[SI-LIST] Re: Rise time bandwidth relation
- From: "Cavanna, Vicente Vaca (Sr. ; ProCurve ASICs)" <vicente.cavanna@xxxxxx>
- To: <david.banas@xxxxxxxxxx>, <rohitsharma@xxxxxxxxxxxxx>, <si-list@xxxxxxxxxxxxx>
- Date: Tue, 8 May 2007 00:27:02 -0000
I will try to find a reference since the results are well known (and
derived in many textbooks) but in the meantime this is what I recall:
1. If the channel has an exponential impulse response, such as would be
obtained from a simple RC low-pass filter, then the 3dB bandwidth (BW)
of that channel is related (exactly) to the 10-90% risetime (t) of the
step response of the same channel by the formula:
t =3D 0.35/BW this is an exact result - no approximation
2. If the channel has a gaussian impulse response, such as would be
obtained from a cascade of a large number of low-pass filters with
arbitrary response but conmensurate bandwidths, then the relationship
between bandwidth and risetime is:
t =3D 0.5/BW this is also an exact result
For other channels the numerator in the above formula tends to be
somewhere in between 0.35 and 0.5.
Vicente
> -----Original Message-----
> From: si-list-bounce@xxxxxxxxxxxxx=20
> [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=20
> the result of an ideal edge having moved through some=20
> composite channel with a Gaussian shaped impulse response,=20
> g(t), with "-3dB" bandwidth, B, then your observed edge,=20
> s(t), will be the integral of that impulse response:
>=20
> s(t) =3D3D Integral{g(t) dt} (1)
>=20
> If:
>=20
> G(w) =3D3D exp{-(a * w)^2} (2)
>=20
> Then:
>=20
> g(t) =3D3D exp{-(t/2a)^2} / 2*sqrt{pi * a} (3)
>=20
> B is found by setting G(w) =3D3D 0.707. We get:
>=20
> B =3D3D .0937 / a. (4)
>=20
> The "error function" (ERF() in Excel) is defined as:
>=20
> ERF(z) =3D3D (2/sqrt(pi)) * Integral_0-z{exp(-u^2) du} (5)
>=20
> and gives the area under the Gaussian curve from -z to +z,=20
> normalized to 1. (That is, ERF(Infinity) =3D3D 1.) ERF() can be=20
> used to find the =3D integral of g(t) by substituting "t/2a"=20
> from (3) in for "u" in (5). The "10-90%"
> rise time of s(t) can be found by setting ERF(z) =3D3D 80%,=20
> solving for =3D "u", and then setting:
>=20
> risetime/2a =3D3D 2u (6)
>=20
> (This works because g(t) is symmetric about its peak, and the=20
> 20% of its area that we're leaving out of the integral is=20
> precisely divided in
> half: 10% before the integrated portion and 10% after,=20
> yielding the 10% and 90% points in s(t), respectively. We=20
> have to double "u", due to the way ERF() is defined. (i.e. -=20
> ERF(z) integrates over the range
> [-z,+z].)) We get:
>=20
> u =3D3D 0.9 (7)
>=20
> rise-time =3D3D 2u * 2a =3D3D 3.6 * a (8)
>=20
> and, using (4),
>=20
> rise-time =3D3D 0.337 / B (9)
>=20
> -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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