How Can A Fanned Dipole Only Use One Feedline?
How does your radio signal know which part of the dipole to use if you've only got 1 coax wire going to the antenna? If the dipole supports 4 bands, let's say, and suppose you key up on 20 meters. How does that 20 meter transmission "know" to only go to the 20 meter length of wire and not the other lengths so as not to have a wacky SWR?
Your signal in the coax looks for the path of least resistance! In this case, the antenna that represents the best match for a 50 ohm load will get the job done. The antennas that are not a match will present an improper impedance and won't be a problem as they will get ignored by the RF.
If I test copper wire on a multimeter, it will usually show almost 0 ohms of resistance and the length of wire makes very little difference. So how can it be 50 ohms on a radio? Or does the resistance change when you change frequencies on the radio?
I guess I just don't understand how if you hookup several wires of different lengths to the same coax line, how the radio signal "knows" to only use one of those wires and not the others. Won't the radio signal just go into all the wires?
It's not resistance but IMPEDANCE that matters! RF is AC, alternating current. The impedance of the dipole that is a half wave for frequency in question is (more or less) 50 ohms. The impedance of the other dipoles will be higher or lower(usually by a lot), and will hence be mismatched. VERY little RF will enter these dipoles.
You might want to go back to basic electricity and review impedance and resistance. I know I had to many times!
Good luck and have fun learning - that's a big part of ham radio.
The radio doesn't "know" anything. It's called "resonance". The length of wire that is a resonant length for the band being transmitted on is the element that will radiate the signal. The other adjacent elements will also modify the pattern due to their proximity to the radiating element, but not enough to cause much distortion of the pattern.
Don't confuse resistance with impedance. Both are measured in ohms but are different entities. A dipole is an antenna that when cut to resonance using the formula length (in feet)=468/(freq. in mHz) has an impedance close to 72 ohms in free space. Now, antennas in free space don't actually exist, except in theory. As it turns out, your dipole sits above ground, which has an affect on the antenna's impedance. Turns out, that causes the impedance at resonance to drop a bit (the antenna is "detuned" due to proximity to the Earth!) so that 50 ohm coax makes a good feed line for dipoles.
Originally Posted by KT4JX
Almost all rigs made today use 50 ohm impedance connectors for the antenna. The rig is designed to output its RF into a 50 ohm load. If it sees that 50 ohm load, that is where the RF will go!
These are things I never really understood in the exams. It's been a long time since the exams and I've forgotten almost everything. So the word "ohm" can mean either resistance or impedance?
Here is a good explanation;
Originally Posted by KT4JX
"Oh what a tangled web we weave when first we practice to receive."
-Otto Watt Sept. 5 1925
Assuming the next question will be: What causes the feedpoint impedances? ... Hold on to your hat!
Originally Posted by KT4JX
To completely answer your question one needs to look beyond the impedances involved because they are virtual impedances caused by the superposition of forward and reflected waves on the wires of a standing-wave antenna. Let's take a simple case of one 1/2WL dipole in parallel with a 1WL dipole. (A single #14 horizontal wire, 30 feet in the air, has a characteristic impedance very close to 600 ohms.)
During the transient time after an RF signal is applied at the feedpoints of the above dipoles, the initial transient currents will split equally in the two dipoles because both dipoles exhibit an initial impedance of 600 ohms. The forward current in the 1/2WL dipole will be reflected from the ends of the 1/2WL dipole and will arrive back at the feedpoint in phase with the forward current on the 1/2WL dipole so the forward and reflected currents on the 1/2WL dipole will add together to a high value. It's called constructive interference. The forward current in the 1WL dipole will be reflected from the ends of the 1WL dipole and will arrive back at the feedpoint 180 degrees out of phase with the forward current on the 1WL dipole so the forward and reflected currents on the 1WL dipole will subtract and tend to cancel to a low value. It's called destructive interference. The overall steady-state result, after the initial transient condition, is a lot of current flowing in the 1/2WL dipole and very little current flowing in the 1WL dipole. Both of these dipoles are standing-wave antennas.
What we call resonance on a standing-wave antenna, e.g. a dipole, is simply a length chosen for the dipole that causes the reflected current from the ends of the dipole to arrive back at the feedpoint in phase with the forward current just being delivered to the feedpoint from the source. There is no resistor at the feedpoint of a dipole. It is a lossless virtual resistance caused by the total voltage divided by the total current resulting in an impedance phase angle of zero. The equation for the feedpoint impedance of a resonant standing-wave antenna is:
Zfp = (Vfor-Vref)/(Ifor+Iref)
This is also the equation for a resonant stub.
I apologize if that is more than anyone wanted to know.
73, Cecil, www.w5dxp.com
The maximum power transfer theorem works just as well for a non-resonant antenna as it does for a resonant antenna.
The height of a half-wave dipole above ground does affect its impedance. Now, height is measured in wavelengths of the radiated signal.
Look at the graph of height versus impedance which is attached.