[SI-LIST] Re: Rise time bandwidth relation
- From: "David Banas" <david.banas@xxxxxxxxxx>
- To: <rohitsharma@xxxxxxxxxxxxx>, <si-list@xxxxxxxxxxxxx>
- Date: Mon, 7 May 2007 15:25:29 -0700
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:
s(t) =3D Integral{g(t) dt} (1)
If:
G(w) =3D exp{-(a * w)^2} (2)
Then:
g(t) =3D exp{-(t/2a)^2} / 2*sqrt{pi * a} (3)
B is found by setting G(w) =3D 0.707. We get:
B =3D .0937 / a. (4)
The "error function" (ERF() in Excel) is defined as:
ERF(z) =3D (2/sqrt(pi)) * Integral_0-z{exp(-u^2) du} (5)
and gives the area under the Gaussian curve from -z to +z, normalized to
1. (That is, ERF(Infinity) =3D 1.) ERF() can be used to find the =
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) =3D 80%, solving for =
"u",
and then setting:
risetime/2a =3D 2u (6)
(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:
u =3D 0.9 (7)
rise-time =3D 2u * 2a =3D 3.6 * a (8)
and, using (4),
rise-time =3D 0.337 / B (9)
-db
> -----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
>=20
> Hi All,
> Can somebody help me in understanding the derivation of the
following
> formula for digital signals:-
> Bandwidth =3D 0.35/Rise time.
>=20
> Regards,
> Rohit
>=20
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