Why the frequency of carrier wave is always kept at least $10$ times higher than the highest frequency of base band signal? Can you explain me this with a practical example?
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2$\begingroup$ Do you have a reference for this statement? $\endgroup$ – Matt L. Mar 31 '15 at 10:18
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$\begingroup$ The statement that you're asking about isn't true; hence, I think this question should probably be closed. $\endgroup$ – Jason R Mar 31 '15 at 13:22
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$\begingroup$ Although "is it true that...as stated in [ref]" could be a valid and potentially interesting question. $\endgroup$ – Deve Mar 31 '15 at 13:57
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Assuming your baseband signal $b(t)$ has bandwidth $B$ and the carrier frequency is $f_c$, the upconverted signal $$s(t)=b(t)\cos(2\pi f_ct)$$ has bandwidth $2B$, extending from $f_c-B$ to $f_c+B$. You need $f_c-B>0$, which implies $f_c>B$. Theoretically, this is the only rule you need to make sure to follow.
Some specific receivers may need a higher $f_c$; for example, a simple envelope detector for amplitude demodulation will work better (and be easier to design) if you use a higher $f_c$.
Also, some textbooks assume $f_c>>B$ because that allows for some mathematical simplifications, such as assuming two up-converted pulses are orthogonal when they are not quite so.
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There is no hard "10 times" rule.
However, depending on the communication medium, there may be problems when a carrier is modulated with a signal of too high a frequency.
A little animation of amplitude modulation on Wikipedia illustrates this. The red line is the original carrier. The blue wave is the signal, and the green lines are the side-bands produced by modulating the carrier with the signal. Notice how the side bands move outwards in frequency as the signal frequency increases.
In radio communications, modulating with too high a frequency could cause two problems:-
- Interference with other people's signals on nearby frequencies.
- Difficulty in producing a tuned circuit that can accept the whole of the resulting frequency range.
