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Re: Fwd: New Sherwood report

Posted by Jan Erik Holm on Dec 03, 2010; 8:11pm
URL: http://elecraft.85.s1.nabble.com/Fwd-New-Sherwood-report-tp5793377p5801336.html

Yes logical at least to me. Well English is my second language
but still makes it difficult.

Yes but if the fall time is longer the discontinuity will be
less abrupt and by so make it easier for a bad regulating device,
if you understand what I mean.
We want to shape the first part of the envelope on the fall side,
not being to sharp of a knee, i e the first 1 ms from full power
is the most important part. If you increase the over all fall time
that transition period will also be longer, i e make it easier for
a system with not enough dynamic range.

Oh well once upon a moon I could do Fourier analysis on this but it
was 35 years ago and I just have forgot it all. The teacher I had
was a ham but he is an SK now.

/ Jim SM2EKM
--------------
On 2010-12-03 20:36, Kok Chen wrote:

>
> On Dec 3, 2010, at 11:06 AM, Jan Erik Holm wrote:
>
>> Yes 99% of all CW stations clicks at the "break" side.
>
> That makes perfect sense, Jan.
>
> If they are using simple IIR filters, the slope discontinuity is worse at the onset of switching than when it is at the end of the leading edge or trailing edge (visualize the RC constant diagrams that we learn in school :-).
>
> Now, the onset of switching states happens to occur at low power (zero power) on the rising edge of a CW pulse, so even if it is dirty, you are not putting out energy.  But it the worst part of an RC filter occurs at the highest power location at the trailing edge of a CW pulse.
>
> If the click spectra is mostly coming from slope discontinuities, the energy from the clicks are going to be much greater when the key is breaking than when the key is making.
>
> This is why a couple of us has said that it is not the "rise time" (or fall time) that is important, it is the n-th order discontinuities (slope discontinuity contributes more than higher order ones obviously, when you look at it as a Fourier series).
>
> Someone else can probably explain better than I can.  I can't do it without using equations.  English is my third language :-).
>
> 73
> Chen, W7AY
>
>
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