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Beam solid angle of antenna

  1. Aug 2, 2011 #1
    The definition of Beam solid angle: For an antenna with single main lobe, the Beam solid angle defines as:

    The solid angle [itex]\; \Omega_A\;[/itex] where all the radiated power would flow with radiation intensity equal to maximum and constant inside the Beam solid angle [itex]\; \Omega_A\;[/itex].

    The book gave:

    [tex]\Omega_A\;=\; \int F(\theta,\phi) d \Omega \;\hbox{ where }\; F(\theta,\phi)=\frac { P_{av}(\theta, \phi)}{P_{max}}[/tex]

    But from the definition, the [itex]\; \Omega_A\;[/itex] depend on the total power radiated divid by the [itex]\;P_{max}\;[/itex] times the total steradians of the sphere.

    Total power radiated = [itex] P_{Rad}=\int P_{av}(\theta,\phi) dS= \int R^2\;P_{av}(\theta,\phi)\;d\Omega [/itex]

    [tex]\Omega_A = 4\pi \frac { P_{Rad}(\theta,\phi)}{P_{max}} = 4\pi R^2 \frac{\int P_{av}(\theta,\phi) d\Omega} {P_{max}} = 4\pi R^2 \int \frac { P_{av}(\theta,\phi)}{P_{max}} d \Omega = 4\pi R^2 \int F(\theta,\phi) d\Omega[/tex]

    You see there is a [itex]\;4\pi R^2\;[/itex] difference between the definition and the given formula, please help.

    Thanks

    Alan
     
    Last edited: Aug 2, 2011
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  3. Aug 2, 2011 #2

    marcusl

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    I don't think your second equation is correct. The total radiated power is the integral over the surface S of the Poynting Vector, not Pav. Since Poynting vector is a power flux density, there will be no mysterious factor of R^2.
     
  4. Aug 2, 2011 #3

    I am typing word to work the definition of Beam Solid Angle from the book:

    [Quote Book]

    [tex]\hbox { Directivity }\;D = \frac {P_{max}}{P_{Rad}4\pi R^2} = \frac { 4\pi R^2 P_{max}}{\int P_{av}(\theta,\phi)R^2\sin\theta\;d\theta\;d\phi};=\; \frac {4\pi U(\theta,\phi)}{\int U(\theta,\phi)\sin\theta\;d\theta\;d\phi}\; = \;\frac {4\pi B_0 F(\theta,\phi)}{\int B_0 F(\theta,\phi)\sin\theta\;d\theta\;d\phi}\; =\; \frac {4\pi}{[ \int F(\theta,\phi)\sin\theta\;d\theta\;d\phi\;]/F_{max}}[/tex]

    [tex]\hbox { Dedfine }\;\Omega_A=\frac 1 {F_{max}} \int F(\theta,\phi)\sin\theta\;d\theta\;d\phi[/tex]

    The Beam solid angle [itex]\;\Omega_A\;[/itex] is defined as the solid angle through which all the power of the antenna would flow if it's radiation intensity is constant ( and equal to the maximum value of U) for all angles within [itex]\;\Omega_A\;[/itex].
    [End Book]

    Notice the definition formula and the words from the book don't agree? As I explained, according to the definition in words:

    [tex]\Omega_A=\frac {P_{Rad}}{U_{max}} \times 4\pi\;\hbox { steradians.}[/tex]

    We use [itex] P_{av}\;[/itex] which is the time average of the poynting vector.

    [tex] \vec P_{av} =\int_0^T \vec E \times \vec H^* d\;t[/tex]


    Radiation Intensity defines as [itex]\; U(\theta,\phi)=R^2 P_{av}(\theta,\phi)[/itex]

    [tex]P_{Rad}=\int_0^{2\pi}\int_0^{\pi} P_{av} R^2 \sin\theta\;d\theta\;d\phi\;=\; \int U(\theta,\phi) d \Omega \;\hbox { where }\;d\Omega = \sin\theta \;d \theta\;d\phi\; \hbox { and } dS=R^2 d\Omega[/tex]
     
    Last edited: Aug 3, 2011
  5. Aug 3, 2011 #4

    marcusl

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    The section you quote from the book makes sense and is consistent. What you type after that is not, and your equation for Pav is wrong. You misunderstand Poynting vector and power flux density, which are not the same as power. Please review these concepts in a good E&M text before proceeding. All should then be clear.
     
  6. Aug 3, 2011 #5
    I figure the problem already.

    But my equation of Pav is correct, this is backed up by 3 books including Balanis. Pav is the time average poynting vector and is per meter square. I never said it is power. Prad is total power in Watts.

    I original mistake is that I forget Umax is per steradian and I mistakenly times 4\pi.

    Thanks for you help anyway.

    Alan
     
  7. Aug 3, 2011 #6

    marcusl

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    I am glad you got things to work. Please be careful about Pav, however. In my copy of Balanis (2nd edition), Eq. (2-9)
    [tex]
    P_{av}=\frac{1}{2}\int{Re(\vec{E}\times\vec{H}^*)}\cdot d\vec{s} [/tex]
    and the text accompanying it clearly show that Pav is power, not power density. Note by comparison that your integral is over time.
     
  8. Aug 3, 2011 #7

    You are right, my mistake from just pulling out of my head, I left out the Re[] part. The confusion is I use 3 books and they all have different names. Pav I use is from Chengs. It is actually Wav in Balanis.

    My original confusion is really about the units of the Beam solid angle, I just overlooked the denominator is the power density in steradians, so power divided by (power per steradian ) is simple steradian!!! It's just that simple.

    Is Balanis a good book for antennas? I have both 2nd and 3th edition. I manage to download the solution manual for the second edition so I just bought the 2nd edition dirt cheap for $26!!! I also have Warren Stutzman and Kraus. Which one is the best for self study?

    Thanks for your help.
     
  9. Aug 3, 2011 #8

    marcusl

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    No. That's important also but you are still missing a key point. My integral is over surface area, yours over time.
    Try sticking with one book then. Wav is Balanis's notation for the time-averaged Poynting vector. You must integrate this over area to get the power Pav through a surface.
    I still suggest that you carefully look at and master the concepts of power density, Poynting vector, and power, as well as the relations between them.
    I own a copy of Balanis and think it is good. I once flipped through Kraus and thought it looked good, too. I've had no exposure to Stutzman.
     
  10. Aug 3, 2011 #9
    I really stop and look it up. What I mean you are right is the 1/2Re[], when I use Pav, it meant the time integrate which is the Wav in Balanis, not the spacial integration with dS.

    I think it's useful to read more than one book because they do present the subject a little different, and they are good in different topics. And some do make mistake.
     
  11. Aug 3, 2011 #10

    marcusl

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    Ok, that makes your equations correct.
    No argument either on the use of multiple books, since a different perspective is sometimes very helpful.
     
  12. Aug 4, 2011 #11
    Actually I was still wrong

    [tex] W_{av}= \frac 1 T \int_0^T \vec E \times \vec H \;d\;t \;\hbox { or }\;\tilde W_{av} = \frac 1 2 \Re e [\tilde E \times \tilde H^*][/tex]

    Sometimes actually two books have different assumption and get different answer. Actually today I am verifying the current approximation on the linear dipole antenna of about [itex]\frac {\lambda}{2}[/itex], center fed at z=0 and one side towards +ve z and the other half towards -ve z. Ulaby use [itex]i(t)=I_0 \cos\beta z [/itex] but both Cheng and Balanis use [itex]i(t)=I_0 sin [\beta(h-|z|)] \;\hbox { where the dipole length is } \; 2h\;[/itex].

    The answer is the same at dipole length is [itex] \frac {\lambda} 2 [/itex]. But the calculation is different if not. I know the current is an approximation for linear dipole antenna, but still I would like to find out who is closer!! I am a self studier, I don't have a professor as a fall back. Physics forum is my only source for answer.
     
    Last edited: Aug 4, 2011
  13. Aug 4, 2011 #12

    marcusl

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    The current at the ends of the antenna must be zero. This is true for I=cos(beta*z) only when z is an odd multiple of lambda/4, while the other form is zero at the end of an antenna of any length h. The second may therefore be used for dipoles that are not lambda/2. Note that it is still only an approximation to the actual current.
     
  14. Aug 4, 2011 #13
    Yes, I know it is approximation.

    I see, Ulaby is more of the introduction book and the formula is for [itex]\; \frac {\lambda} 2 \;[/itex]. I was hoping the formula can work for other length but it is not. Well, at least I get to practice my calculus to work out the formula to compare the difference!!!

    Thanks for the second help.

    Alan
     
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