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