How Do Antenna Size and Frequency Affect Power Density Calculations?

axcelenator
Messages
33
Reaction score
0
Antenna is in size of 0.5[m] X 0.5[m], with transmit frequency of 10GHz. In a distance of 100m the electric field is 250 [V/m]. I calculated the density of power with the formula: 0.5*(E^2)/(120*pi) and got: 82.891
[W/m^2].
the question is: can I calculate the density of power again but un distance of 10m? My other question is: If I have a known G (G=a*D) can i find an expression for the total power comes from the antenna?

Can I use the formula: Pt=S*(4*pi*r^2)/G ? where Pt is total power and S is power in distance r. after I get Pt do I need to multiply it by the size of antenna?

Hi, I need the answer as quickly as possible. Thanks!
 
Last edited:
Physics news on Phys.org
You weren't given what type of antenna this is? Each antenna has its own distinct equations for V, A, E, B, S, <S>, and P.

You should be able to find the power density (time averaged Poyting vector) at 10m by solving your equation for E at 100m (using your known value of E = 250) to obtain the value of many of the contants that are specific to your particular antenna. You can then use the value of those constants as a group to solve for the value of the time averaged Poyting vector at 10m.


As far as your other question - I think the equation "Pt=S*(4*pi*r^2)/G" looks fine. Assuming 'G' is the power gain and S is the time averaged Poynting vector we have already been discussing.. Because that means G = \frac{4 \pi r^2 S}{P_{t}} which I think is right. And once you get P_{t} I don't know why you would then want to multiply that by the dimensions of the antenna. P_{t} is the total power radiated by the antenna.

I'm not an antenna expert (I've never dealt with the power gain equations before, only the directivity), so take my post with some suspicion, but nevertheless I hope I helped.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top