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If a receiver tuned to 146.70 MHz receives an intermodulation product signal whenever a nearby transmitter transmits on 146.52, what are the two most likely frequencies for the other interfering signal?
146.34 MHz and 146.61 MHz
Third-order intermodulation products are common: 2A-B or 2B-A. If 146.70 is 2A-B and A=146.52, then B = 2*146.52 - 146.70 = 293.04 - 146.70 = 146.34. If 146.70 is A+B-C and A=146.52, C could be 146.61, then (146.52 + B - 146.61) or (2*146.52-146.34 = 146.70). The second freq is likely involved in 2B-A or A+B-C etc.
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