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 design log periodic antenna

Status
Not open for further replies.

vanishri

Newbie level 2
Newbie level 2
Joined
Nov 17, 2010
Messages
2
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,295
I want to design LPDA at lowest frequency 300MHz and highest 700MHz. can any1 help me.?
 

hope this helps. from the ARRL handbook. LPDA antenna design

**broken link removed**
 
I understood procedure but not still clear about spacing between feeder, whats the equation for spacing to find.?
 

I'm designing a printed LPDA, my question is how to choose the width of the antenna boom? is there any formula or relation between boom width and the widths of the dipoles?
 

this software design LPDA antenna
 

Attachments

  • lpda.exe
    56 KB · Views: 205
this program is for the conventional LPDA antenna. I need some design equations for calculating the widths of dipoles and the boom of the antenna.
 

dear friend
on your design you have toe , sigma and R........toe 0.8-0.95 sigma 0.05-0.2 R 50 or 75 ohms
first- calculate za =60 ln{ (16 k toe sigma)/[pi (1+toe)]}
k is average ratio of elements LENGTH to DIAMETER.for better resault you may slim elements to be constant k.
ln is natural logarithm.
second- x={R(1+toe)/(32 sigma toe za)}
zo=R[x+square (x power2 +1)]
zo is the characteristic impedance of boom line.
if you use round tube diameter d and distance between centers D....zo=276 log (D/d)
if you use square tube side a equivalent diameter is 1.06 a
so if you choose tube you will find separation between them.you are free to ask more. juji tabrizi

---------- Post added at 07:11 ---------- Previous post was at 06:12 ----------

dear friend
on your design you have toe , sigma and R........toe 0.8-0.95 sigma 0.05-0.2 R 50 or 75 ohms
first- calculate za =60 ln{ (16 k toe sigma)/[pi (1+toe)]}
k is average ratio of elements LENGTH to DIAMETER.for better resault you may slim elements to be constant k.
ln is natural logarithm.
second- x={R(1+toe)/(32 sigma toe za)}
zo=R[x+square (x power2 +1)]
zo is the characteristic impedance of boom line.
if you use round tube diameter d and distance between centers D....zo=276 log (D/d)
if you use square tube side a equivalent diameter is 1.06 a
so if you choose tube you will find separation between them.you are free to ask more. juji tabrizi
 
  • Like
Reactions: amihomo

    amihomo

    Points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top