power delivered to antenna from network analyzer

[windell]

Hi,
Here is my setup info
I'm gathering data on the radiation pattern of a dipole I recently built. My test setup uses two antennas in which a calibrated horn is transmitting to my dipole. Both transmit and receive antennas are connected to a network analyzer with the S-parameter test set. I set the transmit to 0dBm or 1mW. I am reading log plot s12 measurements at 434MHz.

I was reading that at best only half my transmit power will radiate from my antenna due to the 50ohm to 50ohm match. This makes sense to me, however, I am not sure for certain if the 1mW network anlyzer setting compensates for the 50% loss of power and transmits 2mW instead so 0dBm is what is radiating from the antenna.

So my question is...with 1mW set for the transmitting power, if my antenna was a perfect 50ohm impedance, would the radiated power from the antenna be 1/2mW or 1mW?

Also would this mean that the best reception of power at the receiver be 0.25mW if all of the radiated power was received?

Thanks!
Windell

 

[VEBill]

Any instrument worthy of the name will provide an actual output that closely matches the specified output. If it says 0dBm at 50 ohms, then it'll provide 0dBm at 50 ohms (plus or minus some small tolerance).

 

[windell]

Thanks for the reply VE1BLL! Please forgive me if I'm looking into it way too much, I guess I'm still not sure if I'm getting you. For example, the way I'm seeing it is that the 0dBm is inserted right before two 50-ohm loads. The first one being the output impedance of the network analyzer and the second being the antenna's (including the coax and stuff) load. So if the two loads were matched in thise case, only 1/2 the transmit power would radiate out of the antenna.

I guess another way I could look at it is that the power transmitted is 0dBm after the output impedance so 1mW is delivered to the antenna. Is this the correct way to see it?

Much appreciated!

 

[VEBill]

The designer of the instrument would have made the necessary adjustments in internal amplitude levels to provide a calibrated output *** at the output connector ***.

And there are other internal 'Z-out' solutions that he might have used instead of the one you assume. It's not unknown to have a moderate value in-line attenuator to keep things stable no matter what.

 

[Higgler]

I think the full power would get to the antenna, not half.

These aren't resistors and the RF Source impedance isn't 50 ohms.
This is 0 dBm power traveling down a coaxial impedance of 50 ohms, when it reaches a 50 ohm coax. matched antenna, all that power exits the antenna (if it's 100% efficient of course).

Another way to look at it is two coaxes hooked together, if the loss in a cable is very low, adding another 50 ohm coax. to an original 50 ohm coax. doesn't lose any power. Imaging putting 5 short coaxes together, you don't lose half the power at every coax. connection.

kch

 

[VEBill]

"...and the RF Source impedance isn't 50 ohms."

Actually instrument-grade signal sources (such as the O.P.'s "network analyzer") should have Z-out pretty much exactly 50 ohms (or whatever it is specified to be).

Sometimes the architecture will be an expensive low-Z amplifier with literally a lovely, expensive-looking 50-ohm series resistor. And I have seen at least one with a 6dB pad on the way to the output.

One flea market Tracking Generator I have is specified as being Z-out = 75-ohm. The internal circuit is the 50-ohm version of the same product but with literally a 25-ohm resistor in series with the output. The cable and BNC jack after the resistor are all 75-ohm versions as would be expected.

And as already mentioned, the actual final output into the specified load should be as is specified. If it says 0 dBm then the actual output is 0 dBm.

This Z-out assumption should be true for any non-rubbish test equipment, but may be an unsafe assumption anywhere else.

 

[Higgler]

RF source impedance typing oops.........I think (could be true though on second thought),

To generate all those frequencies with diode multiplier rf sources, the actual source impedance of the real source (diode junction) isn't 50 ohms, but forced to 50 ohms by adding attenuators and amps after the diode outputs.

k