The part that I found surprising was that the power is not a sine wave. But
I guess I shouldn't have found that surprising since that's also true of
normal AC power distribution; with a purely resistive load, the power is
proportional to sin^2(ωt).
I only just learned (well, started learning about) radiation resistance
earlier today. I'm clearly a long way from fully understanding this
stuff. The introductory E&M class I took about 7 years ago definitely has
helped, though I don't remember enough of it.
On Thu, May 4, 2017 at 3:18 PM, Eric Schneider <erictschneider@xxxxxxxxx>
Not dumb questions at all!
I'll add to Darrell's accurate response. The key being "proportional".
Electrical power is _always_ the product of current and voltage (at an
instant, not average).
V=I*R -> I=V/R, P=I*V -> P=(V/R)*V=(V^2)/R
For AC, the voltage and current can be out of phase, and the power
transfered is a function of this phase difference (called "power factor").
In the case of a 90 degree difference (purely capacitive or inductive),
there is no power transfer.
So if you tune your antenna perfectly (so it is purely resistive), then
v_peak*sin^2(wt)/R will be accurate. The R (radiation resistance) will be
dependent on the characteristics of the antenna (probably most importantly,
length vs wavelength).
On 05/04/2017 02:20 PM, Eric Smith wrote:
On Thu, May 4, 2017 at 1:07 PM, Darrell Harmon <
Yes, transmitted power is proportional to voltage squared (antenna
input impedance is constant with input power variation).
The electric and magnetic fields are proportional to voltage and
Thanks for the quick answer!