# How to design log periodic antenna

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#### vanishri

##### Newbie level 2 I want to design LPDA at lowest frequency 300MHz and highest 700MHz. can any1 help me.?

#### bbgil

##### Full Member level 2 • ### vanishri

Points: 2

Points: 2

#### vanishri

##### Newbie level 2 I understood procedure but not still clear about spacing between feeder, whats the equation for spacing to find.?

#### amihomo

##### Full Member level 5 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?

#### ferdows this software design LPDA antenna

#### Attachments

• lpda.exe
56 KB · Views: 42
• roni 1980

### roni 1980

Points: 2

#### amihomo

##### Full Member level 5 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.

#### ferdows this software calculate the boom and widths of antenna

• ### roni 1980

Points: 2

Points: 2

#### juji

##### Newbie 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

• amihomo

Points: 2