[Lowfer] Epson Question

[Scott Tilley]

The wonders of Integral calculus...

Why not just use the general case and solve for whatever the situation is
and let'm know where the formula came from?

If you can write an equation for what your seeing on your scope (or make a
general approximation based on observation) one can write:

I^2(rms) = a ~ b (~ is an ascii integration operator) f^2(t) dt / (b-a)

where,

a = start of period (in some time unit)
b = end of period (in some time unit)
f(t) = some time based function describing your waveform...

In most cases you can simply write the area under the curve from common
knowledge (like a square or rectangle) and plug that in rather then doing
the integration... There are sites on the web that can do the integration
for you or you can use Maple or an TI-89 type calc. to do the work and prove
to yourself where this stuff comes from...

Scott, VE7TIL

----- Original Message ----- 
From: "Bill de Carle" <ve2iq@magma.ca>
To: <lowfer@mailman.qth.net>
Sent: Sunday, April 18, 2004 5:40 PM
Subject: RE: [Lowfer] Epson Question


> At 04:41 PM 4/18/2004 -0700, Stewart, KK7KA wrote:
> >Hi all,
> >
> >There is no need to argue -- both Bills are right, but are describing
> >different quantities.  Bill de Carle's formula of V^2/2*R gives the total
> >power delivered to a load, when the signal switches between V and ground,
> >with a 50% duty cycle.  However, half of that power is DC, and would not
> >contribute to radiation, drive power for an amp, etc.  Bill Ashlock's
> >formula of V^2/4*R gives the total AC power in the square wave.  However,
> >that also includes the harmonics, which are not useful HiFER output.
IIRC,
> >the p-p amplitude of the fundamental is ~1.275 times the p-p of a square
> >wave.  Then, applying V^2/8*R, where R is the effective load resistance
at
> >the fundamental frequency, should give a reasonably accurate result.
>
> Very clearly stated, Stewart!
> I was trying to get some discussion (argument?) going to liven things
> up a bit.  We often fail to define precisely what the conditions are
> when tossing formulas out, which can lead to erroneous assumptions.
> Bill was no doubt right in using the term square wave to describe a
> signal that is a limiting case for a clipped sinewave, i.e. one which
> spends half its time above ground, the other half below ground, with
> the load connected to ground.  My beef was that there are other cases
> where the driving signal is a "square" wave, shape-wise, that don't
> have the same power delivered.
>
> Bill
>
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