how to measure power accepted into antenna and power radiated out of antenna in lab?

in a lab setting, how do we measure power accepted into antenna and power radiated out of antenna? What are the tools and procedures to do that?

HFSS has a radiation boundary box that is necessary for measuring the power radiated, accepted, etc....what would be the equivalent in a lab setting?

One more question, is power radiated only measured in far field?

Thank you.

Use a power meter or a spectrum analyzer to measure the power at the antenna connector.
For accurate radiated power measurement you need an anechoic chamber or an open-site setup.
Use a reference antenna, and a calibrated dipole antenna for substitution with your antenna. In this way you get your antenna gain in dBd (referenced to a dipole).
Knowing the power into the antenna and the antenna gain you can calculate the radiated power.

vfone said:

Knowing the power into the antenna and the antenna gain you can calculate the radiated power.

Click to expand...

What is the equation for above statement? P(radiated) = P(gain)*P(incident)? Also, do you assume P(incident) = P(accepted)?

Also, how do you measure the antenna gain? using spectrum analyzer?

It seems to me, that the thread is mainly replaying the discussion from your previous thread. https://www.edaboard.com/threads/247709/

I previously mentioned

Forward power to the antenna is either reflected, radiated or dissipated in the antenna structure (according to conservation of energy law)

Click to expand...

Accepted power is simply input (forward) power minus reflected power. The latter can be precisely determined with a directional coupler respectively a network analyzer.

If the antenna is (known to be) lossless, all accepted power is radiated. In this case, you won't need to measure the radiated field, unless you want to know the radiation pattern.

The term "gain" is related to the radiation pattern, it's describing the field strength of the "beam" compared to a reference antenna, either an hypothetic isotropic or a dipole antenna.

An antenna radiation measurement has to be performed under free and effectively far field conditions, e.g. in anechoic chamber of suffcient size. Network analyzers, power meters or spectrum analyzers can be used, receiver antenna calibration is a critical point.

Thanks FvM for putting up with my silly questions. Your answer makes sense.

Let me be more specific about my problem:
I have a dipole antenna that is impedance matched, so P(radiated)= P(accepted)
Then, I introduce a lossy dielectric object, consisting mainly from fat and muscle, to the simulated model and put it adjacent to the dipole antenna (5cm away). The simulation results show a decrease in P(radiated) mainly because lossy dielectric object absorbs some of the P(radiated).

Note: I did not impedance match the antenna in the presence of the lossy dielectric object, only by itself.

Question is: Is it fair to say that P(absorbed by the lossy dielectric object) = P(accepted) - P(radiated)

Now back to my S11 thread: The reason I had a whole post for that is because I read few papers in the literature using S11 measurements to detect the presence of tumors in the breast since tumors have different dielectric properties from fat and more electric field waves would be reflected when tumors are present. My whole understanding of S11 is like what you explained (reflections would be from the antenna structure only), and I am not totally convinced it includes reflections from adjacent lossy dielectric objects, or does it?

Vielen Dank.

There's no doubt, that objects in the antenna proximity change it's impedance respectively S11, either lossy or lossless objects. The observed S11 number doesn't distinguish between a change of antenna parameters itself or the effect of a backscattered wave from the enviroment.

I can agree to P(absorbed by the lossy dielectric object) = P(accepted) - P(radiated)

But the expression doesn't tell about the quantity you're interested in: The change of S11.

kae_jolie said:

Thanks FvM for putting up with my silly questions. Your answer makes sense.

Let me be more specific about my problem:
I have a dipole antenna that is impedance matched, so P(radiated)= P(accepted)

Click to expand...

That's only true for a loss-less antenna. From a practical point of view, a decent antenna will have high efficiency and radiate most of the RF energy fed to it as RF energy at the same frequency. But some will be converted to heat in the antenna. That can be lost via conduction, convection or radiation (mainly IR)

Note, that the fact the antenna is matched is not a requirement for that equation to be true. If the antenna is lossless, but poorly matched, it might reflect 40% of the incident power. So if you apply 1 W, 0.4W is reflected, and 0.6 W accepted. The 0.6W will be radiated.

kae_jolie said:

Then, I introduce a lossy dielectric object, consisting mainly from fat and muscle, to the simulated model and put it adjacent to the dipole antenna (5cm away). The simulation results show a decrease in P(radiated) mainly because lossy dielectric object absorbs some of the P(radiated).

Click to expand...

Where is there a decrease in power radiated? Are you simply saying the presence of the lossy dielectric causes RF to be reflected by the antenna, so the RF power the antenna radiates is less?

Note the same will happen for a good dielectric like PTFE. Stick a lump of that around an antenna and you will change the resonate frequency. That will cause RF to be reflected from the antenna, and for a constant incident power will mean the antenna radiates less power.

kae_jolie said:

Note: I did not impedance match the antenna in the presence of the lossy dielectric object, only by itself.

Question is: Is it fair to say that P(absorbed by the lossy dielectric object) = P(accepted) - P(radiated)

Click to expand...

I disagree with FvM and say that the equation is not true.

As you sated earlier, P(radiated)= P(accepted), which I stated is only true for a lossless antenna. So assuming a lossless antenna, the right hand side of the above equation is zero. The left hand side however is not zero, since the lossy dielectric will heat up. So the equation can't be true. I think the equation you are looking for is:

P(absorbed by the lossy dielectric object) + P(radiated elsewhere) = P(accepted by the antenna) = P(radiated by the lossless antenna)

If the antenna is not lossless, but has an efficiency η, then

P(absorbed by the lossy dielectric object) + P(radiated elsewhere) = P(radiated by the antenna) = η * P(accepted by the antenna)

(When I talk of radiated power, I am assuming you mean immediate RF power - not heat radiated some time later, even after the excitation is removed.)

kae_jolie said:

Now back to my S11 thread: The reason I had a whole post for that is because I read few papers in the literature using S11 measurements to detect the presence of tumors in the breast since tumors have different dielectric properties from fat and more electric field waves would be reflected when tumors are present. My whole understanding of S11 is like what you explained (reflections would be from the antenna structure only), and I am not totally convinced it includes reflections from adjacent lossy dielectric objects, or does it?

Vielen Dank.

Click to expand...

The S11 of an antenna in free space is not the same as the S11 you will measure on a network analyzer with items near the antenna. You can measure S11 on many things - mixers, dummy loads etc. If you put something near an antenna, the free space properties of the antenna wont change, but the environment the antenna is in has changed, so S11 will change.

A word of caution about your research, from someone who has spent a long time working in medical physics, although I no longer do.

I used to work in the optics group in the Department of Medical Physics, UCL.

https://www.ucl.ac.uk/medphys/

In fact my PhD is in medical physics, and I was measuring the optical properties of tissue.

**broken link removed**

Some in the group were using the change of scattering coefficient to try to detect tumor tissue. Tumor tissue scatters light more than normal tissue.

https://www.ucl.ac.uk/medphys/research/borl/imaging/monstir/breast

It used to, and looking at the web site, still does, get lots of funding, as one can say you are not using x-rays. But I know spacial resolutions were of the order of cm, and personally I don't feel it is ever going to improve considerably because human tissue is highly scattering to light. The usual technique is to time how long the light takes to get from A to B in the tissue, so then you know how far it has gone. Trouble is, you don't know what path the light took, so building images is very hard.

I was chatting to someone in the X-ray group in the Medical Physics department, who assured me the level of x-rays given for mammograms is very low, but it's possible to detect abnormalities as small as 100 µm. His point was that by the time the optical methods could detect anything reliably, the person (usually female) would be dead. So despite millions of pounds being spent on it, the technique is never going to be of any practical use. I spent more than 10 years working in that group, and believe it will never be of any practical use. It's almost 10 years since I left the group, and the images look no better today than they did 10 years ago.

I rather suspect your RF techniques will suffer the same practical problems as the optical techniques, although for different reasons. Although tumor tissue has different dielectric properties to normal tissue, it is not going to be so dramatically different that you are going to be able to tell them apart with a device suitable for breast screening. If you match your antenna away in free space, and put it near a human breast, there will be a dramatic change in S11. I don't believe you are going to be able to distinguish the types of tissue in such a way to make it practical to produce a clinical instrument on which clinical decisions will be based.


Dave

Your discussion of the P(absorbed...) expressions points to some implicite unclear definitions in the setup. I admit, that I was too lazy to mention them in detail. I also don't see much sense in discussing the (as I think) still existing problems with your revised expressions. In a first place, you would need to clearly specify the objects of interest and their connecting ports (what is antenna, what is enviroment, what is far field). I'm sure, that we will easily agree about a model, once it has been clearly defined.

The term radiated power in the original post e.g, should be better named as radiated into far field respectively radiated elsewhere in your words.

The main point is in my last statement. I'm convinced, that the discussion of this terms leads to almost nothing. You are effectively unable to determine radiated power in the discussed setup. So at the end of the day, in-situ S11 change is the only information you get. Either if you analyze the effect as backscatter of radiated power or change of antenna impedance itself, it won't allow you unequivocal conclusions about the surrounding objects.

Your description of the expectable diagnostic benefit sounds simply realistic under this conditions.

As FvM is stating, we do need a precise definition of the terms here.

The physics of a lot of this is very simple - mainly dictated by the conservation of energy.

Again definitions are necessary, but I don't think it will be true to say

P(radiated elsewhere) = P(radiated into free space).

since you no longer have free space once there is this lossy dielectric around. Note also that once you have a lossy dielectric around, it will change how the RF propogates close to it, but not actually in it. The equations for that have been worked out theoertically for a sphere, and possibly some other simple objects, although not the human breast! It depends on the sizes of the wavelength and object.

I'd be intersted in what frequency range you are using, and the size of the dipole. That will dictate whether you are in the far field, or the near field where things get even more complicated.

If this is Ph.D project for kae_jolie, then good luck to you. I'm sure you can get a Ph.D out of it.

However, if you are a post-doc, looking to get funding for a reseach project, I think I'd look at something else, as personally I would not try to base a career around this work.

That said, I know several professors who have got chairs as a result of publications for work which is never likely to be of any practical use. I think that's acceptable for theoetical physics, but this sounds very much like what I'd call "medical physics", where you apply physics and engineering skills to solve medical problems. The case for doing this in a medical physics department is very questionable.

Dave

---------- Post added at 11:05 ---------- Previous post was at 10:09 ----------

With a second antenna close to the first one, there will be coupling between them. The breast will change that coupling, so you would have extra information, having S11, S22 and S21. But I still think it will not lead to anything useful. I only mention it for academic interest.

Dave

drkirkby said:

Where is there a decrease in power radiated? Are you simply saying the presence of the lossy dielectric causes RF to be reflected by the antenna, so the RF power the antenna radiates is less?

Click to expand...

drkirkby, thank you for your invaluable comments and insight. To answer your question, yes there was a decrease in power radiated measurements in HFSS when the lossy dielectric object was introduced, and as the size of the dielectric object increased, the power radiated decreased more indicating more power is being absorbed by dielectric object.

Here is how I derive my equation: P(absorbed by lossy dielectric) = P(accepted by antenna) - P(radiated in far field)

From conservation of energy law:

P(incident on antenna) = P(reflected by antenna structure) + P(absorbed by lossy dielectric) + P(radiated in far field) + P(reflected by lossy dielectric) [Equation 1]

Definition of P(accepted) is: P(accepted by antenna) = P(incident on antenna) - P(reflected by antenna structure) [Equation 2]

Substitute Equation 2 into Equation 1:

P(accepted by antenna) = P(absorbed by lossy dielectric) + P(radiated in far field) + P(reflected by lossy dielectric) [Equation 3]

After re-organizing Equation 3:

P(absorbed by lossy dielectric) = P(accepted by antenna) - [ P(radiated in far field) + P(reflected by lossy dielectric)] [Equation 4]

There is an assumption I make in Equation 4, which is P(reflected by lossy dielectric) gets radiated in far field, hence Equation 4 becomes:

P(absorbed by lossy dielectric) = P(accepted by antenna) - P(radiated in far field) [Equation 5]

Is this a correct assumption?


Thank you for your input.

---------- Post added at 23:27 ---------- Previous post was at 22:58 ----------

drkirkby said:

I'd be intersted in what frequency range you are using, and the size of the dipole. That will dictate whether you are in the far field, or the near field where things get even more complicated.

Click to expand...

I am using 915 MHz frequency.
The size of dipole is 0.475*lambda = 155.6mm
The lossy dielectric object is placed 5mm away from antenna (in near field of antenna)
The radiation air box (in HFSS to measure the power radiated) is placed lambda/4 away from antenna+dielectric structure in each direction. (in far field of antenna)

Regarding the S11 discussion, please review the attached example diagrams for S11 and S21 and let me know if I made any mistakes. These diagrams convey my understanding of S11 and S21. Also, I have two questions on the bottom two diagrams if you could address them please.

Thanks.

**broken link removed**