Attenuation between two whip antennae? Calling all Hams.

Factual Questions
1

How do I calculate the attenuation I’d expect in a radio link?

More specifically, I have two 802.11b wireless access points (Linksys WAP11) that have whip antennae on the back. They are about 5" long (though they’re plastic and I can’t see how long the wire inside actually is, or how far the shielding extends). I think they are fed with a 50 ohm impedance. The simplest thing they could have done is make a quarter wave whip antenna, maybe extending the shield a little ways up the plastic body of this thing so it has a more symmetric radiation pattern despite the mounting point on the back of their little box.

I am replacing the antenna-to-antenna link with a coax cable, and it’s pretty obvious the receive antenna does not expect to receive nearly as much power as the transmit antenna is transmitting. So, in the antenna-to-antenna mode, these things are designed anticipating a range of attenuation. I don’t want to blow out the receive circuit by connecting them with a cable that has practically none of the attenuation they’d expect.

So I think I need a formula for antenna to antenna link, probably quarter wave whip to quarter wave whip - anybody know?

2

I would do it this way.

Compute the power density at the receiving antenna = transmitted powergain of the transmitting antenna/4pi*d[sup]2[/sup] were d is the distance between the antennas.

Compute the capture area of the receiving antenna which is:

Antenna gainwavelength[sup]2[/sup]/4pi

If these two are multiplied together you get the power received by the receiving antenna. Wavelength and distance must be in the same units.

Knowing the power received in your present setup allows you to adjust the power received by the cable to approximately the same level. I don’t think the match between power now and power with the cable needs to be exact because the AGC in your receiver should take care of quite a difference in received power.

This should do until someone comes along with a simpler method.

3

Hey, I can see where that would work - should have thought of it myself, except I didn’t know how to calculate the capture area of the antenna, and might not have realized it would be calculable so I would never have looked.

Yes, I think that’d work quite nicely! Thanks!!!

4

I have to ask why you’d want to do this? It seems that it would be easier to install a Cat-5 cable between 2 switches (or hubs) would be a whole lot easier. (or fiber for a REALLY long run)

If you’ve a good reason, I’d love to know it… who knows, I may have the same need or a customer with that need!

-Butler

5

butler1850,

I am immensely pleased that you asked. I do have a good reason to do this weird thing, I think.

Several years ago I installed satellite TV service, with the dish on the back exterior wall of my barn, which is about 150’ from the house, because the barn is the only structure with a view of the southern sky. I buried two CATV 75 ohm cables, for two television sets in the house. Doing this ditch in my incredibly hard soil took several hours with a very big trenching machine (it was certainly longer than a VW beetle though neither as tall or as wide).

Then I decided to get broadband internet service, and while there are zero options in my house per se, I could install a two way satellite dish on the barn. It uses a different satellite, but only about ten degrees from the television satellite. My internet access is via a satellite 22,000 miles directly above the Galapagos Islands.

So, how to connect the barn and the house? My house is already wired with Cat 5 twisted pair ethernet cable, and that would be the obvious thing to bury if I was going that route (through several years of landscaping).

But better, we thought, to lose one of the television sets (we hardly watch the second one). Then I could use the 75 ohm CATV cable if I could find a way to do it.

I explored getting a converter to coaxial type ethernet first, but that’s become an obscure item, apparently. Then it occurred to me that these $50 wireless access points were, in effect, twisted pair to coax translation boxes.

You’re not going to believe this worked: I stripped several inches of the CATV at each end, and tied it to the antenna on each wireless access point. Or perhaps I should say to the antenna on each end of the wireless bridge (a bridge acts as a link in the network, and doesn’t provide wireless access for other devices - it’s a separate mode specific to this model and you enable it through the config web page). I also laid a few inches of the shield close to the body of each end of the bridge. This stupid setup worked! And, yes, if I break the coax cable somewhere else it all completely dies.

But it’s also just sort of tied together, and the barn end is just sitting on an unused table between the drop spreader and boxes of Christmas lights. Besides, I am still radiating some sort of signal, and the fact that every month there’s some new wireless security hole announced just makes me edgy, is all.

So I found companies like Minicircuits and JEFA that sell the requisite bewildering variety of adapters, attenuators, and impedance matches to connect 50 ohm RPTNC to 75 ohm F with your choice of attenuation, and I’m figuring out exactly what to buy.

Sound like a good reason to you?

6

David Simmons,

How do I get the gain of these antennae? And, you mean some sort of absolute gain, not normalized to the highest gain, right?

I got a copy of the ARRL Antenna Handbook which is loaded with antenna gains. Trouble is, they are all normalized to have a maximum of 0. That is, what they show is graphic representation of the pattern of sensitivity, with the greatest sensitivity set at 0 dB and lower values off to the sides and rear.

I think I need some kind of absolute value, like watts received per watts per square meter signal strength “in the air”.

Though I have looked around, I haven’t found anything that does not appear to be normalized.

Is there another name for what I need?

Thanks!

7

Well, you said that the antennas were quarter wave stubs on some sort of ground plane. If that is true then they act like half-wave dipole antennas because the ground plane reflects the quarter wave stub and the result is a half-wave pattern.

An ordinary half-wave antenna has a pattern that resembles a horizontal doughnut encircline the antenna which pokes up through the hole. The gain of such an antenna over an omnidirectional antenna is about 3 dB, or a factor or 2 if I recall correctly. However, since the bottom half of the antenna is just an image reflected in the ground plane, all the energy that would go in the bottom half of the doughnut is reflected into the top half so there is an additional gain of a factor of 2. The overall gain in the plane at right angles to the antenna is then 6 dB or a factor of 4. This is the maximum gain of the antenna if your receiver is located in a plane at right angles to the axis of the stub antenna and at the same elevation. However, small departures from this gain should easilty be taken care of the the automatic gain control system of the receiver.

The formula for power density at a distance, d, assumes that the power from the antenna is distributed equaly on the surface of a sphere having the radius d. So the formula for that distribution of power is simply the power divided by the area of a sphere with radius d which is 4pid[sup]2[/sup]. However, your antenna isn’t omnidirectional (there isn’t really any such thing) but puts out more power in the horizontal direction than in the vertical direction so the power density in the preferred direction is increased. In the case of your antennas by a factor of 4.

Since your receiving antenna is just like the transmitting antenna its gain should also be 4.