[SI-LIST] Re: Rise time bandwidth relation
- From: "Dennis Han" <Dennis.Han@xxxxxxxxxxxxxxx>
- To: <si-list@xxxxxxxxxxxxx>
- Date: Tue, 8 May 2007 07:46:55 -0700
The exact formula is ln(upper rise time limit / lower rise time limit) /=0D=
=0A
(2 * pi)=0D=0A
=0D=0A
ln(0.9/0.1)/2/pi =3D 0.35014 for 10% to 90% rise time=0D=0A
ln(0.8/0.2)/2/pi =3D 0.21972 for 20% to 80% rise time =0D=0A
=0D=0A
=0D=0A
Dennis=0D=0A
=0D=0A
-----Original Message-----=0D=0A
From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx]=0D=
=0A
On Behalf Of Cavanna, Vicente Vaca (Sr. ; ProCurve ASICs)=0D=0A
Sent: Monday, May 07, 2007 7:27 PM=0D=0A
To: david.banas@xxxxxxxxxx; rohitsharma@xxxxxxxxxxxxx;=0D=0A
si-list@xxxxxxxxxxxxx=0D=0A
Subject: [SI-LIST] Re: Rise time bandwidth relation=0D=0A
=0D=0A
I will try to find a reference since the results are well known (and=0D=0A
derived in many textbooks) but in the meantime this is what I recall:=0D=0A
=0D=0A
1. If the channel has an exponential impulse response, such as would be=0D=
=0A
obtained from a simple RC low-pass filter, then the 3dB bandwidth (BW)=0D=
=0A
of that channel is related (exactly) to the 10-90% risetime (t) of the=0D=
=0A
step response of the same channel by the formula:=0D=0A
=0D=0A
t =3D3D 0.35/BW this is an exact result - no approximation=0D=0A
=0D=0A
2. If the channel has a gaussian impulse response, such as would be=0D=0A
obtained from a cascade of a large number of low-pass filters with=0D=0A
arbitrary response but conmensurate bandwidths, then the relationship=0D=0A
between bandwidth and risetime is:=0D=0A
=0D=0A
t =3D3D 0.5/BW this is also an exact result=0D=0A
=0D=0A
For other channels the numerator in the above formula tends to be=0D=0A
somewhere in between 0.35 and 0.5.=0D=0A
=0D=0A
=0D=0A
Vicente=0D=0A
=0D=0A
> -----Original Message-----=0D=0A
> From: si-list-bounce@xxxxxxxxxxxxx=3D20=0D=0A
> [mailto:si-list-bounce@xxxxxxxxxxxxx] On Behalf Of David Banas=0D=0A
> Sent: Monday, May 07, 2007 3:25 PM=0D=0A
> To: rohitsharma@xxxxxxxxxxxxx; si-list@xxxxxxxxxxxxx=0D=0A
> Subject: [SI-LIST] Re: Rise time bandwidth relation =3D20 If you start=
=0D=0A
>with the assumption that your observed edge is=3D20 the result of an =0D=
=0A
>ideal edge having moved through some=3D20 composite channel with a =0D=0A
>Gaussian shaped impulse response,=3D20 g(t), with "-3dB" bandwidth, B, =
=0D=0A
>then your observed edge,=3D20 s(t), will be the integral of that impuls=
e=0D=0A
=0D=0A
>response:=0D=0A
>=3D20=0D=0A
> s(t) =3D3D3D Integral{g(t) dt} (1)=0D=0A
>=3D20=0D=0A
> If:=0D=0A
>=3D20=0D=0A
> G(w) =3D3D3D exp{-(a * w)^2} (2)=0D=0A
>=3D20=0D=0A
> Then:=0D=0A
>=3D20=0D=0A
> g(t) =3D3D3D exp{-(t/2a)^2} / 2*sqrt{pi * a} (3)=0D=0A
>=3D20=0D=0A
> B is found by setting G(w) =3D3D3D 0.707. We get:=0D=0A
>=3D20=0D=0A
> B =3D3D3D .0937 / a. (4)=0D=0A
>=3D20=0D=0A
> The "error function" (ERF() in Excel) is defined as:=0D=0A
>=3D20=0D=0A
> ERF(z) =3D3D3D (2/sqrt(pi)) * Integral_0-z{exp(-u^2) du} (5)=0D=0A
>=3D20=0D=0A
> and gives the area under the Gaussian curve from -z to +z,=3D20 =0D=0A
>normalized to 1. (That is, ERF(Infinity) =3D3D3D 1.) ERF() can be=3D20 =
=0D=0A
>used to find the =3D3D integral of g(t) by substituting "t/2a"=3D20 fro=
m =0D=0A
>(3) in for "u" in (5). The "10-90%"=0D=0A
> rise time of s(t) can be found by setting ERF(z) =3D3D3D 80%,=3D20 =0D=
=0A
>solving for =3D3D "u", and then setting:=0D=0A
>=3D20=0D=0A
> risetime/2a =3D3D3D 2u (6)=0D=0A
>=3D20=0D=0A
> (This works because g(t) is symmetric about its peak, and the=3D20 20%=
=0D=0A
>of its area that we're leaving out of the integral is=3D20 precisely =0D=
=0A
>divided in=0D=0A
> half: 10% before the integrated portion and 10% after,=3D20 yielding =0D=
=0A
>the 10% and 90% points in s(t), respectively. We=3D20 have to double =0D=
=0A
>"u", due to the way ERF() is defined. (i.e. -=3D20=0D=0A
> ERF(z) integrates over the range=0D=0A
> [-z,+z].)) We get:=0D=0A
>=3D20=0D=0A
> u =3D3D3D 0.9 (7)=0D=0A
>=3D20=0D=0A
> rise-time =3D3D3D 2u * 2a =3D3D3D 3.6 * a=0D=0A
(8)=0D=0A
>=3D20=0D=0A
> and, using (4),=0D=0A
>=3D20=0D=0A
> rise-time =3D3D3D 0.337 / B (9)=0D=0A
>=3D20=0D=0A
> -db=0D=0A
>=3D20=0D=0A
> > -----Original Message-----=0D=0A
> > From: si-list-bounce@xxxxxxxxxxxxx=0D=0A
> [mailto:si-list-bounce@xxxxxxxxxxxxx]=0D=0A
> > On Behalf Of Rohit Sharma=0D=0A
> > Sent: Monday, May 07, 2007 8:01 AM=0D=0A
> > To: si-list@xxxxxxxxxxxxx=0D=0A
> > Subject: [SI-LIST] Re: Rise time bandwidth relation =3D3D20 Hi All,=0D=
=0A
> > Can somebody help me in understanding the derivation of the=0D=0A
> following=0D=0A
> > formula for digital signals:-=0D=0A
> > Bandwidth =3D3D3D 0.35/Rise time.=0D=0A
> >=3D3D20=0D=0A
> > Regards,=0D=0A
> > Rohit=0D=0A
> >=3D3D20=0D=0A
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