pulsewidth and bandwidth
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From: Fan.Zhang on 7 Jul 2010 05:43 Hi Experts, I know bandwidth = 1/pulsewidth. After reconsider the pulse shape and ISI, this question comes to my mind.But how it comes in this way? Thanks, F
From: Clay on 7 Jul 2010 10:32 On Jul 7, 5:43 am, "Fan.Zhang" <zf624(a)n_o_s_p_a_m.sina.com> wrote: > Hi Experts, > > I know bandwidth = 1/pulsewidth. > > After reconsider the pulse shape and ISI, this question comes to my > mind.But how it comes in this way? > > Thanks, > > F In mathematical terms, look at the fourier integral and simply apply time dilation to your pulse function. I.e., replace "t" with " a*t " and then define a new variable "T = a*t " and find the jacobian. You will end up with a omega/a term. And that is all there is to that. Thus you have the reciprocal relationship between time and frequenction dilation. IHTH, Clay
From: Fred Marshall on 7 Jul 2010 13:08 Fan.Zhang wrote: > Hi Experts, > > I know bandwidth = 1/pulsewidth. > Well, that's a rather gross approximation. Not hugely useful when considering ISI I shouldn't think. But maybe..... Think of it this way: Construct a rectangular pulse and run it through a lowpass filter. What happens as you decrease the filter bandwidth? - First, the edges of the pulse start to spread out. - Then, the rise and fall of the pulse spread so much that they meet somewhere near the end of the pulse. - Then, pulse peak amplitude drops as the rise time is longer than ~1/2 the pulse width. - Eventually the pulse doesn't get through the filter "at all". Maybe bandwidth = 1/pulsewidth is somewhere around where the filter bandwidth is 1/pulsewidth. You can try this out with something like Matlab or just using simple math with an assumed RC filter. The output of the filter is just: A *(1-e^-t/RC) A is the amplitude of a rectangular pulse. t is time R is resistance C is capacitance. The bigger RC, the lower the filter bandwidth. So, when t = the pulse width and RC is such that e^-t/RC is around 0.02 i.e. 2% when t/RC=3.9 (or you pick the number) then we might say that the pulse amplitude is reached well enough at the output. If the RC product is great enough, then the output can't reach the selected value. As I recall from the dim past, the "bandwidth" of such a simple filter is 1/RC radians per second or 1/2piRC Hz. And that isn't going to match up exactly with the temporal analysis above that yields an RC perhaps more related to the need. But, I think you'll find that it's close (i.e. within 50%). Fred
From: robert bristow-johnson on 8 Jul 2010 15:09 On Jul 7, 5:43 am, "Fan.Zhang" <zf624(a)n_o_s_p_a_m.sina.com> 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. r b-j
From: Greg Heath on 12 Jul 2010 04:05
On Jul 8, 3:09 pm, robert bristow-johnson <r...(a)audioimagination.com> wrote: > On Jul 7, 5:43 am, "Fan.Zhang" <zf624(a)n_o_s_p_a_m.sina.com> 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. Hope this helps. Greg |