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I encountered the following question on a test, and I'm not sure of the theory behind it. Could someone please explain?

When a signal with spectrum ranging from 13 MHz to 15 MHz is mixed with (i.e., multiplied by) a 14-MHz sine wave, which of the following frequency ranges does NOT contain frequencies of the resulting mixed signal?

  • (A) From 0 to 1 MHz
  • (B) From 13 to 14 MHz
  • (C) From 27 to 28 MHz
  • (D) From 28 to 29 MHz
  • (E) All of the above ranges contain frequencies of the resulting signal
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    $\begingroup$ Can you elaborate why you are posting a multiple-choice question to this site ? Do you expect us to simply answer this ? Or maybe you do not understand the theory, in which case you could rewrite your question about what it is you do not understand, give more background what you have studied and researched about it and narrow it down so we can formulate an appropriate answer. As it stands it just looks like a question for some sort of test.... $\endgroup$
    – Edwin van Mierlo
    Jan 15, 2019 at 7:41
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    $\begingroup$ yep, I actually don't understand what you're asking here, so I voted that this question be closed. Please stop me by explaining exactly what you need help with. $\endgroup$
    – Marcus Müller
    Jan 15, 2019 at 12:45

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When two sinusoidal signals are multiplied through a non-linear mixer circuit, it results in the sum and difference of the mixed frequencies:

$$f_s=f_1+f_2 \tag 1$$ $$f_d=f_1-f_2 \tag 2$$

Of course, negative frequencies are not possible so $f_1$ should be the higher frequency in equation 2.

This multiplying or mixing process is also called heterodyning and is the basis for most modern, analog radios.

So now you can work through the formulas to determine your homework answer. Take note that the question asks for a range that would not be produced by the mixing action. If you care to work through the answer as a comment to this answer, I would be happy to comment on your conclusion.

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    $\begingroup$ Also (though depending on the context) it's often safe to assume that the original frequencies also appear in the output of a mixer. $\endgroup$
    – natevw - AF7TB
    Jan 15, 2019 at 18:33
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    $\begingroup$ @natevw-AF7TB True! An unbalanced mixer or a single balanced mixer are examples of that. Since the question is academic, I went with the ideal case. $\endgroup$
    – Glenn W9IQ
    Jan 15, 2019 at 18:50
  • $\begingroup$ base on the explanation above: fs=13+14=27 MHZ . fd=15-14=1 MHZ please advise me @GlennW9IQ $\endgroup$
    – natasha
    Jan 16, 2019 at 5:42
  • $\begingroup$ @natasha You are close but since you have a range of frequencies you need to add and subtract the upper and lower values of the range. The result will be two ranges of frequencies. Hint: one of those ranges is split into two of the answers. $\endgroup$
    – Glenn W9IQ
    Jan 16, 2019 at 11:34
  • $\begingroup$ @natasha I think Glenn's answer was on-point and answers the question. What further advise do you need, specifically? (Note: please learn to ask precise questions. That's how this website works.) $\endgroup$
    – Marcus Müller
    Jan 16, 2019 at 11:35

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