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 
  250 length_metres = 50
  260 l = (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 " reactance resonant impedance = ";reactance_resonant_impedance;"ohms"
  330 PRINT " inductance of wire / coil = ";l*1E6;"uH"
  340 PRINT " length of equivalent wire = ";INT(length_feet);"feet or ";length_metres;"metres"
  350 
  360 PRINT
  370 
  380 FOR f = 0.1 TO 1 STEP 0.1
  390   
  400   REM inductive reactance
  410   XL= (2*PI*(f*1E6)*l)
  420   
  430   REM RL is low to 50ohms, thus the wire is capacitive and needs inductive loading
  440   IF XL <= reactance_resonant_impedance THEN PROC_low
  450   
  460   REM RL is high to 50ohms, thus thwe wire is inductive and needs capacitive loading
  470   IF XL > reactance_resonant_impedance THEN PROC_high
  480   
  490 NEXT f
  500 
  510 
  520 FOR f = 1 TO 30  STEP 1
  530   
  540   REM inductive reactance
  550   XL= (2*PI*(f*1E6)*l)
  560   
  570   REM RL is low to 50ohms, thus the wire is capacitive and needs inductive loading
  580   IF XL <= reactance_resonant_impedance THEN PROC_low
  590   
  600   REM RL is high to 50ohms, thus thwe wire is inductive and needs capacitive loading
  610   IF XL > reactance_resonant_impedance THEN PROC_high
  620   
  630 NEXT f
  640 END
  650 
  660 
  670 
  680 
  690 REM RL is low to 50ohms, thus the wire is short and capacitive thus needs inductive loading
  700 DEF PROC_low
  710 Xc_low = (reactance_resonant_impedance - XL)
  720 REM Xc_low used as XL, as opposite reactance required
  730 XLoad = Xc_low/((2*(f*1E6)))
  740 PRINT TAB(1);" freq= ";f;"MHz";TAB(19);" XL=";XL;TAB(35);" Xload=";XLoad*1E6;"uH"
  750 ENDPROC
  760 
  770 
  780 REM RL is high to 50ohms, thus the wire is long and inductive thus needs capacitive loading
  790 DEF PROC_high
  800 XL_high = (XL - reactance_resonant_impedance)
  810 REM XL used as Xc, as opposite reactance required
  820 Xcload = 1/((2*PI*(f*1E6)*XL_high))
  830 PRINT TAB(1);" freq= ";f;"MHz";TAB(19);" XL=";XL;TAB(35);" Cload=";Xcload*1E12;"pF"
  840 ENDPROC