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. The detector uses two identical electrical oscillators in the form of LC circuits having resonant frequencies of 715 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 1.490%, while that of the second is unchanged. Determine the beat frequency that would be detected in this situation.

_____ Khz

Explanation / Answer

beat frequency is given by the difference between the two oscillators frequency

Intial frequency = f1 = 715 kHz

f = resonance frequency = 1/(2*pi*sqrt (LC))

Now when inducntance is increased by 1.490%, new frequency will be:

f2 = [1/sqrt (1.0149)]*f1

f2 = 715*[1/sqrt 1.0149] = 709.73 kHz

Beat frequency = 715 - 709.73 = 5.27 kHz

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