# Where does radiated power come from?

I've had electromagnetism classes back in uni, but I'm having a hard time linking them with that matter:

1. Take a car antenna -which I assume is simply a wire, coiled or not-, and apply a voltage across it. The way I see it the current will only be limited by the resistance of the wire, which also means all the power is lost via Joule dissipation. Yet, electromagnetic waves are emitted, which represents radiated power (it has to be nonzero as we can capture it on the receiver end). So how can we account for that power in the power budget?
2. The underlying problem that made me realise this is: how can I calculate a rough estimate of the radiated power, or at least a maximum value, of the radiated power if I know how much electric power I put in? This is to make sure it won't disturb cellphones even though I'm pretty sure it won't - stepper square pulses, 28V 50W switched @ ~10kHz [rise time 500us], cables in single turn loop of ~15cm in diameter.
• What is the rise time on your 10kHz signal? – Matt Young Oct 17 '14 at 17:28
• They are stepper commands, 120ohms 20mH, which means a 95% rise time of 500us. – Mister Mystère Oct 17 '14 at 18:02
• Check your math, your 95% rise time is 5x the period of your input signal. – Matt Young Oct 17 '14 at 19:31
• Good catch, it's supposed to be 99%. – Mister Mystère Oct 17 '14 at 22:53

## 2 Answers

How can I calculate a rough estimate of the radiated power, or at least a maximum value, of the radiated power if I know how much electric power I put in (assume an RLC circuit)?

If you use a resonant RLC circuit (basically a magnetic loop antenna) and you believe you know the value of R, you might still not get an accurate answer. This is because R is really hard to compute. It is it hard to compute because the electric and magnetic fields in the close vicinity of the loop antenna may be producing eddy currents and dielectric heating in things that are reasonably close by. How would you account for such things? It's still going to look like a real power loss and you just can't tell one power from another. simulate this circuit – Schematic created using CircuitLab

In the close vicinity of the loop (or in fact any antenna used for EM transmission), the localized electric and magnetic fields are not sufficiently aligned to form an EM wave and those losses (I mentioned above) are just straight forward local power losses and nothing to do with the emission of an EM wave.

• Good point. So how can I estimate a cap to the radiated power then? Even a rough guesstimate would do, it's really to compare it with the radiated power of cellphones for example, with numbers rather than "I have no clue but that should do". – Mister Mystère Oct 17 '14 at 22:57
• @MisterMystère it's really difficult to estimate (well for me it is). The induced eddy current and DC copper power losses could be a hundred times what is emitted by an EM wave. It all depends on the practical realization of the RLC. It's like asking "how big is a 1uH inductor" - it could be an 0603 component size or, be a 50cm x 20 cm loop of wire (which I know to be about 1uH). – Andy aka Oct 18 '14 at 0:42

When you look at the input impedance of an antenna, there will be a real component, indicating that electrical power is delivered to the antenna.

Part of the real component is due to resistive loss (electrical power transformed to heat), and part is due to radiated loss (electrical power transformed into radiated em waves).

Antennas are inherently not lumped circuit elements, so you can't use all the analysis techniques of lumped circuit theory when analyzing them.

• Thanks for your answer. That clarifies what I thought, but how can I estimate those radiated losses even if it means making lots of assumptions? For example, one loop of known radius through which a known electrical power is forced at a certain frequency. I suggested example figures in my post. – Mister Mystère Oct 17 '14 at 18:00