Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

how to find separation between antenna

Status
Not open for further replies.

ashokone

Member level 3
Joined
Sep 30, 2009
Messages
56
Helped
1
Reputation
2
Reaction score
1
Trophy points
1,288
Location
banglore
Activity points
1,659
what is the maximum separtion between the antennas?

how to calculate it?
 

Ejem.
Using the info you are providing its's more or less 46.5 billion light-years (the radius of the observable universe) :)
Seriously. Try to be be more precise: application, frequency, etc...
Good luck
 

Minimum 20dB is a good number in antenna separation. This isolation is fine for getting at least minimum performances for diversity or for a two-antenna duplex system.
 

suppose frequency of operation is 20GHz. then how to find the separation between antennas ex dipole antenna?
sum paper i reffered that it should be greater then lamda..is it true r not i don't know

plz help
 

Yes, distance greater than λ is fine, but is good to look also for isolation in dB at operating frequency to eliminate any kind of interaction between antennas.
 

https://www.cisco.com/en/US/tech/tk722/tk809/technologies_tech_note09186a008019f646.shtml

above link gives the distance of separation between antenas

in this link it says that
The 2.4 GHz wavelength is approximately 4.92 inches. Therefore, to support diversity on a 2.4 GHz radio with two separate antennas, the antennas should be spaced approximately 5 inches apart. The antenna pair could also be spaced at multiples of 5 inches, but the distance between should not exceed 4 multiples: reflected waves farther apart than that are likely to be so distorted and different in delay spread that the radio could not work with them.

how did he got 4.92 inches?

plz help
 

@ashokone,
Did you calculated for λ pls?
K.
 

i am not getting how he was represented the wavelength in inches?

anyone have an idea regarding this plz help
 

Use your very, very elementary equations. lambda=c/f. Then convert that to inches.
 

i did already, getting lamda=0.125 and if we convert it into inches multiply it by 2.54

gives the lamda in inches=0.3175 is the answer i am getting
 

Uh, that's not correct. Inches -> cm is *2.54. cm -> inches is /2.54.
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top