Antenna and Freespace Impedance

[likephysics]

Why isn't the antenna impedance (usually 50 ohms) matched with free space impedance (377 ohms)?
Won't the mismatch result in some loss.

 

[abluphoton]

i want to know this..

 

[sophiecentaur]

Simple answer is that the dipople acts as a transformer, which, at resonance, happens to transform the 377 ohms to the 73 ohms of a half wave dipole.
How like a transformer? Waving arms around I could say that the varying fields created around the dipole ( E and B fields in quadrature) as current sloshes back and forth along the wire manage to induce E and B fields in space, at a distance, with a ratio of 377. All I can say is that it's not surprising that the actual resistance values are different!
I must say I find the change from quadrature to in-phase a bit weird but the dipole is actually resonating so you have a standing wave on the structure but a progressive wave in space.

 

[jsgruszynski]

Why isn't the antenna impedance (usually 50 ohms) matched with free space impedance (377 ohms)?
Won't the mismatch result in some loss.

First: first, 377 ohms simply "is" the free-space impedance a result of physical constants of nature like the speed of light, etc.

Second: think of an antenna as a 2-port impedance matching filter between a transmission line and free space.

Third: remember that 50 ohms is arbitrary in the sense that it simply happens to be one of two optimal solutions (50 ohms or 75 ohms) commonly used for a coaxial transmission lines. The sweet spots for balanced lines are 300 ohm, hence that's the "nominal value" you see for that transmission line geometry.

Optimal in what sense?

• For the materials involved (per the periodic table)
• With using the coaxial characteristic impedance equation
• For values of conductor and insulator thickness that make the cables human-friendly (not too heavy or inflexible
• For the resource cost of materials minization
  

For these, you get optimal points (defined by arbitrary trade-offs) which are 50 ohms is trade-off for power transfer loss vs. mismatch while 75 ohms is optimize for mismatch loss ignoring power transfer loss.

http://www.microwaves101.com/encyclopedia/why50ohms.cfm

The latter is why CATV uses 75 ohms - you can get sloppier with your installations without having to worry so much about effects of mismatch. It's all \muW of power so power handling is a non-issue.

50 ohms gets used for everything else but especially for military radar originally where power transfer was most important and specialize/expensive matching could be used without much additional cost or problem - it was also a complicated business. The actual peak power optimal impedance is 30 ohms!

 

[sophiecentaur]

One thing worth pointing out is that most antennae, in practice, aren't half wave dipoles. So they don't have a 73 ohm impedance. For instance, a monopole mast would often be 1/4 wave high and an impedance of 39ohms (ish). Yagi antennae, with all those parasitic elements, have much higher impedance. Thus, you nearly always need an impedance matching network at the feedpoint.

 

[wbeaty]

Why isn't the antenna impedance (usually 50 ohms) matched with free space impedance (377 ohms)?
Won't the mismatch result in some loss.

A dipole antenna doesn't exactly "have" a single unique impedance. Yes, if we cut a half-wave conductor in two, then connect a feedline at the break, we'll find a low Z at that spot.

But if we go and google up some "delta match" articles, we get our nose rubbed in the true nature of antennas... try leaving your half-wave wire continuous and unbroken. Then gradually spread apart your feedline conductors, and solder them to points on the antenna far from the center:

http://www.ycars.org/EFRA/Module%20C/AntMatch.htm
http://www.g4nsj.co.uk/delta.shtml

Obviously the dipole antenna "has" much higher impedance at points farther from the midpoint. This isn't so mysterious, since the signal at the exact center of the dipole is basically zero impedance "all current," while the signal at the tips is infinite impedance "all voltage."

As I understand it, the Delta, the spread feedline, acts very much like a conical microwave waveguide, but for longwave HF radio frequencies.

Also try looking up longwave "waveguide style" antennas called Beverage, Rhombic, and Vee or "V antenna" types. Weird stuff to bend one's brain around (especially that strange little resistor on the Rhombic.)

Also visualize employing a delta-match on a VHF loop antenna: keep your resonant metal ring unbroken, then connect your feedline to the loop via a sloping, spread-apart conductor-pair.

 

[wbeaty]

I must say I find the change from quadrature to in-phase a bit weird but the dipole is actually resonating so you have a standing wave on the structure but a progressive wave in space.

To teach such things for 8.02 class at MIT, John Belcher developed the TEAL project: a set of mpeg field animations with a moving wood-grain effect. In this one below you can see the nearfield pattern oscillating in and out, while the farfield peels off and flies away. (It's a video loop, so click "play" over and over.)

http://web.mit.edu/viz/EM/visualizations/light/dipoleRadiationReversing/DipoleRadiationReversing.htm

The site doesn't have every single possible situation of course. I'd love to also see some animations for pure quadrature nearfield, and also for pure radiation with nearfield waves removed.

More info:
http://web.mit.edu/viz/EM/visualizations/light/index.htm
http://web.mit.edu/jbelcher/www/TEALref/fnlEditedLinks.pdf