Free and Forced Oscillations in 

 Radio Telegraphy 



By John Vincent 



:^ 



Fig. 1. 



Simple 

 loaded 

 antenna 



THE February article of this series 

 pointed out how closely all oscil- 

 lation circuits resembled each 

 other, whether or not they contained 

 spark-gaps and whether they were open 

 antennas or closed condenser-circuits. 

 Not all of the similarities were brought 

 out, however, and it is interesting to note 

 that for all practical purposes the rule 

 last given, for finding the time period 

 of an oscillating spark-circuit, is the 

 same as that for determing the resonant 

 wavelength of an antenna. 

 The simplest way to work 

 this out is to compute the ^ 



period and wavelength of ^^ 



an aerial, such as shown in ^ 



Fig. 1, accordin^ to each of 1,^ 



the rules, and then to com- i 1 



pare the results. =^, 



Suppose the antenna 

 system of this diagram has the constants 

 given in the fourth example of the No- 

 vember article. The aerial itself will 

 then be of 0.0012 microfarad capacity 

 and 0.023 millihenry inductance, and the 

 loading-coil will have 0.35 millihenry in- 

 ductance. This last named figure is the 

 sum of the inductances of secondary 

 and loading-coil in the earlier example; 

 the total is taken because in 

 Fig. 1 only a single coil is 

 shown. 



Applying the rule for 

 finding resonant wavelength, 

 when capacity and induct- 

 ance are known, the steps are: (1) mul- 

 tiply the total inductance by the total 

 capacitv (0.0012 microfarad times 0.373 

 millihenry = 0.000447), (2) take the 

 square root of this number, which equals 

 0.0213, (3) multiply this result by times this cycle is performed in on 



17 

 T 



Fig. 2, Spark-gap 

 circuit 



that would be set up when currents of 

 a definite frequency surged back and 

 forth in the antenna, and that the fre- 

 quency could be found by dividing the 

 wavelength in meters into the number 

 300,000,000 (which is the velocity of 

 electromagnetic waves in meters per 

 second). By performing this operation, 

 it is found that the frequency of the 

 1,270-meter wave is 300,000,000 /1, 270 

 =236,000 cycles per second. This is 

 the resonant frequency of a circuit hav- 

 ing the inductance and ca- 

 pacity above stated, or, in 

 other words, the frequency 

 of exciting alternating volt- 

 age which will produce the 

 largest current in that cir- 

 cuit. At that frequency the 

 current will be the greatest 

 possible for any given volt- 

 age, because the circuit has minimum 

 impedance when the capacity and in- 

 ductance neutralize each other's reactive 

 eflfects. as was also explained in Jan- 

 uary. 



Now, taking the same antenna circuit 

 of Fig. 1, and assuming the same values 

 of inductance and capacity, the time 

 period of natural oscillation may be 



_^j^ found by applying the rule 



^ stated last month. This time 

 period is the number of 

 seconds which it takes fcr 

 the alternating" current in 

 the circuit to pass through 

 a complete cycle, i. e., to start from zero, 

 reach a maximum in one direction, re- 

 verse, pass through zero and reach a 

 maximum in the other direction, reverse 

 again, and reach zero. The number of 



-xMW 



60.000 (0.0213 times 60.000=1,270 

 meters) and thus obtain the answer re- 

 quired. Thus the tuned wavelength of 

 the antenna with loading-coil is found to 

 be 1.270 meters. From the January 

 article it appeared that this corre- 

 sponded to the length of the ether-wave 



second is the frequency of the current, 

 and is the numerical reciprocal of the 

 time period. Since one millihenry is 

 one-thousandth of a henry, the value of 

 inductance may be given in either unit. 

 Since for this antenna it is 0.373 milli- 

 henry, in henrys it is one-thousandth of 



464 



