# Understanding the physics of antennas

1. Aug 9, 2013

### harts

I'm currently preparing for a challenge exam, and I'm having trouble understanding antennas. My physics book is good, but it is very brief in its section on antennas. I know there are more complicated types of antennas, but let's just stick with the "simple" half-wave antenna (or dipole antenna, or whatever you want to call it).

As I understand it now, this type of antenna is essentially 2 metal rods with an alternating voltage source in the middle. My book tells me that because the current in this rod is constantly changing, it will emit electromagnetic energy.

How do common antennas flip the charges on the two rods? The book mentions an LC oscillator - how do those work? Are there other kinds of oscillators?

I think where I'm having the most trouble is in understanding how exactly energy is radiated by the antenna.

Here's what my book says: "because $\vec{}E$ and $\vec{}B$ are 90° out of phase at points near the dipole, the net energy flow is zero. From this fact, you might conclude (incorrectly) that no energy is radiated by the dipole."

Is this phase difference because of the oscillator in the middle? Is it trying to tell me that if I integrate the Poynting vector over a full current/charge cycle I would get 0?

It goes on to tell me that at great distances from the antenna, the dipole fields become negligible (since they are proportional to 1/R3). Does that mean that antennas don't work when they are close to the receiver?

Here's the strangest part of the book for me: "At these great distances, something else causes a type of radiation different from that close to the antenna. The source of radiation is the continuous induction of an electric field by the time-varying magnetic field and the induction of a magnetic field by the time-varying electric field". According to the book, these fields ARE in phase. So, where did they come from, and where were they at the closer distances? Were they overruled by the earlier dipole fields? How are the properties of this wave determined? How could I find the power and frequency of said wave?

What if I had two antennas with the same power and frequency? Would it make the strength of my signal in the transmission area better or worse?

Thanks (EDIT: I don't really know how to do vectors properly on physics forums. Sorry about that.)

2. Aug 9, 2013

### davenn

Hi harts

altho I know the basic answers to your questions, I suspect you are looking for some deeper answers.... that i will leave to others :)

but this question needs clarifying....
2 antennas fed from the same RF source ? that is a single coax that is then split to feed 12 separate antennas ? or were you thinking of something different

in the case of 2 antennas and one source say the antennas were Yagi type and 10 dBd ( dBd = dB over a dipole) gain. If you are feeding 10Watts into just one Yagi you have 10 W x 10 gain or an ERP (Effective Radiated Power) of 100W (minus some small losses that we wont go into here)

if hnow you split the coax and feed 2 identical Yagi's with that 10 Watts, you double your power, that's an additional 3 dB .... so your total gain is now 13dBd
no, it's not 10dBd + 10 dBd = 20dBd .... many years ago, it took me a while to grasp that fact

You also have to remember that since you split that 10 Watts between the 2 antennas, each antenna is only seeing 5W from the transmitter

cheers
Dave

3. Aug 9, 2013

### yungman

First of all, below is only for short dipole where current is assumed constant along the rod. It will be too complicate to assume sinusoidal. The idea is the same.

(1) As differential voltage driving into the both rods, current driving into the rod.
$$\tilde A=\frac {\mu_0}{4\pi}\int_{v'}\tilde J \frac{e^{-jkr}}{r}d\vec l'\;\Rightarrow\; \mu_0\tilde H=\nabla\times \tilde A, \;and\; \tilde E=\frac{1}{j\omega \epsilon_0}\nabla\times \tilde H$$
So from current, you get E and H.

$$P=\frac {1}{2}\int_{v'}\tilde E\times \tilde H^* dv'=P_{rad}+P_{img}$$
At close to antenna ( near field), because the $P_{img}$>>$P_{rad}$. Most power is imaginary, that's the reason the real radiation power is like zero. It is too long to write out the formulas. But if you use the formulas given from (1) you'll get
$$E_r=K_1\left(1+\frac{1}{jkr}\right), \;E_{\theta}=K_2\left[1+\frac{1}{jkr}+\frac{1}{(jkr)^2}\right]$$
As you can see, if $kr$<<1 $E_{\theta}$ dominates and is complex. Same goes to H. That's the reason the book say radiating power is zero and close distance ( near field). You calculate the power and you'll see.

(3) It is not correct to say dipole field goes to zero at large distance. Look at (2) above, $E_r\propto\frac{1}{r^2}$ but $E_{\theta}\propto\frac {1}{r}$. Therefore $E_r$ assume to be zero and only $E_{\theta}$ remains at large distance from antenna.

(4) If you refer back to (2) where I explain
$$E_r=K_1\left(1+\frac{1}{jkr}\right), \;E_{\theta}=K_2\left[1+\frac{1}{jkr}+\frac{1}{(jkr)^2}\right]$$
H is the same way. The poynting vector is too complicate to calculate using all the terms. So you make assumption for a given distance. For example if kr<<1,$\Rightarrow \frac{1}{(jkr)^2}$>>$\frac{1}{(jkr)}$>>1. So you ignore the smaller terms. So you will see the fields looks different at different distance. It is just an assumption but it is good enough.

This is a big subject. I am studying antenna right now. No body can explain in a post here. What I give you is just a starting point in trying to answer your question. Remember, the antenna theory book is thicker than the electromagnetic book!!!

If you are going to study antennas and EM, you better learn Latex. This and other forums all use Latex to write formulas. Without that, good luck in even asking question!!! I am getting to the point I can just type as if it's English!!!

Last edited: Aug 9, 2013
4. Aug 10, 2013

### yungman

Look at the formulas of $\vec E$ and $\vec H$ in this article. Make assumption according to the distance and calculate the Poynting vector. You'll see this a lot clearer.

http://en.wikipedia.org/wiki/Dipole_antenna

In the article, it also show you the far field assumption. Do it on the near field assumption, you can see it become imaginary and does not contribute to real radiating power.