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From: "Mike Engelhardt"
Subject: Re: LT SwitcherCAD Current Source Help
Date: 13 Sep 2002 21:28:46 GMT
References: <3D809F7A.firstname.lastname@example.org> <5Pgg9.email@example.com>
Reply-To: "Mike Engelhardt"
X-Newsreader: Microsoft Outlook Express 6.00.2600.0000
The implementation of the voltage controlled switch
in Berkeley SPICE has a continuous 1st derivative,
though it has a discontinuity in value. The
continuous 1st derivative helps it converge even
when its right at the edge of switching between the
two conductivity values.
The only improvement is to offer a switch like that
in LTspice which is everywhere continuous in value
"Kevin Aylward" wrote in message
> "Mike Engelhardt" wrote in message
> > Kevin wrote:
> > > > > ...The spice SW voltage controlled switch is not discontinues.
> > > > > It varies from its min value to its max value. You can not
> > > > > make it infinite...
> > > >
> > > > Err, well, yes, that is discontinuous. You can make switches
> > > > continuous and smoothly
> > >
> > > Er... Nope...don't know what you mean at all.
> > >
> > > If something various from a to b with no missing bits in-between its
> > > continuous, and that is what the spice switch does. Maybe you are
> > > confusing continuity with differentiability. For example, a triangle
> > > wave is continuous, but it is not differentiable at its turning
> point as
> > > its derivative depends on the direction of approach.
> > I've always been of the persuasion that a triangular wave is
> > continuous but has a discontinuous 1st derivative. A square
> > wave has a continuous 1st derivative(always zero) but the
> > function itself is discontinuous.
> There is a subtlety here that makes the square wave derivative
> technically discontinuous, but I can understand the rational for saying
> that it might be continuous.
> A definition of continuous is if and only if:
> f(xo) = f(x)|x->xo
> That is, the limit of a function as x->xo must equal that function at
> xo, independant of .
> In posh talk it would be:
> |f(x)-f(a)| < epsilon for all x with |x-a|< delta,
> A square wave function would have to be so defined as to be single
> so that only on one side of the jump is it say -0 is 0 and at 0+ it is 1
> with f(0)=1.
> Calculating the derivative on the positive side:
> f(x+dx)-f(x)/dx |dx->0
> or (1 - 1)/dx
> On the negative side:
> (0 - 1)/dx
> These cannot be equal, hence its derivative does not exist at x=0.
> Or, in more simple terms, differentiating an ideal square gives infinite
> spikes at the discontinuity, and it is positive or negative, depending
> on the direction, hence it clearly has no derivative at that point.
> However, there are other functions that are discontinuous at a point xo,
> yet still have a derivative at that point.
> Kevin Aylward
> SuperSpice, a very affordable Mixed-Mode
> Windows Simulator with Schematic Capture,
> Waveform Display, FFT's and Filter Design.
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