Practical_Antenna_Handbook_0071639586

186 p a r t I I I : h i g h - F r e q u e n c y B u i l d i n g - B l o c k A n t e n n a s
4. Frequency difference: 4000 kHz – 3500 kHz = 500 kHz
5. 5. Calculate frequency
unit length :
500kHz
16.7ft
= 30kHz/ft or 2.5kHz/in
For 15 m (21.0 – 21.45 MHz in the United States):
468
1.
L
UPPER
= = 21.82 ft
21.45
468
2.
L
22.29 LOWER
= = ft
21
3. Difference in length: 22.29 ft – 21.82 ft = 0.47 ft
4. Frequency difference: 21,450 kHz – 21,000 kHz = 450 kHz
5. 6. Calculate frequency
unit length : 450
0.47
= 960 kHz/ft or 80kHz/in
At 80 m the frequency change per foot is small, but at 15 m small changes in a
dipole’s length can result in very large shifts in its resonant frequency.
Using the preceding steps, you can determine approximately how much to add (or
subtract) from an antenna under test. If, for example, you design an antenna for 21,390
kHz but find the actual resonant point is 21,150 kHz, the frequency shift required is
21,390 – 21,150, or 240 kHz. To determine how much to add or subtract (as a first guess):
1. The factor for 15 m is 80 kHz/in, which is the same as saying 1 in/80 kHz.
2. The required frequency shift is 240 kHz.
3. Therefore:
1 in
Change in length = 240kHz ×
80kHz
= 3 in
Each side of the antenna must be changed by half of the length calculated, or 1.5 in.
Because you wish to raise the resonant frequency of the dipole, each side should be
shortened 1.5 in. Once the length is correct, as determined by new VSWR measurements,
the connections at the center insulator are soldered and made permanent, and the antenna
is hoisted back up to its original height.
Note This is an excellent application for an SWR analyzer or antenna analyzer (see Chap. 27),
especially if the original point of minimum SWR seems to be outside the frequency range
you are authorized to transmit in.
Impedance Matching
While the procedures detailed here may help you place the frequency of minimum
SWR where you want it, they do not directly address the elimination of the reactive part