SU63984A1 - Method of measuring modulation index - Google Patents

Description

Before)

, t - f - sin o t

I In cos

dl dt

tea is equal to - L

Detection is best done with a mixing lamp, in such a way that voltage amplification occurs at the same time.

Integration is necessary to produce, passing a current through a capacitor. The voltage on the capacitor will be:

and Dsh I cos Q t dt, -. - -X

.

Xsin Q i.

The circuit of the device performed by the proposed method (see drawing) shows a mixing lamp L, at the input of which there is an inductor L and resistance Ri for differentiating the signal, and in the output circuit a capacitor C for integration, resistance R and cathode voltmeter KB for measurements.

To measure the modulus index

was found

tion - how

above, it is necessary to measure the voltage in capacitance C.

If the ratio between R and C

chosen so that R (, p-, topo-miu cathode voltmeter will be proportional to:

 . ° C 0) 0 s 1

Therefore, if WQ or Q is known (it is usually easy to know WQ), then using the readings of the cathode voltmeter, you can calculate the modulation index

- (since DtUo and C are constant and known). Coefficient D equals

-2-SnL2I f

where S is the steepness of the lamp.

Subject invention

1. A method for measuring the modulation index, characterized in that for judging the magnitude of the latter, the variable component of the detector current is electrically integrated, to the input of which the alternating voltage of the signal previously subjected to electrical differentiation is applied.

2. A device for implementing the method of clause 1, characterized in that a capacitor is turned on at the output of the frequency modulation detector, the voltage on which is measured with the help of a lamp voltmeter.