FreeLists / si-list / [SI-LIST] Re: Rise time bandwidth relation

["David Banas" <david.banas@xxxxxxxxxx>]

Yes. In fact, I was really just paraphrasing what Howard Johnson and
Martin Graham relate in Appendix B of their book, "High-Speed Digital
Design - A Handbook of Black Magic" (Prentice Hall, New Jersey, 1993).

Sorry, Drs., I should have given you credit originally.

-db


> -----Original Message-----
> From: Asbenson, Lyndell L [mailto:lyndell.l.asbenson@xxxxxxxxx]
> Sent: Monday, May 07, 2007 5:41 PM
> To: David Banas
> Subject: RE: [SI-LIST] Re: Rise time bandwidth relation
>=20
> WOW, Is there any book that describes this? Thanks Lyndell
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> -----Original Message-----
> From: si-list-bounce@xxxxxxxxxxxxx
[mailto:si-list-bounce@xxxxxxxxxxxxx]
> On Behalf Of David Banas
> Sent: Monday, May 07, 2007 3:25 PM
> To: rohitsharma@xxxxxxxxxxxxx; si-list@xxxxxxxxxxxxx
> Subject: [SI-LIST] Re: Rise time bandwidth relation
>=20
> If you start with the assumption that your observed edge is the result
> of an ideal edge having moved through some composite channel with a
> Gaussian shaped impulse response, g(t), with "-3dB" bandwidth, B, then
> your observed edge, s(t), will be the integral of that impulse
response:
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>       s(t) =3D3D Integral{g(t) dt}                              (1)
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> If:
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>       G(w) =3D3D exp{-(a * w)^2}                                (2)
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> Then:
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>       g(t) =3D3D exp{-(t/2a)^2} / 2*sqrt{pi * a}                (3)
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> B is found by setting G(w) =3D3D 0.707. We get:
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>       B =3D3D .0937 / a.                                        (4)
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> The "error function" (ERF() in Excel) is defined as:
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>       ERF(z) =3D3D (2/sqrt(pi)) * Integral_0-z{exp(-u^2) du}    (5)
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> and gives the area under the Gaussian curve from -z to +z, normalized
to
> 1. (That is, ERF(Infinity) =3D3D 1.) ERF() can be used to find the =3D
> integral
> of g(t) by substituting "t/2a" from (3) in for "u" in (5). The
"10-90%"
> rise time of s(t) can be found by setting ERF(z) =3D3D 80%, solving =
for
=3D
> "u",
> and then setting:
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>       risetime/2a =3D3D 2u                                      (6)
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> (This works because g(t) is symmetric about its peak, and the 20% of
its
> area that we're leaving out of the integral is precisely divided in
> half: 10% before the integrated portion and 10% after, yielding the
10%
> and 90% points in s(t), respectively. We have to double "u", due to
the
> way ERF() is defined. (i.e. - ERF(z) integrates over the range
> [-z,+z].)) We get:
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>       u =3D3D 0.9                                               (7)
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>       rise-time =3D3D 2u * 2a =3D3D 3.6 * a                         (8)
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> and, using (4),
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>       rise-time =3D3D 0.337 / B                                 (9)
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> -db
>=20
> > -----Original Message-----
> > From: si-list-bounce@xxxxxxxxxxxxx
> [mailto:si-list-bounce@xxxxxxxxxxxxx]
> > On Behalf Of Rohit Sharma
> > Sent: Monday, May 07, 2007 8:01 AM
> > To: si-list@xxxxxxxxxxxxx
> > Subject: [SI-LIST] Re: Rise time bandwidth relation
> >=3D20
> > Hi All,
> >    Can somebody help me in understanding the derivation of the
> following
> > formula for digital signals:-
> >  Bandwidth =3D3D 0.35/Rise time.
> >=3D20
> > Regards,
> > Rohit
> >=3D20
> >=3D20
> >=3D20
> > --
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