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Let's take an example.

Suppose that the intermediate frequency is 10,7 MHz (FM).

The local oscillator works on 110,7 MHz.

First case: You receive a signal of 100 MHz, the mixer will generate a frequency of 110,7 + 100 = 210,7 MHz, which will be rejected by the band-pass filter. The difference of the two frequencies is 110,7 - 100 = 10,7 MHz (desired one).

Second case: You receive a signal of 121,4 MHz. The sum of that frequency and the local oscillator is 232,1 MHz, which will be rejected. The difference is 121,4 - 110,7 = 10,7 MHz. So the image frequency in that case is going to be 121,4 MHz.

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12y ago

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