RE: [SI-LIST] : Antenna Problem on the Board

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From: Chris Rokusek ([email protected])
Date: Mon Jun 04 2001 - 18:13:01 PDT


I like your power analogy except that it assumes each I(f) is halved which
is not the case. If a driver of 20 ohms feeds a 50 ohm line with/without a
30 ohm series terminator, the current is reduced to .71 (not .5) its
original value. Squaring that is .5 which is multipled by twice the time
ending up back at one. Seems like we agree to me!

For reference:

           k * I(f) * f^2 * A
   E(f) = --------------------

   where k is a constant,
         I(f) if current at a given frequency
         f is frequency,
         A is loop area (sep distance times length)
         r is antenna distance.

Now in attempting to map the series/parallel question into this formula I
reasoned to double the area in the sense that the wave (albreit with less
current than parallel case) travels down and then back. It happens to be at
the same location. I don't see how this thinking is drastically different
than creating a an actual route with twice the length (and loop area) that
folds back and runs along side itself to terminate near its source (with
adjusted feed current and end termination). This reasoning implies that the
reduction in current must outweigh the increase in loop area to be

If you don't like messing with "A", then another way to map this problem
into the simple formula is to look at the Fourier equivalent I(f) of the
signal resulting from each of the cases assuming some periodic switching
frequency (the formula is for sine wave of freq "f"). In the parallel case,
the signal content is that of a gorgeous trapezoidal waveform--each segment
sees a trapezoidal blip travel down it per cycle. However in the series
termination, it looks much different. The energy is not distributed among
harmonics in the same proportions, each small segment sees something
different than the others over a given cycle. Aren't the odds spectacular
that some of these small segments will sing in chorus at some frequency at
some angle and some distance away (at your FCC test).

This is all just hearsay anyway until someone publishes some results.

Best Regards,

Chris Rokusek

> -----Original Message-----
> From: [email protected]
> [mailto:[email protected]]On Behalf Of S. Weir
> Sent: Monday, June 04, 2001 1:52 PM
> To: [email protected]
> Subject: RE: [SI-LIST] : Antenna Problem on the Board
> Chris,
> I really don't follow your reasoning:
> At 11:16 AM 6/4/01 -0700, you wrote:
> >Richard,
> >
> >Your first paragraph sounds good to me.
> >
> >Something else that to consider is that with parallel
> termination, the wave
> >flows down the line without a reflection but with source termination the
> >wave has to travel _twice_ as far before it is absorbed. This
> seems loosely
> >like doubling the loop area. Sounds like a good case for
> >simulation/measurement.
> >
> >Chris Rokusek
> >Innoveda
> The current imparted is one half for each direction, Vdelta/(Rt + Zl)
> versus Vdelta/(Zl) so we have one quarter the power for twice the
> time. This says one-half the total energy, and the peak amplitude is one
> quarter. With the same geometries for each, what condition would
> ever give
> rise to the series termination radiating more than the parallel
> termination?
> Regards,
> Steve.

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