Free and Forced Oscillations in Radio Telegraphy
��By John Vincent
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��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
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