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Radiated Power

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Hi,

Does 0.5*real(emw.Vport_1*conj(emw.Iport_1)) expression really give the radiated power claimed as in https://www.comsol.com/model/download/822281/models.rf.patch_antenna.pdf

Thanks


6 Replies Last Post 24 janv. 2021, 12:34 UTC−5
Robert Koslover Certified Consultant

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Posted: 3 years ago 20 janv. 2021, 18:16 UTC−5

That expression should give the power that goes into port #1, just as the document says. I would not assume that to be identical to the radiated power. First, the document actually refers to a patch with two ports and the expression you cited is for only one of them, and second, there is efficiency to consider as well.

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Scientific Applications & Research Associates (SARA) Inc.
www.comsol.com/partners-consultants/certified-consultants/sara
That expression should give the power that goes into port #1, just as the document says. I would not assume that to be identical to the radiated power. First, the document actually refers to a patch with *two ports* and the expression you cited is for only one of them, and second, there is efficiency to consider as well.

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Posted: 3 years ago 21 janv. 2021, 10:10 UTC−5
Updated: 3 years ago 21 janv. 2021, 10:13 UTC−5

Thank you Robert,

In the document, it is said that "the antenna efficiency η is defined as the fraction of the theoretical max power that actually radiates out of the antenna" on page 5 and the antenna efficiency is calculated by:

where and are the net power flow through ports 1 and 2 respectively. and are calculated in the following way in the same document:

= 0.5*real(emw.Vport_1*conj(emw.Iport_1))

= 0.5*real(emw.Vport_2*conj(emw.Iport_2))

They imply that the radiated power is equal to the net power flow through the ports. I think they neglect the conductive and dielectric losses. Actually It is understandable.

I would like to modify my question in this form:

If we have a single port, the expression P = 0.5*real(emw.Vport_1*conj(emw.Iport_1)) is the power consumed by the antenna, which is not returned from it? This power indicates the radiated power and losess (conduction and dielectric) in antena. Can we say this?

Thank you.

Thank you Robert, In the document, it is said that "*the antenna efficiency η is defined as the fraction of the theoretical max power that actually radiates out of the antenna*" on page 5 and the antenna efficiency is calculated by: ![](https://i.ibb.co/QpXqJn9/chrome-Do-J52-AVtto.png) where P_1 and P_2 are the net power flow through ports 1 and 2 respectively. P_1 and P_2 are calculated in the following way in the same document: P_1 = 0.5\*real(emw.Vport_1\*conj(emw.Iport_1)) P_2 = 0.5\*real(emw.Vport_2\*conj(emw.Iport_2)) They imply that the radiated power is equal to the net power flow through the ports. I think they neglect the conductive and dielectric losses. Actually It is understandable. I would like to modify my question in this form: If we have a single port, the expression *P = 0.5\*real(emw.Vport_1\*conj(emw.Iport_1))* is the power consumed by the antenna, which is not returned from it? This power indicates the radiated power and losess (conduction and dielectric) in antena. Can we say this? Thank you.

Robert Koslover Certified Consultant

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Posted: 3 years ago 21 janv. 2021, 11:18 UTC−5
Updated: 3 years ago 21 janv. 2021, 11:22 UTC−5
  1. You said "They imply that the radiated power is equal to the net power flow through the ports." No. The "radiated power" is the RF power radiated. The net power flow through the ports includes the radiated power + losses (aka, heat) in the antenna.

  2. You said "I think they neglect the conductive and dielectric losses." No. Power_in = Power_into_heat + Power_radiated.

  3. You said "If we have a single port, the expression P = 0.5real(emw.Vport_1conj(emw.Iport_1)) is the power consumed by the antenna, which is not returned from it? This power indicates the radiated power and losess (conduction and dielectric) in antena. Can we say this?" Well, your use of the term "consumed" could be confusing here, but you appear to be referring to the net power entering the antenna via that port. In that sense, you are correct. But don't forget that this "consumed" power is going two places: (a) RF radiation, and (b) heat (aka, conductive and dielectric losses).

Perhaps it will help if you more carefully identify the various terms. When you specify a port boundary condition, you usually specify as an input the forward power. But some power will be reflected. Thus, the actual power accepted (the net input power, or what you are calling consumed) propagates into the antenna (so to speak). Some of this power is then dissipated into heat (aka, losses) and some of it radiates. Confusion arises when using terms like "input" power carelessly (and you aren't the only one to do this), because there is ambiguity there. Just make sure you know exactly what each term means and you will be fine.

-------------------
Scientific Applications & Research Associates (SARA) Inc.
www.comsol.com/partners-consultants/certified-consultants/sara
1. You said "They imply that the radiated power is equal to the net power flow through the ports." No. The "radiated power" is the RF power radiated. The net power flow through the ports includes the radiated power + losses (aka, heat) in the antenna. 2. You said "I think they neglect the conductive and dielectric losses." No. Power_in = Power_into_heat + Power_radiated. 3. You said "If we have a single port, the expression P = 0.5*real(emw.Vport_1*conj(emw.Iport_1)) is the power consumed by the antenna, which is not returned from it? This power indicates the radiated power and losess (conduction and dielectric) in antena. Can we say this?" Well, your use of the term "consumed" could be confusing here, but you appear to be referring to the net power entering the antenna via that port. In that sense, you are correct. But don't forget that this "consumed" power is going two places: (a) RF radiation, and (b) heat (aka, conductive and dielectric losses). Perhaps it will help if you more carefully identify the various terms. When you specify a port boundary condition, you usually specify as an input the *forward* power. But some power will be reflected. Thus, the actual power accepted (the net input power, or what you are calling *consumed*) propagates into the antenna (so to speak). Some of this power is then dissipated into heat (aka, losses) and some of it radiates. Confusion arises when using terms like "input" power carelessly (and you aren't the only one to do this), because there is ambiguity there. Just make sure you know exactly what each term means and you will be fine.

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Posted: 3 years ago 22 janv. 2021, 02:28 UTC−5
Updated: 3 years ago 22 janv. 2021, 02:29 UTC−5

The power accepted by the antenna is converted into two things: RF radiation and (conductive and dielectric) losses. We agree on it.

The estimations that you answered "No" are not mine. If you look at page 5 of the document, they (the people who prepare the document) accept the net power flow into the antenna as radiated power. From my understanding, It can be only possible if the losses are neglected.

Maybe I don't understand what I read in the document. It is perfectly possible. That is why I asked the question on this platform. There is no direct efficiency calculation in Comsol and, most probably, the only official document that mentions antenna efficiency confuses the concept without any solid explanation. I like Comsol but it needs to be improved in this direction, too.

Anyway, thank you for your time and answers.

The power accepted by the antenna is converted into two things: RF radiation and (conductive and dielectric) losses. We agree on it. The estimations that you answered "No" are not mine. If you look at page 5 of the document, they (the people who prepare the document) accept the net power flow into the antenna as radiated power. From my understanding, It can be only possible if the losses are neglected. Maybe I don't understand what I read in the document. It is perfectly possible. That is why I asked the question on this platform. There is no direct efficiency calculation in Comsol and, most probably, the only official document that mentions antenna efficiency confuses the concept without any solid explanation. I like Comsol but it needs to be improved in this direction, too. Anyway, thank you for your time and answers.

Robert Koslover Certified Consultant

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Posted: 3 years ago 22 janv. 2021, 11:52 UTC−5

Suad,

Ah, I apologize -- I should have taken a closer/longer look at that Comsol document. Upon a second look, I see now that they used only PECs and lossless dielectrics. So, of course, there are no heating terms! Don't blame Comsol (or me) too much. There is some ambiguity in how the term "efficiency" is used casually within the antenna community. It helps to add qualifiers, such as "radiation efficiency" or "total efficiency." See, for example, the discussion at: https://antenna-theory.com/basics/efficiency.php It seems that the efficiency being discussed in the Comsol document was actually the impedance mismatch efficiency. Again, I am sorry about adding to the confusion.

-------------------
Scientific Applications & Research Associates (SARA) Inc.
www.comsol.com/partners-consultants/certified-consultants/sara
Suad, Ah, I apologize -- I should have taken a closer/longer look at that Comsol document. Upon a second look, I see now that they used *only* PECs and *lossless* dielectrics. So, of course, there are no heating terms! Don't blame Comsol (or me) too much. There is some ambiguity in how the term "efficiency" is used casually within the antenna community. It helps to add qualifiers, such as "radiation efficiency" or "total efficiency." See, for example, the discussion at: https://antenna-theory.com/basics/efficiency.php It seems that the efficiency being discussed in the Comsol document was actually the impedance *mismatch* efficiency. Again, I am sorry about adding to the confusion.

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Posted: 3 years ago 24 janv. 2021, 12:34 UTC−5

Oh, thank you for your kind answer, no problem. I have been learning many things from your posts in this forum. But this time, it happened in a different way. I had to check the basic sources again and again to be sure. Thanks again.

Oh, thank you for your kind answer, no problem. I have been learning many things from your posts in this forum. But this time, it happened in a different way. I had to check the basic sources again and again to be sure. Thanks again.

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