Isotropic antenna Transmit and Receive power

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 3K views
SPYazdani
Messages
22
Reaction score
0

Homework Statement


Plot and compare the path loss (dB) for the free-space and plane-Earth models at 800MHz vs distance on a logarithmic scale for distances from 1m to 40Km. Assume that the antennas are isotropic and have a height of 10m


Homework Equations



Free space: [itex]P_R=\frac{P_T G_T G_R}{L_P}[/itex]

Plane Earth: [itex]P_R=P_TG_TG_R(\frac{h_Th_R}{R^2})^2[/itex]

Two isotropic antennas separated by a distance [itex]R\epsilon[1m,40km][/itex] at frequency [itex]f=800MHz[/itex].




The Attempt at a Solution


Isotropic antennae have [itex]G_T=G_R=1[/itex]. So That simplifies [itex]P_R=\frac{P_T G_T G_R}{L_P}[/itex] = [itex]P_R=\frac{P_T}{L_P}[/itex]
[itex]L_P=(\frac{R4\pi}{\lambda})^2[/itex].

I'm solving the question for 1m for the free space model, then once I have that, plotting it is easy in Excel.

I'm stuck on finding [itex]P_T[/itex]. I tried deriving an equation for [itex]P_T[/itex]by substituting [itex]L_P=(\frac{R4\pi}{\lambda})^2[/itex] into [itex]P_R=\frac{P_T}{L_P}[/itex] but that lead me nowhere. At least I don't know what the answer means.

Here's what happened.

[itex]P_R=\frac{P_T}{R^24\pi}A_e[/itex]
[itex]P_R=\frac{P_T}{L_P}[/itex]
[itex]L_P=(\frac{R4\pi}{\lambda})^2[/itex]
[itex]\frac{P_T}{R^24\pi}A_e=\frac{P_T}{(\frac{R4\pi}{λ})^2}[/itex]
Then a bunch of cancellation on both sides and finally
[itex]Ae 4\pi = \lambda^2[/itex]

Help! I don't know how to find [itex]P_T[/itex]
 
Physics news on Phys.org
marcusl said:
What is Lp? How does it relate to what you are asked for?

For reliable communication, Lp is the minimum signal level required at the receiving antenna. It's a ratio of [itex]\frac{P_T (mW)}{P_R(mW)}[/itex]. The distance [itex]R = \frac{\lambda\sqrt{L_P}}{4\pi}[/itex]. Rearranging and solving for [itex]L_P = (\frac{4R\pi}{\lambda})^2[/itex] implies the loss is related to the distance separated by the antennas as well as the wavelength of the transmitted signal.

Thanks for pointing that out. I can now solve my problem :D