Re: NST power rating -- another perspective

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br> 

Tesla list wrote:
 >
 > Original poster: "Gerry Reynolds" <gerryreynolds-at-earthlink-dot-net>

 > I think ^^ is a C language notation because ^ is a bitwise operator for
 > exclusive or.

My C compilers don't have exponentiation, but in Fortran exponentiation
is "**". In BASIC it is "^".

 >  >  > the power across RL is:
 >  >  > PL = Vsec(oc)^^2  x   RL / [(sL)^^2 + sL(R + RL) + (R + RL)^^2]
 >  >
 >  > Power is not calculable so directly in this way. A product in the time
 >  > domain is not a product of transforms.
 >
 > My VL, IL, and PL should have been expressed as VL(s), IL(s), and PL(s) and
 > Vsec(oc) is assumed to be the transformed Vsec(oc).  I meant to keep the
 > expressions in frequency domain. The power in frequency domain, I believe,
 > will be the product of VL(s) and IL(s).  One can then take the inverse
 > Laplace transform to convert back to time domain (for the general
 > case).....or, one can do what you suggest below.

To see that the general expression doesn't work, try to compute the
power when Vsec(s)=V/s, a step function (or DC), and L=0:
PL=(V/s)^2*RL/(R+RL)^2
The inverse Laplace transform of this is a ramp, not a constant as it
should be.
Note that to obtain the correct value making s=jw you have to consider
Vsec as a constant, a phasor, not the Laplace transform of Vsec(s)
when s=jw. A product of Laplace transforms is a convolution in time,
not a product. Power is a nonlinear function of the voltage, and can't
be calculated directly from these transforms, although, as you observed,
some simple tricks give the correct answers.

Antonio Carlos M. de Queiroz