coax hidden antenna load

Many radio hams would perhaps use a quarter wave line or vertical for their antenna installations. However looking into the hidden factor of the coax hiding the true antenna load,

However, before embarking on the determination of the hidden antenna impedance due to the coax masking over the antenna true value, it may well be born in mind that a wire antenna is a tuned up inductor. This is to say that a single metre of wire has a 300nH inductance, found from the characteristic impedance equation for a coax line:- " Z = SQR(inductance/capacitance) ", such that a matched wire line or rod antenna is a inductive reactance impedance matched load to the coax line characteristic impedance. Also bearing in mind that an inductance reactance is at right angles to a pure resistance. That being said, a inverse Tan of the angle to a (50ohms antenna / 50 ohms coax) would result in a impedance phase match of 45 degrees to a tuned antenna load to the coax line impedance, thus the antenna / coax SWR = 1:1. It could then be perhaps noted, that a resonant antenna impedance is only resonant at its designed impedance, be this 50ohm or maybe 300ohms, matching thus then a 50ohm or a 300ohm coax line.

If one folds down the antenna impedance reactance down at the right angle focal point, the antenna vector would mask over the coax vector, thus would illustrate an antenna impedance would match to the coax line vector impedance, meaning a SWR of 1:1. However any long then the antenna vector than the coax vector would mean an antenna vector greater than the coax line impedance, any antenna vector line shorter than the coax line vector, the antenna impedance would be shorter than the coax line vector impedance.

Now if the one metre line was a full wave length wire antenna, then the frequency of the RF carrier would be 300MHz. From the equation for an inductive reactance, the full wave wire would have an inductive reactance of some 550ohms, irrespective the frequency provided the wire line is a full wave length. To put another way, a 10m line would have an equivelant inductance of 3uH, a 10m line would be fullwave at 30MHz, which in turn would again have an inductive reactance of 565ohms. A half wave line would be a 282ohms inductive reactance, making thus a 1/4 wave line a 141ohms inductive reactance. A 50ohm stub antenna would be a wire or rod length of only 8·85% of a full wavelength.

Now as a full wave wire or rod would have a full wave coupling to the etha, the surrounding air space, a 1/4 wire would be 25% efficient with the signal radiation, thus 25Watt from a 100Watt transmitter, compared to the 100% signal radiation with a full wave line or a full wave antenna. By this comparison, a 50ohms stub antenna found on many hand held radios, the 50ohm stub antenna would only be 8·85% efficient, meaning that a 5Watt transmitter would only radiate an E.R.P. of 440mW. The received signal on a 50ohm stub antenna would like wise be only 8·85% efficient. This thus equates to a 1uV signal found at the radio terminal connection, the connection the antenna is connected too, would require a signal field strength of 11uV signal for a 1uV terminal signal strength input from the 50ohm stub antenna.

Much has been said how the coax line hides the true identity of a antenna. I came across this equation shown below, while finding out how an equation functions that would determine the true antenna impedance from the coax feed point of the radio set. I found that "c" is the speed of light, "300E6", and "L" is the length of the coax used. URL: http://www.antenna-theory.com/basics/impedance.php

The results shown below illustrate the 1/4 wave vertical appearance to the radio as rather a different thing. The frequency of the band in use also plays apart of the results calculated from the above equation. For tuning a 1/4 wave for 2m or 70cm bands, the results shown give some reason why the matching of one 1/4 wave design is some what problematic. As the coax length increases, so does the offset antenna load reduce in deviation from the good Swr match.

However, if your wire antenna on HF is short on length, say tuning up a 20m 1/4 wave say for 80m band, then its impedance will short than 50 ohms. With the radio viewing the 20m 1/4 wave through a 10 metre coax on the 80m band, the following calculations show the apparent resulting values for the antenna looking as a 35 ohm load.

Now consider the an antenna system where a 40m long wire is interfaced via either a 4:1 or a 9:1 balum.

The first table below shows a 4:1 balum attached to a 40m wire.

Now the "ATU load" is the terminal impedance seen by the FT450d internal ATU. The "antenna erp" relates to the antenna wire length used relative to a its band full wavelength measurement. For the 80m band, a 40m wire is half wavelength and would thus have 50% of the coupling efficiency to a full wavelength antenna to the surrounding etha.

The next diagram table relates to a 9:1 balum used for the same 40m wire.

On the 4:1 balum, to 40m wire on the 10m band shows a 546ohm impedance, while for the 9:1 balum, the impedance calculates to 242ohms in value. However, due to the coaxial cable transformer effect, the antenna impedance experienced and the radio set end would be different.

For a 10m length, 50ohm coax cable, the impedance experienced at the radio end using a 4:1 balum antenna circuit is some 12ohms, while for the 9:1 balum antenna circuit, the radio would experience a terminal impedance of 17ohms.

Provided your radio set internal ATU can matched to either a 12ohms for a 4:1 balum circuit, or matched to 17ohms for the 9:1 balum circuit, the radio would matched to the radio set end of the coax, while the antenna end of the coax would be matched to the antenna loading. In other words, the antenna circuit would be matched through-out its length of the antenna system circuit.

The BBC Basic code program shown below, in each text window, the 1st lists the hidden antenna load program, while the second listing calculates the antenna balum circuit.

Text Box

10 REM calculating the antenna load with coax length against frequency

20 light = 300E6

30 imped_ant_Za = 150

40 imped_line_Zo = 50

50 PRINT " antenna load = "; imped_ant_Za;" ohms : line Coax = ";imped_line_Zo;" ohms "

60 length = 5

70 PRINT " coax length = ";length;" metres"

71 PRINT

72 PRINT

73 freq = 136E3

74 PROC_active_load

76 freq = 457E3

78 PROC_active_load

80 freq = 1.9E6

90 PROC_active_load

100 freq = 3.7E6

110 PROC_active_load

120 freq = 5.2E6

130 PROC_active_load

140 freq = 7.1E6

150 PROC_active_load

160 freq = 10.05E6

170 PROC_active_load

180 freq = 14.1E6

190 PROC_active_load

195 freq = 18.1E6

197 PROC_active_load

200 freq = 21E6

210 PROC_active_load

220 freq = 24.95E6

230 PROC_active_load

240 freq = 29E6

250 PROC_active_load

260 freq = 51E6

270 PROC_active_load

280 freq = 71E6

290 PROC_active_load

300 freq = 145E6

310 PROC_active_load

320 freq = 225E6

350 PROC_active_load

360 freq = 435E6

370 PROC_active_load

375 freq = 1275E6

377 PROC_active_load

380

390 END

400

410 DEF PROC_active_load

420 top = imped_ant_Za + ((imped_line_Zo * TAN((2*PI*freq )/light)*length))

430 bot = imped_line_Zo + ((imped_ant_Za * TAN((2*PI*freq )/light)*length))

440 Z_in = imped_line_Zo *(top/bot)

450 PRINT TAB(10);" freq = ";freq/1E6;"MHz";TAB(30);"found load = ";INT(Z_in*100)/100;" ohms"

460 ENDPROC

Text Box

10 REM calculated load capacitance for long wire antenna though the HF bands

20 REM inductance to wire is 300nH / metre length

30 REM 15metres of wire is half wavelenght at 10MHz, the wire has an inductance of 4.5uH equivalent coil

40 REM reason : 1 metre of wire is equal to 300nH

50

60 REM the Zo=SQR(RL^2 - XL^2) equation is the add on to the co-axial cable, the antenna end

70 REM Thus so, with XL = 50, then the above equation then equals zero, and hence does add to the

80 REM antenna with an antenna reactive loading.

90 REM

100

110 REM The value of reference in the Zo=SQR() equation is the cable impedance, as XL =50,

120 REM the added load of the antenna to the cable equates to zero offset.

130

140 REM By using the cable impedance, the wire is matched to the cable characteristic value.

150

160 REM However if a 1/4 wavelength wire inductive resonance reactance value is used as RL, RL = 141ohms,

170 REM then the wire is matched to the 1/4 wave length terminal impedance.

180

190 REM for the 1/4 wave length match to be used, the ATU would have an input PI section Low Pass Filter, Fc = 30MHz

200 REM the input of the LPF would be 50ohms to match the RF co-axial cable, but the LPF output would be 141ohms.

210

220

230 REM 15metres of wire

240 balam = 9

250 length_metres = 80

260 inductance = (length_metres*(300E-9))

270 length_feet = ((100/2.54)*length_metres)/12

280

290 REM in this example the cable impedance is used as the reference for RL

300 reactance_resonant_impedance = 50

310

320 PRINT " terminal R.F. antenna impedance = ";reactance_resonant_impedance;"ohms"

330 REM PRINT " inductance of wire / coil = ";l*1E6;"uH"

340 PRINT " test radio - Yaesu FT450d @ 100watts Tx"

350 PRINT " eff. length of wire = ";INT(length_feet);"feet or ";INT(length_metres);"metres, equal to ";INT(10 * inductance*1E6)/10;"uH coil inductance"

360 PRINT " ATU matching circuit fed through a impedance step-up ";balam;":1 balam to wire antenna connection"

370 PRINT

380 PRINT TAB(4);"RF band";TAB(20);"ATU load";TAB(36);"ATU match balam i/p";TAB(59);"balam o/p load";TAB(79);"antenna ERP"

390 PRINT

400

410

420 f = 136E3

430 PROC_calculate

440 f = 450E3

450 PROC_calculate

460 PRINT

470 f = 1.9E6

480 PROC_calculate

490 f = 3.65E6

500 PROC_calculate

510 f = 5.2E6

520 PROC_calculate

530 f = 7.1E6

540 PROC_calculate

550 f = 10E6

560 PROC_calculate

570 f = 14.15E6

580 PROC_calculate

590 f = 18E6

600 PROC_calculate

610 f = 21.2E6

620 PROC_calculate

630 f = 24.7E6

640 PROC_calculate

650 f = 29E6

660 PROC_calculate

670 PRINT

680 f = 51E6

690 PROC_calculate

700 f = 71E6

710 PROC_calculate

720 END

730

740

750 DEF PROC_calculate

760 REM inductive reactance

770 XL= (2*PI*(f)*inductance)

780 antenna_load = XL

790 XL = XL/balam

795

800 REM RL is low to 50ohms, thus the wire is capacitive and needs inductive loading

810 IF XL <= reactance_resonant_impedance THEN PROC_low

820

830 REM RL is high to 50ohms, thus thwe wire is inductive and needs capacitive loading

840 IF XL > reactance_resonant_impedance THEN PROC_high

850 ENDPROC

860

870 REM PRINT TAB(4);"RF band";TAB(20);"ATU load";TAB(38);"match i/p balam";TAB(59);"balam o/p Ant. load";TAB(79);"antenna ERP"

880

890 REM RL is low to 50ohms, thus the wire is short and capacitive thus needs inductive loading

900 DEF PROC_low

910 Xc_low = (reactance_resonant_impedance - XL)

920 REM Xc_low used as XL, as opposite reactance required

930 ind_Load = INT((Xc_low/((2*(f)))*1E6)*10)

940 PRINT TAB(1);" freq= ";f/1E6;"MHz";TAB(19);" XL=";INT(XL * 10)/10;" ohms";TAB(37);"series_ind=";ind_Load/10;"uH";TAB(60);INT(antenna_load);"ohms";TAB(80);INT((antenna_load/550)*100);" Watts"

950 ENDPROC

960

970

980 REM RL is high to 50ohms, thus the wire is long and inductive thus needs capacitive loading

990 DEF PROC_high

1000 XL_high = (XL - reactance_resonant_impedance)

1010 REM XL used as Xc, as opposite reactance required

1020 cap_load = INT((1/((2*PI*(f)*XL_high))*1E12)*10)

1030 PRINT TAB(1);" freq= ";f/1E6;"MHz";TAB(19);" XL=";INT(XL * 10 )/10;" ohms";TAB(37);"series_cap=";cap_load/10;"pF";TAB(60);INT(antenna_load);"ohms";TAB(80);INT((antenna_load/550)*100);" Watts"

1040 ENDPROC