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: