From: Rich Peyton (email@example.com)
Date: Tue May 29 2001 - 04:05:09 PDT
I think there something on knee frequency ( beginning of book ) in Dr.
Howard's book. I do not have the book right now but it's in the
beginning of the book. One of my references ( white pages ) also has
> Hi All,
> The expression that Rich mentions can easily be
> derived. To be precise, one will get
> f = ln9/ (PI x Tr) = 0.349/Tr
> I do find some authors using the following expression:
> f = 0.5/Tr
> The frequency is referred to as Knee frequency.
> I'm curious to know how this expression is derived.
> --- Yibing Tang <firstname.lastname@example.org> wrote:
> > Hi, Rich,
> > The formula is correct.
> > Freq here doesn't mean bit rate you want to
> > transmit, it means the bandwidth
> > needed to transmit the signal at certain quality.
> > For drivers, rise time( edge rate) is not
> > neccessarily to be consistent with
> > bit rate (your frequency). Even at very low speed,
> > sometimes it requires
> > very fast rise time. In some situation, faster rise
> > time doesn't mean better
> > result.
> > Yibing
> > -----Original Message-----
> > From: email@example.com
> > [mailto:firstname.lastname@example.org]On Behalf Of
> > richard hill
> > Sent: 2001 05 24 10:32
> > To: email@example.com
> > Subject: [SI-LIST] : Frequency based on rise time
> > for drivers
> > I have often seen people estimating the frequency of
> > a
> > driver based on the rise time for the driver using
> > the
> > formula Freq = .35/Tr (This seems to be true for a
> > perfect sinusoid). There can be two different
> > drivers outputing signal at the same frequency but
> > different edge rates, won't this assumption be
> > invalid. Some times there is a big difference in
> > the
> > edge rates of drivers running at the same frequency.
> > If we use the above formula and estimate the edge
> > rates
> > using the frequency for the two drivers we will get
> > similar no.s.
> > Any comments.
> > Thanks in advance,
> > Rich
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