pulsewidth and bandwidth

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

On Jul 7, 5:43&#4294967295;am, "Fan.Zhang" <zf624@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

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

On Jul 7, 5:43&#4294967295;am, "Fan.Zhang" <zf624@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

On Jul 8, 3:09&#4294967295;pm, robert bristow-johnson <r...@audioimagination.com>
wrote:
> On Jul 7, 5:43&#4294967295;am, "Fan.Zhang" <zf624@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. &#4294967295;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

On Jul 12, 4:05&#4294967295;am, Greg Heath <he...@alumni.brown.edu> wrote:
> On Jul 8, 3:09&#4294967295;pm, robert bristow-johnson <r...@audioimagination.com>
> wrote:
>
> > On Jul 7, 5:43&#4294967295;am, "Fan.Zhang" <zf624@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. &#4294967295;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

On Jul 12, 12:31�pm, robert bristow-johnson
<r...@audioimagination.com> wrote:

>
> &#4294967295; &#4294967295; x(t) = exp(-pi*(t/t0)^2)
>
> which has Fourier Transform of
>
> &#4294967295; &#4294967295; 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

On Jul 12, 2:14&#4294967295;pm, dvsarwate <dvsarw...@gmail.com> wrote:
> On Jul 12, 12:31&#4294967295;pm, robert bristow-johnson
>
> <r...@audioimagination.com> wrote:
>
> > &#4294967295; &#4294967295; x(t) = exp(-pi*(t/t0)^2)
>
> > which has Fourier Transform of
>
> > &#4294967295; &#4294967295; 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? &#4294967295;It is true that exp(-pi*(t)^2) has
> area 1, but what about exp(-pi*(t/t0)^2)? &#4294967295;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

On Jul 7, 2:43&#4294967295;am, "Fan.Zhang" <zf624@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

Jitendra


>
> Thanks,
>
> F

Strictly speaking th

I don't understand what Jitendra Rayala <jray...@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.

--Dilip Sarwate