t/a)^2) and the rectangular time pulse > x2 = step(t)-step(t-T), you can see how a and T > determine the width of the main spectral lobe. a definition that i like to use for the width of a pulse is the width about 92.4% of the entire area of the pulse. the number is motivated by x(t) = exp(-pi*(t/t0)^2) which has Fourier Transform of X(f) = exp(-pi*(f/f0)^2) where the height of both pulses is 1, total area of either pulse is 1, and the pulsewidths in the two domains (t0 and f0) multiply to 1. if you take either pulse, square it (to get power or energy) and integrate it from -t0/2 to +t0/2 (or -f0/2 to +f0/2), you'll get about 0.924 of the total squared area, i think. in the sense of the guassian pulse, if you define pulsewidth that way, then the relationship bandwidth = 1/pulsewidth is exactly correct, and i wonder if it's close to the case for many other classes of signals. it might be interesting to see how that definition compares to the rectangular and sinc() pulses. r b-j From: dvsarwate on 12 Jul 2010 14:14 On Jul 12, 12:31 pm, robert bristow-johnson wrote: > >     x(t) = exp(-pi*(t/t0)^2) > > which has Fourier Transform of > >     X(f) = exp(-pi*(f/f0)^2) > > where the height of both pulses is 1, total area of either pulse is 1, > and the pulsewidths in the two domains (t0 and f0) multiply to 1. Both x(t) = exp(-pi*(t/t0)^2) and X(f) = exp(-pi*(f/f0)^2) have value 1 at the origin, but why is the area 1 regardless of the values of t0 and f0? It is true that exp(-pi*(t)^2) has area 1, but what about exp(-pi*(t/t0)^2)? Surely stretching (or contracting) the time axis by a factor t0 should affect the area? Similarly for the Fourier transform.... --Dilip Sarwate From: robert bristow-johnson on 12 Jul 2010 14:56 On Jul 12, 2:14 pm, dvsarwate wrote:> On Jul 12, 12:31 pm, robert bristow-johnson > > wrote: > > >     x(t) = exp(-pi*(t/t0)^2) > > > which has Fourier Transform of > > >     X(f) = exp(-pi*(f/f0)^2) > > > where the height of both pulses is 1, total area of either pulse is 1, > > and the pulsewidths in the two domains (t0 and f0) multiply to 1. > > Both x(t) = exp(-pi*(t/t0)^2) and X(f) = exp(-pi*(f/f0)^2) > have value 1 at the origin, but why is the area 1 regardless > of the values of t0 and f0?  It is true that exp(-pi*(t)^2) has > area 1, but what about exp(-pi*(t/t0)^2)?  Surely stretching > (or contracting) the time axis by a factor t0 should affect > the area? oops. i overlooked the necessary scaling factor. we need to put 1/t0 and 1/f0 in front of both to get the unity area thing. but then they are not exactly FT of each other so x(t) = 1/t0 * exp(-pi*(t/t0)^2) has area of 1 and FT of X(f) = exp(-pi*(f/f0)^2) and X(f) = 1/f0 * exp(-pi*(f/f0)^2) has area of 1 and an iFT of x(t) = exp(-pi*(t/t0)^2) where in all cases t0*f0 = 1. and because of that, i think it's natural to define t0 as the "pulsewidth" and f0 as the "bandwidth". thanks for keeping me honest. r b-j From: Jitendra Rayala on 12 Jul 2010 20:26 On Jul 7, 2:43 am, "Fan.Zhang" 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 Jitendra > > Thanks, > > F Strictly speaking th From: dvsarwate on 13 Jul 2010 08:53 I don't understand what Jitendra Rayala 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. --Dilip Sarwate First  |  Prev  |  Next  |  Last Pages: 1 2 3 4 5 6 Prev: Floating point DSPsNext: MATLAB indexing issue