From: Fan.Zhang on
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
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
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
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
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
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