Two electrical oscillators are used in a heterodyne metal detector to detect bur

Question

Two electrical oscillators are used in a heterodyne metal detector to detect buried metal objects (see the figure). The detector uses two identical electrical oscillators in the form of LC circuits having resonant frequencies of 690 kHz. When the signals from the two oscillating circuits are combined, the beat frequency is zero because each has the same resonant frequency. However, when the coil of one circuit encounters a buried metal object, the inductance of this circuit increases by 1.150%, while that of the second is unchanged. Determine the beat frequency that would be detected in this situation.
1 kHz

Explanation / Answer

Initial resonant frequency of both oscillators = 1/{2*pi*sqrt(LC)} = 690 kHz

Changed Resonant frequency of the first coil = 1/{2*pi*sqrt(1.0115LC)} = 0.9943 / {2*pi*sqrt(LC)} = 0.9943*690 = 686 kHz

Resonant frequency of second coil remains at 690 kHz.

So, Beat Frequency = 690 - 686 = 4 kHz

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