Hi All,

I am trying to understand the calculations on a LPDA (Logper dipole array).

-edit- there is a typo on the following web page, therefore i din't understand the calculations. I have found it, so it's solved

I have found the following webpage:

The Secrets of the Log Periodic Antenna.

They use the following formulas. The below formulas use the following symbols:

Zb: Impedance of the boom.

Zi: Impedance of the balanced feed line

Zd: Impedance of the largest dipole

ln: length of the largest dipole

dn: diameter of longest dipole

They have used sigma of 0.12 and tau of 0.8. This yields sigma' = 0.134 :

sigma 0.12 0.12

sigma' = -------------- = ------------- = ------- = 0.134

sqrt (tau) sqrt (0.8) 0.894

They have choosen 0.25 inch rods, the longest is 29.52 inch long

(labda for 200MHz = 1.5 meter, 0.5 labda = 75cm = 29.53 inch)

this yields impedance for the longest dipole = 302.4 Ohm:

ln 29.52

Zd = 120 * ( ln (----) -2.25) = 120 * ( ln ( ------- ) -2.25 ) = 120 * ( ln ( 118.08 ) -2.25 ) = 120 * (4.77 - 2.25) = 120 * 2.52 = 302.4 Ohm

dn 0.25

They have choosen Zi= 200 Ohms (impedance of the balanced feed line). This yields Zb (impedance for the boom) = 358.38 Ohm:

sq (Zi) Zi

Zb = ---------------- + Zi*sqrt ( sq(-----------------)+1)

8 sigma' Zd 8 sigma' Zd

sq (200) 200

Zb = ---------------------- + 200*sqrt ( sq(-----------------------)+1)

8 * 0.134 * 302.4 8 * 0.134 * 302.4

40000 200

Zb = ------------- + 200*sqrt ( sq(------------)+1)

324.2 324.2

Zb = 123.38 + 200*sqrt ( sq( 0.617)+1)

Zb = 123.38 + 200*sqrt ( 0.38+1)

Zb = 123.38 + 200*sqrt ( 1.38)

Zb = 123.38 + 200*1.175

Zb = 123.38 + 235 = 358.38 Ohm

Next, they make a typo in the following formula: Zi should have been Zb

S: the spacing between the 2 booms

diam: the diameter of the 2 booms they choose 0.25 inch

Zb: the impedance of the 2 booms

Zb 358.38

----- ---------

diam 276 0.25 276 1.298

S = ------- * 10 = ------- * 10 = 0.125 * 10 = 0.125 * 19.86 = 2.48 inch

2 2

Now I can re-calculate the above for my own logper, but i will do that in a following post.

Best regards,

Cedric

I am trying to understand the calculations on a LPDA (Logper dipole array).

-edit- there is a typo on the following web page, therefore i din't understand the calculations. I have found it, so it's solved

I have found the following webpage:

The Secrets of the Log Periodic Antenna.

They use the following formulas. The below formulas use the following symbols:

Zb: Impedance of the boom.

Zi: Impedance of the balanced feed line

Zd: Impedance of the largest dipole

ln: length of the largest dipole

dn: diameter of longest dipole

They have used sigma of 0.12 and tau of 0.8. This yields sigma' = 0.134 :

sigma 0.12 0.12

sigma' = -------------- = ------------- = ------- = 0.134

sqrt (tau) sqrt (0.8) 0.894

They have choosen 0.25 inch rods, the longest is 29.52 inch long

(labda for 200MHz = 1.5 meter, 0.5 labda = 75cm = 29.53 inch)

this yields impedance for the longest dipole = 302.4 Ohm:

ln 29.52

Zd = 120 * ( ln (----) -2.25) = 120 * ( ln ( ------- ) -2.25 ) = 120 * ( ln ( 118.08 ) -2.25 ) = 120 * (4.77 - 2.25) = 120 * 2.52 = 302.4 Ohm

dn 0.25

They have choosen Zi= 200 Ohms (impedance of the balanced feed line). This yields Zb (impedance for the boom) = 358.38 Ohm:

sq (Zi) Zi

Zb = ---------------- + Zi*sqrt ( sq(-----------------)+1)

8 sigma' Zd 8 sigma' Zd

sq (200) 200

Zb = ---------------------- + 200*sqrt ( sq(-----------------------)+1)

8 * 0.134 * 302.4 8 * 0.134 * 302.4

40000 200

Zb = ------------- + 200*sqrt ( sq(------------)+1)

324.2 324.2

Zb = 123.38 + 200*sqrt ( sq( 0.617)+1)

Zb = 123.38 + 200*sqrt ( 0.38+1)

Zb = 123.38 + 200*sqrt ( 1.38)

Zb = 123.38 + 200*1.175

Zb = 123.38 + 235 = 358.38 Ohm

Next, they make a typo in the following formula: Zi should have been Zb

S: the spacing between the 2 booms

diam: the diameter of the 2 booms they choose 0.25 inch

Zb: the impedance of the 2 booms

Zb 358.38

----- ---------

diam 276 0.25 276 1.298

S = ------- * 10 = ------- * 10 = 0.125 * 10 = 0.125 * 19.86 = 2.48 inch

2 2

Now I can re-calculate the above for my own logper, but i will do that in a following post.

Best regards,

Cedric

Last edited: