4 ms·
There will be losses in the dipole conductor and resonator dielectric, so you have a finite input impedance, but that’s where the energy should go; into heat.
by madengr 5y ago
There will be losses in the dipole conductor and resonator dielectric, so you have a finite input impedance, but that’s where the energy should go; into heat. You could use it as a dummy load over very narrow bandwidth.